Unit 2 — Compound Structure and Properties

Every lesson, checkpoint, FRQ practice set and test of this unit, in course order. Answer keys are marked ✓ KEY. Feedback, model responses and rubrics are shown in full.

Contents — 16 sequences, 151 lessons

U2-S01 — Bond categories and bond polarity CSP-001 CSP-002 CSP-003 CSP-004 CSP-005 CSP-006 CSP-007 CSP-008 CSP-009 CSP-010 CSP-011

U2-S02 — Ionic and metallic particulate models CSP-012 CSP-013 CSP-014 CSP-015 CSP-016 CSP-017 CSP-018 CSP-019 CSP-020 CSP-021

U2-S03 — Alloy models CSP-022 CSP-023 CSP-024 CSP-025 CSP-026

U2-S04 — Property vocabulary and bonding evidence CSP-027 CSP-028 CSP-029 CSP-030 CSP-031 CSP-032 CSP-032A CSP-033 CSP-033A CSP-034 CSP-034A CSP-035 CSP-035A CSP-036 CSP-036A CSP-037 CSP-037A CSP-038 CSP-039 CSP-039A CSP-040 U2-G1 checkpoint

U2-S05 — Potential-energy curves CSP-041 CSP-042 CSP-043 CSP-044 CSP-045 CSP-046 CSP-047 CSP-048 CSP-049 CSP-050 CSP-051

U2-S06 — Bond order, length, and energy CSP-052 CSP-053 CSP-054 CSP-055 CSP-056 CSP-057 CSP-058 CSP-059

U2-S07 — Ionic interaction strength CSP-060 CSP-061 U2-G2 checkpoint

U2-S08 — Lewis-model foundations and boundaries CSP-062 CSP-063 CSP-064 CSP-065 CSP-066 CSP-067 CSP-068

U2-S09 — Lewis construction routine CSP-069 CSP-070 CSP-071 CSP-072 CSP-073 CSP-074 CSP-075 CSP-076 CSP-077 CSP-078

U2-S10 — Formal charge and model selection CSP-079 CSP-080 CSP-081 CSP-082 CSP-083 CSP-084 CSP-085

U2-S11 — Resonance CSP-086 CSP-087 CSP-088 CSP-089 CSP-090 CSP-091 CSP-092 U2-G3 checkpoint

U2-S12 — VSEPR and molecular geometry CSP-093 CSP-094 CSP-095 CSP-096 CSP-097 CSP-098 CSP-099 CSP-100 CSP-101 CSP-102 CSP-103 CSP-104 CSP-105 CSP-106 CSP-107

U2-S13 — Bond angles and VSEPR synthesis CSP-108 CSP-109 CSP-110 CSP-111 CSP-112 CSP-113 CSP-114 CSP-115 CSP-116 CSP-117 CSP-118 U2-G4 checkpoint

U2-S14 — Hybridization CSP-119 CSP-120 CSP-121 CSP-122 CSP-123

U2-S15 — Orbital overlap and geometric isomerism CSP-124 CSP-125 CSP-126 CSP-127 CSP-128 CSP-129 CSP-130 CSP-131 CSP-132 CSP-133 CSP-134 CSP-135 CSP-136 CSP-137 U2-G5 checkpoint

U2-S16 — Molecular polarity CSP-138 CSP-139 CSP-140 CSP-141 CSP-142 CSP-143 CSP-144 U2-G6 checkpoint Unit test

U2-S01 — Bond categories and bond polarity

11 lessons

U2-S01 · LESSON 1 · CSP-001

Metals and nonmetals on the periodic table

Outcome: Classify a main-group element as a metal or a nonmetal from its position in a supplied periodic table.

Activation question — asked before the lesson opens

Activates: Unit 1: the periodic table's layout follows from electron configurations

Elements in the same group (column) of the periodic table behave in similar ways. What feature of their atoms causes this similar behavior?

AThey have the same number of outer-shell (valence) electrons. ✓ KEY — this is the correct answer

Why: A group shares its outer-shell count — the table's layout follows the elements' electron configurations.

BThey have the same number of protons.

Why: The proton count is the atomic number, and it is different for every element in the group.

CThey have the same total number of electrons.

Why: A neutral atom's total electron count equals its proton count, so no two elements share it.

DThey have the same number of neutrons.

Why: Neutron counts vary even between isotopes of one element; the table is not arranged by them.

VIDEO SLOT: APCHEM-U02-L01 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Two families of elements

You’ve already seen that the periodic table’s layout follows from the elements’ electron configurations.

Here is a periodic table with a heavy stair-step line drawn toward its right side.

Elements on the left of the stair-step line are metals — most of the table, including Na, Mg, Ca, and Al.

Elements on the right of the stair-step line are nonmetals — including C, N, O, S, and Cl.

Hydrogen is the exception: it sits on the left, but it is a nonmetal.

Elements touching the line itself behave partly like both families; every element you are asked to classify in this course sits clearly on one side.

What you are expected to be able to do

Classify an element as a metal or a nonmetal from its position in a supplied periodic table.

Part 2Check what you just learned

Check — multiple choice

Using the periodic table shown, which of these elements is a nonmetal?

ACa

Feedback if chosen: This option is wrong — you classified calcium against its position. Ca sits well to the left of the stair-step line, so it is a metal.

BAl

Feedback if chosen: This option is wrong — aluminum sits on the left of the stair-step line, so it is a metal.

Feedback if chosen: This option is wrong — potassium sits at the far left of the table, deep on the metal side of the stair-step line.

DS ✓ KEY — this is the correct answer

Feedback if chosen: Correct. S sits to the right of the stair-step line, so it is a nonmetal.

Check — multiple choice

Using the periodic table shown, which of these elements is a metal?

Feedback if chosen: This option is wrong — phosphorus sits to the right of the stair-step line, so it is a nonmetal.

BMg ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Mg sits on the left of the stair-step line, so it is a metal.

CBr

Feedback if chosen: This option is wrong — bromine sits to the right of the stair-step line, with the other halogens, so it is a nonmetal.

DSe

Feedback if chosen: This option is wrong — selenium sits to the right of the stair-step line, below sulfur, so it is a nonmetal.

Check — multiple choice

Hydrogen sits on the left side of the table shown. How is it classified?

AA metal, because it sits on the left of the stair-step line

Feedback if chosen: This option is wrong — you applied the left-side rule to the one element it does not cover. Hydrogen sits on the left but is a nonmetal, the marked exception.

BA metal, because it sits in the same column as Na and K

Feedback if chosen: This option is wrong — you classified hydrogen by its column neighbors. H shares a column with the Na and K metals, but it is the marked exception: a nonmetal.

CA nonmetal, because it sits to the right of the stair-step line

Feedback if chosen: This option is wrong — you placed hydrogen on the wrong side. Check the table: H sits on the left of the stair-step line, and it is a nonmetal only because it is the marked exception.

DA nonmetal — hydrogen is the marked exception to the left-side pattern ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Hydrogen sits on the left, but it is a nonmetal — the one exception you need to remember.

U2-S01 · LESSON 2 · CSP-002

What makes a bond covalent

Outcome: State that a covalent bond is one or more pairs of valence electrons shared between two atoms, and recognize whether a drawn pairing is covalent.

Activation question — asked before the lesson opens

Activates: Unit 1: ion charges — Na forms Na⁺ and Cl forms Cl⁻ (by electron transfer)

When sodium and chlorine form NaCl, what happens to sodium's single outer electron?

AIt stays in sodium's outer shell, and no ions form.

Why: Sodium's outer electron does leave — NaCl is made of Na⁺ and Cl⁻, and the transfer is what creates them.

BIt moves down into one of sodium's inner shells.

Why: Sodium's inner shells are already full; the electron leaves the atom entirely, onto chlorine.

CIt transfers onto the chlorine atom, forming Na⁺ and Cl⁻. ✓ KEY — this is the correct answer

Why: The transfer creates two oppositely charged ions, and their attraction is the ionic bond.

DIt is destroyed as the two ions form.

Why: Electrons are never destroyed — the electron still exists, now on the chlorine atom.

VIDEO SLOT: APCHEM-U02-L02-SCRIPT — quote this id when requesting a video edit — Video, part A — an alternative to reading this lesson

VIDEO SLOT: APCHEM-U02-L02B-SCRIPT — quote this id when requesting a video edit — Video, part B

Part 1Shared or not shared

You’ve already seen that an atom’s valence electrons are the electrons in its outermost shell.

You’ve already seen how to predict the charge a main-group atom’s ion typically carries — Na forms Na⁺, and Cl forms Cl⁻.

In the drawings that follow, dots stand for valence electrons, and a line between two atom symbols stands for a pair of valence electrons the two atoms share.

Start with sodium and chlorine: Na’s single outer electron has moved onto Cl, leaving two separate ions.

No pair of electrons sits between the two ions — nothing is shared, so this is not a covalent bond.

Now two chlorine atoms: each contributes one valence electron to a pair that sits between the two, drawn as the line joining them.

That shared pair is a covalent bond.

Swap one chlorine for fluorine — two different elements now — and the pair between them is still shared.

Different elements can bond covalently — a bond is covalent because the atoms share a pair, not because the elements match.

Now two oxygen atoms sharing two pairs at once, drawn as two lines.

A covalent bond can be more than one shared pair — two here, and some pairs of atoms share three.

Finally magnesium and oxygen: Mg’s two outer electrons have moved onto O, leaving two separate ions again.

Nothing sits between the ions to be shared — not a covalent bond.

Part 2The rule

A covalent bond consists of one or more pairs of valence electrons shared between two atoms.

To decide whether a bond is covalent, ask one question: is there a pair of electrons that both atoms hold together?

What you are expected to be able to do

State that a covalent bond is one or more pairs of valence electrons shared between two atoms, and recognize whether a drawn pairing is covalent.

Part 3Check what you just learned

Check — multiple choice

The diagram shows two bromine atoms in Br₂. What makes the Br–Br bond covalent?

AThe two atoms share a pair of valence electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The line between the two Br atoms is a pair of valence electrons that both atoms hold together — that sharing is the covalent bond.

BOne Br atom has transferred an electron to the other

Feedback if chosen: This option is wrong — you read the drawing as a transfer. The pair sits between the two atoms and is held by both; that is sharing, not transfer.

CThe two atoms are the same element

Feedback if chosen: This option is wrong — you tied covalent to matching elements. F–Cl is covalent too; what makes a bond covalent is the shared pair, not the elements matching.

DEach atom keeps all of its own valence electrons to itself

Feedback if chosen: This option is wrong — you described atoms with nothing shared. If each atom kept every electron to itself there would be no bond; the pair between the atoms is shared.

Check — multiple choice

The diagram shows N₂ with three pairs of electrons between the two atoms. Is the N–N bond covalent?

ANo — a covalent bond is exactly one shared pair

Feedback if chosen: This option is wrong — you turned the first example into the whole rule. A covalent bond is one or more shared pairs; three shared pairs still count.

BNo — covalent bonds form only between different elements

Feedback if chosen: This option is wrong — you tied covalent bonding to the elements being different. Cl–Cl already showed identical atoms sharing a pair; the sharing makes a bond covalent.

CYes — the atoms share pairs of valence electrons, and a covalent bond can be one, two, or three shared pairs ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Three pairs held between the two N atoms is still sharing — a covalent bond is one or more shared pairs of valence electrons.

DYes — because each N atom also keeps a pair of unshared electrons

Feedback if chosen: This option is wrong — you pointed at the unshared dots. Each unshared pair sits on one atom only; it is the three pairs between the atoms that make the bond covalent.

Check — multiple choice

In the KF drawing shown, K’s outer electron has moved onto F. Is this a covalent bond?

AYes — the electron pairs now on F count as shared pairs

Feedback if chosen: This option is wrong — you counted F’s own electrons as shared. Every dot in the drawing sits on F alone; no pair sits between the two atoms.

BNo — the electrons are not shared between the two atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Nothing sits between K⁺ and F⁻ for the two to hold together — the electrons are not shared, so this is not a covalent bond.

CNo — K and F are different elements

Feedback if chosen: This option is wrong — you gave the right verdict for the wrong reason. F–Cl showed that different elements can share a pair; KF fails because nothing is shared, not because the elements differ.

DYes — any two atoms that attract each other share electrons

Feedback if chosen: This option is wrong — you treated attraction as sharing. The two ions do attract each other, but no pair of electrons sits between them, so the bond is not covalent.

U2-S01 · LESSON 3 · CSP-003

Classifying bonds from the elements involved

Outcome: Classify a bond as generally ionic for a metal–nonmetal pair or covalent for a nonmetal–nonmetal pair.

Activation question — asked before the lesson opens

Activates: CSP-001: classifying an element from its position against the stair-step line

Calcium sits on the left of the periodic table's stair-step line. How is calcium classified?

AAs a nonmetal

Why: Nonmetals sit on the right of the stair-step line.

BAs a metal ✓ KEY — this is the correct answer

Why: Left of the stair-step line means metal — that is the classification rule.

CAs a metalloid

Why: The metalloids sit along the line itself; calcium is well to the left of it.

DIt cannot be classified from its position alone.

Why: Position against the line is exactly how the classification works — hydrogen is the one exception.

VIDEO SLOT: APCHEM-U02-L03 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Sorting the pairings by element type

You’ve already seen how to classify an element as a metal or a nonmetal from its position in the periodic table.

You’ve already seen that a covalent bond is one or more pairs of valence electrons shared between two atoms.

You’ve already seen ionic bonding: oppositely charged ions held together by electrostatic attraction.

Line the pairings we’ve seen up against the periodic table, and a pattern appears.

NaCl pairs a metal with a nonmetal, and it forms oppositely charged ions that attract — ionic bonding.

MgO also pairs a metal with a nonmetal — ionic again.

Cl₂, FCl, and O₂ each pair a nonmetal with a nonmetal, and each shares electron pairs — covalent bonds.

Part 2The rule

A metal bonded to a nonmetal: classify the bond as generally ionic.

A nonmetal bonded to a nonmetal: classify the bond as covalent.

The word generally matters — element types give a reliable starting classification, but a few metal–nonmetal compounds break the pattern.

What you are expected to be able to do

Classify a bond as generally ionic for a metal–nonmetal pair, or covalent for a nonmetal–nonmetal pair.

Part 3Check what you just learned

Check — multiple choice

Using the periodic table shown, classify the bond in KBr.

ACovalent — K and Br are both nonmetals

Feedback if chosen: This option is wrong — you placed K on the nonmetal side. K sits at the far left of the table, deep on the metal side of the stair-step line.

BGenerally ionic — K and Br are both metals

Feedback if chosen: This option is wrong — you placed Br on the metal side. Br sits to the right of the stair-step line, a nonmetal — so KBr is a metal–nonmetal pair, generally ionic.

CGenerally ionic — K is a metal and Br is a nonmetal ✓ KEY — this is the correct answer

Feedback if chosen: Correct. K sits left of the stair-step line (metal) and Br sits right of it (nonmetal); a metal–nonmetal pair is generally ionic.

DCovalent — K and Br sit in the same period

Feedback if chosen: This option is wrong — you classified by period. K and Br do share a period, but the rule sorts by metal versus nonmetal, and a metal–nonmetal pair is generally ionic.

Check — multiple choice

Using the periodic table shown, classify the bond in HBr.

AGenerally ionic — H sits on the left of the table, so it is a metal

Feedback if chosen: This option is wrong — you classified hydrogen by its position. H is the marked exception: a nonmetal despite sitting on the left, so HBr pairs two nonmetals.

BCovalent — H and Br are both nonmetals ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Hydrogen is the exception — a nonmetal despite its left-side position — and Br is a nonmetal, so HBr is a nonmetal–nonmetal pair: covalent.

CGenerally ionic — Br is a nonmetal

Feedback if chosen: This option is wrong — you applied half the rule. One nonmetal is not enough for an ionic classification; the pairing must be metal with nonmetal, and H is a nonmetal too.

DGenerally ionic — H and Br sit far apart on the table

Feedback if chosen: This option is wrong — you classified by distance across the table. The rule sorts by metal versus nonmetal, and both H and Br are nonmetals: covalent.

Check — multiple choice

Using the periodic table shown, which of these bonds is covalent?

ACa–O

Feedback if chosen: This option is wrong — Ca is a metal and O is a nonmetal, so Ca–O is a metal–nonmetal pair: generally ionic.

BK–I

Feedback if chosen: This option is wrong — K is a metal and I is a nonmetal, so K–I is a metal–nonmetal pair: generally ionic.

CLi–F

Feedback if chosen: This option is wrong — Li is a metal and F is a nonmetal, so Li–F is a metal–nonmetal pair: generally ionic.

DN–O ✓ KEY — this is the correct answer

Feedback if chosen: Correct. N and O both sit right of the stair-step line — two nonmetals, so the bond is covalent.

U2-S01 · LESSON 4 · CSP-004

Why measured properties give the best evidence

Outcome: Explain why measured properties of a compound are better evidence for its bonding than element types or electronegativity alone.

Activation question — asked before the lesson opens

Activates: Unit 1: the electronegativity definition

What does an atom's electronegativity measure?

AThe energy needed to remove one of its electrons

Why: That is ionization energy — a different periodic property.

BThe number of valence electrons it has

Why: The valence count comes from the group number; electronegativity is about pull on shared electrons.

CThe overall electric charge that the atom carries

Why: A neutral atom carries no overall charge; electronegativity is an attraction, not a charge.

DIts tendency to attract shared electrons toward itself ✓ KEY — this is the correct answer

Why: That is the definition — the pull an atom exerts on electrons it shares.

VIDEO SLOT: APCHEM-U02-L04 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Predictions and measurements

You’ve already seen the element-type rule: metal–nonmetal pairs are generally ionic, nonmetal–nonmetal pairs covalent.

You’ve already seen that electronegativity measures an atom’s tendency to attract shared electrons toward itself.

Both of those come from the periodic table alone — they are predictions made before anyone touches the compound.

The compound itself can be measured: at what temperature it melts, whether it conducts electricity as a solid and again when molten, whether it shatters or bends when struck.

Measured properties come from the actual substance, so they are the best evidence for characterizing its bonding.

When a prediction from the elements and a measurement of the compound disagree, the measurement wins.

It happens: SnCl₄ pairs a metal with a nonmetal, yet it is a liquid at room temperature — measuring the real compound, not the element-type rule, is what reveals its bonding.

An unknown salt is characterized the same way — for example, by testing whether it conducts electricity as a solid and again when molten.

What you are expected to be able to do

Explain that element types and electronegativity give an initial bond classification, and that measured properties of the compound give the best evidence.

Part 2Check what you just learned

Check — multiple choice

A chemist wants the most reliable characterization of the bonding in a new compound. Which evidence is best?

AWhether the elements it contains are metals or nonmetals on the periodic table

Feedback if chosen: This option is wrong — you stopped at the starting classification. Element types predict the bonding from the periodic table alone; the compound itself has not been examined.

BThe electronegativity difference between its elements, read from a table

Feedback if chosen: This option is wrong — electronegativity is also read off the periodic table, so it is still a prediction about the compound, not evidence from it.

CThe compound’s chemical formula, showing which elements combine in it

Feedback if chosen: This option is wrong — a formula names which elements are present, which is the same element-identity information the starting classification already uses.

DMeasured properties of the compound, such as melting behavior and conductivity ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Measured properties come from the actual compound, so they are the best evidence for characterizing its bonding.

Check — multiple choice

Element types predict that compound Z, made of a metal and a nonmetal, is ionic. Careful measurements of Z point the other way. What should the chemist conclude?

ATrust the measurements — they come from the actual compound, and the rule is only a starting classification ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The element-type rule is a prediction and is generally right — but measured properties of the real compound are the best evidence, so they carry the day when the two disagree.

BTrust the element-type rule — the periodic table is more fundamental than any measurement

Feedback if chosen: This option is wrong — you ranked the prediction above the evidence. The rule says generally ionic because compounds like Z exist; the measurement of the actual compound wins.

CRe-check the electronegativities — electronegativity evidence outranks measurements

Feedback if chosen: This option is wrong — electronegativity is another prediction from the elements, not evidence from the compound, so it cannot outrank the measurements.

DConclude that the measurements must have been performed incorrectly, since the rule cannot fail

Feedback if chosen: This option is wrong — you dismissed evidence to save a prediction. Careful measurements of the actual compound are the best evidence there is about its bonding.

U2-S01 · LESSON 5 · CSP-005

Polar and effectively nonpolar bonds

Outcome: Classify a covalent bond as polar or effectively nonpolar from the electronegativities of its atoms, without using a numerical cutoff.

Activation question — asked before the lesson opens

Activates: Unit 1: the electronegativity trends

How does electronegativity change across the periodic table?

AIt decreases across a period and decreases up a group.

Why: Both directions are reversed — the trends run toward fluorine, not away from it.

BIt increases across a period and increases up a group. ✓ KEY — this is the correct answer

Why: Both trends point the same way — toward fluorine, the most electronegative element.

CIt increases across a period but decreases up a group.

Why: The group trend is the other way: electronegativity increases going up.

DIt decreases across a period but increases up a group.

Why: The period trend is the other way: electronegativity increases going across.

VIDEO SLOT: APCHEM-U02-L05 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Equal pull, unequal pull

You’ve already seen that electronegativity is a measure of an atom’s tendency to attract shared electrons toward itself within a covalent bond.

You’ve already seen the trends: electronegativity increases across a period and up a group.

Here are the electronegativities of the four elements in this lesson’s examples.

Electronegativities
Element	Electronegativity
H	2.20
C	2.55
Cl	3.16
F	3.98

Start with Cl–Cl: the two atoms are identical, so they attract the shared pair equally, and the electron density sits evenly between them.

A bond with evenly shared electron density is nonpolar.

F–F is the same story: identical atoms, equal pull, evenly shared — nonpolar.

Now H–Cl: two different atoms, and the table shows their electronegativities are clearly unequal — Cl attracts the shared pair much more strongly than H.

Because Cl pulls harder, the shared electron density shifts toward Cl — the bond is polar.

Now the C–H bonds in methane: two different elements again — but the table shows their electronegativities are nearly the same.

The pull is nearly equal, so the shared electron density barely shifts: C–H is effectively nonpolar.

Part 2The rule

Atoms of similar electronegativity — identical atoms included — share nearly equally: classify the bond as effectively nonpolar.

Atoms of clearly unequal electronegativity share unevenly: classify the bond as polar.

There is no numerical cutoff to memorize — compare the two electronegativities and judge whether the difference is small or large.

What you are expected to be able to do

Classify a covalent bond as polar when its atoms’ electronegativities are clearly unequal, and as effectively nonpolar when they are similar — no numerical cutoff.

Part 3Check what you just learned

Check — multiple choice

Classify the I–I bond in I₂, shown in the diagram.

APolar

Feedback if chosen: This option is wrong — you treated identical atoms as pulling unequally. Two I atoms attract the shared pair equally, so the electron density does not shift.

BEffectively nonpolar ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Identical atoms have identical electronegativities — equal pull, evenly shared, effectively nonpolar.

CIonic

Feedback if chosen: This option is wrong — you classified a shared pair as ionic. I₂ is two nonmetals sharing a pair, a covalent bond; the question is only whether that bond is polar.

DIt cannot be classified without a numerical cutoff

Feedback if chosen: This option is wrong — you reached for a cutoff that does not exist. Identical atoms settle it directly: equal pull, effectively nonpolar.

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
P	2.19
Br	2.96
F	3.98

Using the table shown, which of these bonds is polar?

AF–F

Feedback if chosen: This option is wrong — the two F atoms are identical, so they pull the shared pair equally: effectively nonpolar.

BBr–Br

Feedback if chosen: This option is wrong — the two Br atoms are identical, so they pull the shared pair equally: effectively nonpolar.

CH–F ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The table shows H and F clearly unequal (2.20 against 3.98), so the shared electron density shifts strongly toward F: polar.

DP–H

Feedback if chosen: This option is wrong — you treated different elements as automatically polar. The table shows P and H nearly the same (2.19 against 2.20), so P–H is effectively nonpolar.

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
P	2.19

Using the table shown, classify the P–H bond in PH₃.

APolar — P and H are different elements

Feedback if chosen: This option is wrong — you classified by element identity. Different elements can still pull nearly equally; the table shows 2.19 against 2.20.

BPolar — P is a larger atom than H

Feedback if chosen: This option is wrong — you classified by atom size. Polarity follows the electronegativities, and the table shows P and H nearly equal.

CIonic — H acts as a metal here

Feedback if chosen: This option is wrong — hydrogen is a nonmetal, and P–H is a shared pair between two nonmetals: a covalent bond, to be classified as polar or effectively nonpolar.

DEffectively nonpolar — the two electronegativities are nearly equal ✓ KEY — this is the correct answer

Feedback if chosen: Correct. P (2.19) and H (2.20) pull almost identically, so the shared electron density barely shifts: effectively nonpolar.

U2-S01 · LESSON 6 · CSP-006

Bond dipoles and partial charges

Outcome: State that a bond dipole is a separation of partial charge across a polar bond, with δ+ on the electron-poor end and δ− on the electron-rich end.

Activation question — asked before the lesson opens

Activates: CSP-005: the DIRECTION of the shift — toward the more electronegative atom (δ placement builds on it; direction is not tested by CSP-005's checks)

In H–F, fluorine is the more electronegative atom. Where does the shared electron density sit?

AShifted toward the fluorine atom ✓ KEY — this is the correct answer

Why: The shared density shifts toward the more electronegative atom — fluorine.

BShifted toward the hydrogen atom

Why: Hydrogen is the less electronegative atom, so the density shifts away from it.

CExactly halfway between the two atoms

Why: Halfway is even sharing, and even sharing needs two atoms that pull equally; H and F pull unequally.

DEntirely on the fluorine atom, with hydrogen losing it completely

Why: That would be a transfer, forming ions; in a polar bond the pair is still shared, only shifted.

VIDEO SLOT: APCHEM-U02-L06-SCRIPT — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Two ends of a polar bond

You’ve already seen that in a polar bond the shared electron density shifts toward the more electronegative atom.

Here is H–Cl again, now with its two ends labeled.

Remember, electrons are negatively charged.

So the Cl end, where the shared electron density shifts to, becomes slightly electron-rich — slightly negative.

The H end, left with less of the shared electron density, becomes slightly electron-poor — slightly positive.

Slightly is the point: each end carries only a fraction of a full charge, called a partial charge.

Partial charges are written with a lowercase delta: δ− on the electron-rich end, δ+ on the electron-poor end.

This separation of partial charge across a polar bond is called a bond dipole.

A partial charge is not an ion’s charge: the H in H–Cl carries δ+, but it has not lost its electron and has not become an ion.

What you are expected to be able to do

State that a bond dipole is a separation of partial charge across a polar bond, with δ+ on the electron-poor end and δ− on the electron-rich end.

Part 2Check what you just learned

Check — multiple choice

The H–F bond is polar, with the shared electron density shifted toward F. Which labeling is correct?

AH(δ−)–F(δ+)

Feedback if chosen: This option is wrong — you swapped the two ends. The electron density shifted toward F, so F is the electron-rich end: δ− belongs on F.

BH(δ+)–F(δ−) ✓ KEY — this is the correct answer

Feedback if chosen: Correct. F holds more of the shared electron density, so it is electron-rich (δ−), and H is left electron-poor (δ+).

CH(+1)–F(−1)

Feedback if chosen: This option is wrong — you wrote full ion charges. Neither atom has gained or lost an electron; each end carries only a fraction of a charge, written δ+ and δ−.

DNo charge labels — the molecule is neutral overall

Feedback if chosen: This option is wrong — you let the molecule’s overall neutrality erase the bond dipole. HF is neutral overall and its two ends still carry opposite partial charges.

Check — multiple choice

In N–H bonds, N is the more electronegative atom. What does the δ− written beside N represent?

AN has gained H’s electron completely, becoming an ion with a 1− charge

Feedback if chosen: This option is wrong — you read δ− as an ion’s charge. No electron has moved completely; the shared density has only shifted, leaving a fraction of a charge.

BThe whole molecule carries a negative charge

Feedback if chosen: This option is wrong — you moved the charge from one end of the bond to the whole molecule. δ− marks one end of one bond; the molecule stays neutral overall.

CN’s end of the bond is slightly electron-rich — a partial negative charge ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The shared electron density sits closer to N, so N’s end is slightly electron-rich — a partial negative charge, written δ−.

DN has more unshared electron pairs than H

Feedback if chosen: This option is wrong — you counted dots instead of following the shared density. δ− marks where the shared electron density has shifted, not how many unshared pairs an atom has.

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
C	2.55
I	2.66
Cl	3.16
F	3.98

Using the table shown, which of these bonds has a bond dipole?

AF–F

Feedback if chosen: This option is wrong — F–F joins identical atoms, so the sharing is even: no separation of partial charge, no bond dipole.

BC–H

Feedback if chosen: This option is wrong — C and H pull nearly equally, so C–H is effectively nonpolar: the electron density barely shifts, and there is no meaningful bond dipole.

CI–I

Feedback if chosen: This option is wrong — I–I joins identical atoms, so the sharing is even and no partial charges separate.

DH–Cl ✓ KEY — this is the correct answer

Feedback if chosen: Correct. H–Cl is polar: the shared electron density shifts toward Cl, separating δ+ on H from δ− on Cl — a bond dipole.

U2-S01 · LESSON 7 · CSP-007

Which way a bond dipole points

Outcome: Predict that a bond-dipole arrow points toward the more electronegative atom.

No activation question for this lesson — recap (δ+/δ− placement) is exactly what CSP-006's checks just tested.

VIDEO SLOT: APCHEM-U02-L07 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1The dipole drawn as an arrow

You’ve already seen that a polar bond carries a bond dipole: δ+ on the electron-poor end and δ− on the electron-rich end.

Here is H–Cl with its bond dipole drawn as an arrow.

Because Cl is the more electronegative atom, the shared electron density shifts toward Cl — and the arrow points the same way.

The arrow points toward the more electronegative atom: the δ− end of the bond.

To predict the direction for any polar bond, compare the two electronegativities and point the arrow at the more electronegative atom.

What you are expected to be able to do

Predict that a bond-dipole arrow points toward the more electronegative atom.

Part 2Check what you just learned

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
Br	2.96

Using the table shown, which statement gives the direction of the H–Br bond-dipole arrow and the reason?

AToward Br, because Br is more electronegative than H ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The table shows Br (2.96) above H (2.20), so the shared electron density shifts toward Br and the arrow points toward Br.

BToward H, because H is more electronegative than Br

Feedback if chosen: This option is wrong — you ranked H above Br. Check the table: Br is 2.96 and H is 2.20, so Br is the more electronegative atom.

CToward Br, because Br is the larger atom

Feedback if chosen: This option is wrong — you reasoned from atom size. The arrow does point toward Br, but because Br is more electronegative, not because it is larger.

DToward H, because the arrow points at the δ+ end

Feedback if chosen: This option is wrong — you aimed the arrow at the wrong end. The arrow points at the δ− end, the more electronegative atom — here, Br.

Check — multiple choice

Electronegativities
Element	Electronegativity
Cl	3.16
F	3.98
H	2.20
N	3.04
O	3.44

Using the table shown, in which bond does the bond-dipole arrow point toward the N atom?

AN–F

Feedback if chosen: This option is wrong — the table puts F (3.98) above N (3.04), so in N–F the arrow points toward F, away from N.

BN–H ✓ KEY — this is the correct answer

Feedback if chosen: Correct. N (3.04) is more electronegative than H (2.20), so in N–H the arrow points toward N.

CN–O

Feedback if chosen: This option is wrong — the table puts O (3.44) above N (3.04), so in N–O the arrow points toward O, away from N.

DN–Cl

Feedback if chosen: This option is wrong — you judged by position instead of the supplied values. The table shows Cl at 3.16 and N at 3.04, so Cl pulls the shared electrons harder and the arrow points toward Cl, not N.

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
C	2.55
N	3.04
O	3.44
Cl	3.16
F	3.98

Using the table shown, in which of these bonds does the bond-dipole arrow point toward the atom written first?

AC–F

Feedback if chosen: This option is wrong — F (3.98) outpulls C (2.55), so the arrow points toward F, the atom written second.

BH–Cl

Feedback if chosen: This option is wrong — Cl (3.16) outpulls H (2.20), so the arrow points toward Cl, the atom written second.

CO–H ✓ KEY — this is the correct answer

Feedback if chosen: Correct. O (3.44) is more electronegative than H (2.20), so the arrow points toward O — the atom written first.

DN–F

Feedback if chosen: This option is wrong — F (3.98) outpulls N (3.04), so the arrow points toward F, the atom written second.

U2-S01 · LESSON 8 · CSP-008

Electronegativity difference and dipole size

Outcome: State that, for comparable single bonds, a larger electronegativity difference produces a larger bond dipole.

No activation question for this lesson — recap (a polar bond carries a bond dipole) is exercised by every CSP-007 check.

VIDEO SLOT: APCHEM-U02-L08 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Comparing two dipoles

You’ve already seen that a polar bond carries a bond dipole — a separation of partial charge across the bond.

Compare two bonds built on the same H atom: H–F and H–Cl.

Electronegativities
Element	Electronegativity
H	2.20
Cl	3.16
F	3.98

F attracts the shared electron density more strongly than Cl does — its electronegativity is higher.

So the electronegativity difference across H–F is larger than the electronegativity difference across H–Cl.

As the electronegativity difference gets larger, the bond dipole gets larger: H–F has the larger bond dipole.

The comparison is fair because the bonds are comparable — both are single bonds (bonds sharing one pair), and both share the same H atom.

What you are expected to be able to do

State that, for comparable single bonds, a larger electronegativity difference produces a larger bond dipole.

Part 2Check what you just learned

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
N	3.04
O	3.44

Using the table shown, which bond has the larger bond dipole: N–H or O–H?

AO–H ✓ KEY — this is the correct answer

Feedback if chosen: Correct. O–H spans the larger electronegativity difference (3.44 against 2.20, versus 3.04 against 2.20 for N–H), and a larger difference produces a larger bond dipole.

BN–H

Feedback if chosen: This option is wrong — you picked the smaller electronegativity difference. N (3.04) sits closer to H (2.20) than O (3.44) does, so N–H has the smaller dipole.

CThey are equal, because both bonds contain H

Feedback if chosen: This option is wrong — sharing an atom does not equalize the dipoles. The other atom differs, so the electronegativity differences differ, and the dipoles differ with them.

DIt cannot be decided without a numerical cutoff

Feedback if chosen: This option is wrong — no cutoff is needed to compare. Just compare the two electronegativity differences: the larger one belongs to O–H.

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
Br	2.96
F	3.98

Using the table shown, which statement correctly compares the bond dipoles of H–F and H–Br?

AH–Br has the larger dipole, because Br is the larger and heavier atom

Feedback if chosen: This option is wrong — you ranked by atom size. Dipole size follows the electronegativity difference, and F sits much further from H than Br does.

BH–Br has the larger dipole, because its atoms' electronegativities are more closely matched

Feedback if chosen: This option is wrong — you reversed the rule. Closely matched electronegativities mean more even sharing and a smaller dipole; the larger difference makes the larger dipole.

CH–F has the larger dipole, because its electronegativity difference is larger ✓ KEY — this is the correct answer

Feedback if chosen: Correct. F (3.98) sits much further from H (2.20) than Br (2.96) does, so H–F spans the larger electronegativity difference and has the larger bond dipole.

DThey are equal, because both are single bonds from hydrogen to a halogen

Feedback if chosen: This option is wrong — being the same kind of bond makes them comparable, not equal. Their electronegativity differences still differ, so their dipoles do too.

U2-S01 · LESSON 9 · CSP-009

Why a larger difference makes a larger dipole

Outcome: Explain that a larger electronegativity difference produces a larger bond dipole because the shared electron density is distributed more unevenly.

No activation question for this lesson — recap (larger difference → larger dipole) is exactly what CSP-008's checks just tested.

VIDEO SLOT: APCHEM-U02-L09 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1What the difference does to the shared electrons

You’ve already seen that, for comparable single bonds, a larger electronegativity difference produces a larger bond dipole.

Here is why. Both atoms attract the shared electron density, and the more electronegative atom attracts it more strongly.

When the two electronegativities are close, the contest is nearly even, and the shared electron density shifts only slightly.

As the electronegativity difference grows, the contest becomes more one-sided, and the shared electron density is distributed more and more unevenly.

A more uneven distribution piles more negative charge at one end and leaves less at the other — larger partial charges, and so a larger bond dipole.

That is the H–F versus H–Cl comparison: F shifts the shared electron density toward itself more strongly than Cl does, so H–F ends up with the larger bond dipole.

What you are expected to be able to do

Explain that a larger electronegativity difference produces a larger bond dipole because the shared electron density is distributed more unevenly.

Part 2Check what you just learned

Check — multiple choice

Why does H–F have a larger bond dipole than H–Cl?

AThe H–F bond is shorter, and shorter bonds always have larger dipoles

Feedback if chosen: This option is wrong — you reasoned from bond length. The dipole is larger because the shared electron density is pulled more unevenly toward F, not because of how long the bond is.

BF attracts the shared electron density more strongly than Cl does, so the sharing in H–F is more uneven ✓ KEY — this is the correct answer

Feedback if chosen: Correct. F outpulls H by more than Cl does, so the shared electron density in H–F is distributed more unevenly — larger partial charges, larger bond dipole.

CF has more valence electrons than Cl

Feedback if chosen: This option is wrong — F and Cl sit in the same group and have the same number of valence electrons. The difference is how strongly each attracts the shared density.

DThe H atom in H–F holds its own electron more tightly than the H atom in H–Cl

Feedback if chosen: This option is wrong — you put the cause on H. H is the same atom in both bonds; what changes is how strongly the other atom pulls the shared electron density away.

Check — multiple choice

Two single bonds are compared: X–Y with a small electronegativity difference, and X–Z with a large one. Which statement describes the shared electron density?

AIn X–Z it is distributed more unevenly, giving X–Z the larger bond dipole ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The larger electronegativity difference in X–Z makes the pull more one-sided, so its shared electron density is distributed more unevenly — the larger bond dipole.

BIn X–Y it is distributed more unevenly, because similar atoms compete harder for the electrons

Feedback if chosen: This option is wrong — you inverted the relationship. When the atoms pull almost equally, the density ends up nearly centered; it is the unequal contest that shifts it.

CIt is distributed evenly in both bonds, because every covalent bond shares equally

Feedback if chosen: This option is wrong — covalent bonds share, but not always equally. Unequal electronegativities distribute the shared density unevenly, which is exactly what a bond dipole is.

DIn X–Z it leaves the bond entirely and transfers to Z

Feedback if chosen: This option is wrong — you turned a shift into a transfer. In a polar bond the electron density is still shared; it has moved toward Z, not onto Z.

U2-S01 · LESSON 10 · CSP-010

Ranking bond dipoles

Outcome: Rank comparable single bonds by bond-dipole size using supplied electronegativities.

No activation question for this lesson — recap (same rule) is exercised by CSP-009's checks.

VIDEO SLOT: APCHEM-U02-L10 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1From comparing two to ranking several

You’ve already seen that, for comparable single bonds, a larger electronegativity difference produces a larger bond dipole.

Here are supplied electronegativities for ranking three bonds: H–F, H–Cl, and H–Br.

Electronegativities
Element	Electronegativity
H	2.20
Br	2.96
Cl	3.16
F	3.98

Work out each electronegativity difference in turn: H–F differs by 1.78, H–Cl by 0.96, H–Br by 0.76.

As the differences decrease from H–F to H–Cl to H–Br, the bond dipoles decrease in the same order: H–F > H–Cl > H–Br.

The greater-than signs compare bond-dipole size, largest first.

What you are expected to be able to do

Rank comparable single bonds by bond-dipole size using supplied electronegativities.

Part 2Check what you just learned

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
N	3.04
O	3.44
F	3.98

Using the table shown, rank N–H, O–H, and F–H from largest bond dipole to smallest.

AN–H > O–H > F–H

Feedback if chosen: This option is wrong — you ranked the smallest electronegativity difference as the largest dipole. The differences are F–H 1.78, O–H 1.24, N–H 0.84: largest difference, largest dipole.

BF–H > O–H > N–H ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The differences are F–H 1.78, O–H 1.24, and N–H 0.84 — the dipoles rank in the same order.

CF–H > N–H > O–H

Feedback if chosen: This option is wrong — you swapped N and O. Check the table: O (3.44) sits above N (3.04), so O–H spans the larger difference and outranks N–H.

DO–H > F–H > N–H

Feedback if chosen: This option is wrong — you ranked O above F. Check the table: F (3.98) sits above O (3.44), so F–H spans the largest difference of the three.

Check — multiple choice

Electronegativities
Element	Electronegativity
P	2.19
H	2.20
S	2.58
N	3.04

Using the table shown, rank N–H, S–H, and P–H from largest bond dipole to smallest.

AP–H > S–H > N–H

Feedback if chosen: This option is wrong — you ranked by atom size, largest atom first. Dipole size follows the electronegativity difference: N–H 0.84, S–H 0.38, P–H 0.01.

BS–H > P–H > N–H

Feedback if chosen: This option is wrong — you ranked by electron count, heaviest atom first. Dipole size follows the electronegativity difference: N–H 0.84, S–H 0.38, P–H 0.01.

CN–H > S–H > P–H ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The differences are N–H 0.84, S–H 0.38, and P–H 0.01 — the dipoles rank in the same order.

DThey tie, because each is a single bond from hydrogen to a nonmetal

Feedback if chosen: This option is wrong — being the same kind of bond makes them comparable, not equal. Their electronegativity differences differ, so their dipoles rank apart.

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
I	2.66
Br	2.96
Cl	3.16

Using the table shown, rank H–Cl, H–Br, and H–I from largest bond dipole to smallest.

AH–Cl > H–Br > H–I ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The differences are H–Cl 0.96, H–Br 0.76, and H–I 0.46 — the dipoles rank in the same order.

BH–I > H–Br > H–Cl

Feedback if chosen: This option is wrong — you ranked by atom size, largest halogen first. The largest dipole belongs to the largest electronegativity difference, which is H–Cl at 0.96.

CH–Br > H–Cl > H–I

Feedback if chosen: This option is wrong — you ranked Br above Cl. Check the table: Cl (3.16) sits above Br (2.96), so H–Cl spans the larger difference.

DThey cannot be ranked without measuring each dipole

Feedback if chosen: This option is wrong — no measurement is needed. The supplied electronegativities rank the differences directly: 0.96, 0.76, 0.46.

U2-S01 · LESSON 11 · CSP-011

The continuum between covalent and ionic bonding

Outcome: Explain that polar bonds have some ionic character, and that ionic and covalent bonding form a continuum with no sharp boundary.

Activation question — asked before the lesson opens

Activates: CSP-005/CSP-006: evenly vs unevenly shared electron density

In Cl–Cl the shared electron density sits evenly between the two atoms; in H–Cl it shifts toward Cl. How are the two bonds classified?

ACl–Cl is polar, and H–Cl is effectively nonpolar.

Why: It is the shifted density that makes a bond polar — that is H–Cl, not Cl–Cl.

BBoth bonds are polar.

Why: Cl–Cl's two identical atoms pull equally, so its density sits evenly — not polar.

CCl–Cl is effectively nonpolar, and H–Cl is polar. ✓ KEY — this is the correct answer

Why: Even sharing is effectively nonpolar; a shifted share is polar.

DBoth bonds are effectively nonpolar.

Why: H–Cl's density has shifted toward Cl, and a shifted share is what polar means.

VIDEO SLOT: APCHEM-U02-L11 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Between two ends

You’ve already seen the two ways a shared pair can sit: evenly shared in an effectively nonpolar bond, unevenly shared in a polar bond.

Whether two elements bond in a more ionic or a more covalent way follows from how the electron density between them is distributed.

Here are four bonds placed on one line, from even sharing at one end to a nearly complete handover of the electron density at the other.

Cl–Cl sits at the covalent end: the electron density is shared evenly.

H–Cl and H–F sit partway along: their shared electron density has shifted toward Cl or F — partly even sharing, partly the beginnings of ionic charge separation.

That partial charge separation is described as ionic character, and every polar bond has some.

As the electronegativity difference between the atoms grows, the electron density shifts further, and the bond has more ionic character: H–F has more ionic character than H–Cl.

NaCl sits near the ionic end — but even there the handover is not total; no bond is perfectly ionic.

So ionic and covalent are not two boxes separated by a sharp boundary — they are the two ends of a continuum.

What you are expected to be able to do

Explain that polar bonds have some ionic character and that ionic and covalent bonding form a continuum — compare bonds as having more or less ionic character, never with a cutoff.

Part 2Check what you just learned

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
Br	2.96
Cl	3.16

Using the table shown, which bond has more ionic character: H–Cl or H–Br?

AH–Br, because Br is the larger atom with the bigger electron cloud

Feedback if chosen: This option is wrong — you ranked by atom size. Ionic character follows how far the electron density has shifted, which follows the electronegativity difference.

BNeither — covalent bonds have no ionic character

Feedback if chosen: This option is wrong — you kept covalent and ionic in separate boxes. Both bonds are polar, and every polar bond carries some ionic character.

CBoth are fully ionic, because both bonds are polar with a δ+ and a δ−

Feedback if chosen: This option is wrong — you pushed polar all the way to ionic. A polar bond still shares its electron density; it has some ionic character, not full ionic bonding.

DH–Cl, because its electronegativity difference is larger ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Cl (3.16) sits further from H (2.20) than Br (2.96) does, so H–Cl’s electron density shifts further — more ionic character.

Check — multiple choice

Which statement describes how ionic and covalent bonding are related?

AEvery bond is either purely ionic or purely covalent, with nothing in between the two types

Feedback if chosen: This option is wrong — you sorted every bond into two pure boxes. Real bonds sit along a continuum, and most are somewhere between the two ends.

BA bond becomes ionic once the electronegativity difference passes a set cutoff value

Feedback if chosen: This option is wrong — you drew a sharp boundary where there is none. No cutoff value separates ionic from covalent; bonds simply have more or less ionic character.

CThey are two ends of a continuum: bonds differ in how unevenly the electron density is shared, with no sharp boundary ✓ KEY — this is the correct answer

Feedback if chosen: Correct. From even sharing to nearly complete handover, bonding shifts by degree — a continuum with no sharp boundary between covalent and ionic.

DA bond’s place between ionic and covalent depends only on whether a metal is present

Feedback if chosen: This option is wrong — you promoted the element-type rule into the full story. Metal–nonmetal pairs are generally more ionic, but the bond’s place on the continuum follows the electron-density distribution.

Check — multiple choice

HF and HCl are both polar. Which sentence uses the continuum idea correctly?

AHF is a fully ionic compound

Feedback if chosen: This option is wrong — you pushed a polar bond to the ionic end. HF’s electron density is strongly shifted toward F, but it is still shared: more ionic character, not fully ionic.

BHF has more ionic character than HCl ✓ KEY — this is the correct answer

Feedback if chosen: Correct. HF’s larger electronegativity difference shifts its electron density further, so HF sits closer to the ionic end than HCl — more ionic character, with neither bond fully ionic.

CHCl has no ionic character, because it is covalent

Feedback if chosen: This option is wrong — you treated covalent as excluding ionic character. HCl is polar, and every polar bond carries some ionic character.

DHF and HCl have identical ionic character, because both contain H

Feedback if chosen: This option is wrong — sharing an atom does not equalize the bonds. The other atom differs, so the electron-density shift differs, and so does the ionic character.

U2-S02 — Ionic and metallic particulate models

10 lessons

U2-S02 · LESSON 12 · CSP-012

Cations and anions

Outcome: Classify an ion as a cation or an anion from the sign of its charge.

Activation question — asked before the lesson opens

Activates: Unit 1: an atom becomes an ion by losing or gaining electrons (the recap's Na → Na⁺, Cl → Cl⁻ point; CSP-011's checks tested the ionic–covalent continuum, not ion formation)

A neutral fluorine atom gains one electron. What does the atom become?

AAn F⁺ ion

Why: The atom gained a negative electron, so its charge has gone down, not up — a positive ion forms when an atom loses electrons.

BA fluorine atom that is still neutral

Why: A neutral atom has equal numbers of electrons and protons; with one extra electron the counts no longer match, so the atom is now charged — an ion.

CA neon atom

Why: The element is fixed by the number of protons, and that has not changed; the extra electron changes the atom's charge, not its identity.

DAn F⁻ ion ✓ KEY — this is the correct answer

Why: With one more electron than protons, the atom carries a single negative charge: it has become the F⁻ ion.

VIDEO SLOT: APCHEM-U02-S02-L01 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Two names for ions

You’ve already seen that an atom becomes an ion when it loses or gains electrons — Na loses one electron to become Na⁺, and Cl gains one to become Cl⁻.

Chemists sort ions into two groups by the sign of the charge, and each group has a name.

A positively charged ion is a cation. Na⁺ is a cation.

A negatively charged ion is an anion. Cl⁻ is an anion.

Part 2The sign decides

Ca²⁺ carries a 2+ charge. The charge is larger than 1+, but it is still positive, so Ca²⁺ is still a cation.

O²⁻ carries a 2− charge. Negative, so O²⁻ is an anion.

H⁺ is a cation too. Hydrogen is a nonmetal, but the name follows the ion’s charge, not the element.

A neutral atom such as Ne carries no charge, so it is not an ion at all — neither name applies.

What you are expected to be able to do

Classify any positively charged ion as a cation and any negatively charged ion as an anion, using only the sign of the charge.

Part 3Check what you just learned

Check — multiple choice

Which one of these ions is a cation?

AS²⁻

Feedback if chosen: This option is wrong — you picked a negatively charged ion. S²⁻ carries a negative charge, so it is an anion; a cation carries a positive charge.

BBr⁻

Feedback if chosen: This option is wrong — you picked a negatively charged ion. Br⁻ carries a negative charge, so it is an anion.

CAl³⁺ ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Al³⁺ carries a positive charge, and a positively charged ion is a cation.

DO²⁻

Feedback if chosen: This option is wrong — you picked a negatively charged ion. O²⁻ carries a negative charge, so it is an anion.

Check — multiple choice

A student writes down four ions: K⁺, Mg²⁺, N³⁻, and Li⁺. How many of them are anions?

Feedback if chosen: This option is wrong — you missed the negative charge on N³⁻. An anion is a negatively charged ion, and N³⁻ is one.

B1 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Only N³⁻ carries a negative charge, so exactly one of the four is an anion; K⁺, Mg²⁺, and Li⁺ are cations.

Feedback if chosen: This option is wrong — you counted a cation as an anion. Only N³⁻ carries a negative charge.

Feedback if chosen: This option is wrong — you counted the cations. Three of these ions are positively charged, which makes them cations, not anions.

Check — multiple choice

Which one of these ions is an anion?

ABa²⁺

Feedback if chosen: This option is wrong — you picked a positive ion. The 2+ charge makes Ba²⁺ a cation; an anion carries a negative charge.

BH⁺

Feedback if chosen: This option is wrong — you may have sorted by element type. Hydrogen is a nonmetal, but the name follows the charge, and H⁺ is positive: a cation.

CN³⁻ ✓ KEY — this is the correct answer

Feedback if chosen: Correct. N³⁻ carries a negative charge, and every negatively charged ion is an anion — whatever the size of the charge.

DAg⁺

Feedback if chosen: This option is wrong — you picked a positive ion. Ag⁺ is a cation; anions are the negatively charged ions.

U2-S02 · LESSON 13 · CSP-013

Particulate models

Outcome: State what a particulate model shows — particle identities, positions, and charges — and read a labeled circle as one specific ion.

No activation question for this lesson — no 'You've already seen' recap — the lesson opens fresh, with the grain-of-salt problem.

VIDEO SLOT: APCHEM-U02-S02-L02 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Too small to see, too many to draw

A single grain of salt contains more ions than you could draw in a lifetime, and each one is far too small to see.

To show what a solid is made of, chemists draw a tiny fragment of it, particle by particle.

Here is such a drawing for solid sodium chloride.

Each circle stands for one particle — here, one ion.

The label on a circle tells you the particle’s identity and its charge: Na⁺ is a sodium cation, Cl⁻ is a chloride anion.

The circles’ positions show how the particles are arranged — which particles sit next to which.

A drawing like this is called a particulate model.

Part 2What the model does not show

The model is drawn at the scale of the particles, not at a scale you can see — the real grain contains vastly more ions than any drawing could hold.

It also does not show the solid’s appearance — no color, no shine.

What you are expected to be able to do

State that a particulate model represents the identities, positions, and charges of particles rather than showing a substance at visible scale, and read a labeled circle as one specific ion.

Part 3Check what you just learned

Check — multiple choice

In the model of solid potassium bromide shown, what does one circle labeled Br⁻ represent?

AOne pair of potassium and bromide ions

Feedback if chosen: This option is wrong — you merged two particles into one circle. One circle is one particle: here, a single Br⁻ ion.

BOne grain of solid potassium bromide

Feedback if chosen: This option is wrong — you read the model at visible scale. A grain contains vastly more ions than the model shows; each circle is a single ion.

COne molecule of KBr

Feedback if chosen: This option is wrong — you read the circle as a molecule. Each circle is one particle, and in this solid the particles are single ions.

DOne bromide ion ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each circle is one particle — this one is a single Br⁻ anion.

Check — multiple choice

The model of solid magnesium oxide shown has 12 circles. How many ions does a real speck of the solid contain?

AExactly the 12 shown in the model

Feedback if chosen: This option is wrong — you read the model as a complete picture. It is a fragment drawn at particle scale; the real speck holds vastly more ions.

BVastly more — the model draws only a tiny fragment ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A particulate model shows a tiny fragment at particle scale — the real solid contains far more ions than any drawing could hold.

CExactly 24 — each circle stands for a pair of ions, one Mg²⁺ and one O²⁻

Feedback if chosen: This option is wrong — twice over. Each circle stands for one ion, not a pair; and the model draws only a tiny fragment, so the real speck holds vastly more ions than any count of circles.

DNone — the circles stand for something other than ions

Feedback if chosen: This option is wrong — you took the circles to be decoration. In a particulate model each circle stands for one ion; the model's limitation is its tiny sample size, not its symbols.

Check — multiple choice

Which information can you read from the model of solid lithium fluoride shown?

AThe exact number of ions in the whole solid

Feedback if chosen: This option is wrong — you treated the fragment as a census. The model shows a small sample of the arrangement; the real solid contains vastly more ions than it draws.

BThe color of the solid and how brightly it shines

Feedback if chosen: This option is wrong — you asked the model for appearance. A particulate model shows particles and their arrangement, not what the solid looks like to the eye.

CThe size of the solid as it would sit on your bench

Feedback if chosen: This option is wrong — you read the model at visible scale. It is drawn at particle scale and shows only a tiny fragment.

DThe ions present, their charges, and their arrangement ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A particulate model carries exactly this: which particles are present (from the labels), their charges (from the labels), and how they sit relative to one another (from the positions).

U2-S02 · LESSON 14 · CSP-014

The crystal lattice

Outcome: State that a crystal lattice is a regular, repeating, three-dimensional arrangement of particles.

No activation question for this lesson — no 'You've already seen' recap — the lesson opens with its own question about the NaCl model.

VIDEO SLOT: APCHEM-U02-S02-L03 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1A pattern, not a pile

Inside solid sodium chloride, are the ions piled up at random?

Here is the particulate model of solid sodium chloride again.

The arrangement is regular: the same alternation of Na⁺ and Cl⁻ repeats, row after row and column after column.

The real solid keeps this repetition going in three dimensions — the model shows one flat layer of it.

A regular, repeating, three-dimensional arrangement of particles is called a crystal lattice.

Part 2No separate molecules

Here is a drawing that gets solid sodium chloride wrong: it shows the solid as separate Na⁺Cl⁻ pairs.

Solid NaCl contains no such pairs — no Na⁺ belongs to any single Cl⁻.

The lattice simply repeats: there is no separate NaCl molecule anywhere in the solid.

What you are expected to be able to do

State that a crystal lattice is a regular, repeating, three-dimensional arrangement of particles, and that an ionic solid such as NaCl repeats throughout with no separate molecules.

Part 3Check what you just learned

Check — multiple choice

Which arrangement of particles is a crystal lattice?

AA single giant molecule with a fixed number of atoms

Feedback if chosen: This option is wrong — you turned the solid into a molecule. A lattice has no fixed particle count; the pattern simply repeats for as far as the solid extends.

BA random jumble packed as tightly as possible

Feedback if chosen: This option is wrong — you kept the packing but dropped the pattern. A lattice is not a jumble: the same arrangement repeats regularly throughout.

CSeparate two-particle pairs scattered through the solid

Feedback if chosen: This option is wrong — you broke the solid into pairs. A lattice is one continuous repeating arrangement; no particle belongs to a separate pair.

DThe same pattern repeating in three dimensions ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A crystal lattice is regular and repeating, and the repetition extends through the solid in three dimensions.

Check — multiple choice

Solid lithium fluoride has the structure shown. Why is “a molecule of LiF” the wrong way to describe any part of this solid?

AMolecules exist only in liquids and gases, never in solids

Feedback if chosen: This option is wrong — you made it a matter of state. Plenty of solids are built of molecules; an ionic solid is not, because its lattice repeats without separate units.

BNo Li⁺ is attached to just one F⁻ — the pattern repeats without separate LiF units ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The lattice repeats throughout: every Li⁺ sits among several F⁻ neighbors, and none of them is its private partner, so there is no separate LiF unit to call a molecule.

CThe solid as a whole is too large to be a molecule, but each Li⁺F⁻ pair inside it is one

Feedback if chosen: This option is wrong — you kept the pair idea. No Li⁺ in the lattice belongs to one particular F⁻; each ion sits among several oppositely charged neighbors, so the lattice contains no Li⁺F⁻ pair.

DLiF would need three or more atoms to count as a molecule

Feedback if chosen: This option is wrong — you made molecule a matter of size. Two atoms can form a molecule (Cl₂ does); the reason there is no LiF molecule here is that the lattice has no separate LiF units at all.

U2-S02 · LESSON 15 · CSP-015

Drawing an ionic solid

Outcome: Draw a particulate model of an ionic solid as a repeating arrangement of alternating, labeled ions.

Activation question — asked before the lesson opens

Activates: Unit 1 (Coulomb's law): opposite charges attract, like charges repel — the recap's second 'You've also seen' line; the lattice line itself is exactly what CSP-014's checks just tested

Solid potassium bromide contains K⁺ ions and Br⁻ ions. Which of the ions attract each other?

AOnly the oppositely charged ions: each K⁺ with the Br⁻ ions around it ✓ KEY — this is the correct answer

Why: Opposite charges attract — each positive K⁺ pulls on the negative Br⁻ ions beside it, and that attraction holds the solid together.

BOnly the like-charged neighbors: K⁺ with K⁺, and Br⁻ with Br⁻

Why: Two K⁺ ions carry like charges, and like charges repel; the same goes for two Br⁻ ions.

CAll of the ions: every ion attracts every other ion in the solid

Why: Two K⁺ ions repel each other, and two Br⁻ ions repel each other — like charges never attract.

DNone of the neighbors: the ions exert no forces on each other

Why: Charged particles always exert forces on one another: opposite charges attract, and like charges repel.

VIDEO SLOT: APCHEM-U02-S02-L04 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1The rules of the drawing

You’ve already seen the crystal lattice of solid sodium chloride — now you learn to draw one.

You’ve also seen that opposite charges attract — that attraction is what holds the solid together, so the drawing must put each ion among oppositely charged neighbors.

Rule one: alternate the ions in both directions, so that no two like charges sit side by side.

Rule two: label every ion with its symbol and its charge — Na⁺, Cl⁻.

Rule three: draw each ion at its relative size. In every compound here, the anion really is the larger ion, so draw the anions larger.

Rule four: draw at least three rows and three columns, so the repeating pattern is visible and could clearly continue in every direction.

Keep the grid continuous — no gaps that split the drawing into separate pairs.

The real lattice continues in three dimensions; your flat drawing shows one layer of it, and that is the accepted way to draw it.

What you are expected to be able to do

Draw a particulate model of an ionic solid: a repeating arrangement that places each ion among oppositely charged neighbors, with no two like charges side by side, every ion labeled, and each ion drawn at its relative size — here, anions larger.

Part 2Check what you just learned

Check — drawing task (pen and paper, self-marked)

Solid potassium bromide is made of K⁺ and Br⁻ ions. Take a pen and paper and draw a particulate model of a fragment of solid KBr — at least three rows and three columns of ions. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows K⁺ and Br⁻ alternating in both directions — no two like charges sit side by side.
Your drawing shows every ion labeled with its symbol and charge: K⁺ or Br⁻.
Your drawing shows the Br⁻ ions larger than the K⁺ ions.
Your drawing shows at least three rows and three columns, in one continuous grid — no separate pairs, no gaps.

Check — drawing task (pen and paper, self-marked)

Solid magnesium oxide is made of Mg²⁺ and O²⁻ ions. Take a pen and paper and draw a particulate model of a fragment of solid MgO — at least three rows and three columns of ions. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows Mg²⁺ and O²⁻ alternating in both directions — no two like charges sit side by side.
Your drawing shows every ion labeled with its symbol and charge: Mg²⁺ or O²⁻ — the 2+ and 2− charges change the labels, not the pattern.
Your drawing shows the O²⁻ ions larger than the Mg²⁺ ions.
Your drawing shows at least three rows and three columns, in one continuous grid — no separate pairs, no gaps.

Check — multiple choice

A classmate drew this model of solid lithium fluoride. What must they fix before the drawing is a correct ionic-solid model?

AThe ions carry written charges — the labels must show symbols only

Feedback if chosen: This option is wrong — you flagged a correct feature. Every ion must be labeled with both its symbol and its charge.

BThe F⁻ ions are drawn larger than the Li⁺ ions — the circles must be equal in size

Feedback if chosen: This option is wrong — you flagged a correct feature. In LiF the F⁻ ion really is larger than the Li⁺ ion, so the F⁻ circles should be drawn larger.

CThe drawing shows only a small fragment — it must show every ion in the solid

Feedback if chosen: This option is wrong — you asked the drawing to do the impossible. A model always shows a small fragment; it only has to make the repeating pattern clear.

DLike-charged ions sit side by side — every ion must sit among oppositely charged neighbors ✓ KEY — this is the correct answer

Feedback if chosen: Correct. One Li⁺ and one F⁻ have swapped places in the top row, so like charges sit as nearest neighbors; alternating the ions in both directions removes that.

U2-S02 · LESSON 16 · CSP-016

Attractions and repulsions in an ionic solid

Outcome: Identify how an ionic-solid model maximizes opposite-charge attractions and minimizes like-charge repulsions.

Activation question — asked before the lesson opens

Activates: Unit 1 (Coulomb's law): as the distance between two charges decreases, the force increases — the distance–force part of the recap; CSP-015's checks exercised only the alternation rule

A Na⁺ ion and a Cl⁻ ion attract each other. The two ions drift closer together. What happens to the attraction between them?

AIt gets weaker.

Why: Coulomb's law says the opposite: as the distance between two charges decreases, the force between them increases.

BIt gets stronger. ✓ KEY — this is the correct answer

Why: As the distance between two charges decreases, the force between them increases — so the attraction strengthens as the ions move closer.

CIt stays the same.

Why: The force between two charges depends on the distance between them, so a change in the distance changes the force.

DIt becomes a repulsion.

Why: The direction of the force follows the signs of the charges, and the signs have not changed — opposite charges attract at any distance.

VIDEO SLOT: APCHEM-U02-S02-L05 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1Reading the arrangement

Here is the particulate model of solid sodium chloride.

You’ve already seen Coulomb’s law: opposite charges attract, like charges repel, and as the distance between two charges decreases, the force between them increases.

Pick any Na⁺ in the model. Every one of its nearest neighbors is a Cl⁻.

So the closest pairs — where the force is strongest — are all oppositely charged: the strongest forces in the solid are attractions.

The like-charged ions, Na⁺ and Na⁺ or Cl⁻ and Cl⁻, are never nearest neighbors — they sit farther apart, where the force between them is weaker.

This is what the alternating arrangement achieves: it maximizes the attractions and minimizes the repulsions.

You’ve already seen that ionic bonding is the attraction between oppositely charged ions — and the model shows that each ion is attracted to all of its oppositely charged neighbors, not to one partner.

What you are expected to be able to do

Identify, in a supplied ionic-solid model, how the arrangement maximizes opposite-charge attractions and minimizes like-charge repulsions.

Part 2Check what you just learned

Check — multiple choice

In the model of solid lithium fluoride shown, which ions are the nearest neighbors of each Li⁺ inside the model?

AWhichever ions happened to land beside it

Feedback if chosen: This option is wrong — you read the arrangement as chance. The lattice is a regular repeating pattern, and the pattern always puts oppositely charged ions side by side.

BOther Li⁺ ions only

Feedback if chosen: This option is wrong — you put like charges side by side. Like charges repel; the lattice keeps them apart, and each Li⁺ sits among F⁻ neighbors instead.

CA mix of Li⁺ and F⁻ ions

Feedback if chosen: This option is wrong — you let some like charges in close. In a correct lattice the alternation is complete: every nearest neighbor of a Li⁺ is an F⁻.

DF⁻ ions only ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The alternation puts only oppositely charged ions side by side, so every nearest neighbor of a Li⁺ is an F⁻ — the closest, strongest forces are all attractions.

Check — multiple choice

In the same arrangement, why are the repulsions in the solid weak?

ALike charges are never nearest neighbors, so they sit farther apart, where the force is weaker ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Like charges are never nearest neighbors, so every repulsion acts across a larger distance than the attractions do — and as the distance between two charges increases, the force between them decreases.

BLike-charged ions stop repelling each other once they are locked inside a rigid solid

Feedback if chosen: This option is wrong — you switched Coulomb’s law off inside the solid. Like charges always repel; the lattice just keeps them farther apart, where the force is weaker.

CThe charges of neighboring ions cancel each other out, leaving no force to repel

Feedback if chosen: This option is wrong — you canceled the charges. Each ion keeps its full charge; the arrangement doesn’t erase forces, it makes the attractions closer and stronger than the repulsions.

DThe larger F⁻ ions sit between the Li⁺ ions and absorb the repulsions between them

Feedback if chosen: This option is wrong — you gave the anions a shielding job. Ions don’t absorb forces; the repulsions are weak simply because like charges sit farther apart.

Check — multiple choice

Solid potassium bromide has the arrangement shown. What holds the solid together?

AEach ion is attracted to all of the oppositely charged ions around it ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Ionic bonding is the attraction between oppositely charged ions, and in the lattice every ion is attracted to all of its oppositely charged neighbors — the attraction extends through the whole solid.

BEach K⁺ is bonded only to the single Br⁻ ion it gave its own electron to

Feedback if chosen: This option is wrong — you made the electron transfer the bond. Losing and gaining electrons is how the ions formed; the bonding itself is the attraction between opposite charges, and it ties each ion to all of its neighbors, not to one partner.

CNeighboring ions share pairs of valence electrons between them

Feedback if chosen: This option is wrong — you described a covalent bond. In an ionic solid the electrons are not shared between neighbors; the ions attract each other as whole charged particles.

DThe large Br⁻ ions hold the small K⁺ ions in place by surrounding them

Feedback if chosen: This option is wrong — you made it mechanical. Size doesn’t hold anything in place; the solid is held by the attractions between the opposite charges.

U2-S02 · LESSON 17 · CSP-017

Spotting the flaw in an ionic-solid model

Outcome: Judge whether an ionic-solid model is consistent with a repeating lattice, and name the specific flaw when it is not.

Activation question — asked before the lesson opens

Activates: CSP-013/CSP-015: every circle in a particulate model of an ionic solid carries its label (the piece of the checklist no preceding check keys)

In a particulate model of an ionic solid, what must the label on each circle show?

AThe ion's symbol only

Why: The symbol alone hides the charge, and the charges are what hold the solid together.

BThe ion's charge only

Why: A bare charge does not say which ion it is — Na⁺ and K⁺ would look the same.

CNothing — the circle sizes identify the ions

Why: Sizes alone are ambiguous in a drawing; the model is read from its labels.

DThe ion's symbol and its charge ✓ KEY — this is the correct answer

Why: Each circle is named in full — symbol and charge — so the model can be read without guessing.

VIDEO SLOT: APCHEM-U02-S02-L06 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1What a correct model must show

You’ve already seen what a correct ionic-solid model shows: a repeating pattern, each ion among oppositely charged neighbors, no two like charges side by side, no separate pairs, and every ion labeled.

To judge a model, check it against that list — and when it fails, name the one specific thing that is wrong.

Here is a flawed model of solid sodium chloride.

Check the neighbors: in the top row, one Na⁺ and one Cl⁻ have swapped places — two Na⁺ ions now sit side by side, and so do two Cl⁻ ions.

Like charges repel, and nearest neighbors are where the forces are strongest — so this arrangement adds strong repulsion where the solid should have attraction.

That is the specific flaw: like charges as nearest neighbors. Everything else about the drawing may be fine, but this one swapped pair breaks the model.

What you are expected to be able to do

Evaluate whether a supplied ionic-solid model shows a repeating lattice that maximizes attractions and minimizes repulsions, and identify one specific flaw when it does not.

Part 2Check what you just learned

Check — multiple choice

A student drew this model of solid potassium bromide. What is the specific flaw?

AThe model contains exactly as many K⁺ ions as Br⁻ ions

Feedback if chosen: This option is wrong — you flagged a correct feature. KBr contains one K⁺ for every Br⁻, so equal numbers are exactly right.

BThe Br⁻ ions are drawn larger than the K⁺ ions throughout the model

Feedback if chosen: This option is wrong — you flagged a correct feature. In KBr the Br⁻ ion really is larger than the K⁺ ion, so Br⁻ should be drawn larger.

CThe ions are labeled with their charges as well as their symbols

Feedback if chosen: This option is wrong — you flagged a correct feature. Every ion in a particulate model must be labeled with its symbol and its charge.

DThe ions sit in separate K⁺Br⁻ pairs instead of one continuous repeating array ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Solid KBr has no separate pairs — no K⁺ belongs to one particular Br⁻. The lattice repeats continuously, each ion among several oppositely charged neighbors.

Check — multiple choice

What is the specific flaw in this model of solid sodium fluoride?

ALike-charged ions sit side by side, adding strong repulsion where the solid needs attraction ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Nearest neighbors are where the forces are strongest, so a correct lattice never places like charges side by side.

BThe F⁻ ions are drawn larger than the Na⁺ ions in every row

Feedback if chosen: This option is wrong — you flagged a correct feature. In NaF the F⁻ ion really is larger than the Na⁺ ion, so F⁻ should be drawn larger.

CThe same alternating pattern repeats over and over across the drawing

Feedback if chosen: This option is wrong — you flagged the repetition itself. Repeating is what a lattice does; the flaw here is which ions the pattern puts next to each other.

DThe drawing shows a single flat layer rather than a three-dimensional block

Feedback if chosen: This option is wrong — you flagged the accepted drawing convention. A flat layer is how a three-dimensional lattice is drawn on paper; it is not a flaw.

Check — multiple choice

A student drew this model of solid calcium oxide. Which judgment is right?

AThe model is consistent — the ions alternate, and no two like charges are nearest neighbors ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The drawing passes every check: a continuous repeating pattern, every nearest-neighbor pair oppositely charged, labels on every ion, anions larger.

BThe model is flawed — the O²⁻ ions are drawn larger than the Ca²⁺ ions

Feedback if chosen: This option is wrong — you flagged a correct feature. In CaO the O²⁻ ion really is larger than the Ca²⁺ ion, so O²⁻ should be drawn larger.

CThe model is flawed — a real solid would not keep such a regular pattern

Feedback if chosen: This option is wrong — you doubted the lattice itself. A crystal lattice is exactly that regular: the same arrangement repeating in three dimensions.

DThe model is flawed — ions with 2+ and 2− charges are too strongly charged to sit side by side

Feedback if chosen: This option is wrong — you turned the larger charges into a problem. Larger opposite charges attract more strongly — placing them side by side is the point of the arrangement.

U2-S02 · LESSON 18 · CSP-018

Metallic bonding

Outcome: State what a metallic bond is — the attraction between positive metal ions and delocalized valence electrons.

Activation question — asked before the lesson opens

Activates: Unit 1: valence electrons are an atom's outermost electrons (CSP-017's checks tested flawed ionic models, not valence electrons)

A magnesium atom has 12 electrons: 10 in its inner shells and 2 in its outermost shell. Which of these electrons are its valence electrons?

AAll 12 of its electrons

Why: Only the outermost electrons count as valence electrons; the ten inner electrons are the atom's core.

BThe 10 electrons in its inner shells

Why: The inner-shell electrons are the core electrons; the valence electrons are the ones in the outermost shell.

CThe 2 electrons in its outermost shell ✓ KEY — this is the correct answer

Why: Valence electrons are the electrons in an atom's outermost shell — for magnesium, its two outer electrons.

DThe 2 electrons closest to the nucleus

Why: The electrons closest to the nucleus sit in the innermost shell; the valence electrons sit at the opposite end, in the outermost shell.

VIDEO SLOT: APCHEM-U02-S02-L07 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1What holds a metal together

A copper wire is one solid piece of metal — yet it contains only copper. There are no oppositely charged partners in sight, so what holds it together?

You’ve already seen that an atom’s outermost electrons are its valence electrons.

Here is the particulate model of solid copper.

In the solid, each copper atom releases its valence electrons.

What remains of each atom is a positive copper ion — the circles in the model — sitting in a regular repeating array.

The released electrons spread through the whole solid: they belong to no single ion. Electrons spread out like this are called delocalized.

In the model, the delocalized electrons are the continuous shaded region labeled e⁻ that fills the space between the ions.

The metallic bond is the attraction between the positive metal ions and the delocalized electrons around them.

Notice what the model does not contain: no negative ions anywhere, and no electron pair fixed between two particular atoms.

What you are expected to be able to do

State that a metallic bond is the attraction between positive metal ions and valence electrons that are delocalized across the whole solid rather than assigned to one atom.

Part 2Check what you just learned

Check — multiple choice

Solid magnesium has the model shown. What is the metallic bond in this solid?

AThe attraction between the neighboring positive ions packed side by side

Feedback if chosen: This option is wrong — you attracted like charges to each other. The positive ions repel one another; it is the delocalized electrons between them that they are attracted to.

BThe attraction between the positive ions and the delocalized electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The positive Mg²⁺ ions are attracted to the delocalized electrons spread around them — that attraction is the metallic bond.

CA pair of electrons shared between two particular ions

Feedback if chosen: This option is wrong — you described a covalent bond. In a metal no electron pair sits between two particular ions; the electrons are delocalized across the whole solid.

DThe pull of each positive ion on the single electron that it released

Feedback if chosen: This option is wrong — you kept each electron tied to its own atom. A released electron belongs to no single ion; every ion is attracted to the delocalized electrons around it.

Check — multiple choice

Which statement describes the delocalized electrons in the model of solid sodium shown?

AThey move throughout the whole solid and belong to no single ion ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Delocalized means exactly this: the valence electrons are spread across the whole solid, not assigned to any one atom.

BEach stays beside the ion that released it

Feedback if chosen: This option is wrong — you kept each electron with its own atom. Once released, an electron is delocalized: it belongs to the whole solid, not to the ion it came from.

CThey sit in fixed pairs between particular ions

Feedback if chosen: This option is wrong — you localized the electrons into covalent-style pairs. The shaded region is continuous because the electrons are spread out, not parked between particular ions.

DThey have left the solid, which is why the ions are positive

Feedback if chosen: This option is wrong — you sent the electrons away. They are still inside the solid — the ions are positive because each atom released its valence electrons into the shared region, not because electrons left the metal.

Check — multiple choice

Where did the delocalized electrons in solid copper come from?

AThey are the valence electrons the copper atoms released ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each copper atom releases its valence electrons into the shared region; the core electrons stay with their ion.

BThey are the core electrons of the copper atoms

Feedback if chosen: This option is wrong — you released the wrong electrons. The core electrons stay bound inside each ion; it is the outermost, valence electrons that become delocalized.

CThey were donated by a nonmetal bonded to the copper

Feedback if chosen: This option is wrong — you brought in a second element. Solid copper contains only copper; the delocalized electrons are copper’s own valence electrons.

DThey flowed into the metal from the air around it

Feedback if chosen: This option is wrong — you gave the electrons an outside source. The delocalized electrons are the metal's own valence electrons, released by the copper atoms themselves.

U2-S02 · LESSON 19 · CSP-019

Drawing a metallic solid

Outcome: Draw a particulate model of a metallic solid — a repeating array of positive ions within one shared region of delocalized electrons.

No activation question for this lesson — recap (the metallic-solid particulate model) is exactly what CSP-018's three checks just tested.

VIDEO SLOT: APCHEM-U02-S02-L08 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1The rules of the drawing

You’ve already seen the particulate model of a metallic solid — now you learn to draw one.

Rule one: draw the positive metal ions as a regular repeating array — evenly spaced rows, at least three rows and three columns.

Rule two: label every ion with its symbol and positive charge.

Rule three: draw one continuous region filling the space between all the ions, and label it e⁻ — these are the delocalized electrons.

Do not draw any electron as belonging to a particular ion — no dots or pairs parked beside individual circles.

And there are no negative ions anywhere in a metal — the only negative charge is the delocalized electrons.

What you are expected to be able to do

Draw a particulate model of a metallic solid: a repeating array of labeled positive metal ions within one shared, continuous region of delocalized valence electrons.

Part 2Check what you just learned

Check — drawing task (pen and paper, self-marked)

In solid iron, each atom releases two electrons, leaving Fe²⁺ ions. Take a pen and paper and draw a particulate model of solid iron — at least three rows and three columns of ions. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows the ions in a regular repeating array — evenly spaced, at least three rows and three columns.
Your drawing shows every ion labeled Fe²⁺ — the positive charge is written, not implied.
Your drawing shows one continuous region, labeled e⁻, filling the space between all the ions.
Your drawing shows no electrons parked beside individual ions — no fixed dots or pairs.
Your drawing shows no negative ions anywhere.

Check — multiple choice

A classmate drew this model of solid silver. What must they change before it is a correct metallic-solid model?

AMake some of the ions negative so the charges in the drawing balance

Feedback if chosen: This option is wrong — you added negative ions. A metal has none: the delocalized electrons are the negative charge that balances the positive ions.

BScatter the ions randomly — a real metal’s ions do not line up in a regular array

Feedback if chosen: This option is wrong — you broke the array. The positive ions of a metallic solid do sit in a regular repeating arrangement; the flaw here is the electrons, not the ions.

CReplace the fixed electron dots beside each ion with one continuous shared region labeled e⁻ ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The drawing assigns each electron pair to one ion, but the valence electrons in a metal are delocalized — one shared region spanning the whole solid.

DRemove the charge labels — metal ions are drawn without charges

Feedback if chosen: This option is wrong — you removed a required label. The circles are positive ions, and the model must say so: symbol and charge on every ion.

U2-S02 · LESSON 20 · CSP-020

Reading a metallic-solid model

Outcome: Identify the positive metal ions and the delocalized electron region in a metallic-solid model.

No activation question for this lesson — no 'You've already seen' recap — the lesson reopens the copper model directly.

VIDEO SLOT: APCHEM-U02-S02-L09 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1The two parts of the model

Here is the particulate model of solid copper again.

Every metallic-solid model has exactly two parts, and you should be able to name both.

The labeled circles are the positive metal ions — each one a copper atom that has released its valence electrons.

The continuous region between the circles, labeled e⁻, is the delocalized electrons — the released valence electrons, spread through the whole solid and free to move.

The region is drawn continuous on purpose: it says the electrons belong to the whole solid, not to any one ion.

What you are expected to be able to do

Identify, in a supplied metallic-solid model, which part represents the positive metal ions and which part represents the delocalized valence electrons.

Part 2Check what you just learned

Check — multiple choice

In the model of solid zinc shown, what does the shaded region between the circles represent?

AThe delocalized valence electrons, spread through the solid ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The continuous region labeled e⁻ is the delocalized valence electrons — released by the atoms and free to move through the whole solid.

BEmpty space left between the particles when the solid formed

Feedback if chosen: This option is wrong — you read the shading as nothing. The region is labeled e⁻: it is filled with the delocalized electrons that hold the solid together.

CNegatively charged zinc ions packed between the positive ones

Feedback if chosen: This option is wrong — you turned the electrons into anions. A metal contains no negative ions; the negative charge is the delocalized electrons themselves.

DA fixed pair of electrons owned by each ion and held beside it

Feedback if chosen: This option is wrong — you split the region into per-ion pairs. It is drawn as one continuous region precisely because the electrons belong to the whole solid, not to individual ions.

Check — multiple choice

In the model of solid silver shown, what does each circle labeled Ag⁺ represent?

AOne positive silver ion ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each circle is a silver atom that has released its valence electron, leaving a positive ion — that is what the ⁺ in the label says.

BOne neutral silver atom

Feedback if chosen: This option is wrong — you left the atom neutral. The label says Ag⁺: in the solid, each atom has released its valence electron and is a positive ion.

COne molecule of silver

Feedback if chosen: This option is wrong — you read the circle as a molecule. Solid silver contains no molecules; each circle is a single positive ion.

DThe delocalized electrons of one atom

Feedback if chosen: This option is wrong — you swapped the two parts of the model. The electrons are the shaded region labeled e⁻; the circles are the positive ions.

Check — multiple choice

Which feature of the magnesium model shown tells you the valence electrons are delocalized?

AThe circles are all drawn with exactly the same size

Feedback if chosen: This option is wrong — you pointed at the drawing style. Equal circle sizes just mean the ions are identical; the delocalization is carried by the continuous e⁻ region.

BThe ions sit in a regular, evenly spaced repeating pattern

Feedback if chosen: This option is wrong — you pointed at the ions. The regular array only describes where the ions sit; it does not describe the electrons at all.

CEvery ion in the model carries the same positive charge

Feedback if chosen: This option is wrong — you pointed at the charges. Equal charges tell you each atom released the same number of electrons, not where those electrons are now.

DThe e⁻ region is one continuous span across the whole model ✓ KEY — this is the correct answer

Feedback if chosen: Correct. One unbroken region shared by all the ions is the model’s way of saying the electrons belong to the whole solid, not to any single atom.

U2-S02 · LESSON 21 · CSP-021

Spotting the flaw in a metallic-solid model

Outcome: Judge whether a metallic-solid model shows positive ions within delocalized electrons, and name the specific flaw when it does not.

No activation question for this lesson — recap (the correct metallic-model checklist) is exercised across CSP-020's checks — the delocalized e⁻ region and the positive-ion circles are the keys of all three, and no-fixed-pairs / no-negative-ions are confronted in those checks' options; an item on them would duplicate CSP-021's own checks.

VIDEO SLOT: APCHEM-U02-S02-L10 — quote this id when requesting a video edit — Video — an alternative to reading this lesson

Part 1What a correct model must show

You’ve already seen what a correct metallic-solid model shows: a regular array of labeled positive ions, one continuous region of delocalized electrons, no fixed electron pairs, and no negative ions.

To judge a model, check it against that list — and when it fails, name the one specific thing that is wrong.

Here is a flawed model of solid copper.

Check the electrons: each ion has its own little pair of dots parked beside it.

That says every electron belongs to one particular ion — but the valence electrons in a metal are delocalized, shared across the whole solid.

That is the specific flaw: localized electrons where the model needs one shared, continuous electron region.

What you are expected to be able to do

Evaluate whether a supplied metallic-solid model shows positive metal ions within delocalized electrons, and identify one specific flaw when it does not.

Part 2Check what you just learned

Check — multiple choice

A student drew this model of solid zinc. What is the specific flaw?

AEach ion is given its own fixed electrons instead of one shared region across the solid ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Parking electrons beside individual ions makes them localized; in a metal the valence electrons are delocalized — one continuous region shared by every ion.

BThe ions are drawn in a regular array — a real metal’s ions are scattered randomly

Feedback if chosen: This option is wrong — you broke the array. The positive ions of a metallic solid really do sit in a regular repeating arrangement.

CThe ions are labeled with positive charges — metal ions should be drawn neutral

Feedback if chosen: This option is wrong — you neutralized the ions. Each atom has released its valence electrons, so the circles are positive ions and must be labeled that way.

DThe drawing shows too few ions to stand for a real piece of metal

Feedback if chosen: This option is wrong — you asked the drawing to do the impossible. A model always shows a tiny fragment; it only has to show the arrangement correctly.

Check — multiple choice

A student drew this model of solid copper. What is the specific flaw?

AHalf of the copper is drawn as negative ions — a metal contains no negative ions ✓ KEY — this is the correct answer

Feedback if chosen: Correct. There are no Cu⁻ ions in copper. The student drew the solid like an ionic lattice; the negative charge in a metal is the delocalized valence electrons, drawn as one continuous region around the positive ions.

BThe circles alternate in a repeating pattern — a metal’s ions should not be ordered

Feedback if chosen: This option is wrong — you rejected the ordering. A metal’s ions do sit in a regular repeating array; the flaw is what the student put between them.

CThe charges are written on the particles — charge labels do not belong in a metallic model

Feedback if chosen: This option is wrong — you rejected the labels. Charges belong on every ion in a particulate model; the problem is that half of these charges are negative, and a metal has no negative ions.

DThe two kinds of circle are drawn with two different sizes

Feedback if chosen: This option is wrong — you flagged the drawing style. The sizes follow from what the student drew; the real error is drawing Cu⁻ ions that do not exist in a metal.

Check — multiple choice

A student drew this model of solid magnesium. Which judgment is right?

AThe model is flawed — the electrons should be drawn in pairs beside each ion

Feedback if chosen: This option is wrong — you localized the electrons. Fixed per-ion pairs are the classic flaw; the shared continuous region is what a correct metallic model shows.

BThe model is consistent — positive ions in a repeating array within shared delocalized electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The drawing passes every check: a regular array of labeled positive ions, one continuous e⁻ region, no fixed pairs, no negative ions.

CThe model is flawed — half of the ions should be negative to balance the charge

Feedback if chosen: This option is wrong — you added negative ions. A metal has none: the delocalized electrons already balance the positive ions.

DThe model is flawed — the ions should touch, with no room for a region between them

Feedback if chosen: This option is wrong — you squeezed out the electrons. The region between the ions is exactly where the model shows the delocalized electrons that hold the solid together.

U2-S03 — Alloy models

5 lessons

U2-S03 · LESSON 22 · CSP-022

What an alloy is

Outcome: State what an alloy is: a metallic material containing a mixture of a metal with one or more other elements.

No activation question for this lesson — no 'You've already seen' recap — the lesson opens fresh, with the brass-is-not-on-the-periodic-table problem (immediately preceding lesson is CSP-021 in U2-S02, but the rule fires only on a recap opening).

VIDEO SLOT: APCHEM-U02-S03-L01 — quote this id when requesting a video edit — Video — an alternative to reading this lesson (render pending)

Part 1A metal that is not on the periodic table

Brass is a metal you have handled: door handles, keys, trumpets.

Look for brass on the periodic table, though, and it is not there.

Copper is there. Zinc is there. Brass is not an element.

Here is a particulate model of brass: it is copper with zinc mixed all through it — one metallic solid made of two elements.

A metallic material made by mixing a metal with one or more other elements is called an alloy.

The other elements do not have to be metals.

Steel is an alloy of iron mixed with a small amount of carbon — a nonmetal — and steel is still a metallic material.

Because an alloy is a mixture, its proportions can vary: brass can contain more zinc or less.

An element on its own is not an alloy: pure copper is just copper, with nothing mixed into it.

A material that is not metallic is not an alloy either: sodium chloride contains the metal sodium, but it is a crystal lattice of Na⁺ and Cl⁻ ions, not a metallic material.

What you are expected to be able to do

State that an alloy is a metallic material containing a mixture of a metal with one or more other elements, and recognize whether a given material is an alloy.

Part 2Check what you just learned

Check — multiple choice

Which of these materials is an alloy?

ABronze — a metallic material of copper mixed with tin ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Bronze has everything the definition asks for: it is a metallic material, and it is a mixture of a metal (copper) with another element (tin).

BPure aluminum — a single metallic element

Feedback if chosen: This option is wrong — you picked a metallic material with no mixture in it. Pure aluminum is a single element; an alloy must be a metal mixed with at least one other element.

CSodium chloride — a crystal lattice of Na⁺ and Cl⁻ ions containing the metal sodium

Feedback if chosen: This option is wrong — you counted any material containing a metal as an alloy. Sodium chloride is a lattice of ions, not a metallic material, so it is not an alloy.

DAir — a mixture of nitrogen and oxygen gases

Feedback if chosen: This option is wrong — you counted any mixture as an alloy. Air is a mixture, but it is not a metallic material and contains no metal.

Check — multiple choice

Cast iron is a metallic material of iron mixed with carbon, a nonmetal. Is cast iron an alloy?

ANo — an alloy must be a mixture of two or more metals

Feedback if chosen: This option is wrong — you added a requirement the definition does not have. The elements mixed with the metal can be metals or nonmetals; the material just has to stay metallic.

BYes — any material that contains iron is an alloy

Feedback if chosen: This option is wrong — you reached the right verdict for the wrong reason. Containing iron is not enough: rust contains iron and is not an alloy. Cast iron qualifies because it is a metallic material of a metal mixed with another element.

CNo — mixing in a nonmetal means the material cannot be metallic

Feedback if chosen: This option is wrong — you assumed a nonmetal destroys the metallic character of the mixture. The stem tells you cast iron is a metallic material; a small amount of carbon mixed into iron leaves it metallic.

DYes — it is a metallic material containing a metal mixed with another element ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Cast iron is a metallic material, and it is a mixture of a metal (iron) with another element (carbon). That is the whole definition — the other element does not have to be a metal.

Check — multiple choice

Which of these materials is NOT an alloy?

AStainless steel — a metallic material of iron mixed with chromium and nickel

Feedback if chosen: This option is wrong — you treated a three-element mixture as if it could not be an alloy. An alloy is a metal mixed with one or more other elements, so stainless steel is an alloy.

BPewter — a metallic material of tin mixed with copper and antimony

Feedback if chosen: This option is wrong — you treated a three-element mixture as if it could not be an alloy. Pewter is a metallic material of tin mixed with two other elements, which is exactly what an alloy is.

CPure zinc — a single metallic element ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Pure zinc is a single element with nothing mixed into it. An alloy must be a mixture of a metal with one or more other elements.

DSolder — a metallic material of tin mixed with silver and copper

Feedback if chosen: This option is wrong — you treated a three-element mixture as if it could not be an alloy. Solder is a metallic material of a metal mixed with other elements, so it is an alloy.

U2-S03 · LESSON 23 · CSP-023

Interstitial and substitutional alloys

Outcome: Classify an alloy as interstitial or substitutional from a description of where the atoms of the other element sit.

No activation question for this lesson — recap (an alloy is a metallic material containing a mixture of a metal with one or more other elements) is exactly what CSP-022's three checks just keyed — bronze, cast iron and pure zinc all key on that definition.

VIDEO SLOT: APCHEM-U02-S03-L02A — quote this id when requesting a video edit — Video, part A — an alternative to reading this lesson (render pending)

VIDEO SLOT: APCHEM-U02-S03-L02B — quote this id when requesting a video edit — Video, part B — an alternative to reading this lesson (render pending)

Part 1Two ways to mix atoms into a metal

You've already seen that an alloy is a metallic material containing a mixture of a metal with one or more other elements.

Mix the atoms of the other element into a metal and there are only two places they can end up. Steel shows the first; brass shows the second.

Here is a particulate model of steel: iron mixed with a small amount of carbon.

The carbon atoms are much smaller than the iron atoms.

Because they are much smaller, the carbon atoms can fit into the spaces between the iron atoms.

Every iron atom keeps its position — none has been replaced.

The spaces between the metal atoms are called interstices.

An alloy in which smaller atoms occupy these spaces is called an interstitial alloy.

Now here is a particulate model of brass: copper mixed with zinc.

The zinc atoms are of comparable size to the copper atoms.

Because they are of comparable size, the zinc atoms cannot fit into the spaces between the copper atoms.

Instead, each zinc atom takes the place of a copper atom in the array.

An alloy in which atoms of comparable size — a little smaller, equal, or a little larger — take the places of metal atoms is called a substitutional alloy.

Both labels answer one question about the atoms of the other element: do they sit in the spaces between the metal atoms (interstitial), or do they take the places of metal atoms (substitutional)?

What you are expected to be able to do

Classify an alloy as interstitial or substitutional from its description: smaller atoms occupying the spaces between the metal atoms make it interstitial; comparable-size atoms taking the places of metal atoms make it substitutional.

Part 2Check what you just learned

Check — multiple choice

Palladium metal absorbs hydrogen. In the resulting alloy, small hydrogen atoms sit in the spaces between the palladium atoms, and every palladium atom keeps its position. How is this alloy classified?

ASubstitutional — the hydrogen atoms take the places of palladium atoms

Feedback if chosen: This option is wrong — you misread the arrangement. The description says every palladium atom keeps its position, so nothing has been replaced.

BSubstitutional — the hydrogen atoms sit in the spaces between the palladium atoms

Feedback if chosen: This option is wrong — you matched the right arrangement to the wrong label. Atoms sitting in the spaces between the metal atoms are the mark of an interstitial alloy, not a substitutional one.

CInterstitial — the hydrogen atoms sit in the spaces between the palladium atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Small atoms occupying the spaces between the metal atoms, with no metal atom replaced, is exactly what interstitial means.

DInterstitial — the hydrogen atoms take the places of palladium atoms

Feedback if chosen: This option is wrong — you picked the right label with the wrong reason. Interstitial atoms sit in the spaces between the metal atoms; the description says nothing was replaced.

Check — multiple choice

Sterling silver is silver mixed with copper. The copper atoms are of comparable size to the silver atoms, and each copper atom takes the place of a silver atom in the array. How is this alloy classified?

AInterstitial — the copper atoms take the places of silver atoms

Feedback if chosen: This option is wrong — you matched the right arrangement to the wrong label. Taking the places of metal atoms is the mark of a substitutional alloy; interstitial atoms sit in the spaces between the metal atoms.

BSubstitutional — the copper atoms take the places of silver atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Atoms of comparable size taking the places of metal atoms is exactly what substitutional means.

CInterstitial — the copper atoms sit in the spaces between the silver atoms

Feedback if chosen: This option is wrong — you misread the arrangement and attached the wrong label to it. The description says the copper atoms take the places of silver atoms, and replacement is the mark of a substitutional alloy.

DSubstitutional — the copper atoms sit in the spaces between the silver atoms

Feedback if chosen: This option is wrong — you picked the right label with the wrong reason. The copper atoms take the places of silver atoms; they do not sit in the spaces between them.

Check — multiple choice

Bronze is copper mixed with tin. The tin atoms are slightly larger than the copper atoms, and each tin atom takes the place of a copper atom in the array. How is this alloy classified?

ASubstitutional — the slightly larger tin atoms take the places of copper atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Comparable size includes slightly larger. The tin atoms take the places of copper atoms, and replacement is what substitutional means.

BSubstitutional — the tin atoms sit in the spaces between the copper atoms

Feedback if chosen: This option is wrong — you picked the right label with the wrong reason. The description says each tin atom takes the place of a copper atom; nothing sits in the spaces between them.

CInterstitial — the tin atoms take the places of copper atoms

Feedback if chosen: This option is wrong — you matched replacement to the wrong label. Taking the places of metal atoms is the mark of a substitutional alloy.

DInterstitial — the tin atoms sit in the spaces between the copper atoms

Feedback if chosen: This option is wrong — you misread the arrangement. An atom slightly larger than a copper atom cannot fit into the spaces between copper atoms; the description says the tin atoms take the places of copper atoms.

U2-S03 · LESSON 24 · CSP-024

Predicting alloy type from atomic radii

Outcome: Predict whether an alloy will be interstitial or substitutional from the atomic radii of its elements.

Activation question — asked before the lesson opens

Activates: Unit 1: atomic radius — the measure of an atom's size (the recap's first line; the steel/brass-models recap is exactly what CSP-023's checks just keyed, and no CSP-023 check mentions atomic radius)

What does an atom's atomic radius measure?

AThe number of protons in its nucleus

Why: The proton count is the atomic number; a radius is a distance, not a count.

BThe size of the atom ✓ KEY — this is the correct answer

Why: Atomic radius is the measure of an atom's size: the distance from the nucleus to the atom's outer edge.

CThe mass of the atom

Why: An atom's mass and its size are different properties; a radius is a distance, not a mass.

DThe tendency of the atom to attract shared electrons

Why: That is electronegativity — a different periodic property.

VIDEO SLOT: APCHEM-U02-S03-L03 — quote this id when requesting a video edit — Video — an alternative to reading this lesson (render pending)

Part 1From radii to a prediction

You've already seen atomic radius in Unit 1 — the measure of an atom's size.

You've already seen steel's model: much smaller carbon atoms occupying the spaces between the iron ions — an interstitial alloy.

And brass's model: zinc ions of comparable size taking the places of copper ions — a substitutional alloy.

Given the atomic radii of an alloy's elements, you can predict which type will form before ever seeing a model.

Here are the atomic radii of the four elements in those two alloys.

Atomic radii (Slater empirical values)
Element	Atomic radius (pm)
Fe	140
C	70
Cu	135
Zn	135

Start with steel. Carbon's radius, 70 pm, is half of iron's 140 pm — the carbon atoms are much smaller than the iron atoms.

Because they are much smaller, the carbon atoms can fit into the spaces between the iron atoms — predict interstitial. Steel is indeed interstitial.

Now brass. Zinc's radius, 135 pm, equals copper's — the zinc atoms are of comparable size to the copper atoms.

Because they are of comparable size, the zinc atoms cannot fit into the spaces between the copper atoms; they can only take the places of copper atoms — predict substitutional. Brass is indeed substitutional.

The prediction rule: when the atoms of the other element are much smaller than the metal atoms, predict interstitial; when the two radii are comparable — a little smaller, equal, or a little larger — predict substitutional.

What you are expected to be able to do

Predict whether an alloy will be interstitial or substitutional by comparing the atomic radii of its elements.

Part 2Check what you just learned

Check — multiple choice

Atomic radii (Slater empirical values)
Element	Atomic radius (pm)
Fe	140
N	65

Nitrogen atoms can be mixed into solid iron. Use the atomic radii shown. Which type of alloy do you predict?

AInterstitial — the nitrogen atoms are much smaller than the iron atoms, so they can fit into the spaces between them ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Nitrogen's 65 pm is less than half of iron's 140 pm, so the nitrogen atoms are much smaller and can fit into the spaces between the iron atoms.

BSubstitutional — the nitrogen atoms are of comparable size to the iron atoms, so they take the places of iron atoms

Feedback if chosen: This option is wrong — you read the radii as comparable. Nitrogen's 65 pm is less than half of iron's 140 pm; that is a much smaller atom, and much smaller atoms fit into the spaces between the metal atoms.

CSubstitutional — the nitrogen atoms are much smaller than the iron atoms, so they take the places of iron atoms

Feedback if chosen: This option is wrong — you compared the radii correctly but attached the wrong prediction. Much smaller atoms do not take the places of metal atoms; they fit into the spaces between them, which is interstitial.

DInterstitial — the nitrogen atoms are of comparable size to the iron atoms, so they can fit into the spaces between them

Feedback if chosen: This option is wrong — you picked the right label with a wrong reading of the table. Nitrogen's 65 pm is far from comparable to iron's 140 pm; it is the much smaller size that lets the atoms fit into the spaces.

Check — multiple choice

Atomic radii (Slater empirical values)
Element	Atomic radius (pm)
Cu	135
Ni	135

Cupronickel, the metallic material used in many coins, is copper mixed with nickel. Use the atomic radii shown. Which type of alloy do you predict?

AInterstitial — the nickel atoms can fit into the spaces between the copper atoms

Feedback if chosen: This option is wrong — you predicted spaces-between placement for atoms that are not small enough. At 135 pm, a nickel atom is exactly as large as a copper atom and cannot fit into the spaces between copper atoms.

BInterstitial — the nickel atoms take the places of copper atoms

Feedback if chosen: This option is wrong — you paired the substitutional arrangement (taking the places of metal atoms) with the interstitial label.

CSubstitutional — the nickel atoms are of comparable size to the copper atoms, so they take the places of copper atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The two radii are equal, so the nickel atoms are of comparable size and can only take the places of copper atoms.

DSubstitutional — the nickel atoms are much smaller than the copper atoms

Feedback if chosen: This option is wrong — you reached the right label from a wrong reading of the table. The radii are equal — comparable size, not much smaller — and comparable size is why the alloy is substitutional.

Check — multiple choice

Atomic radii (Slater empirical values)
Element	Atomic radius (pm)
Ti	140
O	60

Titanium metal can absorb oxygen atoms into the solid. Use the atomic radii shown. Which type of alloy do you predict?

ASubstitutional — the oxygen atoms take the places of titanium atoms

Feedback if chosen: This option is wrong — you predicted replacement for atoms that do not need it. At 60 pm against titanium's 140 pm, the oxygen atoms are much smaller and fit into the spaces between the titanium atoms.

BInterstitial — the oxygen atoms are of comparable size to the titanium atoms

Feedback if chosen: This option is wrong — you picked the right label with a wrong reading of the table. Oxygen's 60 pm is far from comparable to titanium's 140 pm; it is the much smaller size that lets the atoms fit into the spaces.

CSubstitutional — oxygen is a nonmetal, and nonmetal atoms take the places of metal atoms

Feedback if chosen: This option is wrong — you classified by element type instead of by radius. The prediction comes from comparing atomic radii, and 60 pm against 140 pm predicts interstitial. Steel's carbon is also a nonmetal, and it sits in the spaces between the iron atoms.

DInterstitial — the oxygen atoms are much smaller than the titanium atoms, so they can fit into the spaces between them ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Oxygen's 60 pm is less than half of titanium's 140 pm — much smaller — so the oxygen atoms fit into the spaces between the titanium atoms.

U2-S03 · LESSON 25 · CSP-025

Reading an alloy model

Outcome: Identify an alloy particulate model as interstitial or substitutional from the positions and relative sizes of its atoms.

No activation question for this lesson — recap (the steel model — interstitial; the brass model — substitutional) is keyed by all three CSP-024 checks: each key states the size-and-placement → label pairing (much smaller atoms in the spaces → interstitial; comparable-size atoms taking places → substitutional).

VIDEO SLOT: APCHEM-U02-S03-L04 — quote this id when requesting a video edit — Video — an alternative to reading this lesson (render pending)

Part 1What to look for in the model

You've already seen this particulate model of steel — an interstitial alloy.

And this model of brass — a substitutional alloy.

A model like these answers the classification question by itself — you read two things from it.

First, the sizes: are the atoms of the other element much smaller than the metal atoms, or of comparable size?

Then, the positions: do those atoms sit in the spaces between the metal atoms, or do they take the places of metal atoms in the array?

Much smaller atoms sitting in the spaces between the metal atoms — interstitial. Comparable-size atoms taking the places of metal atoms — substitutional.

What you are expected to be able to do

Identify a supplied alloy particulate model as interstitial or substitutional from the positions and relative sizes of its atoms.

Part 2Check what you just learned

Check — multiple choice

The model shows an alloy of silver and gold. Which classification does the model support?

AInterstitial — the gold atoms sit in the spaces between the silver atoms

Feedback if chosen: This option is wrong — you misread the positions. The gold atoms sit at positions in the array where silver atoms would otherwise be, not in the spaces between them.

BSubstitutional — the gold atoms are of comparable size and take the places of silver atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The model shows comparable-size gold atoms taking the places of silver atoms — the two marks of a substitutional alloy.

CInterstitial — the gold atoms are much smaller than the silver atoms

Feedback if chosen: This option is wrong — you misread the sizes. The gold atoms in the model are of comparable size to the silver atoms, and comparable-size atoms taking the places of metal atoms make a substitutional alloy.

DSubstitutional — the gold atoms sit in the spaces between the silver atoms

Feedback if chosen: This option is wrong — you picked the right label for the wrong feature. The gold atoms take the places of silver atoms; they do not sit in the spaces between them.

Check — multiple choice

The model shows an alloy of iron and boron. Which classification does the model support?

ASubstitutional — the boron atoms take the places of iron atoms

Feedback if chosen: This option is wrong — you misread the positions. Every iron atom in the model keeps its position; the boron atoms sit in the spaces between them, and nothing has been replaced.

BSubstitutional — the boron atoms are much smaller than the iron atoms

Feedback if chosen: This option is wrong — you read the sizes correctly but attached the wrong label. Much smaller atoms sitting in the spaces between the metal atoms are the mark of an interstitial alloy.

CInterstitial — the boron atoms are much smaller and sit in the spaces between the iron atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The model shows much smaller boron atoms in the spaces between the iron atoms, with no iron atom replaced — an interstitial alloy.

DInterstitial — the boron atoms are of comparable size to the iron atoms

Feedback if chosen: This option is wrong — you picked the right label with the wrong reading of the model. The boron atoms are drawn much smaller than the iron atoms; that is what lets them sit in the spaces between them.

Check — multiple choice

The model shows an alloy of copper and tin. Which classification does the model support?

AInterstitial — the tin atoms are larger than the copper atoms

Feedback if chosen: This option is wrong — you classified from a size difference alone. Interstitial means much smaller atoms sitting in the spaces between the metal atoms; these tin atoms are slightly larger and take the places of copper atoms.

BInterstitial — the tin atoms sit in the spaces between the copper atoms

Feedback if chosen: This option is wrong — you misread the positions. An atom slightly larger than a copper atom cannot sit in the spaces between copper atoms, and the model shows the tin atoms at positions where copper atoms would otherwise be.

CSubstitutional — the tin atoms sit in the spaces between the copper atoms

Feedback if chosen: This option is wrong — you picked the right label for the wrong feature. The tin atoms take the places of copper atoms; they do not sit in the spaces between them.

DSubstitutional — the slightly larger tin atoms take the places of copper atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Comparable size includes slightly larger, and the tin atoms take the places of copper atoms in the array — a substitutional alloy.

U2-S03 · LESSON 26 · CSP-026

Drawing an alloy model

Outcome: Draw an interstitial or substitutional alloy model from atomic radii, keeping the delocalized electrons of the metal.

Activation question — asked before the lesson opens

Activates: CSP-020: the pure-metal particulate model — a repeating array of positive ions within one continuous region of delocalized electrons (the recap's 'you've already drawn' lines; CSP-025's checks classify alloy models and never key the array or the electron region)

Which of these describes a correct particulate model of solid nickel, a pure metal?

AA repeating array of neutral nickel atoms, with no electrons shown

Why: Each nickel atom has released its valence electrons, so the particles in the array are positive ions — and the released electrons stay in the model.

BA repeating array of positive nickel ions, each with its own fixed pair of electrons beside it

Why: The valence electrons in a metal are delocalized — shared across the whole solid — so they are drawn as one continuous region, never parked beside individual ions.

CA repeating array of positive nickel ions within one continuous region of delocalized electrons ✓ KEY — this is the correct answer

Why: That is the metallic-solid model: the positive ions sit in a regular repeating array, and the delocalized valence electrons form one continuous region covering the whole array.

DAlternating positive and negative nickel ions in a repeating array

Why: A metal contains no negative ions; the negative charge is the delocalized electrons, drawn as one continuous region around the positive ions.

VIDEO SLOT: APCHEM-U02-S03-L05 — quote this id when requesting a video edit — Video — an alternative to reading this lesson (render pending)

Part 1Everything into one drawing

Here is a complete particulate model of steel — the kind of drawing you will make in this lesson.

The iron ions form a repeating array — you've already drawn arrays like this for a pure metal.

The continuous region of delocalized electrons covers the whole array — you've already drawn this too.

The much smaller carbon atoms in the spaces between the iron ions are the only new feature.

To draw an alloy model from atomic radii, work in three steps.

Step 1 — draw the metal: a repeating array of positive metal ions.

Step 2 — place the atoms of the other element using the radii: much smaller atoms sit in the spaces between the metal ions; comparable-size atoms take the places of metal ions.

Step 3 — keep the alloy metallic: draw the continuous region of delocalized electrons covering the whole array.

Step 3 is the one drawings most often lose. An alloy is still a metallic material, so the delocalized electrons stay in the model.

Here is the same three-step drawing for brass: zinc ions of comparable size take the places of copper ions, and the continuous region of delocalized electrons still covers the whole array.

What you are expected to be able to do

Draw an interstitial or substitutional alloy model from supplied atomic radii, keeping the repeating array of positive metal ions and the continuous region of delocalized electrons.

Part 2Check what you just learned

Check — drawing task (pen and paper, self-marked)

Palladium metal absorbs hydrogen to form an alloy. Atomic radii: Pd 140 pm; H 25 pm. Take a pen and paper and draw a particulate model of this alloy. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows the palladium ions as a repeating array, each marked positive.
Your drawing shows the hydrogen atoms much smaller than the palladium ions — the radii (25 pm against 140 pm) make this an interstitial alloy.
Your drawing shows the hydrogen atoms in the spaces between the palladium ions, with no palladium ion replaced.
Your drawing shows a continuous region of delocalized electrons (for example, shading) covering the whole array — the alloy is still metallic.

Check — drawing task (pen and paper, self-marked)

Rose gold is gold mixed with copper. Atomic radii: Au 135 pm; Cu 135 pm. Take a pen and paper and draw a particulate model of this alloy. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows one repeating array of positive metal ions.
Your drawing shows the copper ions the same size as the gold ions — the equal radii (135 pm and 135 pm) make this a substitutional alloy.
Your drawing shows each copper ion taking the place of a gold ion in the array, not sitting in the spaces between gold ions.
Your drawing shows a continuous region of delocalized electrons covering the whole array — the alloy is still metallic.

Check — multiple choice

A student drew this particulate model of steel, an alloy of iron and carbon. What is the flaw in the model?

AThe carbon atoms should take the places of iron ions rather than sit in the spaces between them

Feedback if chosen: This option is wrong — you moved the flaw to the carbon atoms' positions. Carbon atoms are much smaller than iron atoms, so the spaces between the iron ions are exactly where they belong.

BThe delocalized electrons should be drawn as one continuous region covering the whole array, not as fixed pairs on individual ions ✓ KEY — this is the correct answer

Feedback if chosen: Correct. An alloy is still a metallic material, so its model keeps the delocalized electrons as one continuous region covering the whole array. You've already seen fixed pairs on individual ions rejected in metallic-solid models.

CThe carbon atoms should be drawn the same size as the iron ions

Feedback if chosen: This option is wrong — you resized atoms that are drawn correctly. Carbon atoms really are much smaller than iron atoms; that is why they sit in the spaces between the iron ions.

DNothing — the model shows every required feature

Feedback if chosen: This option is wrong — you accepted a model that has lost the metallic bonding. The electrons here are drawn as fixed pairs on individual ions instead of one continuous region covering the whole array.

U2-S04 — Property vocabulary and bonding evidence

21 lessons

U2-S04 · LESSON 27 · CSP-027

Melting point

Outcome: State what melting point means.

Part 1A temperature with a name

Take an ice cube out of the freezer and leave it on the bench. At 0 °C it starts to turn to liquid water.

Do the same with a crystal of table salt and nothing happens — even in a 250 °C oven it stays solid.

Every solid has a temperature at which it stops being a solid. For the solids in this course, it stops by turning to liquid, and that temperature is called its melting point — a few solids instead break down, or turn straight to gas, when heated.

The melting point of a substance is the temperature at which the solid changes to a liquid.

Pressure shifts this temperature slightly, so a melting point is quoted under a stated pressure — usually standard pressure.

Here are measured melting points for three substances at standard pressure.

Melting points at standard pressure
Substance	Melting point (°C)
Water (ice)	0
Sodium chloride (table salt)	801
Iron	1538

As a solid's temperature rises past its melting point, the solid changes to a liquid. Below its melting point, it stays solid.

What you are expected to be able to do

State that melting point is the temperature at which a solid changes to a liquid, under the stated pressure.

Part 2Check what you just learned

Check — multiple choice

A student heats a sample of solid iodine and records the temperature at which the solid turns to liquid: 114 °C. What has the student measured?

AThe highest temperature iodine can reach

Feedback if chosen: This option is wrong — 114 °C is not a ceiling on iodine's temperature. It is the temperature at which the solid changes to a liquid; the liquid can be heated further.

BThe boiling point of iodine

Feedback if chosen: This option is wrong — boiling is a liquid changing to a gas. The student watched a solid change to a liquid, which is melting.

CThe temperature at which iodine dissolves

Feedback if chosen: This option is wrong — dissolving needs a second substance, such as water, for the solid to mix into. Nothing was added here; the solid changed to a liquid on heating alone.

DThe melting point of iodine ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The solid changed to a liquid at 114 °C, and the temperature at which a solid changes to a liquid is the melting point.

Check — multiple choice

Melting and boiling points at standard pressure
Substance	Melting point (°C)	Boiling point (°C)
Naphthalene	80	218
Bromine	−7	59
Sodium chloride	801	1413
Iron	1538	2861

Use the table of melting and boiling points. Which substance is a liquid at room temperature, 25 °C?

ANaphthalene

Feedback if chosen: This option is wrong — 25 °C is below naphthalene's melting point of 80 °C, so naphthalene has not yet melted and is still a solid.

BIron

Feedback if chosen: This option is wrong — 25 °C is far below iron's melting point of 1538 °C, so it is still a solid.

CSodium chloride

Feedback if chosen: This option is wrong — 25 °C is far below sodium chloride's melting point of 801 °C, so it is still a solid.

DBromine ✓ KEY — this is the correct answer

Feedback if chosen: Correct. 25 °C sits above bromine's melting point of −7 °C and below its boiling point of 59 °C, so bromine has melted but not boiled — a liquid.

U2-S04 · LESSON 28 · CSP-028

Electrical conductivity

Outcome: State what electrical conductivity is.

Part 1What it takes to carry a current

Connect a battery and a bulb with two wires, and touch the free ends to a strip of copper: the bulb lights. Touch them to a strip of plastic: nothing.

The ability of a material to carry an electric current is called its electrical conductivity.

Carrying a current means moving charge through the material from one end to the other.

So a material conducts only if it contains charged particles that are free to move.

Both parts matter. Charged particles that are locked in place cannot carry a current, and particles that are free to move but carry no charge cannot either.

A molten salt conducts because its ions — charged particles — are free to move.

What you are expected to be able to do

State that electrical conductivity is the ability of a material to carry electric current through mobile charged particles.

Part 2Check what you just learned

Check — multiple choice

What must a material contain in order to conduct an electric current?

AMetal atoms

Feedback if chosen: This option is wrong — you tied conduction to one kind of substance. A molten salt conducts with no metal atoms present, because its ions are charged and free to move.

BParticles of any kind that are free to move

Feedback if chosen: This option is wrong — you dropped the charge requirement. Moving particles carry a current only if they are charged; neutral particles move plenty of matter but no charge.

CCharged particles that are free to move ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Conduction is charge moving through the material, so the particles must be charged and free to move — both requirements at once.

DCharged particles, whether or not they can move

Feedback if chosen: This option is wrong — you dropped the movement requirement. Charged particles locked in place cannot carry charge from one end of the material to the other.

Check — multiple choice

Molten potassium bromide carries an electric current. What does this observation tell you about molten potassium bromide?

AAny liquid would conduct, because particles in a liquid are free to move

Feedback if chosen: This option is wrong — you kept only the movement requirement. Particles in every liquid move, but a current flows only if the moving particles are charged.

BMelting turned it into a metal

Feedback if chosen: This option is wrong — melting changes the state, not the substance. The current is evidence about the particles present: they are charged and free to move.

CIts particles are free to move but carry no charge

Feedback if chosen: This option is wrong — particles that carry no charge cannot carry a current, yet this liquid conducts. Its mobile particles must be charged.

DIt contains charged particles that are free to move ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A flowing current means charge is moving through the liquid, so the liquid must contain charged particles that are free to move — here, its ions.

U2-S04 · LESSON 29 · CSP-029

Brittleness

Outcome: State what it means for a solid to be brittle.

Part 1Shattering under stress

Tap a large crystal of rock salt with a hammer. It does not dent and it does not bend — it cracks into smaller pieces.

A solid that fractures like this instead of permanently changing shape is called brittle.

When a brittle solid is stressed — struck, squeezed, or bent — it fractures rather than taking on a new shape and keeping it.

The test is what the stress leaves behind: fragments mean the solid is brittle; a dent or a new shape that stays means it is not.

What you are expected to be able to do

State that a brittle solid fractures rather than permanently changing shape when stressed.

Part 2Check what you just learned

Check — multiple choice

Which behaviour shows that a solid is brittle?

AIt becomes a liquid when heated

Feedback if chosen: This option is wrong — melting is a response to temperature, not to stress. Brittleness is about what happens when the solid is struck, squeezed, or bent.

BIt flattens and stays flattened when struck

Feedback if chosen: This option is wrong — flattening and keeping the new shape is permanent deformation without fracture, the opposite of brittle behaviour.

CIt bends under load and springs back straight when released

Feedback if chosen: This option is wrong — springing back means the solid neither fractured nor kept a new shape, so this observation does not show brittleness.

DIt cracks into fragments when struck ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Fracturing into fragments under stress, instead of permanently changing shape, is exactly what brittle means.

Check — multiple choice

Response of four solids to a sharp hammer blow
Solid	Response when struck
A	dents; the dent remains
B	cracks into several pieces
C	bends, then springs back straight
D	no visible change

Use the table. Which solid is brittle?

ASolid B ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Solid B fractured into pieces instead of permanently changing shape — the defining response of a brittle solid.

BSolid D

Feedback if chosen: This option is wrong — solid D shows no response at all to this blow. Brittleness is shown by fracture, and nothing fractured.

CSolid C

Feedback if chosen: This option is wrong — solid C returned to its original shape, so it neither fractured nor deformed permanently. That is not brittleness.

DSolid A

Feedback if chosen: This option is wrong — solid A took a permanent new shape without fracturing. That is deformation, not the fracture that defines brittleness.

U2-S04 · LESSON 30 · CSP-030

Malleability

Outcome: State what it means for a solid to be malleable.

Part 1Hammered into sheets

Aluminum foil starts as a thick block of metal. Factories roll it thinner and thinner until it is a sheet you can wrap food in — reshaped enormously, and still in one piece.

A solid that can be hammered or rolled into sheets without shattering is called malleable.

You've already seen that a brittle solid fractures when stressed. A malleable solid responds the opposite way: the stress reshapes it, and it holds together.

What you are expected to be able to do

State that a malleable solid can be hammered or rolled into sheets without shattering.

Part 2Check what you just learned

Check — multiple choice

Which observation shows that gold is malleable?

AGold shatters into small fragments when it is struck with a hammer

Feedback if chosen: This option is wrong — shattering under the hammer is brittleness, the opposite of malleability.

BGold keeps a bright shine even after years of handling

Feedback if chosen: This option is wrong — shine is an appearance, not a response to stress. Malleability is shown by reshaping into sheets without shattering.

CGold melts at 1064 °C when it is heated in a furnace

Feedback if chosen: This option is wrong — a melting point describes the response to temperature, not the response to hammering or rolling.

DGold can be hammered into sheets thin enough to let light through ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Being hammered into thin sheets without shattering is exactly what malleable means.

Check — multiple choice

Response of four solids to being rolled under pressure
Solid	Response when rolled
A	cracks into pieces
B	springs back to its original shape
C	flattens into a sheet and holds the shape
D	crumbles to powder

Use the table. Which solid is malleable?

ASolid A

Feedback if chosen: This option is wrong — cracking into pieces is fracture, so solid A is brittle, not malleable.

BSolid B

Feedback if chosen: This option is wrong — springing back means solid B kept no new shape; malleability is taking a new shape as a sheet and holding it together.

CSolid C ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Solid C was rolled into a sheet and held the new shape without shattering — the defining response of a malleable solid.

DSolid D

Feedback if chosen: This option is wrong — crumbling to powder is fracture into many fragments, not reshaping. Solid D failed to hold together.

U2-S04 · LESSON 31 · CSP-031

Ductility

Outcome: State what it means for a solid to be ductile.

Part 1Drawn into wire

Electrical wire is made by pulling a copper rod through a series of smaller and smaller holes, until meter after meter of thin wire comes out the far side — unbroken.

A solid that can be drawn out into a wire without shattering is called ductile.

You've already seen that a malleable solid reshapes into sheets. Ductile describes a different reshaping: being drawn out into a wire.

What you are expected to be able to do

State that a ductile solid can be drawn into a wire without shattering.

Part 2Check what you just learned

Check — multiple choice

Which observation shows that a metal sample is ductile?

AIt can be pulled through a die into a thin wire without breaking ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Being drawn out into a wire without breaking is exactly what ductile means.

BIt can be rolled between heavy rollers into a thin sheet without breaking

Feedback if chosen: This option is wrong — rolling into sheets shows malleability. Ductility is specifically about being drawn out into a wire.

CIt cracks apart as soon as it is pulled through a die

Feedback if chosen: This option is wrong — cracking apart under the pull is fracture, the opposite of surviving the draw.

DIt carries an electric current when connected in a circuit

Feedback if chosen: This option is wrong — carrying a current is electrical conductivity, not a response to being pulled or reshaped.

Check — multiple choice

A jeweler pulls a small bar of silver through smaller and smaller holes until it becomes a fine, unbroken thread. Which term describes the silver?

ABrittle

Feedback if chosen: This option is wrong — a brittle solid fractures under stress. The silver survived the drawing intact, which is the opposite behaviour.

BMalleable

Feedback if chosen: This option is wrong — malleable describes hammering or rolling into sheets. The silver was drawn out into a thread, which is ductility.

CDuctile ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The silver was drawn into a wire without shattering — the defining behaviour of a ductile solid.

DMolten

Feedback if chosen: This option is wrong — molten means melted to a liquid. The silver stayed solid the whole time; it was reshaped, not melted.

U2-S04 · LESSON 32 · CSP-032

High melting points as evidence for ionic bonding

Outcome: Identify a relatively high melting point as evidence consistent with an ionic substance.

Part 1What a high melting point suggests

A jar of white crystals has lost its label. You cannot see the particles or the bonding — but you can measure the substance's properties.

You've already seen that measured properties give the best evidence for characterizing bonding. Melting point is the first property in the kit.

You've already seen that in an ionic solid such as NaCl, cations and anions repeat throughout a crystal lattice, and that a melting point is the temperature at which a solid changes to a liquid.

Here are measured melting points for some ionic substances and some substances made of small molecules.

Measured melting points
Substance	Bonding	Melting point (°C)
Sodium chloride	ionic	801
Potassium bromide	ionic	734
Magnesium oxide	ionic	2852
Water	molecular	0
Bromine	molecular	−7
Naphthalene	molecular	80

The pattern: the ionic substances in the table melt hundreds of degrees higher than the molecular ones.

So when you are deciding between ionic and molecular for an unknown, a relatively high melting point is evidence supporting ionic.

Treat it as support, not proof. One property narrows the options; it does not settle the classification on its own.

The word relatively matters too: judge the melting point against the other substances in the comparison, not against a fixed cutoff.

What you are expected to be able to do

Identify a relatively high melting point as evidence consistent with an ionic substance rather than a molecular one.

Part 2Check what you just learned

Check — multiple choice

Measured melting points of two unknowns
Substance	Melting point (°C)
X	770
Y	−23

Substances X and Y are each either ionic or made of small molecules. Use the measured melting points in the table. Which claim does the evidence support?

AX is the ionic substance and Y is the molecular one ✓ KEY — this is the correct answer

Feedback if chosen: Correct. X melts hundreds of degrees higher than Y, and in an ionic-versus-molecular comparison the relatively high melting point supports ionic.

BY is the ionic substance and X is the molecular one

Feedback if chosen: This option is wrong — you reversed the pattern. In these comparisons the ionic substances sit at the high end of the melting points and the molecular ones at the low end.

CThe melting points prove that X is ionic

Feedback if chosen: This option is wrong — you turned support into proof. The high melting point is evidence consistent with X being ionic; on its own it does not prove it, which is why several properties are weighed together.

DMelting points carry no information about bonding

Feedback if chosen: This option is wrong — you threw away usable evidence. The melting-point pattern you studied separates ionic from molecular substances by hundreds of degrees.

Check — multiple choice

Measured melting points of four unknowns
Substance	Melting point (°C)
W	5.5
X	734
Y	−7
Z	114

Each substance in the table is either ionic or made of small molecules. Which one is most likely ionic?

ASubstance W

Feedback if chosen: This option is wrong — W melts near room temperature, at the low end of this comparison, which supports a molecular substance rather than an ionic one.

BSubstance X ✓ KEY — this is the correct answer

Feedback if chosen: Correct. X melts hundreds of degrees above the others — the relatively high melting point that supports ionic bonding in this comparison.

CSubstance Y

Feedback if chosen: This option is wrong — Y melts below room temperature, the lowest of the set, which supports a molecular substance rather than an ionic one.

DSubstance Z

Feedback if chosen: This option is wrong — Z's 114 °C is modest next to X's 734 °C. Relatively low in this comparison, it supports a molecular substance.

U2-S04 · LESSON 33 · CSP-032A

Why ionic melting points are high

Outcome: Explain why ionic substances have relatively high melting points.

Part 1Back to the lattice

You've already seen that a relatively high melting point — NaCl stays solid until 801 °C — is evidence supporting an ionic substance. Here is the reason, and it starts with the lattice.

Here is the particulate model of solid NaCl: cations and anions repeating throughout a crystal lattice.

You've already seen that opposite charges attract — a Coulombic attraction.

In the lattice, every cation sits among anions and every anion among cations, so these attractions run throughout the whole crystal.

For the solid to melt, the ions must break out of their fixed positions and move past one another.

That takes energy — enough to overcome those strong attractions.

Ionic substances have relatively high melting points because melting requires enough energy to overcome the strong Coulombic attractions holding oppositely charged ions throughout the lattice.

That is why NaCl stays solid until about 801 °C: below that temperature, the ions do not have enough energy to break away from their oppositely charged neighbors.

What you are expected to be able to do

Explain why ionic substances have relatively high melting points, in terms of the strong Coulombic attractions holding oppositely charged ions throughout the lattice.

Part 2Check what you just learned

Check — multiple choice

Why does NaCl stay solid until about 801 °C?

AMelting requires breaking the strong covalent bonds that join the Na and Cl atoms inside each NaCl molecule

Feedback if chosen: This option is wrong — you've already seen that solid NaCl contains no molecules: it is a lattice of Na⁺ and Cl⁻ ions, and the attractions holding it together are ionic, not covalent bonds.

BMelting requires enough energy to overcome the strong Coulombic attractions holding oppositely charged ions throughout the lattice ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Every ion is held by strong Coulombic attractions to its oppositely charged neighbors throughout the lattice, and melting cannot happen until there is enough energy to overcome them.

CThe ions are too heavy to move, and heavy particles need very high temperatures before they can move at all

Feedback if chosen: This option is wrong — you reached for mass, but the ions move plenty below 801 °C; they vibrate in place. What holds them in position is the Coulombic attraction of their oppositely charged neighbors.

DSodium is a metal, and a solid that contains a metal keeps its shape until it reaches a metal's melting temperature

Feedback if chosen: This option is wrong — you reached for the element instead of the bonding, and NaCl's melting behaviour comes from its ionic lattice, not from sodium alone. Metal or not, the question is what holds the particles in place.

Check — multiple choice

Potassium bromide, KBr, is also an ionic solid; it melts at 734 °C. What must the added energy accomplish for KBr to melt?

ABreak each K–Br unit apart into two separate neutral atoms, one K and one Br

Feedback if chosen: This option is wrong — melting does not make neutral atoms. The K⁺ and Br⁻ ions stay ions in the liquid; they just gain the freedom to move past one another.

BBreak the covalent bonds joining each K atom to its neighboring Br atom

Feedback if chosen: This option is wrong — KBr's bonding is ionic, not covalent. What melting must overcome is the Coulombic attraction between the oppositely charged ions.

CFree the ions from the Coulombic attractions of their oppositely charged neighbors ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Melting means the K⁺ and Br⁻ ions break out of their fixed lattice positions, and that requires enough energy to overcome the strong Coulombic attractions of their oppositely charged neighbors.

DStrip electrons away from the Br⁻ anions so that the charges are released

Feedback if chosen: This option is wrong — no electrons are removed when a solid melts. The ions keep their charges; melting only frees them from their fixed positions in the lattice.

U2-S04 · LESSON 34 · CSP-033

Conductivity evidence for ionic bonding

Outcome: Identify the ionic conductivity pattern: no as a solid, yes when molten or dissolved.

Part 1A conductivity pattern

You've already seen that a material conducts only when it contains charged particles that are free to move.

Here is what a conductivity meter shows for sodium chloride in three situations.

Conductivity of sodium chloride in three situations
Sample	Conducts?
Solid NaCl	no
Molten NaCl (above 801 °C)	yes
NaCl dissolved in water	yes

Solid: no. Molten: yes. Dissolved: yes.

This pattern — no conduction as a solid, conduction once molten or dissolved — is evidence consistent with an ionic substance.

For the dissolved case, the substance must actually dissolve and give ions that can move. A substance that does not dissolve gives no reading either way.

As with melting point, treat the pattern as support for the classification, weighed alongside the other measured properties.

What you are expected to be able to do

Identify nonconductivity as a solid, together with conductivity when molten or dissolved, as evidence consistent with an ionic substance.

Part 2Check what you just learned

Check — multiple choice

Conductivity of unknown substance P
Sample	Conducts?
Solid P	no
Molten P	yes

Substance P was tested as shown in the table. Which bonding does this evidence support for P?

AMetallic

Feedback if chosen: This option is wrong — you kept only the molten-yes reading. A metallic solid conducts in the solid state, and solid P does not.

BIonic ✓ KEY — this is the correct answer

Feedback if chosen: Correct. No conduction as a solid but conduction once molten is the ionic pattern — exactly what you saw for NaCl.

CMolecular

Feedback if chosen: This option is wrong — you kept only the solid-no reading. A molecular substance would not conduct once molten either, and molten P does conduct.

DThese measurements cannot identify the bonding type

Feedback if chosen: This option is wrong — this pattern is exactly the evidence the lesson is about: solid-no together with molten-yes supports an ionic substance.

Check — multiple choice

A white solid does not conduct. It dissolves completely in water, and the solution conducts. Which statement reads this evidence correctly?

AThe evidence proves the solid is ionic

Feedback if chosen: This option is wrong — you turned support into proof. The pattern is evidence consistent with an ionic solid; classification still rests on weighing all the measured properties.

BThe evidence supports a metallic solid, because the solution conducts

Feedback if chosen: This option is wrong — a metallic solid conducts before anything is dissolved, and metals do not dissolve in water this way. The solid-no reading counts against metallic.

CThe evidence supports an ionic solid ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Nonconductive as a solid, then conductive once dissolved — the substance dissolved and gave ions that can move. That is the ionic pattern.

DThe evidence supports a molecular substance, because the solid does not conduct

Feedback if chosen: This option is wrong — you kept only the solid-no reading. A molecular substance's solution would not conduct either, and this solution does.

U2-S04 · LESSON 35 · CSP-033A

Why ionic substances conduct only when molten or dissolved

Outcome: Explain why an ionic substance conducts when molten or dissolved but not as a solid.

Part 1Fixed ions, mobile ions

You've already seen the pattern: solid NaCl does not conduct, but molten or dissolved NaCl does. Here is the solid's particulate model again.

The charged particles are there all along — the lattice is packed with Na⁺ and Cl⁻ ions.

But in the solid, each ion is held in a fixed position by its oppositely charged neighbors. Charged, yes; free to move, no. So no current flows.

As the solid melts, the lattice comes apart and the ions become free to move.

Dissolving frees them the same way: the ions separate into the water and can move.

Ionic substances conduct electricity only when molten or dissolved because the ions are fixed in place in the solid lattice and become mobile charge carriers once freed.

What you are expected to be able to do

Explain why an ionic substance conducts when molten or dissolved but not as a solid: the ions are fixed in the solid lattice and become mobile charge carriers once freed.

Part 2Check what you just learned

Check — multiple choice

Why does molten NaCl conduct while solid NaCl does not?

AThe solid contains no charged particles at all; the ions only come into existence when the solid melts

Feedback if chosen: This option is wrong — the ions exist in the solid too; the lattice is made of them. What melting changes is not whether there are ions but whether they can move.

BMelting releases electrons from the ions, and the moving electrons carry the current

Feedback if chosen: This option is wrong — you transplanted the metal's story. In molten NaCl the moving charges are the ions themselves, not freed electrons.

CMelting splits each ion into neutral atoms, and the neutral atoms then carry the current

Feedback if chosen: This option is wrong — melting does not neutralize the ions, and neutral atoms could not carry a current anyway. The ions stay charged and become free to move.

DIn the solid the ions are fixed in the lattice; once molten they are free to move and carry the current ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The charged particles are present in both states; only their freedom changes. Fixed in the solid lattice, the ions carry no current; freed by melting, they become mobile charge carriers.

Check — multiple choice

Solid calcium chloride, CaCl₂, does not conduct, but its solution in water does. What carries the current through the solution?

AThe mobile Ca²⁺ and Cl⁻ ions ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Dissolving freed the Ca²⁺ and Cl⁻ ions from the lattice; now mobile, these charged particles carry the current.

BDelocalized electrons moving through the water

Feedback if chosen: This option is wrong — delocalized electrons are the metal's charge carriers. In a dissolved ionic substance, the mobile charges are the ions themselves.

CNeutral CaCl₂ units drifting through the water

Feedback if chosen: This option is wrong — a neutral unit carries no charge, so its drifting moves no charge. The current is carried by the separated, charged ions.

DThe water molecules themselves

Feedback if chosen: This option is wrong — water molecules are neutral, so their motion carries no charge. The dissolved Ca²⁺ and Cl⁻ ions are what carry the current.

U2-S04 · LESSON 36 · CSP-034

Brittleness as evidence for ionic bonding

Outcome: Identify brittleness as evidence consistent with an ionic solid.

Part 1Struck with a hammer

Strike a copper coin and it dents. Strike a crystal of rock salt the same way and it shatters into fragments.

You've already seen that a solid that fractures rather than permanently changing shape is brittle.

Ionic solids are typically brittle.

So when an unknown solid shatters under stress, that brittleness is evidence supporting ionic bonding rather than metallic bonding.

Support, not proof: some non-ionic solids are brittle too, so brittleness is weighed together with the other measured properties.

What you are expected to be able to do

Identify brittleness as evidence consistent with an ionic solid rather than a metallic one.

Part 2Check what you just learned

Check — multiple choice

An unknown crystalline solid shatters into fragments when struck. Which claim does this evidence support?

AThe solid is ionic rather than metallic ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Shattering is brittleness, and brittleness supports ionic bonding over metallic — a metal typically dents or flattens instead.

BThe solid is metallic rather than ionic

Feedback if chosen: This option is wrong — you attached the behaviour to the wrong side. A metallic solid typically dents or flattens under a blow; fracturing into fragments points toward ionic.

CThe solid is proved to be ionic

Feedback if chosen: This option is wrong — you turned support into proof. Some non-ionic solids shatter too, so brittleness supports ionic bonding but cannot prove it alone.

DShattering cannot identify the bonding type

Feedback if chosen: This option is wrong — you threw away usable evidence. How a solid responds to stress is one of the measured properties used to characterize its bonding.

Check — multiple choice

Which observation about a crystalline solid supports ionic bonding rather than metallic bonding?

AIt flattens under a hammer blow

Feedback if chosen: This option is wrong — flattening without fracture is malleability, which is evidence on the metallic side, not the ionic side.

BIt shatters under a hammer blow ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Shattering is brittleness, and brittleness is the deformation evidence that supports ionic bonding over metallic.

CIt conducts in the solid state

Feedback if chosen: This option is wrong — conduction in the solid state is evidence for metallic bonding; you've already seen that an ionic solid does not conduct until molten or dissolved.

DIt has a shiny surface

Feedback if chosen: This option is wrong — shine is an appearance, not one of the measured behaviours used here: melting point, conductivity, and the response to stress.

U2-S04 · LESSON 37 · CSP-034A

Why ionic solids shatter

Outcome: Explain why ionic solids are brittle.

Part 1Sliding a layer of ions

You've already seen that an NaCl crystal shatters when struck — brittleness that supports ionic bonding. The reason is in the lattice. Here it is before any stress: every ion sits among oppositely charged neighbors.

A hammer blow forces one layer of ions to slide past the layer below.

Here is the same lattice just after the top layer has slid one position: now cations sit alongside cations, and anions alongside anions.

You've already seen that like charges repel. The slid layer is now pushed away by the layer below instead of being held by it.

That repulsion forces the layers apart, and the crystal fractures along the slip.

Ionic solids are brittle because stress that slides one lattice layer brings like-charged ions alongside each other, and the resulting repulsion fractures the crystal.

That is why an NaCl crystal shatters when struck instead of denting.

What you are expected to be able to do

Explain why ionic solids are brittle: sliding a lattice layer brings like-charged ions alongside each other, and the repulsion fractures the crystal.

Part 2Check what you just learned

Check — multiple choice

The model shows a KBr lattice just after a stress has slid the top layer one position. What has the slide changed?

ALike-charged ions now sit alongside each other, so the layers repel ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Before the slide, opposite charges faced each other across the layer boundary; a one-position slide lines up like with like, and like charges repel.

BOppositely charged ions still face each other, so the layers attract as before

Feedback if chosen: This option is wrong — you missed the offset. The one-position slide moved every ion in the top layer over by one place, so like charges now face each other where opposite charges used to.

CThe ions in the top layer have become neutral atoms

Feedback if chosen: This option is wrong — sliding moves the ions; it does not neutralize them. Each ion keeps its charge, which is exactly why the new like-against-like arrangement repels.

DThe ions in the top layer have swapped their charges

Feedback if chosen: This option is wrong — no charges change hands in a slide. The ions keep their identities; only their positions relative to the layer below have shifted.

Check — multiple choice

Why does an ionic crystal shatter when struck, rather than denting?

AThe blow breaks the covalent bonds that run from ion to ion through the crystal, and the crystal falls apart

Feedback if chosen: This option is wrong — an ionic crystal is held together by Coulombic attractions between ions, not covalent bonds. The fracture comes from like-charge repulsion after a layer slides.

BThe blow squeezes the oppositely charged ions closer together, and the strengthened attraction snaps the crystal apart

Feedback if chosen: This option is wrong — you made attraction the culprit. Attraction holds the crystal together; the fracture happens because the slide replaces attraction with repulsion between like-charged ions.

CThe blow slides a lattice layer, bringing like-charged ions alongside each other, and the resulting repulsion fractures the crystal ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The slide lines up like-charged ions across the slip plane, attraction gives way to repulsion, and the repulsion forces the crystal apart.

DThe ions are hard little spheres, and a solid built from hard particles cracks instead of denting

Feedback if chosen: This option is wrong — you reached for hardness, which is a different property. The fracture is electrical in origin: sliding brings like charges alongside each other and they repel.

Check — multiple choice

A student explains: "Ionic crystals shatter because the attractions between the ions are weak." What is wrong with this explanation?

AThe attractions are strong, not weak; the fracture happens because a sliding layer brings like charges alongside each other, and they repel ✓ KEY — this is the correct answer

Feedback if chosen: Correct. You've already seen that the Coulombic attractions in an ionic lattice are strong — they are why the melting point is high. Brittleness comes from a different event: a slide that turns neighbors from opposite charges into like charges, which repel.

BNothing — the attractions between the ions are weak, and weak attractions are why ionic crystals shatter

Feedback if chosen: This option is wrong — you accepted the student's premise. If the attractions were weak, the melting point would not be near 801 °C. The fracture comes from like-charge repulsion after a slide, not from weak attraction.

CThe attractions are weak, but weak attractions only explain why the crystal melts, not why it shatters

Feedback if chosen: This option is wrong — you kept the false premise that the attractions are weak. They are strong in both stories; shattering happens because sliding replaces attraction with like-charge repulsion.

DThe ions lose their charges once the crystal has formed, so neither attraction nor repulsion plays any part in the fracture

Feedback if chosen: This option is wrong — the ions keep their charges permanently in the crystal; the strong attractions between them are what hold the lattice together in the first place.

U2-S04 · LESSON 38 · CSP-035

Low melting points as evidence for molecular substances

Outcome: Identify a relatively low melting point as evidence consistent with a molecular substance.

Part 1What a low melting point suggests

You've already seen that some substances are made of discrete neutral molecules — such as H₂O — rather than a lattice of ions.

Here are measured melting points for some molecular substances, with an ionic solid for comparison.

Measured melting points
Substance	Bonding	Melting point (°C)
Methane	molecular	−182
Ethanol	molecular	−114
Water	molecular	0
Naphthalene	molecular	80
Iodine	molecular	114
Sodium chloride	ionic	801

The molecular substances all melt low — several below room temperature — while the ionic solid melts hundreds of degrees higher.

So a relatively low melting point is evidence supporting a substance made of small molecules rather than an ionic one.

Support, not proof — and relatively low means low against the other candidates in the comparison, not below some fixed number.

What you are expected to be able to do

Identify a relatively low melting point as evidence consistent with a substance made of discrete neutral molecules.

Part 2Check what you just learned

Check — multiple choice

Measured melting points
Substance	Melting point (°C)
Q	−23
Sodium chloride (an ionic solid)	801

Substance Q is either ionic or made of small molecules. An ionic solid is shown for comparison. Which claim does the measured melting point in the table support?

AQ is ionic

Feedback if chosen: This option is wrong — you attached the evidence to the wrong side. Q melts hundreds of degrees below the ionic comparison, and the low end supports molecular, not ionic.

BQ is made of small molecules ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Q melts at −23 °C, hundreds of degrees below the ionic comparison — a relatively low melting point, which supports a substance made of small molecules over an ionic solid.

CQ is proved to be molecular

Feedback if chosen: This option is wrong — you turned support into proof. The low melting point is consistent with a molecular substance, but classification rests on weighing all the measured properties.

DA melting point carries no information about bonding

Feedback if chosen: This option is wrong — you threw away usable evidence. In an ionic-versus-molecular comparison, the melting point separates the two by hundreds of degrees.

Check — multiple choice

Measured melting points of four unknowns
Substance	Melting point (°C)
A	801
B	734
C	44
D	993

Each substance in the table is either ionic or made of small molecules. Which is most likely made of small molecules?

ASubstance A

Feedback if chosen: This option is wrong — A melts at 801 °C, hundreds of degrees above C. In this comparison that is the high end, which supports ionic bonding.

BSubstance B

Feedback if chosen: This option is wrong — B melts at 734 °C, near the high end of the set. A relatively high melting point supports ionic, not molecular.

CSubstance C ✓ KEY — this is the correct answer

Feedback if chosen: Correct. C melts at 44 °C, far below every other substance in the set — the relatively low melting point that supports a molecular substance.

DSubstance D

Feedback if chosen: This option is wrong — D has the highest melting point in the table, and the high end of the comparison supports ionic bonding, not molecular.

U2-S04 · LESSON 39 · CSP-035A

Why molecular melting points are low

Outcome: Explain why molecular substances have relatively low melting points.

Part 1What melting has to separate

You've already seen that a relatively low melting point — ice melts at 0 °C — is evidence supporting a molecular substance.

There is a puzzle in that fact. The O–H covalent bonds inside each water molecule are strong, yet ice melts at a temperature your kitchen freezer sits below.

The resolution: a molecular solid contains two different kinds of attraction, and melting only has to overcome one of them.

Within each molecule, the atoms are joined by covalent bonds — you've already seen these as shared pairs of valence electrons.

Between neighboring molecules there are also attractions — and these are much weaker than the covalent bonds within the molecules.

Substances made of discrete neutral molecules have relatively low melting points because the attractions between molecules are much weaker than the covalent bonds within them, so melting separates whole molecules without breaking any covalent bonds.

When ice melts at 0 °C, whole H₂O molecules separate from one another; every O–H bond stays intact.

Because the attractions between molecules are weak, only a little energy is needed to overcome them — so the temperature at which melting happens is low.

What you are expected to be able to do

Explain why molecular substances have relatively low melting points: the attractions between molecules are much weaker than the covalent bonds within them, so melting separates whole molecules without breaking any covalent bonds.

Part 2Check what you just learned

Check — multiple choice

Why does ice melt at 0 °C even though the O–H covalent bonds in water are strong?

AMelting breaks the O–H covalent bonds one by one, because the bonds are much weaker than they appear

Feedback if chosen: This option is wrong — you sent the melting energy to the wrong attraction. The O–H bonds are strong and stay intact; melting overcomes only the weak attractions between molecules.

BMelting only separates whole molecules, overcoming attractions much weaker than the covalent bonds — no O–H bond breaks ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Melting pulls whole molecules away from each other. The attractions between molecules are much weaker than the covalent bonds within them, so only a little energy — a low temperature — is needed.

CThe O–H covalent bonds break, but only in about half of the water molecules, so less energy is needed

Feedback if chosen: This option is wrong — no O–H bonds break in any of the molecules when ice melts. The molecules separate from one another whole and intact.

DWater molecules are small and light, and a solid made of small, light molecules melts at a low temperature

Feedback if chosen: This option is wrong — size and mass are not the mechanism. The melting point is set by what melting must overcome: the weak attractions between the molecules, and no covalent bond breaks.

Check — multiple choice

Solid iodine, I₂, melts at 114 °C. What happens to the I–I covalent bonds when it melts?

ANothing — whole I₂ molecules separate from one another, and every I–I bond stays intact ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Melting overcomes only the weak attractions between the I₂ molecules; the covalent bond inside each molecule is far stronger and is untouched.

BAll of the I–I bonds break, so the liquid consists of separate iodine atoms rather than molecules

Feedback if chosen: This option is wrong — breaking every I–I bond would take far more energy than 114 °C supplies. Melting separates whole molecules; it does not take them apart.

CThe I–I bonds break in about half of the molecules, and the intact half holds the liquid together

Feedback if chosen: This option is wrong — melting is not a partial bond-breaking. No I–I bonds break; the molecules leave the solid whole, held less tightly to each other but intact inside.

DThe I–I bonds convert into weaker attractions between the molecules as the solid melts

Feedback if chosen: This option is wrong — a covalent bond does not turn into anything when a solid melts. The bond within each molecule and the weak attraction between molecules are different things, and only the weak attractions are overcome.

U2-S04 · LESSON 40 · CSP-036

Nonconductivity as evidence for molecular substances

Outcome: Identify electrical nonconductivity as evidence consistent with a molecular substance.

Part 1No reading in any state

You've already seen that a material conducts only when it contains charged particles that are free to move.

Here is what a conductivity meter shows for some molecular substances, with molten NaCl for comparison.

Conductivity measurements
Sample	Conducts?
Liquid bromine, Br₂	no
Liquid ethanol	no
Molten naphthalene	no
Molten NaCl (ionic)	yes

The molecular substances give no reading — solid or liquid, the answer stays no.

So when a substance conducts in no state, that nonconductivity is evidence supporting a substance made of neutral molecules.

Support, not proof — nonconductivity is one measured property, weighed together with the others.

What you are expected to be able to do

Identify electrical nonconductivity as evidence consistent with a substance composed of neutral molecules.

Part 2Check what you just learned

Check — multiple choice

A pure liquid does not conduct an electric current. Which claim does this evidence support?

AThe liquid is made of neutral molecules ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A liquid's particles are free to move, so a nonconducting liquid must lack mobile charged particles — evidence supporting a substance made of neutral molecules.

BThe liquid is a molten ionic substance

Feedback if chosen: This option is wrong — you've already seen that a molten ionic substance conducts, because melting frees its ions to move. A nonconducting liquid points away from ionic.

CThe liquid is a molten metal

Feedback if chosen: This option is wrong — a metal conducts as a solid and stays conductive when molten. A liquid giving no reading points away from metallic.

DThe liquid is proved to be molecular

Feedback if chosen: This option is wrong — you turned support into proof. Nonconductivity is consistent with a molecular substance, but the classification rests on weighing all the measured properties.

Check — multiple choice

Which conductivity pattern supports a substance made of neutral molecules?

AConducts as a liquid but not as a solid

Feedback if chosen: This option is wrong — conduction only after melting is the ionic pattern: the ions are freed to move when the solid melts.

BConducts both as a solid and as a liquid

Feedback if chosen: This option is wrong — conduction in both states is the metallic pattern, not the molecular one.

CConducts as a solid but not as a liquid

Feedback if chosen: This option is wrong — no bonding type behaves this way: a substance whose charged particles can move in the solid still has them, still mobile, once it melts.

DConducts neither as a solid nor as a liquid ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A molecular substance gives no reading in either state — with no mobile charged particles to be freed, melting changes nothing.

U2-S04 · LESSON 41 · CSP-036A

Why molecular substances do not conduct

Outcome: Explain why molecular substances do not conduct electricity.

Part 1Neutral through and through

You've already seen that molecular substances conduct in no state — pure liquid bromine gives no reading.

The molecules of a liquid are certainly free to move. Movement is not what is missing.

Molecular substances do not conduct electricity because their molecules are electrically neutral, so no mobile charged particles are available to carry current.

Each Br₂ molecule drifts freely through the liquid, but it carries no overall charge — so its motion moves no charge.

Conduction needs particles that are both charged and free to move. Molten NaCl has both. Liquid bromine has movement without charge.

What you are expected to be able to do

Explain why molecular substances do not conduct: the molecules are neutral, so there are no mobile charged particles to carry current.

Part 2Check what you just learned

Check — multiple choice

Why does pure liquid bromine not conduct an electric current?

AIts molecules are locked in place in the liquid and cannot move from one position to another

Feedback if chosen: This option is wrong — the molecules of a liquid move freely. What liquid bromine lacks is not movement but charge on the moving particles.

BIts molecules move from place to place too slowly, even as a liquid, for their motion to carry a current

Feedback if chosen: This option is wrong — you blamed speed, but no speed would help: a particle that carries no charge moves no charge, however fast it travels.

CIts molecules are electrically neutral, so no mobile charged particles are available to carry current ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The Br₂ molecules are free to move but electrically neutral, so their motion carries no charge — and with no mobile charged particles, no current can flow.

DConduction is a property of metals, and bromine sits on the nonmetal side of the periodic table

Feedback if chosen: This option is wrong — you tied conduction to metals, but molten NaCl conducts and it is not a metal. What matters is mobile charged particles, and liquid bromine has none.

Check — multiple choice

Salt water conducts an electric current; sugar water does not. What accounts for the difference?

ASalt dissolves in the water and sugar does not dissolve at all

Feedback if chosen: This option is wrong — both dissolve. The difference is what each releases into the water: salt gives ions, sugar gives neutral molecules.

BSalt dissolves to give mobile ions; sugar dissolves as whole neutral molecules ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Dissolved salt supplies mobile charged particles — ions — that carry the current. Sugar molecules enter the water whole and neutral, so they add nothing that can carry charge.

CSugar's particles are heavier than salt's ions, so they move through the water too slowly

Feedback if chosen: This option is wrong — you blamed speed and mass, but a neutral particle carries no charge at any speed. Sugar's molecules are neutral; that is why they cannot carry the current.

DDissolved sugar gives ions, but they stay fixed in place in the water

Feedback if chosen: This option is wrong — dissolved sugar gives no ions at all. Its molecules stay whole and neutral in the water; nothing charged is added, fixed or free.

U2-S04 · LESSON 42 · CSP-037

Solid-state conductivity as evidence for metallic bonding

Outcome: Identify conduction in the solid state as evidence consistent with metallic bonding.

Part 1Conducting while still solid

Copper wire carries a current exactly as it is — cold, solid metal. No melting, no dissolving.

Here is what a conductivity meter shows for some solids at room temperature.

Conductivity of solids at room temperature
Solid	Conducts?
Copper	yes
Aluminum	yes
Sodium chloride	no
Iodine	no

You've already seen that ionic solids do not conduct until they are molten or dissolved, and that molecular substances do not conduct in any state.

The metals stand apart: they conduct in the solid state.

So conductivity in the solid state is evidence supporting metallic bonding — of the three bonding types you are comparing, only the metallic one conducts as a solid.

Support, not proof — conductivity is weighed together with the other measured properties.

What you are expected to be able to do

Identify electrical conductivity in the solid state as evidence consistent with metallic bonding.

Part 2Check what you just learned

Check — multiple choice

An unknown solid conducts an electric current at room temperature. Which bonding does this evidence support?

AMetallic ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Of the three bonding types being compared, the metallic one is the one that conducts in the solid state.

BIonic

Feedback if chosen: This option is wrong — you've already seen that an ionic solid does not conduct until it is molten or dissolved; a conducting solid points away from ionic.

CMolecular

Feedback if chosen: This option is wrong — a molecular substance does not conduct in any state, so a conducting solid points away from molecular.

DNo conclusion can be drawn from a single reading

Feedback if chosen: This option is wrong — one reading is not proof, but it is usable evidence: among ionic, molecular, and metallic substances, conduction in the solid state supports the metallic one.

Check — multiple choice

Conductivity of four unknown solids
Solid	Conducts as a solid?
W	no
X	no
Y	yes
Z	no

Four unknown solids were tested as shown in the table. Which is most likely a metal?

ASolid W

Feedback if chosen: This option is wrong — W does not conduct as a solid, and a metal would. Its bonding is more likely ionic or molecular.

BSolid X

Feedback if chosen: This option is wrong — X gives no reading as a solid, which points away from metallic bonding.

CSolid Y ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Y is the one solid that conducts in the solid state — the evidence that supports metallic bonding.

DSolid Z

Feedback if chosen: This option is wrong — Z does not conduct as a solid, and a metal would. The conducting solid in the set is Y.

U2-S04 · LESSON 43 · CSP-037A

Why metals conduct as solids

Outcome: Explain why metals conduct electricity in the solid state.

Part 1Charge carriers already in place

You've already seen that solid copper conducts — evidence supporting metallic bonding. Here is the particulate model of a metal again: positive metal ions among valence electrons that are delocalized — shared across the whole solid rather than assigned to one atom.

Metals conduct electricity in the solid state because their delocalized valence electrons are free to move throughout the lattice.

The ions stay in their lattice positions; the delocalized electrons do the moving.

Compare the ionic solid: there, every charged particle is an ion locked in the lattice, so nothing mobile is left to carry current. In the metal, mobile charge is built in.

So a metal needs no melting and no dissolving — its charge carriers are free to move from the start.

What you are expected to be able to do

Explain why metals conduct in the solid state: their delocalized valence electrons are free to move throughout the lattice.

Part 2Check what you just learned

Check — multiple choice

What carries the current when solid copper conducts?

AIts delocalized valence electrons, which are free to move throughout the lattice ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The delocalized valence electrons belong to no single atom and are free to move throughout the lattice — they are the mobile charged particles that carry the current.

BCu ions breaking out of the lattice and moving from one end of the wire to the other

Feedback if chosen: This option is wrong — the positive ions stay in their lattice positions; it is the delocalized electrons that move through the solid.

CNeutral copper atoms bumping into one another and passing the current along the wire

Feedback if chosen: This option is wrong — a neutral particle carries no charge, so neutral atoms could not carry a current. The metal's mobile charges are its delocalized electrons.

DIons that break free from the lattice when the wire warms up

Feedback if chosen: This option is wrong — you imported the ionic story, where charge carriers must be freed. Copper conducts cold and solid because its delocalized electrons are free to move already.

Check — multiple choice

Solid silver conducts an electric current, but solid silver chloride does not. What accounts for the difference?

ASilver's ions are free to slide through the solid and carry its current; silver chloride's ions cannot slide

Feedback if chosen: This option is wrong — in both solids the ions are fixed in their lattice positions. What silver has and silver chloride lacks is mobile charge: delocalized electrons.

BSilver has delocalized valence electrons free to move; in silver chloride every charged particle is an ion fixed in the lattice ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The metal's delocalized valence electrons are mobile charge carriers present in the solid; the ionic solid's charges are all ions, fixed in place until melting or dissolving frees them.

CSilver chloride has no charged particles anywhere in it, so it has nothing that could carry a current

Feedback if chosen: This option is wrong — silver chloride is full of charged particles: Ag⁺ and Cl⁻ ions. They cannot carry a current because they are fixed in the lattice, not because they are absent.

DSilver is the shinier of the two solids, and a bright shine goes together with conduction

Feedback if chosen: This option is wrong — shine is an appearance, not a mechanism. The difference in conduction comes from what can move: delocalized electrons in the metal, nothing in the ionic solid.

U2-S04 · LESSON 44 · CSP-038

Malleability as evidence for metallic bonding

Outcome: Identify malleability as evidence consistent with metallic bonding.

Part 1Rolled thin without cracking

Gold leaf begins as a small bar of gold and is hammered down to a sheet a few hundred atoms thick — reshaped enormously, and still in one piece.

You've already seen that a solid that can be hammered or rolled into sheets without shattering is malleable.

Malleability is typical of metals. You've already seen the contrast: an ionic crystal struck the same way shatters.

So when an unknown solid can be hammered or rolled into a new shape without shattering, that malleability is evidence supporting metallic bonding.

Support, not proof — malleability is weighed together with the other measured properties.

What you are expected to be able to do

Identify malleability as evidence consistent with metallic bonding.

Part 2Check what you just learned

Check — multiple choice

A machine presses an unknown solid into a thin sheet, and the sheet holds together. Which bonding does this evidence support?

AIonic

Feedback if chosen: This option is wrong — you've already seen that ionic solids fracture when stressed. A solid that reshapes into a sheet without shattering points away from ionic.

BMetallic ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Being pressed into a sheet without shattering is malleability, and malleability is evidence supporting metallic bonding.

CMolecular

Feedback if chosen: This option is wrong — malleability is the metal's signature deformation. The evidence used to support a molecular substance is a relatively low melting point or nonconductivity, not reshaping into sheets.

DThe observation cannot identify the bonding type

Feedback if chosen: This option is wrong — you threw away usable evidence. How a solid responds to stress is one of the measured properties used to characterize bonding, and this response supports metallic.

Check — multiple choice

Which observation about a solid supports metallic bonding?

AIt shatters into small fragments when it is rolled

Feedback if chosen: This option is wrong — shattering is brittleness, which is evidence supporting ionic bonding rather than metallic.

BIt melts at a far lower temperature than a typical ionic solid

Feedback if chosen: This option is wrong — a relatively low melting point is at the low end of these comparisons, which is evidence on the molecular side, not the metallic one.

CIt can be rolled into a sheet without shattering ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Rolling into a sheet without shattering is malleability — the deformation evidence that supports metallic bonding.

DIt does not conduct a current in the solid state

Feedback if chosen: This option is wrong — a metal conducts in the solid state, so a nonconducting solid points away from metallic bonding, not toward it.

U2-S04 · LESSON 45 · CSP-039

Ductility as evidence for metallic bonding

Outcome: Identify ductility as evidence consistent with metallic bonding.

Part 1Pulled into wire without snapping

Copper is sold by the kilometer: rods pulled through dies into unbroken electrical wire.

You've already seen that a solid that can be drawn into a wire without shattering is ductile.

Ductility is typical of metals, so when an unknown solid survives being drawn into wire, that ductility is evidence supporting metallic bonding.

Support, not proof — ductility is weighed together with the other measured properties.

What you are expected to be able to do

Identify ductility as evidence consistent with metallic bonding.

Part 2Check what you just learned

Check — multiple choice

An unknown solid can be drawn into a fine wire without breaking. Which bonding does this evidence support?

AMolecular

Feedback if chosen: This option is wrong — being drawn into wire is not the molecular substance's behaviour, and it is not the evidence used for that classification; low melting point and nonconductivity are.

BIonic

Feedback if chosen: This option is wrong — you've already seen that ionic solids fracture when stressed; a solid that survives being drawn out points away from ionic.

CThe observation cannot identify the bonding type

Feedback if chosen: This option is wrong — you threw away usable evidence. The response to stress is one of the measured properties, and surviving the draw is ductility, which supports metallic bonding.

DMetallic ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Being drawn into a wire without shattering is ductility, and ductility is evidence supporting metallic bonding.

Check — multiple choice

Gold can be drawn into wire finer than a human hair. Which statement reads this evidence correctly?

ADuctility of this kind proves gold is metallic

Feedback if chosen: This option is wrong — you turned support into proof. Ductility is evidence consistent with metallic bonding; the classification still rests on weighing all the measured properties.

BDuctility of this kind supports ionic bonding in gold

Feedback if chosen: This option is wrong — you attached the behaviour to the wrong side: an ionic solid fractures under this treatment rather than drawing out.

CDuctility of this kind supports metallic bonding in gold ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Surviving being drawn into fine wire is ductility, and ductility is evidence supporting metallic bonding.

DBeing drawn thin is a change of state, so it is evidence about melting point

Feedback if chosen: This option is wrong — no state changed: the gold stayed solid throughout. Drawing is a response to stress, which is deformation evidence, not melting-point evidence.

U2-S04 · LESSON 46 · CSP-039A

Why metals bend instead of shattering

Outcome: Explain why metals are malleable and ductile.

Part 1Ions that move, bonding that stays

You've already seen two pieces of evidence for metallic bonding: metals roll into sheets, and metals draw into wire. Both survive the same rough treatment — layers of the solid forced into new positions.

Here is the metal's particulate model again: positive metal ions within delocalized valence electrons shared across the whole solid.

The delocalized electrons behave like a sea that the ions sit in: an ion can move within it, and it holds the ions wherever they sit.

Here is what a hammer blow or a drawing die does to the metal: a layer of ions shifts to a new position — and the electron sea still surrounds every ion, exactly as before.

Wherever an ion ends up, it is still surrounded by the shared electrons, so the attraction bonding the solid is still there.

Metals are malleable and ductile because the positive metal ions can shift position within the delocalized electron sea without breaking the metallic bonding.

You've already seen why an ionic crystal fractures instead: sliding its layer swaps oppositely charged neighbors for like-charged ones, and they repel. In the metal the shift changes nothing about each ion's surroundings — every ion is still a positive ion within the electron sea — so the shift creates no new repulsion.

One mechanism, two properties: hammering into sheets and drawing into wire both work because the ions can shift without the bonding breaking.

What you are expected to be able to do

Explain why metals are malleable and ductile: the positive metal ions can shift position within the delocalized electron sea without breaking the metallic bonding.

Part 2Check what you just learned

Check — multiple choice

Why can copper be drawn into wire without shattering?

ADrawing the wire briefly melts the copper, and the liquid metal flows into its new shape before freezing again

Feedback if chosen: This option is wrong — the copper stays solid throughout the draw. The reshaping happens because the ions shift positions within the electron sea, not because anything melts.

BIts ions lose their positive charges while the wire is being drawn, so nothing holds them back

Feedback if chosen: This option is wrong — the ions keep their charges at all times; the positive ions and the delocalized electrons are what the metallic bonding is made of.

CIts positive ions can shift position within the delocalized electron sea without breaking the metallic bonding ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The ions can take up new positions and the delocalized electron sea still surrounds them, so the metallic bonding survives the reshaping.

DThe covalent bonds between the copper atoms stretch as the wire lengthens, without ever breaking

Feedback if chosen: This option is wrong — a metal's bonding is metallic, not covalent: positive ions attracted to delocalized electrons, which keep holding the ions wherever they shift.

Check — multiple choice

A layer of ions shifts when a metal is hammered, and a layer of ions slides when an ionic crystal is struck. Why does the metal hold together while the ionic crystal fractures?

AThe attractions in the metal are far stronger than those in the ionic crystal, so only the metal holds together

Feedback if chosen: This option is wrong — you compared strengths, but the difference is what the shift creates. The ionic slide creates new like-charge repulsion across the slip; the metal's shift leaves each ion's surroundings unchanged, so it creates no new repulsion.

BThe metal's shifted ions are still held by the electron sea around them; the ionic crystal's slide brings like-charged ions alongside each other, and they repel ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The electron sea surrounds the metal's ions in their new positions, so the bonding survives; the ionic slide replaces opposite-charge attraction with like-charge repulsion, which forces the crystal apart.

CThe metal's particles carry no charge at all, so nothing in it can repel; the ionic crystal is full of charged ions that can repel one another

Feedback if chosen: This option is wrong — the metal's ions are positive, not neutral. They avoid fracturing not by lacking charge but by staying surrounded by the delocalized electrons that bond them.

DThe ionic crystal's ions are larger than the metal's ions, and larger particles crack a crystal when forced into new positions

Feedback if chosen: This option is wrong — you reached for size, but the fracture is electrical: the ionic slide lines up like charges, and their repulsion — not their size — splits the crystal.

Check — multiple choice

Which single mechanism explains both malleability and ductility in metals?

ADelocalized electrons carrying current from one end of the solid metal to the other

Feedback if chosen: This option is wrong — you picked the mechanism for a different property: moving electrons explain why metals conduct. Reshaping is about the ions shifting while the bonding holds.

BLayers of the solid sliding until like-charged particles meet each other and repel

Feedback if chosen: This option is wrong — that is the ionic crystal's story, and it explains fracture, not reshaping. In a metal, a shifted layer's ions are still surrounded by the electron sea, so the shift creates no new repulsion.

CWhole molecules separating from one another without any of their covalent bonds breaking

Feedback if chosen: This option is wrong — that is the molecular substance's melting story. A metal has no molecules; its reshaping is ions shifting within the electron sea.

DPositive metal ions shifting within the delocalized electron sea while the metallic bonding stays intact ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Hammering into sheets and drawing into wire are the same event at the particle level: ions shift to new positions, the electron sea still surrounds them, and the metallic bonding never breaks.

U2-S04 · LESSON 47 · CSP-040

Classifying bonding from measured properties

Outcome: Classify an unknown substance as ionic, molecular, or metallic by weighing all the supplied property evidence.

Part 1Weighing all the evidence

You've already seen that measured properties give the best evidence for characterizing bonding.

You've also seen the individual pieces of evidence: melting point, conductivity in each state, and how the solid responds to stress.

Here is the whole set side by side.

Typical property patterns for the three bonding types
Property	Ionic	Molecular	Metallic
Melting point	relatively high	relatively low	varies widely
Conducts as a solid?	no	no	yes
Conducts when molten?	yes	no	yes
Struck with a hammer	shatters	varies; often shatters or crumbles	dents or flattens

No single row settles the classification. Metals melt anywhere from below room temperature to thousands of degrees, and some molecular crystals shatter as convincingly as ionic ones.

The decisive move is to weigh the whole set: find the one classification consistent with every measurement, not just one.

The conductivity pattern usually narrows it fastest: yes as a solid points metallic; no as a solid but yes when molten points ionic; no in every state points molecular. Then check that the melting point and the hammer test agree.

These three are not the only bonding types. In Unit 3 you will meet a fourth kind of solid — covalent-network solids such as diamond — which this comparison does not cover.

What you are expected to be able to do

Classify an unknown substance as ionic, molecular, or metallic by weighing supplied melting-point, conductivity, and deformation evidence together, rather than relying on one cue.

Part 2Check what you just learned

Check — multiple choice

Measured properties of substance X
Property	Measurement
Melting point	770 °C
Conducts as a solid?	no
Conducts when molten?	yes
Struck with a hammer	shatters

Substance X was tested, and the measurements are shown in the table. Weighing all of the evidence, how should X be classified?

AMetallic

Feedback if chosen: This option is wrong — you leaned on the high melting point alone. Some metals do melt near 770 °C, but X does not conduct as a solid, and a metal would; and X shatters where a metal would dent or flatten.

BIonic ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Every measurement agrees: no conduction as a solid but conduction when molten is the ionic pattern, and the relatively high melting point and the shattering are both consistent with it.

CMolecular

Feedback if chosen: This option is wrong — you set the conductivity aside. A molecular substance would not conduct once molten, and X does; 770 °C is also far above the molecular range in these comparisons.

DThe data cannot decide among the three

Feedback if chosen: This option is wrong — the set is decisive. Solid-no together with molten-yes conductivity is the ionic pattern, and both remaining measurements agree with it.

Check — multiple choice

Measured properties of substance Y
Property	Measurement
Melting point	80 °C
Conducts as a solid?	no
Conducts when molten?	no
Struck with a hammer	shatters

Substance Y was tested, and the measurements are shown in the table. Weighing all of the evidence, how should Y be classified?

AIonic

Feedback if chosen: This option is wrong — you leaned on the shattering alone. Brittleness is consistent with ionic, but an ionic substance conducts once molten, and Y does not; and 80 °C sits at the low end of these comparisons.

BMetallic

Feedback if chosen: This option is wrong — a metal conducts in the solid state, and Y does not; and Y shatters where a metal would dent or flatten.

CMolecular ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Y conducts in no state, which points molecular; the relatively low melting point agrees. The shattering shows why one cue is never enough — some molecular crystals are brittle too.

DThe data cannot decide among the three

Feedback if chosen: This option is wrong — the set is decisive. No conduction in either state rules out metallic and ionic, and the low melting point agrees with molecular.

Check — multiple choice

Measured properties of substance Z
Property	Measurement
Melting point	327 °C
Conducts as a solid?	yes
Conducts when molten?	yes
Struck with a hammer	dents and flattens

Substance Z was tested, and the measurements are shown in the table. Weighing all of the evidence, how should Z be classified?

AMetallic ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Z conducts in the solid state, which of the three types only a metal does; the malleable response agrees, and a modest melting point fits — metals' melting points vary widely.

BIonic

Feedback if chosen: This option is wrong — an ionic solid would not conduct until molten, and it would shatter under the hammer. Z conducts as a solid and flattens instead.

CMolecular

Feedback if chosen: This option is wrong — you leaned on the melting point alone. 327 °C looks modest, but a molecular substance conducts in no state, and Z conducts in both; and Z flattens where a molecular crystal would shatter or crumble.

DThe data cannot decide among the three

Feedback if chosen: This option is wrong — the set is decisive. Conduction in the solid state belongs to the metals in this comparison, and the deformation evidence agrees.

AFTER U2-S04 · CHECKPOINT · U2-G1 · 8 items

Checkpoint 1 — Bond types, particulate models, and property evidence

Check — multiple choice

You are asked to draw a particulate model of solid calcium sulfide, CaS, which is made of Ca²⁺ and S²⁻ ions. Which described drawing is a correct ionic-solid model?

ASeparate Ca²⁺S²⁻ pairs, each pair drawn touching, with clear space between one pair and the next

Feedback if chosen: This option is wrong — you drew the solid as separate molecules. An ionic solid is one continuous lattice; the ions do not pair off with space between the pairs.

BRows that alternate: one row of only Ca²⁺ ions, then one row of only S²⁻ ions, repeating

Feedback if chosen: This option is wrong — you separated the charges into stripes. Within each stripe, like charges sit side by side and repel; in a correct lattice every ion's nearest neighbors are oppositely charged.

CA continuous repeating grid of neutral Ca and S atoms, drawn without any charges

Feedback if chosen: This option is wrong — you drew neutral atoms. An ionic solid is built of charged ions, and the model must label the charges that hold the lattice together.

DA continuous repeating grid in which Ca²⁺ and S²⁻ alternate, so every nearest neighbor of each ion is oppositely charged ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Alternating Ca²⁺ and S²⁻ throughout one continuous repeating grid puts every ion among oppositely charged neighbors and keeps like charges apart.

Check — multiple choice

Atomic radii
Atom	Atomic radius (pm)
Fe	140
C	70

Steel is an alloy of iron and carbon. Using the radii shown, which described drawing is the correct particulate model of this alloy?

AC atoms drawn the same size as the Fe ions, replacing Fe ions at some lattice positions, inside the shared electron region

Feedback if chosen: This option is wrong — you drew a substitutional alloy. The radii show C is half the size of Fe, so the C atoms fit into the spaces between the Fe ions instead of replacing them.

BSmall C atoms drawn in the spaces between the Fe ions, with one continuous shared region of delocalized electrons around them all ✓ KEY — this is the correct answer

Feedback if chosen: Correct. C (70 pm) is much smaller than Fe (140 pm), so the C atoms occupy interstices — and the model keeps the essential metallic feature: one continuous region of delocalized electrons.

CSmall C atoms drawn in the spaces between the Fe ions, with a separate fixed pair of electrons drawn on each ion

Feedback if chosen: This option is wrong — you localized the electrons. Assigning a fixed pair to each ion removes the delocalized electron sea that makes the material metallic.

DAlternating Fe⁺ and C⁻ ions in one continuous repeating grid

Feedback if chosen: This option is wrong — you drew an ionic solid. An alloy is a metallic material: positive metal ions within delocalized electrons, not alternating opposite charges.

Check — multiple choice

Measured properties of substance Q
Property	Measurement
Melting point	1455 °C
Conducts as a solid?	yes
Conducts when molten?	yes
Struck with a hammer	dents and flattens

Substance Q was tested, and the measurements are shown in the table. Weighing all of the evidence, how should Q be classified?

AIonic

Feedback if chosen: This option is wrong — you leaned on the high melting point alone. A solid that conducts in the solid state cannot be ionic: an ionic solid's ions are fixed in the lattice until it melts.

BMetallic ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Conduction in the solid state plus denting rather than shattering together identify metallic bonding, and a very high melting point is consistent with a metal.

CMolecular

Feedback if chosen: This option is wrong — a molecular substance would conduct in neither state and would melt far lower than 1455 °C.

DThe data cannot decide among the three

Feedback if chosen: This option is wrong — the evidence set is decisive: solid-state conduction together with deformation without shattering identifies metallic bonding.

Check — multiple choice

Electronegativities
Element	Electronegativity
Si	1.90
Cl	3.16

Using the table shown, which statement gives the direction of the Si–Cl bond-dipole arrow and the reason?

AToward Cl, because Cl is more electronegative than Si ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The arrow points toward the more electronegative atom — Cl (3.16) pulls the shared electrons more strongly than Si (1.90).

BToward Si, because the arrow points at the δ+ end of the bond

Feedback if chosen: This option is wrong — you reversed the convention. The arrow points toward the electron-rich δ− end, which is the more electronegative atom, Cl.

CToward Cl, because Cl atoms are smaller than Si atoms

Feedback if chosen: This option is wrong — you used atomic size. The direction comes from electronegativity: Cl wins the tug on the shared electrons because of its stronger pull, not its size.

DToward Si, because Si is written first in Si–Cl

Feedback if chosen: This option is wrong — you used the written order of the symbols, which carries no meaning. Compare the electronegativities: the arrow points toward Cl.

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
S	2.58
O	3.44

Using the table shown, which statement compares the ionic character of the O–H and S–H bonds correctly?

AS–H has more ionic character, because S is the larger atom

Feedback if chosen: This option is wrong — you used atomic size. Position on the continuum follows the electronegativity difference, and O–H's difference (1.24) is much larger than S–H's (0.38).

BNeither has any ionic character — both are covalent bonds

Feedback if chosen: This option is wrong — you treated covalent and ionic as sealed categories. Polar covalent bonds have some ionic character; the two bond types are ends of one continuum.

CO–H has more ionic character, because its electronegativity difference is larger ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The larger electronegativity difference in O–H (3.44 − 2.20 = 1.24, versus 0.38 for S–H) shifts the shared electrons more unevenly, giving O–H more ionic character.

DBoth sit at the ionic end of the continuum, because both are polar

Feedback if chosen: This option is wrong — you equated polar with ionic. Both bonds are polar covalent; being polar moves a bond along the continuum but does not make it ionic.

Check — multiple choice

A student drew this model of solid strontium oxide. What is the specific flaw?

AThe O²⁻ ions are drawn larger than the Sr²⁺ ions

Feedback if chosen: This option is wrong — you flagged a correct feature. O²⁻ really is the larger ion, so drawing it larger is right.

BThe model contains equal numbers of Sr²⁺ and O²⁻ ions

Feedback if chosen: This option is wrong — you flagged a correct feature. With 2+ and 2− charges, a one-to-one ion ratio is exactly what neutrality requires.

CLike-charged ions sit side by side as nearest neighbors, adding repulsion where the lattice needs attraction ✓ KEY — this is the correct answer

Feedback if chosen: Correct. One Sr²⁺ and one O²⁻ have swapped places, putting like charges as nearest neighbors — strong repulsion where a consistent lattice surrounds every ion with oppositely charged neighbors.

DThe same alternating pattern repeats across the whole model

Feedback if chosen: This option is wrong — you flagged a correct feature. A crystal lattice is a systematic, periodic array, so the repetition is required, not a flaw.

Check — multiple choice

A student drew this model of solid silver and asked for feedback. Which one change must be made before the model shows metallic bonding?

AReplace the fixed pair of electrons on each ion with one continuous shared region of mobile electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The model assigns each ion its own fixed electrons. Metallic bonding needs the valence electrons delocalized — one continuous shared region across the whole solid.

BRedraw half of the Ag⁺ ions as negative ions so that the charges balance

Feedback if chosen: This option is wrong — you imported an ionic picture. A metal contains only positive metal ions; the delocalized electrons, not anions, balance the charge.

CRemove the positive charges — the particles in a metal are neutral atoms

Feedback if chosen: This option is wrong — the positive labels are correct. Each metal atom contributes valence electrons to the delocalized electron region, leaving positive ions behind.

DBreak the regular array into a random scatter — metal ions are not arranged in rows

Feedback if chosen: This option is wrong — the regular array is correct. Metal ions sit in an ordered repeating arrangement; it is the electrons that must be shared, not the ions scattered.

Check — multiple choice

A chemist suspects that solid substance M is ionic, and measures its electrical conductivity as a solid and again when molten. Which pair of results would support ionic bonding?

AConducts as a solid, and conducts when molten

Feedback if chosen: This option is wrong — conduction in the solid state is the metallic pattern. An ionic solid's ions are locked in the lattice, so the solid cannot carry current.

BDoes not conduct as a solid, and conducts when molten ✓ KEY — this is the correct answer

Feedback if chosen: Correct. This is the pattern that supports ionic bonding: the ions are fixed in place in the solid, then become mobile charge carriers once the solid melts.

CDoes not conduct as a solid, and does not conduct when molten

Feedback if chosen: This option is wrong — conducting in neither state is the molecular pattern: neutral molecules give no mobile charged particles in either state.

DConducts as a solid, and does not conduct when molten

Feedback if chosen: This option is wrong — no bonding type gives this pattern. Melting frees particles to move; it does not remove the charge carriers a conducting solid already had.

U2-S05 — Potential-energy curves

11 lessons

U2-S05 · LESSON 48 · CSP-041

Internuclear distance

Outcome: Identify the internuclear distance between two atoms in a diagram.

Part 1Measuring between fuzzy objects

An atom has no hard edge — its electrons form a fuzzy cloud around the nucleus.

So “the distance between two atoms” needs a precise meaning: where exactly do you measure from?

Chemists measure from nucleus to nucleus.

The distance between the nuclei of two atoms is called the internuclear distance.

In H₂, the internuclear distance is the separation between the two H nuclei — not between the edges of the electron clouds.

What you are expected to be able to do

Identify the internuclear distance in a diagram of two atoms as the distance between their nuclei.

Part 2Check what you just learned

Check — multiple choice

Two chlorine atoms are shown with four distances marked A to D. Which marked distance is the internuclear distance?

ADistance A ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Distance A extends from one nucleus to the other nucleus — the internuclear distance.

BDistance B

Feedback if chosen: This option is wrong — you measured between the outer edges of the electron clouds. The clouds are fuzzy, so edges make a poor measuring point; measure between the nuclei.

CDistance C

Feedback if chosen: This option is wrong — you measured from one nucleus to the edge of the other atom. Both ends of the measurement must be nuclei.

DDistance D

Feedback if chosen: This option is wrong — you measured across a single atom. The internuclear distance extends between the nuclei of the two different atoms.

Check — multiple choice

One hydrogen atom and one chlorine atom are shown with four distances marked A to D. Which marked distance is the internuclear distance?

ADistance A

Feedback if chosen: This option is wrong — you measured the gap between the two electron clouds. The internuclear distance extends all the way from nucleus to nucleus, whatever the sizes of the atoms.

BDistance B ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Distance B extends from the H nucleus to the Cl nucleus — the internuclear distance. The two atoms being different sizes changes nothing about where you measure.

CDistance C

Feedback if chosen: This option is wrong — you stopped at the midpoint between the atoms. The internuclear distance is the full distance from one nucleus to the other.

DDistance D

Feedback if chosen: This option is wrong — you measured from the Cl nucleus to its own edge, which is the size of one atom, not a distance between two atoms.

U2-S05 · LESSON 49 · CSP-042

Potential energy of a pair of atoms

Outcome: State what the potential energy of a pair of atoms depends on, and where it is set to zero.

Part 1One number for a pair of atoms

Bring two hydrogen atoms close together and they join into H₂ — and they stay joined.

To describe why, chemists track one number for the pair: its potential energy.

You’ve already seen that each atom is built from charged particles: a positively charged nucleus and negatively charged electrons.

The potential energy of a pair of atoms depends on how all of those charged particles are positioned relative to one another.

Move the atoms, and the charged particles move with them — so the potential energy of the pair changes.

Part 2Where zero is

A potential-energy value only means something compared with an agreed starting point.

The agreed starting point: when the two atoms are too far apart to interact, the potential energy of the pair is set to zero.

So two widely separated hydrogen atoms have zero potential energy.

A potential energy below zero means the pair has less potential energy than it would have with the atoms too far apart to interact.

What you are expected to be able to do

State that the potential energy of a pair of atoms depends on how their charged particles are positioned, and that it is set to zero when the atoms are too far apart to interact.

Part 3Check what you just learned

Check — multiple choice

Two chlorine atoms are so far apart that they no longer interact. What is the potential energy of the pair?

ALarge and positive

Feedback if chosen: This option is wrong — you may be picturing far-apart atoms as high in energy, but this is exactly where the zero is set: atoms too far apart to interact have zero potential energy.

BLarge and negative

Feedback if chosen: This option is wrong — a negative potential energy would put the pair below the starting point, and the starting point is defined at exactly this situation: too far apart to interact means zero.

CImpossible to say without more information

Feedback if chosen: This option is wrong — no more information is needed. Setting the potential energy to zero for atoms too far apart to interact is the agreed convention, whatever the atoms are.

DZero ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The potential energy of a pair is set to zero when the atoms are too far apart to interact — that is the agreed starting point.

Check — multiple choice

What does the potential energy of a pair of atoms depend on?

AThe mass of the two atoms

Feedback if chosen: This option is wrong — mass matters for gravity, which is far too weak to matter here. The potential energy of the pair comes from its charged particles and their positions.

BThe number of neutrons in each nucleus

Feedback if chosen: This option is wrong — neutrons carry no charge, so adding or removing them does not change the interactions that set the pair’s potential energy.

CThe temperature of the sample

Feedback if chosen: This option is wrong — temperature is about how fast particles move. The potential energy of the pair is about where its charged particles are positioned.

DHow the charged particles of the two atoms are positioned ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The nuclei and electrons are charged, and how those charged particles are positioned sets the potential energy of the pair.

Check — multiple choice

The potential energy of a pair of atoms is below zero. What does that tell you?

AThe pair has more potential energy than it would with the atoms too far apart to interact

Feedback if chosen: This option is wrong — you flipped the comparison. Zero belongs to the far-apart atoms, so a value below zero is less than that, not more.

BThe pair has less potential energy than it would with the atoms too far apart to interact ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Zero is set at atoms too far apart to interact, so a potential energy below zero means the pair sits below that starting point.

CThe atoms are not interacting

Feedback if chosen: This option is wrong — atoms too far apart to interact sit exactly at zero. A value below zero means the atoms are close enough to interact.

DThe atoms have lost some of their electrons

Feedback if chosen: This option is wrong — you read the negative sign as a charge. It is an energy value below the agreed zero, not a statement about losing electrons.

U2-S05 · LESSON 50 · CSP-043

Equilibrium bond length

Outcome: Identify the equilibrium bond length as the distance at which a bonded pair has its minimum potential energy.

Part 1The distance where the energy is lowest

Here is a graph of how the potential energy of two hydrogen atoms depends on the distance between them.

You’ve already seen that internuclear distance is the distance between the nuclei of two atoms.

You’ve already seen that the potential energy of the pair is zero when the atoms are too far apart to interact.

The graph has a single lowest point: the minimum.

At the minimum, the pair has its lowest possible potential energy.

The distance between them at the minimum has a name: the equilibrium bond length.

At any other distance between them, the potential energy of the pair is higher.

What you are expected to be able to do

Identify the equilibrium bond length as the distance between the two atoms at which a bonded pair has its minimum potential energy.

Part 2Check what you just learned

Check — multiple choice

On the potential-energy curve shown, four points are marked A to D. At which point is the distance between the atoms equal to the equilibrium bond length?

APoint A

Feedback if chosen: This option is wrong — at point A the distance between them is shorter than the equilibrium bond length, and the potential energy is high, not at its minimum.

BPoint B ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Point B is the minimum — the pair’s lowest potential energy — and the distance between them there is the equilibrium bond length.

CPoint C

Feedback if chosen: This option is wrong — at point C the potential energy is still falling; the pair has not yet reached its lowest value. The equilibrium bond length is at the minimum itself.

DPoint D

Feedback if chosen: This option is wrong — at point D the atoms are too far apart to interact and the potential energy is zero. Zero is not the lowest value on this curve; the minimum sits below it.

Check — multiple choice

At the equilibrium bond length, the potential energy of a bonded pair is:

AZero

Feedback if chosen: This option is wrong — zero is the potential energy of atoms too far apart to interact. At the equilibrium bond length the pair sits at its minimum, which is below zero.

BRising steeply

Feedback if chosen: This option is wrong — the steep rise happens at distances shorter than the equilibrium bond length. At the equilibrium bond length itself, the pair is at the bottom of the curve.

CAt its highest value

Feedback if chosen: This option is wrong — you inverted the definition. The equilibrium bond length is the distance of lowest potential energy, not highest.

DAt its lowest value ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The equilibrium bond length is defined as the distance between the two atoms at which the pair’s potential energy is at its minimum.

Check — multiple choice

For Cl₂, the potential energy of the pair is at its minimum when the distance between the nuclei is 199 pm. What is the equilibrium bond length of Cl–Cl?

A199 pm ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The equilibrium bond length is the distance between the nuclei at minimum potential energy — exactly the 199 pm given.

B99.5 pm

Feedback if chosen: This option is wrong — you halved the distance, as if 199 pm covered both atoms’ shares. The 199 pm is already the full nucleus-to-nucleus distance at the minimum.

C398 pm

Feedback if chosen: This option is wrong — you doubled the distance, treating 199 pm as if it were the size of one atom. It is already the full distance between the two nuclei.

DIt cannot be found without seeing the curve

Feedback if chosen: This option is wrong — no curve is needed here. The equilibrium bond length is defined as the distance at minimum potential energy, and that distance is supplied.

U2-S05 · LESSON 51 · CSP-044

Bond energy

Outcome: State what bond energy measures for a bonded pair of atoms.

Part 1The cost of pulling a pair apart

A bonded pair of hydrogen atoms holds together — the atoms do not drift apart on their own.

Separating them costs energy, and that cost has a name.

You’ve already seen that the potential energy of the pair is zero when the atoms are too far apart to interact.

You’ve already seen that a bonded pair has its lowest potential energy at the equilibrium bond length.

Here is the potential-energy curve for two hydrogen atoms, with the minimum and the zero level marked.

Pulling the bonded pair completely apart means taking it from the minimum all the way back up to the zero level — atoms too far apart to interact.

The energy required to pull a bonded pair completely apart is called the bond energy.

For H–H that climb is 436 kJ/mol, so the bond energy of H–H is 436 kJ/mol.

Bond energy is a positive quantity — it is energy that must be supplied to the pair, not energy the pair gives out.

What you are expected to be able to do

State that bond energy is the energy required to pull a bonded atom pair completely apart, ending with the atoms too far apart to interact.

Part 2Check what you just learned

Check — multiple choice

Bond energy is the energy:

ARequired to push the bonded atoms closer together than the equilibrium bond length

Feedback if chosen: This option is wrong — you picked the wrong direction along the curve. Pushing the atoms closer also raises the energy, but bond energy measures the separation of the pair, out to atoms too far apart to interact.

BReleased when a bonded pair is pulled completely apart

Feedback if chosen: This option is wrong — you flipped the energy flow. Pulling a pair apart requires energy to be supplied; energy is released when the bond forms, not when it is pulled apart.

CRequired to pull a bonded pair completely apart ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Bond energy is the energy required to take a bonded pair from its minimum all the way apart, to where the atoms no longer interact.

DThat a bonded pair has at the equilibrium bond length

Feedback if chosen: This option is wrong — the pair’s potential energy at the equilibrium bond length is a value below zero on the curve. Bond energy is the energy required to climb from there up to zero — a different quantity.

Check — multiple choice

A bonded pair is pulled apart until the atoms no longer interact. What is the potential energy of the pair at the end?

AZero ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Atoms too far apart to interact sit at the zero level — that is where the pull ends.

BAt its lowest value

Feedback if chosen: This option is wrong — the lowest value belongs to the bonded pair at the equilibrium bond length, which is where the pull starts, not where it ends.

CBelow zero but higher than at the equilibrium bond length

Feedback if chosen: This option is wrong — that describes a pair only partway apart. Pulled completely apart means too far apart to interact, and there the potential energy is zero.

DAbove zero

Feedback if chosen: This option is wrong — supplying energy raises the pair only as far as the zero level. Once the atoms are too far apart to interact, the potential energy is zero, not above it.

Check — multiple choice

To pull the bonded pair in HCl completely apart, energy must be:

AReleased by the pair

Feedback if chosen: This option is wrong — you flipped the energy flow. The pair ends higher on the curve than it started, so energy must go in, not come out.

BSupplied to the pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The pair starts at its minimum and ends at zero — higher in energy — so the difference must be supplied to the pair. That supplied amount is the bond energy.

CNeither supplied nor released

Feedback if chosen: This option is wrong — without an energy input the pair would stay at its minimum. Climbing from the minimum to the zero level requires energy to be supplied.

DFirst supplied, then released again as the atoms separate

Feedback if chosen: This option is wrong — the separated atoms keep the higher potential energy; nothing is handed back. The supplied energy stays with the pair as raised potential energy.

U2-S05 · LESSON 52 · CSP-045

The shape of a potential-energy curve

Outcome: Describe the shape of a potential-energy curve from large distance down to short distance.

Part 1Tracing the curve

Here is the potential-energy curve for two hydrogen atoms. It shows how the potential energy of the pair changes with the distance between them.

You’ve already seen that when the two atoms are very far apart, their potential energy is zero.

Start at the right-hand end of the curve, where the distance between them is large: the curve sits flat at zero.

As the distance between them decreases, the potential energy also decreases.

It keeps decreasing down to a single lowest point: the minimum.

You’ve already seen that the distance between them at the minimum is the equilibrium bond length.

As the distance between them decreases below the equilibrium bond length, the potential energy rises again.

And the shorter the distance between them gets, the steeper that rise becomes.

Every potential-energy curve for a bonded pair has this same overall shape: a fall, a single minimum, and a steep rise at short distance.

What you are expected to be able to do

Describe the shape of a supplied potential-energy curve: the potential energy decreases as the distance between them decreases, reaches a single minimum at the equilibrium bond length, and rises increasingly steeply at short distance.

Part 2Check what you just learned

Check — multiple choice

The potential-energy curve for two chlorine atoms is shown. As the distance between them decreases from very far apart, which sequence describes the potential energy?

AIt falls, bottoms out at the equilibrium bond length, then rises sharply ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Fall, single minimum, then a steep rise at short distance — the shape every potential-energy curve for a bonded pair shares.

BIt rises to a single maximum at the equilibrium bond length, then falls steeply

Feedback if chosen: This option is wrong — you flipped the curve upside down. The pair’s energy falls below zero to a minimum; it does not rise to a maximum first.

CIt decreases steadily all the way to the shortest distances

Feedback if chosen: This option is wrong — you missed the turn. The fall stops at the minimum, and below the equilibrium bond length the potential energy rises steeply.

DIt stays at zero until the atoms touch, then rises steeply

Feedback if chosen: This option is wrong — the potential energy starts changing while the atoms are still well separated; the curve leaves zero long before the atoms could be said to touch.

Check — multiple choice

As the distance between two chlorine atoms decreases below the equilibrium bond length, the potential energy:

AKeeps decreasing

Feedback if chosen: This option is wrong — the decrease belongs to the other side of the curve, at distances larger than the equilibrium bond length. Below it, the energy rises.

BStays at the minimum

Feedback if chosen: This option is wrong — the minimum is a single point at the equilibrium bond length, not a flat region. Any shorter distance puts the pair above the minimum.

CRises, more and more steeply ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Below the equilibrium bond length the curve rises, and it gets steeper and steeper as the distance between them shrinks.

DReturns to zero and stays there

Feedback if chosen: This option is wrong — zero belongs to atoms too far apart to interact. At short distance the curve climbs steeply and passes zero without stopping there.

Check — multiple choice

At very large distance between two chlorine atoms, the potential-energy curve:

ASits at the minimum

Feedback if chosen: This option is wrong — the minimum sits at the equilibrium bond length, a short, specific distance. At very large distance the pair is far from the minimum.

BSits flat at zero ✓ KEY — this is the correct answer

Feedback if chosen: Correct. At very large distance the atoms are too far apart to interact, so the curve flattens out at zero.

CRises steeply

Feedback if chosen: This option is wrong — the steep rise belongs to very short distances, at the other end of the curve.

DFalls below zero

Feedback if chosen: This option is wrong — below zero means the atoms are interacting. At very large distance they are not, and the curve sits at zero exactly.

U2-S05 · LESSON 53 · CSP-046

Why the potential energy falls

Outcome: Explain why the potential energy falls as the distance between two separated atoms decreases.

Part 1The falling part of the curve

Here is the potential-energy curve for two hydrogen atoms again.

You’ve already seen its shape: as the distance between them decreases, the potential energy also decreases, down to a minimum, and then rises steeply at short distance.

This lesson explains the falling part — the long slide from zero down to the minimum.

You’ve already seen that opposite charges attract and like charges repel.

Each atom brings charged particles: a positively charged nucleus and negatively charged electrons.

When the two atoms come close enough to interact, new attractions appear: each nucleus attracts the other atom’s electrons.

New repulsions appear too: the two nuclei repel each other, and the two atoms’ electrons repel each other.

On the falling part of the curve, the attractions outweigh the repulsions.

So as the distance between them decreases, the potential energy decreases, because the attractions between each nucleus and the other atom’s electrons dominate over the repulsions.

What you are expected to be able to do

Explain why the potential energy falls as the distance between two separated atoms decreases: in that region, nucleus–electron attractions dominate over the repulsions.

Part 2Check what you just learned

Check — multiple choice

Two chlorine atoms start very far apart. As the distance between them decreases, the potential energy decreases, as the curve shows. Why?

AThe two nuclei attract each other

Feedback if chosen: This option is wrong — both nuclei are positively charged, so they repel each other. The attractions in this pair are between nuclei and electrons.

BThe two atoms’ electrons attract each other

Feedback if chosen: This option is wrong — the electrons of both atoms are negatively charged, so they repel each other. The attractions are between nuclei and electrons.

CThe repulsions dominate over the attractions

Feedback if chosen: This option is wrong — if the repulsions dominated, moving closer would raise the potential energy. The potential energy falls here, so the attractions must be winning.

DThe attractions between each nucleus and the other atom’s electrons dominate over the repulsions ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each nucleus attracts the other atom’s electrons, and in this region those attractions dominate over the repulsions — so the potential energy falls as the distance between them decreases.

Check — multiple choice

On the potential-energy curve shown, four regions of the distance axis are marked A to D. In which region do the attractions between each nucleus and the other atom’s electrons dominate over the repulsions?

ARegion A

Feedback if chosen: This option is wrong — region A is the steep rise at short distance, where the repulsions dominate and the potential energy rises.

BRegion B

Feedback if chosen: This option is wrong — region B surrounds the minimum, where attractions and repulsions balance; neither dominates, which is why the curve is flat there.

CRegion C ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Region C is the falling portion: as the distance between them decreases through this region, the potential energy decreases because the nucleus–electron attractions dominate.

DRegion D

Feedback if chosen: This option is wrong — in region D the atoms are too far apart to interact, so there are no attractions or repulsions to speak of and the potential energy sits at zero.

Check — multiple choice

When two atoms come close enough to interact, which pairs of charged particles attract one another?

AEach nucleus and the other atom’s electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A nucleus is positive and electrons are negative, so each nucleus attracts the other atom’s electrons — these are the attractions that pull the pair’s energy down.

BThe two nuclei

Feedback if chosen: This option is wrong — both nuclei carry positive charge, and like charges repel.

CThe two atoms’ electrons

Feedback if chosen: This option is wrong — both atoms’ electrons carry negative charge, and like charges repel.

DNo pairs — every particle repels every other

Feedback if chosen: This option is wrong — the pair contains opposite charges too. Each positively charged nucleus attracts the other atom’s negatively charged electrons.

U2-S05 · LESSON 54 · CSP-047

Why the potential energy rises steeply

Outcome: Explain why the potential energy rises steeply when the distance between two atoms becomes very short.

Part 1The steep rise at short distance

Here is the potential-energy curve for two hydrogen atoms once more.

You’ve already seen its shape: as the distance between them decreases, the potential energy also decreases, down to a minimum, and then rises steeply at short distance.

This lesson explains that steep rise — the left-hand end of the curve.

You’ve already seen that opposite charges attract and like charges repel.

You’ve already seen that when two atoms interact, each nucleus attracts the other atom’s electrons, while the two nuclei repel each other and the two atoms’ electrons repel each other.

As the distance between them decreases below the equilibrium bond length, the two positively charged nuclei are forced closer and closer together.

The two atoms’ electrons are forced into the same space too.

At very short distance those repulsions — nucleus–nucleus and electron–electron — grow faster than the attractions.

So as the distance between them decreases below the equilibrium bond length, the potential energy rises steeply, because the nucleus–nucleus and electron–electron repulsions dominate.

What you are expected to be able to do

Explain why the potential energy rises steeply at very short distance: the nucleus–nucleus and electron–electron repulsions dominate there.

Part 2Check what you just learned

Check — multiple choice

The curve for one hydrogen atom and one chlorine atom is shown. As the distance between them decreases well below the equilibrium bond length, the potential energy rises steeply. Why?

AThe nuclei and the electrons begin to repel each other

Feedback if chosen: This option is wrong — nuclei and electrons carry opposite charges, and opposite charges always attract; that never switches to repulsion.

BThe nucleus–nucleus and electron–electron repulsions dominate ✓ KEY — this is the correct answer

Feedback if chosen: Correct. At very short distance the two nuclei are forced together and the two atoms’ electrons are forced together — those repulsions dominate, so the potential energy rises steeply.

CThe attractions dominate over the repulsions

Feedback if chosen: This option is wrong — when the attractions dominate, the potential energy falls; that is the falling part of the curve at larger distances, not the steep rise.

DThe atoms stop interacting at short distance

Feedback if chosen: This option is wrong — you have it backwards: the closer the atoms, the more strongly their charged particles interact. Atoms stop interacting at very large distance.

Check — multiple choice

On the potential-energy curve shown, four regions of the distance axis are marked A to D. In which region do the nucleus–nucleus and electron–electron repulsions dominate?

ARegion A ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Region A is the steep rise at short distance: the nuclei and the electrons are forced together, the repulsions dominate, and the potential energy rises steeply.

BRegion B

Feedback if chosen: This option is wrong — region B surrounds the minimum, where attractions and repulsions balance; neither dominates there.

CRegion C

Feedback if chosen: This option is wrong — region C is the falling portion, where the nucleus–electron attractions dominate and the potential energy decreases.

DRegion D

Feedback if chosen: This option is wrong — in region D the atoms are too far apart to interact; nothing dominates and the potential energy sits at zero.

Check — multiple choice

When two atoms are very close together, which pairs of charged particles repel one another?

AEach nucleus and the other atom’s electrons

Feedback if chosen: This option is wrong — a nucleus and electrons carry opposite charges, so these pairs attract, whatever the distance.

BThe two nuclei, but not the two atoms’ electrons

Feedback if chosen: This option is wrong — you left out half the repulsion. The two atoms’ electrons are both negative and repel each other just as the two nuclei do.

CThe two nuclei, and the two atoms’ electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Like charges repel: the two positive nuclei repel each other, and the two atoms’ negative electrons repel each other. At very short distance these repulsions dominate.

DEach atom’s nucleus and its own electrons

Feedback if chosen: This option is wrong — an atom’s own nucleus and electrons carry opposite charges and attract; that attraction is what holds each atom together.

U2-S05 · LESSON 55 · CSP-048

Reading bond length from a potential-energy curve

Outcome: Read the equilibrium bond length of a bonded pair from a potential-energy curve.

Part 1Two features you’ve already seen

Here is the potential-energy curve for two hydrogen atoms.

You’ve already seen how a curve like this behaves. When the two atoms are very far apart, their potential energy is zero.

As the distance between them decreases, the potential energy also decreases, down to a minimum.

As the distance between them decreases further still, the potential energy rises again.

You’ve also seen that the equilibrium bond length is the distance between the two atoms at which the bonded pair has its lowest potential energy.

Part 2Where bond length sits on the curve

The minimum is the bonded pair at its equilibrium bond length.

So the minimum’s position along the distance axis is the equilibrium bond length.

On this curve the minimum sits at 74 pm, so the equilibrium bond length of H–H is 74 pm.

The −436 kJ/mol label marks how far the minimum sits below zero on the energy axis — that is an energy, not a distance. The equilibrium bond length is read from the distance axis.

Read it in two steps every time: find the minimum, then read straight down from the minimum to the distance axis using the labeled guide.

What you are expected to be able to do

Read the equilibrium bond length of a bonded pair from a supplied potential-energy curve, as the position of the minimum on the distance axis.

Part 3Check what you just learned

Check — numeric entry

Use the potential-energy curve shown for two chlorine atoms. What is the equilibrium bond length of Cl–Cl, in pm?

✓ ANSWER KEY: 199 pm

Feedback when correct: Correct. The minimum sits at 199 pm on the distance axis, so the equilibrium bond length of Cl–Cl is 199 pm.

Anticipated wrong answers (feedback shown):

243	This is how far the minimum sits below the zero line, read from the vertical energy axis in kJ/mol. Equilibrium bond length is the minimum’s position on the horizontal distance axis: 199 pm.
-243	This is the energy-axis label at the minimum, in kJ/mol — an energy, not a distance. Read the minimum’s position on the distance axis instead: 199 pm.
0	The zero line is the energy level of the far-apart atoms — it is not a distance. Read the minimum’s position on the distance axis: 199 pm.

Sign-error retry feedback:

-199	Right distance, wrong sign — a distance is positive. The minus sign on this graph belongs to the energy axis, not the distance axis.

Check — numeric entry

Use the potential-energy curve shown for one hydrogen atom and one chlorine atom. What is the equilibrium bond length of H–Cl, in pm?

✓ ANSWER KEY: 127 pm

Feedback when correct: Correct. The minimum sits at 127 pm on the distance axis, so the equilibrium bond length of H–Cl is 127 pm.

Anticipated wrong answers (feedback shown):

431	This is how far the minimum sits below the zero line, read from the vertical energy axis in kJ/mol. Read the minimum’s position on the horizontal distance axis instead: 127 pm.
-431	This is the energy-axis label at the minimum, in kJ/mol — an energy, not a distance. The equilibrium bond length is the minimum’s position on the distance axis: 127 pm.

Sign-error retry feedback:

-127	Right distance, wrong sign — a distance is positive. The minus sign on this graph belongs to the energy axis, not the distance axis.

Check — multiple choice

On the potential-energy curve shown, four features are marked A to D. Which marked feature gives the equilibrium bond length?

AFeature A

Feedback if chosen: This option is wrong — Feature A measures along the distance axis from the minimum out to the flat region: the span the atoms cross while being pulled apart, not the minimum’s own position on the distance axis.

BFeature B

Feedback if chosen: This option is wrong — Feature B is the vertical gap from the minimum up to the zero line. That gap is the bond energy — an energy, not a length.

CFeature C ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Feature C marks the minimum’s position on the distance axis — the distance between the two atoms at which the potential energy is lowest, which is the equilibrium bond length.

DFeature D

Feedback if chosen: This option is wrong — Feature D measures up the steep rise at short distance: a vertical gap on the energy axis, not a distance.

U2-S05 · LESSON 56 · CSP-049

Reading bond energy from a potential-energy curve

Outcome: Read the bond energy of a bonded pair from a potential-energy curve.

Part 1Two features you’ve already seen

Here is the potential-energy curve for two hydrogen atoms.

You’ve already seen how a curve like this behaves. When the two atoms are very far apart, their potential energy is zero.

As the distance between them decreases, the potential energy also decreases, down to a minimum.

As the distance between them decreases further still, the potential energy rises again.

You’ve also seen that bond energy is the energy required to pull a bonded pair apart until the atoms are too far apart to interact.

Part 2Where bond energy sits on the curve

The flat right-hand end of the curve is the zero-energy level: the two atoms very far apart.

The minimum is the bonded pair at its equilibrium distance.

The bond energy is how far the minimum sits below zero.

On this curve the minimum sits at −436 kJ/mol, so separating the bonded atoms requires 436 kJ/mol. The bond energy of H–H is 436 kJ/mol.

Bond energy is a positive quantity — it is the energy required to separate the atoms. The −436 kJ/mol label marks where the minimum sits; the bond energy is the size of that gap, 436 kJ/mol.

Read it in two steps every time: find the minimum, then read the vertical gap from the minimum up to the zero line using the labeled guide.

What you are expected to be able to do

Read the bond energy of a bonded pair from a supplied potential-energy curve, as how far the minimum sits below the zero-energy level of the far-apart atoms.

Part 3Check what you just learned

Check — numeric entry

Use the potential-energy curve shown for two chlorine atoms. What is the bond energy of Cl–Cl, in kJ/mol?

✓ ANSWER KEY: 243 kJ/mol

Feedback when correct: Correct. The minimum sits at −243 kJ/mol relative to the zero line, so separating the pair requires 243 kJ/mol.

Anticipated wrong answers (feedback shown):

199	This is the equilibrium bond length read from the horizontal axis, in pm. Bond energy is the vertical gap between the minimum and the zero line: 243 kJ/mol.
0	The zero line is where the separated atoms sit, not the bond energy. Read how far the minimum sits below it: 243 kJ/mol.

Sign-error retry feedback:

-243	Right gap, wrong sign. Bond energy is the energy required to pull the atoms apart, so it is positive. The minus sign on the axis label marks where the minimum sits relative to zero.

Check — numeric entry

Use the potential-energy curve shown for one hydrogen atom and one chlorine atom. What is the bond energy of H–Cl, in kJ/mol?

✓ ANSWER KEY: 431 kJ/mol

Feedback when correct: Correct. The minimum sits at −431 kJ/mol below the zero line, so the H–Cl bond energy is 431 kJ/mol.

Anticipated wrong answers (feedback shown):

127	This is the equilibrium bond length from the horizontal axis, in pm. Read the vertical gap from the minimum to the zero line instead: 431 kJ/mol.

Sign-error retry feedback:

-431	Right gap, wrong sign. Bond energy is the energy required to pull the atoms apart: 431 kJ/mol.

Check — multiple choice

On the potential-energy curve shown, four features are marked A to D. Which marked feature gives the bond energy?

AFeature A

Feedback if chosen: This option is wrong — Feature A measures along the distance axis, from the minimum out to the flat region. Bond energy is an energy, so it is read as a vertical gap, from the minimum up to the zero line.

BFeature B ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Feature B is the vertical gap from the minimum up to the zero line — the energy required to take the bonded pair from the minimum back up to zero, atoms too far apart to interact.

CFeature C

Feedback if chosen: This option is wrong — Feature C marks the minimum's position on the distance axis, which is the equilibrium bond length, not the energy needed to pull the atoms apart.

DFeature D

Feedback if chosen: This option is wrong — Feature D measures up the steep rise at short distance. Pulling atoms apart moves them to large distance, where the curve flattens to zero — not up the left-hand branch.

U2-S05 · LESSON 57 · CSP-050

Sketching a potential-energy curve

Outcome: Sketch a qualitative potential-energy curve with all of its required features.

Part 1The features every curve needs

Here is a model potential-energy curve for two hydrogen atoms, drawn with every required feature.

You’ve already seen each feature on its own — this lesson puts them together into one sketch.

Feature one: labeled axes — potential energy on the vertical axis, internuclear distance on the horizontal axis.

Feature two: at large distance the curve sits flat at zero, the level of atoms too far apart to interact.

Feature three: a single minimum, marked at the equilibrium bond length on the distance axis.

Feature four: as the distance between them decreases below the equilibrium bond length, the curve rises steeply.

Feature five: the bond energy, marked as a vertical gap from the minimum up to the zero level.

Your sketch is qualitative — no numbers are needed unless the question supplies them.

If the question does supply values, place them where they belong: the equilibrium bond length on the distance axis, and the bond energy on the vertical gap.

What you are expected to be able to do

Sketch a qualitative potential-energy curve with labeled axes, a flat zero level at large distance, a single minimum at the equilibrium bond length, a steep rise at short distance, and the bond energy marked.

Part 2Check what you just learned

Check — drawing task (pen and paper, self-marked)

Take a pen and paper and draw the potential-energy curve for two chlorine atoms as the distance between them changes. Your curve is qualitative — no numbers. Label both axes, mark the equilibrium bond length, and mark the bond energy. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing labels both axes: potential energy on the vertical axis, internuclear distance on the horizontal axis.
Your curve sits flat at zero at large distance.
Your curve has a single minimum, marked at the equilibrium bond length on the distance axis.
Your curve rises steeply at distances shorter than the equilibrium bond length.
Your drawing marks the bond energy as a vertical gap from the minimum up to the zero level.

Check — drawing task (pen and paper, self-marked)

The bonded pair in HCl has an equilibrium bond length of 127 pm and a bond energy of 431 kJ/mol. Take a pen and paper and draw the potential-energy curve for this pair, labeling these two values in the correct places. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing labels both axes and shows the same overall shape: flat at zero at large distance, a single minimum, and a steep rise at short distance.
Your minimum sits at 127 pm on the distance axis.
Your bond energy is marked as a vertical gap of 431 kJ/mol from the minimum up to the zero level.
The minimum’s level on the energy axis reads −431 kJ/mol — below zero, because the zero level belongs to the far-apart atoms.

U2-S05 · LESSON 58 · CSP-051

Checking a potential-energy curve for flaws

Outcome: Evaluate a supplied potential-energy curve and identify the flaw when it has one.

Part 1What a correct curve must show

Here is a correctly drawn potential-energy curve, with every required feature.

You’ve already seen the required features when you sketched a curve yourself: labeled axes, a flat zero level at large distance, a single minimum at the equilibrium bond length, a steep rise at short distance, and the bond energy marked as a vertical gap from the minimum up to the zero level.

Part 2How to check a supplied curve

Check a supplied curve one feature at a time, in the same order every time.

First check the axes and the zero level: both axes labeled, and the curve flat at zero at large distance.

Then check the shape: a single minimum, and a steep rise at short distance.

Last, check the markings: bond length marked on the distance axis, bond energy marked as a vertical gap on the energy axis.

A curve with even one required feature wrong is not acceptable — and your job is to name the specific flaw.

What you are expected to be able to do

Evaluate a supplied potential-energy curve against the required features, and identify the one specific flaw when the curve is not acceptable.

Part 3Check what you just learned

Check — multiple choice

A student’s potential-energy curve for two chlorine atoms is shown. Evaluate it. Which statement is correct?

AThe curve is acceptable as drawn

Feedback if chosen: This option is wrong — you accepted the bond-energy marking without checking which axis it measures along. A horizontal arrow measures a distance, and bond energy is an energy.

BNot acceptable — the curve is missing the single minimum that should sit below the zero level

Feedback if chosen: This option is wrong — the curve does show a single minimum. The flaw is in how the bond energy is marked.

CNot acceptable — a qualitative curve is incomplete unless both of its axes carry numerical scales

Feedback if chosen: This option is wrong — a qualitative curve needs no numbers; numbers are required only when the task supplies data. The real flaw is the horizontal bond-energy arrow.

DNot acceptable — the bond energy must be marked as a vertical gap from the minimum up to the zero level ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The bond energy is the vertical gap between the minimum and the zero level; drawing it as a horizontal arrow marks a distance instead of an energy.

Check — multiple choice

Another potential-energy curve is shown. Evaluate it. Which statement is correct?

AThe curve is acceptable as drawn

Feedback if chosen: This option is wrong — you missed where the flat end sits. It levels off below zero, but the far-apart atoms define the zero level itself.

BNot acceptable — at large distance the curve must flatten at the zero level, not below it ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The flat end at large distance is the pair with its atoms too far apart to interact, and that is exactly where the potential energy is set to zero — the curve must flatten at the zero line.

CNot acceptable — the curve is missing its steep rise at distances below the equilibrium bond length

Feedback if chosen: This option is wrong — the curve does rise steeply at short distance. The flaw is where the curve flattens at large distance.

DNot acceptable — the minimum of the curve must sit exactly at zero on the energy axis

Feedback if chosen: This option is wrong — the minimum belongs below zero; that is what makes the bond energy a climb back up to the zero level. It is the flat end at large distance that must sit at zero.

Check — multiple choice

A third potential-energy curve is shown. Evaluate it. Which statement is correct?

AThe curve is acceptable as drawn

Feedback if chosen: This option is wrong — you accepted a curve with no steep rise. At short distance the repulsions dominate, so the potential energy cannot stay at the minimum.

BNot acceptable — the flat region at large distance must sit below zero, not at the zero level

Feedback if chosen: This option is wrong — the flat region at large distance belongs at zero, and this curve puts it there correctly. The flaw is at the short-distance end.

CNot acceptable — the bond-energy arrow should reach from the minimum down to the distance axis, not up to the zero line

Feedback if chosen: This option is wrong — the bond energy is an energy, so its marker is the vertical gap from the minimum up to the zero line, exactly as this curve draws it. The flaw is at the short-distance end.

DNot acceptable — below the equilibrium bond length the curve must rise steeply instead of staying flat ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Below the equilibrium bond length the potential energy rises steeply — the repulsions dominate there. A curve that stays flat at the minimum is missing the steep rise at short distance.

Check — multiple choice

A curve for the bonded pair in HCl is shown, with its values labeled. Evaluate it. Which statement is correct?

AThe curve is acceptable as drawn ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Every required feature checks out: labeled axes, flat at zero at large distance, a single minimum at 127 pm, a steep rise at short distance, and the bond energy of 431 kJ/mol marked as a vertical gap up to the zero level.

BNot acceptable — the bond energy should be marked as a horizontal arrow along the distance axis

Feedback if chosen: This option is wrong — you demanded the flaw instead of the feature. Bond energy is an energy, so its marker is a vertical gap, exactly as this curve draws it.

CNot acceptable — the minimum should sit at zero on the energy axis

Feedback if chosen: This option is wrong — the minimum belongs below zero; the climb from the minimum up to zero is the bond energy. A minimum at zero would leave no gap to mark.

DNot acceptable — at large distance the curve should keep rising above zero instead of flattening

Feedback if chosen: This option is wrong — at large distance the atoms are too far apart to interact, so the curve must flatten at zero, exactly as this curve does.

U2-S06 — Bond order, length, and energy

8 lessons

U2-S06 · LESSON 59 · CSP-052

Bond order

Outcome: State the order of a covalent bond: 1 for a single bond, 2 for a double bond, 3 for a triple bond.

Part 1Counting shared pairs

You’ve already seen that a covalent bond consists of one or more pairs of valence electrons shared between two atoms.

One or more: some atom pairs share one pair, some share two, and some share three.

Each shared pair is drawn as one line between the element symbols — here are three examples. These drawings show only the shared pairs; the atoms' remaining valence electrons are not drawn, and you will meet the complete drawings later in this unit.

A bond made of one shared pair is a single bond, drawn as one line: F–F.

A bond made of two shared pairs is a double bond, drawn as two lines: O=O.

A bond made of three shared pairs is a triple bond, drawn as three lines: N≡N.

The number of shared pairs in a covalent bond is called its bond order.

A single bond has bond order 1, a double bond has bond order 2, and a triple bond has bond order 3.

What you are expected to be able to do

State that the order of a covalent bond is 1 for a single bond, 2 for a double bond, and 3 for a triple bond.

Part 2Reading bond order from a drawn structure

Here is a drawn structure of carbon dioxide, CO₂.

Each carbon–oxygen bond is drawn with two lines: two shared pairs, so it is a double bond.

The order of each C=O bond in CO₂ is 2.

One structure can hold bonds of different orders — here is an example.

Each carbon–hydrogen bond is one line: bond order 1.

The carbon–oxygen bond is two lines: bond order 2.

Read each bond’s order separately — count the lines of that bond only.

Part 3Check what you just learned

Check — numeric entry

In the structure shown, what is the bond order of the carbon–carbon bond?

✓ ANSWER KEY: 2

Feedback when correct: Correct. The carbon–carbon bond is drawn with two lines — two shared pairs — so its bond order is 2.

Anticipated wrong answers (feedback shown):

1	This counts the carbon–carbon connection as one bond no matter how it is drawn. Bond order counts the shared pairs: the bond is drawn with two lines, so its bond order is 2.
4	This counts the shared electrons. Bond order counts shared pairs, not electrons: two lines mean two shared pairs, so the bond order is 2.
6	This counts every line drawn in the structure. Read one bond at a time: the carbon–carbon bond alone is drawn with two lines, so its bond order is 2.

Check — numeric entry

In the structure shown, what is the bond order of the carbon–nitrogen bond?

✓ ANSWER KEY: 3

Feedback when correct: Correct. The carbon–nitrogen bond is drawn with three lines — three shared pairs — so its bond order is 3.

Anticipated wrong answers (feedback shown):

1	This is the bond order of the hydrogen–carbon bond. The question asks about the carbon–nitrogen bond, which is drawn with three lines: bond order 3.
6	This counts the shared electrons. Bond order counts shared pairs: three lines mean three shared pairs, so the bond order is 3.
4	This counts all four lines drawn in the structure. Read the carbon–nitrogen bond alone: three lines, so its bond order is 3.

Check — multiple choice

The structure shown contains three bonds. Reading from left to right, what are their bond orders?

A1, 2, 1

Feedback if chosen: This option is wrong — you read the carbon–carbon bond as two lines. Look again: it is drawn with three lines, so its bond order is 3.

B1, 1, 1

Feedback if chosen: This option is wrong — you counted each bond as one connection no matter how many lines it is drawn with. The carbon–carbon bond is drawn with three lines, so its bond order is 3.

C3, 3, 3

Feedback if chosen: This option is wrong — you gave every bond the order of the largest bond in the structure. Each bond is read separately: the hydrogen–carbon bonds are one line each, bond order 1.

D1, 3, 1 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The two hydrogen–carbon bonds are drawn with one line each (bond order 1), and the carbon–carbon bond is drawn with three lines (bond order 3).

U2-S06 · LESSON 60 · CSP-053

Bond order and bond length

Outcome: State that, for bonds between the same two elements, higher bond order means shorter bond length.

Part 1Three carbon–carbon bonds

You’ve already seen that bond order counts the shared pairs in a covalent bond: 1 for a single bond, 2 for a double bond, 3 for a triple bond.

You’ve already seen equilibrium bond length: the distance between the two nuclei at which a bonded pair’s potential energy is at its minimum. This lesson calls it bond length.

Carbon–carbon bonds come in all three orders — here are their measured bond lengths.

Measured carbon–carbon bond lengths (pm). Typical values, rounded.

C–C, with bond order 1, is the longest of the three.

C=C, with bond order 2, is shorter.

C≡C, with bond order 3, is the shortest.

For bonds between the same two elements: as bond order increases, bond length decreases.

Each shared pair sits between the two nuclei and attracts both, so a higher bond order pulls the nuclei closer together.

Part 2Where the comparison applies

The comparison needs the same two elements on both sides.

C–C against C≡C is a fair comparison — both bonds are carbon–carbon.

C–C against Si–Si is not covered — the elements changed, not just the bond order.

The two elements do not have to be identical to each other: C–O and C=O are both carbon–oxygen bonds, so they can be compared.

What you are expected to be able to do

State that, for bonds between the same two elements, higher bond order corresponds to shorter bond length.

Part 3Check what you just learned

Check — multiple choice

For bonds between the same two elements, what happens to bond length as bond order increases?

ABond length increases.

Feedback if chosen: This option is wrong — you reversed the direction. As bond order increases, more shared pairs pull the nuclei closer together, so bond length decreases.

BBond length decreases. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. For bonds between the same two elements, as bond order increases, bond length decreases.

CBond length stays the same.

Feedback if chosen: This option is wrong — you treated bond order as having no effect on bond length. More shared pairs pull the nuclei closer together, so bond length decreases.

DBond length changes, but the direction depends on which elements are bonded.

Feedback if chosen: This option is wrong — for bonds between the same two elements, the direction is always the same: as bond order increases, bond length decreases. It is comparisons across different element pairs that the pattern does not cover.

Check — multiple choice

Nitrogen atoms form both N=N and N≡N bonds. Which bond is shorter?

AN=N, because it has the lower bond order

Feedback if chosen: This option is wrong — you reversed the direction. For bonds between the same two elements, the higher bond order gives the shorter bond, so N≡N is shorter.

BThey are the same length, because both bonds are nitrogen–nitrogen

Feedback if chosen: This option is wrong — same two elements makes the bonds comparable, not equal. The bond orders differ, and higher bond order means shorter bond length.

CN≡N, because it has the higher bond order ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both bonds are nitrogen–nitrogen — the same two elements — so the higher bond order, N≡N, is the shorter bond.

DThere is no way to tell without measured bond lengths

Feedback if chosen: This option is wrong — you treated the comparison as needing measured values. Because both bonds are between the same two elements, bond order alone decides it: N≡N is shorter.

Check — multiple choice

Which pair of bonds can be compared by bond order alone?

AC–C and Si–Si

Feedback if chosen: This option is wrong — these bonds have the same bond order and differ in the elements. Bond order alone compares bonds between the same two elements.

BC=C and O=O

Feedback if chosen: This option is wrong — both are double bonds, but carbon–carbon and oxygen–oxygen are different element pairs, so bond order alone cannot compare them.

CC–C and C=O

Feedback if chosen: This option is wrong — C–C is a carbon–carbon bond and C=O is a carbon–oxygen bond. The two elements must be the same on both sides of the comparison.

DC=O and C≡O ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both bonds are carbon–oxygen bonds — the same two elements — so bond order alone compares them: C≡O, with the higher order, is the shorter bond.

U2-S06 · LESSON 61 · CSP-054

Ranking bond lengths by bond order

Outcome: Rank bonds between the same two elements by length, using bond order.

Part 1Writing a length ranking

You’ve already seen that, for bonds between the same two elements, higher bond order means shorter bond length.

To rank such bonds by length, first write down each bond’s order.

Then arrange them: the lower the bond order, the longer the bond.

For the carbon–carbon bonds, the orders are 1, 2, and 3, so the ranking from longest to shortest is C–C > C=C > C≡C.

The measured bond lengths agree with the ranking.

Measured carbon–carbon bond lengths (pm). Typical values, rounded.

Read a ranking like this as: C–C is longer than C=C, which is longer than C≡C.

Part 2A second family: carbon–oxygen bonds

The same two elements do not have to be identical to each other: C–O, C=O, and C≡O are all carbon–oxygen bonds.

Their orders are 1, 2, and 3, so the ranking from longest to shortest is C–O > C=O > C≡O — again the measured bond lengths agree.

Measured carbon–oxygen bond lengths (pm). Typical values, rounded.

What you are expected to be able to do

Rank bonds between the same two elements by length using bond order.

Part 3Check what you just learned

Check — multiple choice

Rank the nitrogen–nitrogen bonds N–N, N=N, and N≡N from longest to shortest.

AN–N > N=N > N≡N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. All three bonds are nitrogen–nitrogen — the same two elements — so the lowest bond order, N–N, is the longest and the highest, N≡N, is the shortest.

BN≡N > N=N > N–N

Feedback if chosen: This option is wrong — you reversed the direction: this is the ranking from shortest to longest. As bond order increases, bond length decreases, so N–N is the longest.

CAll three are about the same length

Feedback if chosen: This option is wrong — you treated bond order as having no effect. The three orders differ, and for the same two elements a higher bond order means a shorter bond.

DThey cannot be ranked without measured bond lengths

Feedback if chosen: This option is wrong — you treated the ranking as needing measured values. Because all three bonds are between the same two elements, bond order alone ranks them.

Check — multiple choice

Two structures are shown, one above the other. Which carbon–carbon bond is shorter?

AThe carbon–carbon bond in the upper structure

Feedback if chosen: This option is wrong — the upper carbon–carbon bond is drawn with two lines (bond order 2) and the lower with three (bond order 3). For bonds between the same two elements, the higher order is the shorter bond.

BThe carbon–carbon bond in the lower structure ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both bonds are carbon–carbon, so bond order decides: the lower structure’s bond has order 3 against the upper’s 2, and the higher order is the shorter bond.

CThe two carbon–carbon bonds are the same length

Feedback if chosen: This option is wrong — being carbon–carbon in both cases makes the bonds comparable, not equal. Their bond orders differ, so their lengths differ.

DThe bonds cannot be compared because they are in different substances

Feedback if chosen: This option is wrong — the comparison needs the same two elements, not the same substance. Both bonds are carbon–carbon, so bond order ranks them across the two structures.

Check — multiple choice

Carbon and nitrogen form C–N, C=N, and C≡N bonds. Rank them from longest to shortest.

AC≡N > C=N > C–N

Feedback if chosen: This option is wrong — you reversed the direction: this is the ranking from shortest to longest. The lowest bond order, C–N, is the longest bond.

BAll three are the same length because each is carbon–nitrogen

Feedback if chosen: This option is wrong — sharing the same two elements makes the bonds comparable, not equal. The bond orders differ, so the lengths differ.

CC–N > C=N > C≡N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. All three are carbon–nitrogen bonds — the same two elements — so the ranking follows bond order: C–N (order 1) longest, then C=N, then C≡N.

DThey cannot be ranked because carbon and nitrogen are different elements

Feedback if chosen: This option is wrong — “the same two elements” means the same pair on both sides of the comparison. Carbon–nitrogen against carbon–nitrogen qualifies; the two elements within a bond need not be identical.

U2-S06 · LESSON 62 · CSP-055

Bond order and bond energy

Outcome: State that, for bonds between the same two elements, higher bond order means greater bond energy.

Part 1Pulling a bonded pair apart

You’ve already seen that bond energy is the energy required to pull a bonded atom pair completely apart.

You’ve already seen that bond order counts the shared pairs in a covalent bond.

Here are the average bond energies of the three carbon–carbon bonds.

Average carbon–carbon bond energies (kJ/mol). Typical values, rounded; bond energies are averages over many substances containing the bond.

C–C, with bond order 1, has the smallest bond energy of the three.

C=C, with bond order 2, has a larger bond energy.

C≡C, with bond order 3, has the largest bond energy.

For bonds between the same two elements: as bond order increases, bond energy increases.

As bond order increases, more shared pairs attract the two nuclei, so pulling the pair completely apart requires more energy.

Part 2Where the comparison applies

As with bond length, the comparison needs the same two elements on both sides.

N=N against N≡N is covered; N≡N against C≡C is not — the elements changed.

What you are expected to be able to do

State that, for bonds between the same two elements, higher bond order corresponds to greater bond energy.

Part 3Check what you just learned

Check — multiple choice

For bonds between the same two elements, what happens to bond energy as bond order increases?

ABond energy stays the same.

Feedback if chosen: This option is wrong — you treated bond order as having no effect on bond energy. More shared pairs attract the nuclei, so a higher bond order takes more energy to pull apart.

BBond energy decreases.

Feedback if chosen: This option is wrong — you reversed the direction. As bond order increases, more shared pairs attract the two nuclei, so bond energy increases.

CBond energy changes, but the direction depends on which elements are bonded.

Feedback if chosen: This option is wrong — for bonds between the same two elements, the direction is always the same: as bond order increases, bond energy increases. It is comparisons across different element pairs that the pattern does not cover.

DBond energy increases. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. For bonds between the same two elements, as bond order increases, bond energy increases.

Check — multiple choice

Oxygen atoms form both O–O and O=O bonds. Which bond requires more energy to pull completely apart?

AO=O, because it has the higher bond order ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both bonds are oxygen–oxygen — the same two elements — so the higher bond order, O=O, has the greater bond energy.

BO–O, because it has the lower bond order

Feedback if chosen: This option is wrong — you reversed the direction. For bonds between the same two elements, higher bond order means greater bond energy, so O=O takes more energy to pull apart.

CBoth require the same energy, because both bonds are oxygen–oxygen

Feedback if chosen: This option is wrong — same two elements makes the bonds comparable, not equal. The bond orders differ, and the higher order has the greater bond energy.

DBond order predicts bond length, so it cannot answer this question

Feedback if chosen: This option is wrong — bond order predicts both. For bonds between the same two elements, a higher bond order means a shorter bond and a greater bond energy.

Check — multiple choice

A student wants to use bond order alone to decide which of two bonds has the greater bond energy. For which pair is that justified?

AN≡N and C≡C

Feedback if chosen: This option is wrong — these bonds have the same order and differ in the elements. Bond order alone compares bonds between the same two elements.

BN–N and N=N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both bonds are nitrogen–nitrogen — the same two elements — so bond order alone decides it: N=N, with the higher order, has the greater bond energy.

CC–C and C–O

Feedback if chosen: This option is wrong — C–C is carbon–carbon and C–O is carbon–oxygen. The two elements must match on both sides of the comparison.

DC=O and O=O

Feedback if chosen: This option is wrong — both are double bonds, but carbon–oxygen and oxygen–oxygen are different element pairs, so bond order alone cannot compare them.

U2-S06 · LESSON 63 · CSP-056

Ranking bond energies by bond order

Outcome: Rank bonds between the same two elements by bond energy, using bond order.

Part 1Writing a bond-energy ranking

You’ve already seen that, for bonds between the same two elements, higher bond order means greater bond energy.

To rank such bonds by bond energy, write down each bond’s order: the higher the order, the greater the bond energy.

For the carbon–carbon bonds, the ranking from greatest bond energy to least is C≡C > C=C > C–C — the average bond energies agree.

Average carbon–carbon bond energies (kJ/mol). Typical values, rounded; bond energies are averages over many substances containing the bond.

Read it as: C≡C has a greater bond energy than C=C, which has a greater bond energy than C–C.

Part 2A second family: carbon–oxygen bonds

C–O, C=O, and C≡O are all carbon–oxygen bonds, so bond order ranks them: C≡O > C=O > C–O.

The average bond energies agree with the ranking.

Average carbon–oxygen bond energies (kJ/mol). Typical values, rounded; bond energies are averages over many substances containing the bond.

What you are expected to be able to do

Rank bonds between the same two elements by bond energy using bond order.

Part 3Check what you just learned

Check — multiple choice

Rank the nitrogen–nitrogen bonds N–N, N=N, and N≡N by bond energy, from greatest to least.

AN–N > N=N > N≡N

Feedback if chosen: This option is wrong — this is the ranking by bond length, longest to shortest, not by bond energy. For the same two elements, the higher bond order has the greater bond energy, so N≡N comes first.

BAll three have about the same bond energy

Feedback if chosen: This option is wrong — you treated bond order as having no effect. The three orders differ, and for the same two elements a higher bond order means a greater bond energy.

CN≡N > N=N > N–N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. All three bonds are nitrogen–nitrogen — the same two elements — so the highest bond order, N≡N, has the greatest bond energy and N–N the least.

DThey cannot be ranked without measured bond energies

Feedback if chosen: This option is wrong — you treated the ranking as needing measured values. Because all three bonds are between the same two elements, bond order alone ranks them.

Check — multiple choice

Rank the carbon–nitrogen bonds C–N, C=N, and C≡N by bond energy, from greatest to least.

AC–N > C=N > C≡N

Feedback if chosen: This option is wrong — you reversed the direction; this is the ranking by bond length. For the same two elements, higher bond order means greater bond energy, so C≡N comes first.

BAll three are equal because each bond is carbon–nitrogen

Feedback if chosen: This option is wrong — sharing the same two elements makes the bonds comparable, not equal. The bond orders differ, so the bond energies differ.

CThey cannot be ranked because carbon and nitrogen are different elements

DC≡N > C=N > C–N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. All three are carbon–nitrogen bonds — the same two elements — so the ranking follows bond order: C≡N (order 3) greatest, then C=N, then C–N.

Check — multiple choice

In the structure shown, a student argues that the carbon–nitrogen bond has a greater bond energy than the hydrogen–carbon bond because its bond order is higher. Why is this use of bond order not justified?

AThe two bonds are not between the same two elements ✓ KEY — this is the correct answer

Feedback if chosen: Correct. One bond is hydrogen–carbon and the other is carbon–nitrogen. The bond-order comparison only covers bonds between the same two elements, so it does not apply here.

BBond order compares bond lengths, not bond energies

Feedback if chosen: This option is wrong — bond order compares both bond length and bond energy. The problem here is the elements: the two bonds are not between the same two elements.

CA bond of order 3 cannot be compared with a bond of order 1

Feedback if chosen: This option is wrong — orders 1 and 3 compare fine when the elements match, as in C–C against C≡C. The problem here is that the element pairs differ.

DBonds within the same substance cannot be compared with each other

Feedback if chosen: This option is wrong — being in the same substance is not the issue. The comparison needs the same two elements on both sides, and these two bonds involve different pairs.

U2-S06 · LESSON 64 · CSP-057

Atomic radius and bond length

Outcome: State that, at the same bond order, bonds between larger atoms are generally longer.

Part 1Same bond order, different atoms

You’ve already seen the atomic radius trend: going down a group, atomic radius increases.

Bond order compared bonds between the same two elements. What if the bond order stays the same and the elements change instead?

C–C and Si–Si are both single bonds — here are their measured bond lengths.

Measured single-bond lengths (pm). Typical values, rounded.

Si sits below C in group 14, so Si atoms have the larger atomic radius.

The bond between the larger atoms is the longer one: Si–Si is longer than C–C.

Part 2A whole group of examples

The halogens all form single bonds with themselves — here are the measured bond lengths down group 17.

Measured halogen single-bond lengths (pm). Typical values, rounded.

Going down the group, atomic radius increases, and bond length increases with it.

At the same bond order, bonds between larger atoms are generally longer.

Generally: the pattern holds across comparisons like these, but it is a trend, not a guarantee in every case.

The comparison needs the same bond order on both sides — it cannot compare C–C against Si=Si.

What you are expected to be able to do

State that, at the same bond order, bonds between larger atoms are generally longer.

Part 3Check what you just learned

Check — multiple choice

Two bonds have the same bond order. As the atomic radius of the bonded atoms increases, what generally happens to bond length?

ABond length decreases.

Feedback if chosen: This option is wrong — you reversed the direction. As the atomic radius of the bonded atoms increases, the nuclei sit farther apart, so bond length generally increases.

BBond length increases. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. At the same bond order, as the atomic radius of the bonded atoms increases, bond length generally increases.

CBond length stays the same.

Feedback if chosen: This option is wrong — you treated atomic radius as having no effect. Larger atoms hold their nuclei farther apart, so the bond is generally longer.

DBond length changes only if the bond order also changes.

Feedback if chosen: This option is wrong — bond order is fixed in this comparison. Atomic radius changes bond length on its own: larger atoms, generally longer bonds.

Check — multiple choice

N–N and P–P are both single bonds, and P sits below N in group 15. Which bond is generally longer?

AN–N, because N has the smaller atomic radius

Feedback if chosen: This option is wrong — you attached the longer bond to the smaller atoms. Bonds between larger atoms are generally longer, so P–P is the longer bond.

BThey are the same length, because both bonds are single bonds

Feedback if chosen: This option is wrong — matching bond orders makes the comparison fair, not the lengths equal. The atomic radii differ, so the bond lengths differ.

CP–P, because P has the larger atomic radius ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both bonds have bond order 1, and P has the larger atomic radius, so P–P is generally the longer bond.

DThere is no way to tell, because the bonds are between different elements

Feedback if chosen: This option is wrong — different elements at the same bond order is exactly the comparison atomic radius covers: larger atoms, generally longer bond.

Check — multiple choice

The atomic-radius pattern compares bonds at the same bond order. Which pair meets that condition?

AC–C and C=C

Feedback if chosen: This option is wrong — these bonds are between the same two elements but their orders are 1 and 2. That comparison belongs to bond order, not atomic radius.

BN≡N and P–P

Feedback if chosen: This option is wrong — the orders are 3 and 1, so the same-order condition fails before atomic radius can be used.

CO=O and Cl–Cl

Feedback if chosen: This option is wrong — the orders are 2 and 1, so the same-order condition fails.

DCl–Cl and Br–Br ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both are single bonds — the same bond order — so the atomic-radius comparison applies: Br atoms are larger, and Br–Br is generally the longer bond.

U2-S06 · LESSON 65 · CSP-058

Ranking bond lengths by atomic radius

Outcome: Rank same-order bonds by length, using the atomic radii of the bonded atoms.

Part 1Writing a radius-based ranking

You’ve already seen that, at the same bond order, bonds between larger atoms are generally longer.

To rank same-order bonds by length, compare the atomic radii of the bonded atoms: the larger the atoms, the longer the bond.

Going down group 17 the atomic radius increases, so the halogen single bonds rank, from longest to shortest: I–I > Br–Br > Cl–Cl > F–F.

The measured bond lengths agree with the ranking.

Measured halogen single-bond lengths (pm). Typical values, rounded.

Part 2One atom in common

The bonded atoms do not both have to change — hold one atom fixed and compare the other.

In the hydrogen–halogen single bonds, only the halogen changes: going down the group its atomic radius increases, and the ranking is H–I > H–Br > H–Cl > H–F.

Measured hydrogen–halogen single-bond lengths (pm). Typical values, rounded.

What you are expected to be able to do

Rank same-order covalent bonds by length using the radii of the bonded atoms.

Part 3Check what you just learned

Check — multiple choice

N–N, P–P, and As–As are all single bonds, and the elements sit in group 15 with N at the top, then P, then As. Rank the bonds from longest to shortest.

AAs–As > P–P > N–N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. All three bonds have order 1, and atomic radius increases down the group, so the bond between the largest atoms, As–As, is generally the longest.

BN–N > P–P > As–As

Feedback if chosen: This option is wrong — you reversed the radius trend. Going down a group, atomic radius increases, so As–As is the longest bond, not the shortest.

CAll three are about the same length because all are single bonds

Feedback if chosen: This option is wrong — matching bond orders makes the comparison fair, not the lengths equal. The atomic radii differ, so the bond lengths differ.

DThey cannot be ranked without measured bond lengths

Feedback if chosen: This option is wrong — you treated the ranking as needing measured values. At the same bond order, the atomic radii of the bonded atoms rank the bonds.

Check — multiple choice

Rank the single bonds C–F, C–Cl, and C–Br from longest to shortest.

AC–F > C–Cl > C–Br

Feedback if chosen: This option is wrong — you reversed the radius trend. Going down group 17 the halogen’s atomic radius increases, so C–Br is the longest bond and C–F the shortest.

BC–Br > C–Cl > C–F ✓ KEY — this is the correct answer

Feedback if chosen: Correct. All three bonds have order 1 and only the halogen changes: its atomic radius increases down the group, so C–Br is generally the longest and C–F the shortest.

CAll three are the same length because each one contains carbon

Feedback if chosen: This option is wrong — sharing one atom does not equalize the lengths. The other atom’s radius still changes, and the bond grows with it.

DThey cannot be ranked because the bonds are between different pairs of elements

Feedback if chosen: This option is wrong — different element pairs at the same bond order is exactly what the atomic-radius comparison covers. Only the halogen changes here, so its radius ranks the bonds.

Check — multiple choice

A student claims Si=Si is longer than C–C because Si atoms have the larger atomic radius. Does the atomic-radius comparison justify the claim?

AYes — bonds between larger atoms are longer in every comparison

Feedback if chosen: This option is wrong — you overgeneralized the pattern. It reads “generally longer,” and it applies at the same bond order; these two bonds have orders 2 and 1.

BNo — Si atoms have a smaller atomic radius than C atoms

Feedback if chosen: This option is wrong — Si sits below C in group 14, so its atomic radius is larger. The real problem is that the bond orders differ.

CNo — the bonds have different bond orders, and the radius comparison applies at the same bond order ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Si=Si has bond order 2 and C–C has bond order 1, so the same-order condition fails and the radius comparison alone cannot justify the claim. Measured lengths do agree with the student — Si=Si, at about 215 pm, is longer than C–C at 154 pm — but only the measurements can show that.

DNo — bonds between different elements can never be compared

Feedback if chosen: This option is wrong — comparing bonds between different elements is exactly what the radius comparison is for, provided the bond orders match. Here they do not.

U2-S06 · LESSON 66 · CSP-059

Ranking bond lengths: bond order and atomic radius

Outcome: Rank covalent bond lengths by applying the bond-order and atomic-radius comparisons separately, then combining them.

Part 1Two comparisons, used separately

You’ve already seen two ways to compare bond lengths.

For bonds between the same two elements: as bond order increases, bond length decreases.

At the same bond order: bonds between larger atoms are generally longer.

Each comparison answers one kind of question — so compare one pair of bonds at a time, and pick the comparison that fits that pair.

Part 2Ranking three bonds

Rank Si–Si, C–C, and C=C from longest to shortest.

Si–Si and C–C: same bond order, different atoms — the atomic-radius comparison applies. Si atoms are larger, so Si–Si is longer than C–C.

C–C and C=C: same two elements, different bond order — the bond-order comparison applies. C=C has the higher order, so C=C is shorter than C–C.

Combine the two comparisons: Si–Si > C–C > C=C.

The measured bond lengths agree.

Measured bond lengths (pm). Typical values, rounded.

Part 3When the comparisons collide

If both the elements and the bond order differ, the two comparisons can pull in opposite directions.

Si=Si against C–C: larger atoms predict a longer bond, but higher bond order predicts a shorter one. The two comparisons disagree, so comparisons alone cannot rank the pair.

When the comparisons pull in opposite directions, the ranking needs measured bond lengths.

What you are expected to be able to do

Rank covalent bond lengths by considering bond order and atomic radius separately before combining them.

Part 4Check what you just learned

Check — multiple choice

Rank P–P, N–N, and N=N from longest to shortest.

AN=N > N–N > P–P

Feedback if chosen: This option is wrong — you reversed both comparisons. P atoms are larger than N atoms, so P–P is the longest single bond, and the higher order N=N is the shortest.

BN–N > P–P > N=N

Feedback if chosen: This option is wrong — you reversed the atomic-radius comparison. P–P and N–N are both single bonds, and P atoms are larger, so P–P is the longer of the two.

CP–P > N=N > N–N

Feedback if chosen: This option is wrong — you reversed the bond-order comparison. N–N and N=N are both nitrogen–nitrogen, and the higher order N=N is the shorter bond, not the longer.

DP–P > N–N > N=N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. P–P and N–N share bond order 1 and P atoms are larger, so P–P is the longest; N–N and N=N share their elements and the higher order N=N is the shortest.

Check — multiple choice

Three measured bond lengths. Typical values, rounded to the nearest picometer.
Measured bond length (pm)
205
148
121

The table lists three measured bond lengths. One belongs to each of S–S, O–O, and O=O. Which assignment is correct?

A205 pm is S–S, 148 pm is O–O, 121 pm is O=O ✓ KEY — this is the correct answer

Feedback if chosen: Correct. S–S and O–O share bond order 1 and S atoms are larger, so S–S is the longest; O–O and O=O share their elements and the higher order O=O is the shortest.

B205 pm is O=O, 148 pm is O–O, 121 pm is S–S

Feedback if chosen: This option is wrong — you reversed both comparisons. S atoms are larger, so the single bond S–S is the longest at 205 pm, and the higher order O=O is the shortest at 121 pm.

C205 pm is S–S, 148 pm is O=O, 121 pm is O–O

Feedback if chosen: This option is wrong — you reversed the bond-order comparison. O–O and O=O are both oxygen–oxygen, and the higher order O=O is the shorter of the two.

D205 pm is O–O, 148 pm is S–S, 121 pm is O=O

Feedback if chosen: This option is wrong — you reversed the atomic-radius comparison. S–S and O–O are both single bonds, and S atoms are larger, so S–S is the longer of the two.

Check — multiple choice

Rank Si–Si, C–N, C=N, and C≡N from longest to shortest.

AC–N > C=N > C≡N > Si–Si

Feedback if chosen: This option is wrong — you kept the bond-order ranking but reversed the atomic-radius comparison. Si atoms are larger than both C and N, so the single bond Si–Si is the longest, not the shortest.

BSi–Si > C–N > C=N > C≡N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Si–Si and C–N share bond order 1 and Si atoms are larger, so Si–Si is the longest; the three carbon–nitrogen bonds then follow bond order, C–N longest of the three down to C≡N.

CSi–Si > C≡N > C=N > C–N

Feedback if chosen: This option is wrong — you reversed the bond-order comparison within the carbon–nitrogen family. The lowest order, C–N, is the longest of the three, and C≡N is the shortest.

DC≡N > C=N > C–N > Si–Si

Feedback if chosen: This option is wrong — you reversed both comparisons. Si–Si, between the largest atoms at order 1, is the longest bond, and the highest order C≡N is the shortest.

Check — multiple choice

Which pair of bonds cannot be ranked using the two comparisons alone?

ASi–Si and C–C

Feedback if chosen: This option is wrong — these bonds share bond order 1, so the atomic-radius comparison decides it: Si atoms are larger, and Si–Si is longer.

BC–C and C=C

Feedback if chosen: This option is wrong — these bonds share their elements, so the bond-order comparison decides it: C=C has the higher order and is shorter.

CSi=Si and C–C ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both the elements and the bond order differ, and the comparisons pull in opposite directions — larger atoms predict longer, higher order predicts shorter — so measured bond lengths are needed.

DSi–Si and Si=Si

Feedback if chosen: This option is wrong — these bonds share their elements, so the bond-order comparison decides it: Si=Si has the higher order and is shorter.

U2-S07 — Ionic interaction strength

2 lessons

U2-S07 · LESSON 67 · CSP-060

Ion size and the distance between the ion centers

Outcome: Predict which pair of oppositely charged ions has the smaller distance between the ion centers, using supplied ionic radii.

Part 1Ion size sets how close the centers can get

Here are two ion pairs drawn to scale: Li⁺ touching F⁻, and K⁺ touching I⁻.

You've already seen that an ion has a size — its ionic radius, which measures how far its electron cloud extends from its center.

You've already seen the distance between the nuclei of two atoms as internuclear distance. For two ions, the same measurement is the distance between the ion centers.

When a cation and an anion attract, they pull together until their electron clouds press against each other — the two ions end up touching.

Once the ions touch, the distance between the ion centers equals the sum of the two ionic radii.

So the smaller the two ions, the smaller that sum, and the closer the centers of the opposite charges can approach.

In the figure, Li⁺ (radius 76 pm) touching F⁻ (radius 133 pm) puts the ion centers 209 pm apart.

K⁺ (radius 138 pm) touching I⁻ (radius 220 pm) puts the ion centers 358 pm apart.

Li⁺ and F⁻ are the smaller ions, so their centers approach more closely.

What you are expected to be able to do

State that smaller ions let the centers of opposite charges approach more closely, and use supplied ionic radii to pick the pair of ions with the smaller distance between the ion centers.

Part 2Check what you just learned

Check — multiple choice

Ionic radii
Ion	Ionic radius (pm)
Na⁺	102
Cl⁻	181
K⁺	138
Br⁻	196

Use the table of ionic radii. In which pair can the centers of the oppositely charged ions approach more closely: Na⁺ with Cl⁻, or K⁺ with Br⁻?

AThe two pairs approach equally closely, because all four ions carry a single charge

Feedback if chosen: This option is wrong — you used the charges to judge the distance. The charges set how strongly the ions pull on each other, not how close the touching ions sit; that distance is the sum of the two ionic radii.

BK⁺ with Br⁻, because larger ions attract each other more strongly

Feedback if chosen: This option is wrong — you connected larger ions with a closer approach. It is the opposite: the distance between the ion centers is the sum of the two radii, so larger ions hold their centers farther apart.

CK⁺ with Br⁻, because ions with more electrons pack together more tightly

Feedback if chosen: This option is wrong — you reasoned from electron count. How closely the centers approach is set by the ionic radii, and K⁺ and Br⁻ each have the larger radius.

DNa⁺ with Cl⁻, because both of its ions have the smaller radius ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Na⁺ (102 pm) is smaller than K⁺ (138 pm) and Cl⁻ (181 pm) is smaller than Br⁻ (196 pm), so the sum of the radii — the distance between the ion centers — is smaller for Na⁺ with Cl⁻: 283 pm against 334 pm.

Check — multiple choice

Ionic radii
Ion	Ionic radius (pm)
Na⁺	102
K⁺	138
F⁻	133
Br⁻	196

Use the table of ionic radii. Which cation–anion pair puts its ion centers closest together when the ions touch?

AK⁺ with Br⁻

Feedback if chosen: This option is wrong — you picked the two largest ions. Their radii sum to 138 + 196 = 334 pm, the largest distance between the ion centers of the four pairs, not the smallest.

BNa⁺ with Br⁻

Feedback if chosen: This option is wrong — you checked only the cation. Na⁺ is the smaller cation, but Br⁻ is the larger anion: 102 + 196 = 298 pm. Na⁺ with F⁻ gives 102 + 133 = 235 pm.

CNa⁺ with F⁻ ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The distance between the ion centers is the sum of the two radii, and Na⁺ and F⁻ are the smallest cation and the smallest anion in the table: 102 + 133 = 235 pm.

DK⁺ with F⁻

Feedback if chosen: This option is wrong — you checked only the anion. F⁻ is the smaller anion, but K⁺ is the larger cation: 138 + 133 = 271 pm. Na⁺ with F⁻ gives 102 + 133 = 235 pm.

Check — multiple choice

Ionic radii
Ion	Ionic radius (pm)
Li⁺	76
Cl⁻	181
K⁺	138
F⁻	133

Use the table of ionic radii. Which pair has the smaller distance between the ion centers: Li⁺ with Cl⁻, or K⁺ with F⁻?

AK⁺ with F⁻, because F⁻ is the smaller of the two anions

Feedback if chosen: This option is wrong — you judged from one ion. Both radii count toward the sum: 138 + 133 = 271 pm for K⁺ with F⁻, but 76 + 181 = 257 pm for Li⁺ with Cl⁻.

BLi⁺ with Cl⁻, because its two radii add up to the smaller sum ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The distance between the ion centers is the sum of the two radii: 76 + 181 = 257 pm for Li⁺ with Cl⁻, against 138 + 133 = 271 pm for K⁺ with F⁻.

CK⁺ with F⁻, because its two ions are more similar in size

Feedback if chosen: This option is wrong — you used how alike the two radii are. Only their sum sets the distance between the ion centers, and 76 + 181 = 257 pm is smaller than 138 + 133 = 271 pm.

DThe two distances are the same, because each pair combines one small ion with one large ion

Feedback if chosen: This option is wrong — you treated small-plus-large as one fixed size. The two sums differ: 76 + 181 = 257 pm against 138 + 133 = 271 pm, so the Li⁺ and Cl⁻ centers sit closer.

U2-S07 · LESSON 68 · CSP-061

Comparing the strength of ionic attractions

Outcome: Predict which of two ion pairs has the stronger attraction by comparing the ion charges and the distances between the ion centers separately.

Part 1Two comparisons, made separately

Some ionic attractions are far stronger than others, and two comparisons tell you which of two ion pairs holds together more strongly.

You've already seen that as two charged particles move closer together, the force between them grows, and as they move apart, it weakens.

You've already seen that when either charge is larger, with the distance unchanged, the force between the charges is stronger.

You've already seen Coulomb's law for the force between two charged particles: |F| ∝ |q₁q₂| / r², with opposite charges attracting.

You've already seen how to use Coulomb's law to judge which of two situations — different charges, different distances — has the stronger force.

You've already seen that smaller ions let the centers of opposite charges approach more closely.

For a pair of oppositely charged ions, q₁ and q₂ are the ion charges, r is the distance between the ion centers, and the force is the attraction holding the pair together.

To compare two ion pairs, make the two comparisons separately: first compare the ion charges, then compare the distances between the ion centers.

When one comparison favors a pair and the other is even — or both favor the same pair — you can call the stronger attraction directly.

Treat distances that differ by only a few percent as even: a difference that small cannot stand against a charge comparison.

When the two comparisons clearly favor different pairs, the effects compete, and you need supplied values to decide.

Part 2When the charges differ

Compare MgO with NaCl, using the data shown. In MgO the ion charges are 2+ and 2−; in NaCl they are 1+ and 1−.

Ion charges and distances between the ion centers
Pair	Ion charges	Distance between the ion centers (pm)
MgO	2+ and 2−	212
NaCl	1+ and 1−	283

Here both comparisons favor MgO: its charges are larger, and its distance between the ion centers is smaller — 212 pm against 283 pm.

Because each MgO charge is twice as large, the charge product q₁q₂ — the two charges multiplied together — is four times larger for MgO.

With the smaller distance and a four-times-larger charge product, both comparisons agree: the attraction between Mg²⁺ and O²⁻ is much stronger than the attraction between Na⁺ and Cl⁻.

Part 3When the charges are the same

Compare LiF with KBr, using the data shown.

Ion charges and distances between the ion centers
Pair	Ion charges	Distance between the ion centers (pm)
LiF	1+ and 1−	209
KBr	1+ and 1−	334

The ion charges are the same in both pairs: 1+ and 1−. The charge comparison is even.

Li⁺ and F⁻ are smaller ions than K⁺ and Br⁻, so the distance between the ion centers is smaller in LiF: 209 pm against 334 pm.

Because the distance between the ion centers is smaller in LiF, the attraction between Li⁺ and F⁻ is stronger than the attraction between K⁺ and Br⁻.

Part 4When the two effects compete

Compare CaO with LiF, using the data shown.

Ion charges and distances between the ion centers
Pair	Ion charges	Distance between the ion centers (pm)
CaO	2+ and 2−	240
LiF	1+ and 1−	209

The charge comparison favors CaO: its charge product is four times larger. The distance comparison favors LiF: its distance between the ion centers is smaller.

The two effects compete, so neither comparison alone can decide — weigh the supplied values.

The charge product is 4 times larger for CaO, while the squared distance is only about 1.3 times larger — and dividing by 1.3 cannot undo multiplying by 4.

So the attraction between Ca²⁺ and O²⁻ is stronger, even though LiF has the smaller distance between the ion centers.

What you are expected to be able to do

Predict which of two ion pairs has the stronger attraction by comparing the ion charges and the distances between the ion centers separately, using supplied data whenever the two effects compete.

Part 5Check what you just learned

Check — multiple choice

Ion charges and distances between the ion centers
Pair	Ion charges	Distance between the ion centers (pm)
NaF	1+ and 1−	235
CaO	2+ and 2−	240

Use the data shown. Which solid has the stronger attraction between its ions: NaF or CaO?

ANaF, because the distance between the ion centers is smaller in NaF

Feedback if chosen: This option is wrong — you let a 5 pm difference decide. The two distances between the ion centers are nearly equal (235 pm and 240 pm), while the charge product is four times larger for CaO, so the charges decide here.

BCaO, because its ion charges are twice as large while the two distances are nearly equal ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The distances between the ion centers are nearly equal, so the comparison comes down to the charges: 2+ with 2− gives a charge product four times larger than 1+ with 1−, and the attraction in CaO is stronger.

CNaF, because 1+ and 1− charges match each other more exactly than 2+ and 2−

Feedback if chosen: This option is wrong — you treated matching charges as a source of strength. Every cation–anion pair has charges that attract; what sets the strength is how large the charges are, and CaO's are twice as large.

DNeither — the larger charges in CaO and the smaller distance between the ion centers in NaF offset each other

Feedback if chosen: This option is wrong — you assumed the two effects cancel. Compare them: the charge product is four times larger for CaO, while its distance between the ion centers is larger by only 5 pm in 235 — nowhere near enough to offset a factor of four.

Check — multiple choice

Ion charges and distances between the ion centers
Pair	Ion charges	Distance between the ion centers (pm)
MgO	2+ and 2−	212
CaS	2+ and 2−	284

Use the data shown. Which solid has the stronger attraction between its ions: MgO or CaS?

ACaS, because larger ions attract each other more strongly

Feedback if chosen: This option is wrong — you connected larger ions with stronger attraction. Larger ions hold the ion centers farther apart, and a larger distance between the ion centers weakens the attraction.

BNeither — the attractions are equal, because the ion charges are the same in both solids

Feedback if chosen: This option is wrong — you stopped after the charge comparison. The charges are equal, so the decision passes to the distances between the ion centers, and those differ: 212 pm against 284 pm.

CCaS, because the attraction between the ions grows as the distance between the ion centers grows

Feedback if chosen: This option is wrong — you reversed the distance effect. As the distance between the ion centers grows, the attraction weakens, so the 284 pm pair is the weaker one.

DMgO, because the charges are equal and its distance between the ion centers is smaller ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The ion charges are the same in both solids, so the distances decide: 212 pm in MgO against 284 pm in CaS, and the smaller distance between the ion centers makes the attraction in MgO stronger.

Check — multiple choice

Ion charges and distances between the ion centers
Pair	Ion charges	Distance between the ion centers (pm)
MgS	2+ and 2−	256
NaF	1+ and 1−	235

Use the data shown. Which solid has the stronger attraction between its ions: MgS or NaF?

AMgS — the charge product is 4 times larger, while the squared distance is only about 1.2 times larger ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The effects compete — MgS has the larger charges, NaF the smaller distance between the ion centers — so weigh the values: the charge product is 4 times larger for MgS, while the squared distance is only about 1.2 times larger. The factor of 4 wins.

BNaF — its distance between the ion centers is smaller, and the distance comparison outweighs the charge comparison

Feedback if chosen: This option is wrong — you gave the distance comparison automatic priority. Weigh it: squaring 256/235 gives only about 1.2, while the charge product for MgS is 4 times larger, so the charges win here.

CNeither — the larger charges and the smaller distance offset each other exactly

Feedback if chosen: This option is wrong — you assumed the competing effects cancel. Check with the values: a factor of 4 from the charges stands against only about 1.2 from the squared distance — far from an exact offset.

DMgS — S²⁻ is a large ion, and larger ions pull harder

Feedback if chosen: This option is wrong — you credited the large S²⁻ ion with extra pull. A larger ion increases the distance between the ion centers, which weakens the attraction; MgS wins despite that, because of its larger charges.

Check — multiple choice

Ion charges and distances between the ion centers
Pair	Ion charges	Distance between the ion centers (pm)
MgO	2+ and 2−	212
NaCl	1+ and 1−	283

Use the data shown. One of these two solids melts at 801 °C and the other at 2852 °C. Which solid melts at 2852 °C, and why?

ANaCl — its ions are larger, and larger ions attract each other more strongly, so more heat is needed to pull them apart

Feedback if chosen: This option is wrong — you connected larger ions with stronger attraction. Larger ions hold the ion centers farther apart — 283 pm against 212 pm — and a larger distance between the ion centers weakens the attraction, so NaCl is the 801 °C solid.

BMgO — because its ions are smaller; the ion charges play no part in a melting point, which follows ion size alone

Feedback if chosen: This option is wrong — you dropped the charge comparison. Ion size matters only through the distance between the ion centers, and the charges matter too: 2+ with 2− gives MgO a charge product four times larger, the bigger reason its attraction is stronger.

CNeither can be named from the data — Coulomb's law describes attractions between ions, not a measured melting point

Feedback if chosen: This option is wrong — you treated melting as unrelated to the attraction. You've already seen that melting an ionic solid means giving the ions enough energy to overcome the attractions holding them, so the stronger attraction sets the higher melting point.

DMgO — larger charges and a smaller distance make the attraction between its ions stronger, so pulling its ions apart takes more energy ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both comparisons favor MgO: its charge product is four times larger and its distance between the ion centers is smaller. Melting must pull the ions apart against their attraction, so the stronger attraction in MgO sets the higher melting point — MgO is the 2852 °C solid.

Check — multiple choice

Two ion pairs are being compared. In which situation do you need to weigh supplied values before you can say which attraction is stronger?

AThe ion charges are the same in both pairs, and the distance between the ion centers is smaller in one of the two pairs

Feedback if chosen: This option is wrong — you called an even charge comparison a conflict. With the charges the same, the distance comparison decides on its own: the pair with the smaller distance between the ion centers has the stronger attraction.

BThe distances between the ion centers differ by only a few percent, and the ion charges are larger in one pair

Feedback if chosen: This option is wrong — you called nearly even distances a conflict. Distances within a few percent are treated as even, so the charge comparison decides on its own: the pair with the larger charges has the stronger attraction.

CThe ion charges are larger in one pair, and the distance between the ion centers is much smaller in the other pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. One pair clearly wins on charges and the other clearly wins on distance, so the effects compete — only the supplied values can tell you which effect is bigger.

DThe ion charges are the same in both pairs, and the distances between the ion centers are also the same

Feedback if chosen: This option is wrong — you saw a decision where nothing differs. With both comparisons even, the attractions are comparable, and there is nothing left to weigh.

AFTER U2-S07 · CHECKPOINT · U2-G2 · 7 items

Checkpoint 2 — Potential-energy curves, bond trends, and ionic attractions

Check — multiple choice

A student sketched this potential-energy curve for two nitrogen atoms. The axes are labeled correctly. What must the student fix before the sketch is acceptable?

AThe minimum must sit below the zero line — the bonded pair has lower potential energy than the separated atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The zero level belongs to the far-apart atoms, and forming a bond lowers the potential energy, so the minimum must dip below zero — not float above it.

BThe flat region at large distance must be lowered so that it sits level with the minimum

Feedback if chosen: This option is wrong — you moved the wrong feature. The flat region is correctly at zero (atoms too far apart to interact); it is the minimum that must sit below that level.

CThe steep rise at short distance must be removed — the curve should keep falling as the atoms approach

Feedback if chosen: This option is wrong — you removed a required feature. Below the equilibrium bond length the curve must rise increasingly steeply as repulsions dominate.

DNothing — the sketch is acceptable as drawn

Feedback if chosen: This option is wrong — you accepted a curve whose minimum sits above zero, which would mean the bonded pair has more energy than the separated atoms and the bond would not hold together.

Check — multiple choice

Rank the bonds Si–O, C–O, and C=O from longest to shortest.

AC=O > C–O > Si–O

Feedback if chosen: This option is wrong — you reversed both comparisons. Higher bond order pulls the atoms closer (C=O is the shortest), and the larger Si atom sits farther from O (Si–O is the longest).

BC–O > Si–O > C=O

Feedback if chosen: This option is wrong — you ranked C–O above Si–O. At the same bond order, the bond with the larger atom is longer, and Si is larger than C.

CSi–O > C–O > C=O ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Size first: Si is larger than C, so Si–O is longer than C–O. Order second: the double bond pulls the same two atoms closer, so C=O is shorter than C–O.

DSi–O > C=O > C–O

Feedback if chosen: This option is wrong — you ranked C=O above C–O. Between the same two elements, higher bond order means a shorter bond, so the double bond is the shorter of the two.

Check — multiple choice

Ion charges and distances between the ion centers
Pair	Ion charges	Distance between the ion centers (pm)
BaO	2+ and 2−	275
KF	1+ and 1−	271

Use the data shown. Which solid has the stronger attraction between its ions: BaO or KF?

AKF, because its distance between the ion centers is slightly smaller

Feedback if chosen: This option is wrong — you let a 4 pm difference decide. Weigh each factor separately: the distances are nearly equal, while BaO's charge product is four times KF's.

BBaO, because its ion charges are twice as large while the two distances are nearly the same ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Apply the two factors separately: the distances (275 pm and 271 pm) are nearly identical, so the charges decide — 2+ with 2− gives a charge product four times that of 1+ with 1−.

CNeither — BaO's larger charges and KF's smaller distance offset each other exactly

Feedback if chosen: This option is wrong — you assumed the effects cancel without weighing them. A fourfold charge product against a distance difference of about 1.5% is not a standoff.

DBaO, because Ba²⁺ is the largest ion of the four, and larger ions attract more strongly

Feedback if chosen: This option is wrong — right solid, wrong reason. A larger ion matters only by increasing the distance between the ion centers, which weakens attraction; BaO wins on charges.

Check — numeric entry

Use the potential-energy curve shown for two bromine atoms. What is the equilibrium bond length of Br–Br, in pm?

✓ ANSWER KEY: 228 pm

Feedback when correct: Correct. The equilibrium bond length is the horizontal coordinate of the minimum, and the dashed guide marks it at 228 pm.

Anticipated wrong answers (feedback shown):

193	This is the depth of the minimum read from the energy axis, in kJ/mol. Equilibrium bond length is read from the horizontal distance axis: 228 pm.
-193	This is the energy-axis label at the minimum, in kJ/mol. Equilibrium bond length is the minimum's position along the horizontal distance axis: 228 pm.

Check — numeric entry

Use the potential-energy curve shown for two fluorine atoms. What is the bond energy of F–F, in kJ/mol?

✓ ANSWER KEY: 159 kJ/mol

Feedback when correct: Correct. The minimum sits at −159 kJ/mol relative to the zero line, so separating the bonded pair requires 159 kJ/mol.

Anticipated wrong answers (feedback shown):

142	This is the equilibrium bond length read from the horizontal axis, in pm. Bond energy is the vertical gap between the minimum and the zero line: 159 kJ/mol.
0	The zero line is where the separated atoms sit, not the bond energy. Read how far the minimum sits below it: 159 kJ/mol.

Sign-error retry feedback:

-159	Right gap, wrong sign. Bond energy is the energy required to pull the atoms apart, so it is positive: 159 kJ/mol. The minus sign on the axis label marks where the minimum sits relative to zero.

Check — multiple choice

Carbon and oxygen form C–O, C=O, and C≡O bonds. Rank them from longest to shortest.

AC≡O > C=O > C–O

Feedback if chosen: This option is wrong — you ranked more shared pairs as longer. More shared pairs pull the two nuclei closer together, so higher bond order means a shorter bond.

BC–O > C=O > C≡O ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Between the same two elements, bond length falls as bond order rises: the single bond is the longest and the triple bond the shortest.

CAll three are about the same length, because each is a carbon–oxygen bond

Feedback if chosen: This option is wrong — you ignored bond order. The number of shared pairs changes how tightly the same two atoms are pulled together, so the three lengths differ.

DThey cannot be ranked without measured bond lengths

Feedback if chosen: This option is wrong — no measurements are needed here. Because all three bonds join the same two elements, bond order alone ranks them.

Check — multiple choice

Rank the carbon–carbon bonds C–C, C=C, and C≡C by bond energy, from greatest to least.

AAll three have about the same bond energy, because each is carbon–carbon

Feedback if chosen: This option is wrong — you ignored bond order. Extra shared pairs make the same two atoms harder to pull apart, so the three bond energies differ.

BC–C > C=C > C≡C

Feedback if chosen: This option is wrong — you ranked fewer shared pairs as stronger. More shared pairs bind the pair more tightly, so higher bond order means greater bond energy.

CThey cannot be ranked without measured bond energies

Feedback if chosen: This option is wrong — no measurements are needed. All three bonds join the same two elements, so bond order alone ranks their energies.

DC≡C > C=C > C–C ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Between the same two elements, bond energy rises with bond order: the triple bond takes the most energy to pull apart, the single bond the least.

U2-S08 — Lewis-model foundations and boundaries

7 lessons

U2-S08 · LESSON 69 · CSP-062

What a Lewis diagram shows

Outcome: State what the lines and the dots in a Lewis diagram represent.

Part 1Lines and dots

You’ve already seen that a covalent bond is one or more pairs of valence electrons shared between two atoms.

Chemists record where a molecule’s valence electrons are with a drawing built from atom symbols, lines, and dots — a Lewis diagram.

Here is the Lewis diagram of H₂.

Each atom symbol stands for the atom itself: its nucleus and its inner electrons.

The line between the two H symbols is a shared pair of valence electrons — the pair that forms the covalent bond.

Here is the Lewis diagram of H₂O.

Each O–H line is a shared pair of valence electrons.

The two pairs of dots on O are valence electrons that are not shared — they stay with O.

Every valence electron of the molecule appears in the diagram: shared pairs as lines, unshared electrons as dots.

What you are expected to be able to do

State what the lines and the dots in a supplied Lewis diagram represent.

Part 2Check what you just learned

Check — multiple choice

In the Lewis diagram of F₂ shown, what does the line between the two F symbols represent?

AOne shared valence electron

Feedback if chosen: This option is wrong — you read the line as a single electron. One line stands for a pair: two shared valence electrons.

BThe attraction between two oppositely charged ions

Feedback if chosen: This option is wrong — you read a covalent diagram as an ionic one. The two F atoms are neutral and share electrons; the line is that shared pair, not an attraction between ions.

CAll fourteen valence electrons of the molecule

Feedback if chosen: This option is wrong — you assigned all the molecule’s electrons to the line. The line shows only the one shared pair; the other twelve valence electrons appear as the dots.

DOne shared pair of valence electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. One line between two atom symbols stands for one shared pair of valence electrons.

Check — multiple choice

In the Lewis diagram of F₂ shown, what do the pairs of dots next to each F represent?

AInner-shell electrons of the F atom

Feedback if chosen: This option is wrong — you counted the dots as inner electrons. Inner electrons are not drawn at all — the atom symbol stands for them; every dot in the diagram is a valence electron.

BValence electrons that are not shared ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Dots show valence electrons that are not shared — they stay with their own atom.

CElectrons shared between the two F atoms

Feedback if chosen: This option is wrong — you swapped the two marks. Shared electrons are drawn as the line between the symbols; dots sit next to one atom because they are not shared.

DElectrons each F gained to become an ion

Feedback if chosen: This option is wrong — you read the molecule as ions. F₂ is two neutral atoms sharing one pair; no electrons were gained or lost.

Check — numeric entry

How many shared pairs of valence electrons does the Lewis diagram of NH₃ show?

✓ ANSWER KEY: 3

Feedback when correct: Correct. Each of the three N–H lines is one shared pair; the pair of dots on N is not shared.

Anticipated wrong answers (feedback shown):

4	This answer counts the pair of dots on N as well. Dots show electrons that are not shared; only the three lines are shared pairs.
6	This answer counts electrons, not pairs. Each line is one shared pair of two electrons, and there are three lines: three shared pairs.
1	This answer counts the pair of dots on N. That pair is not shared — the shared pairs are the three lines, so the count is 3.

U2-S08 · LESSON 70 · CSP-063

Bonding pairs and lone pairs

Outcome: Classify each valence-electron pair in a Lewis diagram as a bonding pair or a lone pair.

Part 1Two jobs for a pair

You’ve already seen the Lewis diagram of H₂O: lines are shared pairs of valence electrons, and pairs of dots are valence electrons that are not shared.

A pair shared between two atoms holds those atoms together — it is called a bonding pair.

A pair that stays with one atom is called a lone pair.

In H₂O, each O–H line is a bonding pair: two bonding pairs in all.

The two pairs of dots on O are lone pairs: they belong to O alone.

Here is the Lewis diagram of Cl₂.

The line is a bonding pair, shared by the two Cl atoms.

Each Cl also keeps three pairs of dots: six lone pairs in the molecule.

The test is sharing: a pair between two atoms is a bonding pair; a pair with one atom is a lone pair.

What you are expected to be able to do

Classify each valence-electron pair in a supplied Lewis diagram as a bonding pair or a lone pair.

Part 2Check what you just learned

Check — multiple choice

In the Lewis diagram of HCl shown, which statement classifies the electron pairs correctly?

AThe line is a lone pair, and each pair of dots on Cl is a bonding pair.

Feedback if chosen: This option is wrong — you swapped the two labels. The shared pair drawn as the line is the bonding pair; the unshared pairs of dots are the lone pairs.

BAll four pairs are bonding pairs, because they all belong to the same molecule.

Feedback if chosen: This option is wrong — you classified by molecule instead of by sharing. Belonging to the molecule is not the test: a bonding pair is shared between two atoms, and only the line is.

CAll four pairs are lone pairs, because each pair sits closest to Cl.

Feedback if chosen: This option is wrong — you classified by position on the page. The test is sharing, not nearness: the line is shared between H and Cl, so it is a bonding pair.

DThe line is a bonding pair, and each pair of dots on Cl is a lone pair. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The line is shared between H and Cl, so it is a bonding pair; each pair of dots stays with Cl, so those are lone pairs.

Check — numeric entry

How many lone pairs does the Lewis diagram of NH₃ show?

✓ ANSWER KEY: 1

Feedback when correct: Correct. The pair of dots on N stays with N alone — one lone pair. The three lines are bonding pairs.

Anticipated wrong answers (feedback shown):

3	This answer counts the bonding pairs. The three lines are each shared between N and an H; the lone pair is the pair of dots on N.
4	This answer counts every pair. Three of the four are bonding pairs (the lines); only the pair of dots on N is a lone pair.

Check — numeric entry

How many lone pairs does the Lewis diagram of CH₄ show?

✓ ANSWER KEY: 0

Feedback when correct: Correct. All four pairs are lines shared between C and an H — bonding pairs. No pair stays with one atom, so there are no lone pairs.

Anticipated wrong answers (feedback shown):

4	This answer counts the bonding pairs. All four lines are shared between C and an H; a lone pair would be a pair of dots staying with one atom, and CH₄ has none.
8	This answer counts electrons, and all eight are in bonding pairs anyway. CH₄ has no pairs of dots, so it has zero lone pairs.

U2-S08 · LESSON 71 · CSP-064

The octet rule

Outcome: State the octet rule and count the electrons around an atom in a Lewis diagram.

Part 1A tendency toward eight

You’ve already seen that an atom’s valence electrons are the electrons in its outermost shell.

Across the main groups, atoms tend to gain, lose, or share electrons until eight valence electrons surround them.

A set of eight is called an octet, and the tendency toward it is called the octet rule.

The octet rule is a working rule of the Lewis model — a strong tendency, not a law every molecule obeys.

Part 2Counting to eight in a diagram

Here is the Lewis diagram of CO₂.

To count the electrons around an atom, add its lone-pair dots to every electron in its bonds.

Both electrons of a bonding pair count for each of the two atoms the line joins — sharing lets one pair serve both.

Count around the left O: two lone pairs hold 4 electrons, and the double bond holds 4 more — 8 in total.

Count around C: two double bonds hold 8 electrons, and C has no lone pairs — 8 again.

Every atom in CO₂ has an octet.

What you are expected to be able to do

State the octet rule, and count the electrons around an atom in a supplied Lewis diagram.

Part 3Check what you just learned

Check — multiple choice

According to the octet rule, what do main-group atoms tend to gain, lose, or share electrons to obtain?

AEight valence electrons around the atom ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The tendency is toward eight valence electrons around the atom — counted from its lone pairs and its bonds in a Lewis diagram.

BEight electrons in total, counting the inner shells

Feedback if chosen: This option is wrong — you counted inner-shell electrons. The octet is counted among valence electrons only; inner electrons are not part of it, and a Lewis diagram never draws them.

CEight bonds to other atoms

Feedback if chosen: This option is wrong — you counted bonds instead of electrons. The octet is eight electrons, and they can come from lone pairs and bonds together.

DEight pairs of valence electrons

Feedback if chosen: This option is wrong — you counted pairs instead of electrons. Eight pairs would be sixteen electrons; an octet is eight electrons — four pairs.

Check — numeric entry

How many electrons are counted around each Cl atom in the Cl₂ Lewis diagram shown?

✓ ANSWER KEY: 8

Feedback when correct: Correct. Three lone pairs hold 6 electrons, and the bonding pair adds 2 more: 8 — an octet.

Anticipated wrong answers (feedback shown):

6	This answer counts only the lone-pair dots. The bonding pair counts too, and both of its electrons count for each atom the line joins: 6 + 2 = 8.
7	This answer credits each Cl with only one electron of the bonding pair. Both shared electrons count for each atom the line joins: 6 + 2 = 8.
14	This answer counts every electron in the diagram. Count only the electrons around one Cl: three lone pairs plus the bonding pair, 6 + 2 = 8.

Check — numeric entry

How many electrons are counted around each N atom in the N₂ Lewis diagram shown?

✓ ANSWER KEY: 8

Feedback when correct: Correct. The triple bond holds 6 electrons and the lone pair holds 2: 6 + 2 = 8.

Anticipated wrong answers (feedback shown):

4	This answer counts the triple bond as one shared pair. Three lines are three pairs — 6 electrons — plus the lone pair’s 2: 8.
5	This answer counts each line as one electron. Each line is a pair, so the triple bond holds 6 electrons; add the lone pair’s 2: 8.
10	This answer counts every electron in the diagram. Count only the electrons around one N: the triple bond’s 6 plus its own lone pair’s 2 = 8.

U2-S08 · LESSON 72 · CSP-065

The hydrogen duet

Outcome: State that bonded hydrogen is complete with two valence electrons rather than eight.

Part 1Two is complete for hydrogen

You’ve already seen the octet rule: main-group atoms tend to gain, lose, or share electrons until eight valence electrons surround them.

Hydrogen is the built-in exception.

Hydrogen’s valence shell is the first shell, and the first shell holds at most two electrons.

So bonded hydrogen is complete with two valence electrons — one bonding pair — rather than eight.

A completed set of two is called a duet, matching the octet’s set of eight.

Part 2Counting the duet

Here is the Lewis diagram of H₂O.

Count around either H: its one bonding pair holds 2 electrons, and that is the complete count.

In a correct Lewis diagram, bonded H always looks like this: one bonding pair, no lone pairs, never more.

What you are expected to be able to do

State that bonded hydrogen is complete with two valence electrons rather than eight, and count them in a supplied Lewis diagram.

Part 3Check what you just learned

Check — numeric entry

How many electrons are counted around each H atom in the CH₄ Lewis diagram shown?

✓ ANSWER KEY: 2

Feedback when correct: Correct. Each H shares one bonding pair with C — 2 electrons — and bonded hydrogen is complete with two.

Anticipated wrong answers (feedback shown):

8	This answer applies the octet to H. Hydrogen’s valence shell is the first shell, which holds at most two electrons — bonded H is complete with 2.
1	This answer counts only the electron H brought with it. Both electrons of the bonding pair count for each atom the line joins, so each H has 2.

Check — multiple choice

Why is bonded hydrogen complete with two valence electrons rather than eight?

AH is physically too small for eight electrons to fit around it

Feedback if chosen: This option is wrong — you reasoned from the atom’s size. The limit is the shell, not the space: H’s valence shell is the first shell, and it holds at most two electrons.

BH’s valence shell is the first shell, and the first shell holds at most two electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The first shell holds at most two electrons, so two electrons fill hydrogen’s valence shell completely.

CBonded H really has eight — the other six are simply not drawn

Feedback if chosen: This option is wrong — you assumed hidden electrons. A Lewis diagram shows every valence electron; bonded H has exactly the two in its bonding pair.

DThe octet rule applies only to atoms that form more than one bond

Feedback if chosen: This option is wrong — you tied the rule to bond count. The octet rule is about reaching eight valence electrons; hydrogen is an exception because its first shell fills at two, not because of how many bonds it forms.

Check — multiple choice

In a correct Lewis diagram, how does bonded hydrogen always appear?

AWith one bonding pair and one lone pair

Feedback if chosen: This option is wrong — you gave H a lone pair. That would put four electrons around H, and its first shell holds at most two: the bonding pair is everything.

BWith two bonding pairs

Feedback if chosen: This option is wrong — you gave H two bonds. Two bonding pairs would be four electrons, more than the first shell holds. Bonded H has exactly one bonding pair.

CWith one bonding pair and no lone pairs ✓ KEY — this is the correct answer

Feedback if chosen: Correct. One bonding pair — two electrons — completes hydrogen, and it carries no lone pairs.

DWith eight electrons in four pairs

Feedback if chosen: This option is wrong — you applied the octet to H. Eight electrons would overfill hydrogen’s first shell; bonded H is complete with two.

U2-S08 · LESSON 73 · CSP-066

Odd-electron molecules

Outcome: Explain why a species with an odd number of valence electrons cannot give every atom a paired octet in one Lewis diagram.

Part 1Every mark holds a pair

You’ve already seen Lewis diagrams like this one of N₂: lines are bonding pairs, and dots come in lone pairs.

Every mark holds an even number of electrons: a line holds 2, and a lone pair holds 2.

N₂ has 10 valence electrons — 5 from each N — and the diagram places all 10 in pairs: three bonding pairs and two lone pairs.

When an atom’s octet has every electron sitting in a pair, call it a paired octet — each N here has one.

Part 2Eleven electrons cannot all pair

NO — nitrogen monoxide — is a real molecule, a gas found in car exhaust.

N brings 5 valence electrons and O brings 6, so NO has 11.

Eleven is odd: ten of the electrons can sit in five pairs, and one electron is always left over, unpaired.

No choice of lines and lone pairs changes that, because every mark holds an even number of electrons.

Here is the accepted Lewis diagram of NO: the unpaired electron appears as a single dot on N.

Count around O: the double bond holds 4 and the two lone pairs hold 4 — a paired octet of 8.

Count around N: the double bond holds 4, the lone pair holds 2, and the single dot adds 1 — seven electrons, one of them unpaired.

So one Lewis diagram of NO cannot give every atom a paired octet: the odd total guarantees one atom is left short.

A species with an odd number of valence electrons is called an odd-electron molecule — a known limit of the octet rule, not a mistake in the diagram.

What you are expected to be able to do

Explain why a species with an odd number of valence electrons cannot give every atom a paired octet in one Lewis diagram.

Part 3Check what you just learned

Check — multiple choice

NO₂ is a real molecule with 17 valence electrons. What does the odd total mean for its Lewis diagram?

AOne electron must be left unpaired, so not every atom can have a paired octet. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Lines and lone pairs each hold 2 electrons, so 16 of the 17 can pair; one electron is left unpaired, and the atom carrying it cannot have a paired octet.

BAll 17 electrons can be paired if enough double bonds are drawn.

Feedback if chosen: This option is wrong — you expected bonds to change the arithmetic. A double bond still holds an even number of electrons (4); no arrangement of even-sized marks can absorb an odd total.

CA Lewis diagram of NO₂ cannot be drawn at all.

Feedback if chosen: This option is wrong — you turned a limit into an impossibility. The diagram can be drawn — it simply shows one unpaired electron as a single dot.

DOne electron must be left out of the diagram to make the total even.

Feedback if chosen: This option is wrong — you dropped an electron. A Lewis diagram shows every valence electron the species has; the odd one is drawn as a single dot, not removed.

Check — multiple choice

Total valence electrons of four molecules
Molecule	Total valence electrons
CO	10
SO₂	18
ClO₂	19
N₂O	16

Use the table shown. Which molecule cannot give every atom a paired octet in one Lewis diagram?

ACO

Feedback if chosen: This option is wrong — you picked an even total. CO’s 10 electrons can all sit in pairs, and its diagram gives both atoms a paired octet; the impossibility comes only from an odd total.

BSO₂

Feedback if chosen: This option is wrong — 18 is even, so all of SO₂’s electrons can sit in pairs. Look for the total that cannot be split into pairs.

CClO₂ ✓ KEY — this is the correct answer

Feedback if chosen: Correct. ClO₂’s total is 19 — odd — so one electron must remain unpaired, and the atom carrying it cannot have a paired octet.

DN₂O

Feedback if chosen: This option is wrong — 16 is even, so N₂O’s electrons can all be paired. Only an odd total forces an unpaired electron.

Check — numeric entry

A molecule has 15 valence electrons. In any Lewis diagram of it, how many electrons must be left unpaired?

✓ ANSWER KEY: 1

Feedback when correct: Correct. Fourteen of the 15 electrons fill seven pairs; the one left over cannot pair, whatever the diagram looks like.

Anticipated wrong answers (feedback shown):

0	This answer assumes all 15 can pair. Every mark in a Lewis diagram holds 2 electrons, so marks always hold an even total — 15 leaves exactly one electron unpaired.
7	Seven is the number of complete pairs (14 electrons). The question asks how many electrons are left unpaired: 15 − 14 = 1.
15	Only the leftover electron is unpaired. Fourteen of the 15 sit in seven pairs; 1 remains unpaired.

U2-S08 · LESSON 74 · CSP-067

Central atoms with fewer than eight electrons

Outcome: Recognize valid Lewis diagrams in which the central atom has fewer than eight electrons.

Part 1Valid with fewer than eight

You’ve already seen that the octet rule is a strong tendency, not a law every molecule obeys.

Some molecules pair every electron and still leave the central atom short of eight.

Here is the accepted Lewis diagram of BF₃.

Count around each F: three lone pairs hold 6 electrons and the bonding pair holds 2 — an octet.

Count around B: three bonding pairs and no lone pairs — 6 electrons.

This diagram is valid: BF₃ is a real, stable molecule, and six electrons around B is how the Lewis model draws it.

A central atom with fewer than eight electrons is called electron-deficient; Be and B are the atoms most often drawn this way.

This is another honest limit of the octet rule: for a few central atoms, the completed diagram stops short of eight.

What you are expected to be able to do

Accept and read Lewis diagrams in which the central atom has fewer than eight electrons.

Part 2Check what you just learned

Check — numeric entry

How many electrons are counted around B in the BCl₃ Lewis diagram shown?

✓ ANSWER KEY: 6

Feedback when correct: Correct. B has three bonding pairs and no lone pairs: 3 × 2 = 6 electrons.

Anticipated wrong answers (feedback shown):

8	This answer states the octet the rule predicts instead of counting the diagram. B has three bonding pairs and no lone pairs: 6 electrons.
3	This answer counts bonds, not electrons. Each bonding pair holds 2 electrons, so three bonds hold 6.
24	This answer counts every electron in the diagram. Count only the electrons around B: three bonding pairs, 6 electrons.

Check — multiple choice

A student says the BCl₃ Lewis diagram shown must be wrong, because B does not have an octet. Which response is correct?

AThe student is right — every atom in a valid Lewis diagram needs an octet.

Feedback if chosen: This option is wrong — you treated the octet rule as a law. It is a tendency: B is one of the atoms accepted with fewer than eight, and BCl₃ is a real, stable molecule.

BThe diagram is valid — some central atoms, like B, are accepted with fewer than eight electrons. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Electron-deficient central atoms such as Be and B are accepted with fewer than eight electrons; the diagram is complete as drawn.

CThe student is right — a lone pair should be added to B to reach eight.

Feedback if chosen: This option is wrong — you added electrons the molecule does not have. A Lewis diagram shows exactly the valence electrons the atoms supply; there are no spare electrons to give B a lone pair.

DThe diagram is valid, but only because Cl atoms do not need octets.

Feedback if chosen: This option is wrong — you moved the exception to the wrong atom. Each Cl here has a full octet — three lone pairs plus a bonding pair; the atom drawn with fewer than eight is B.

Check — numeric entry

How many electrons are counted around Be in the BeCl₂ Lewis diagram shown?

✓ ANSWER KEY: 4

Feedback when correct: Correct. Be has two bonding pairs and no lone pairs: 2 × 2 = 4 electrons — another electron-deficient central atom.

Anticipated wrong answers (feedback shown):

8	This answer states an octet instead of counting the diagram. Be has two bonding pairs and no lone pairs: 4 electrons.
2	This answer counts bonds, not electrons. Each bonding pair holds 2 electrons, so two bonds hold 4.
16	This answer counts every electron in the diagram. Count only the electrons around Be: two bonding pairs, 4 electrons.

U2-S08 · LESSON 75 · CSP-068

Central atoms with more than eight electrons

Outcome: Recognize that a central atom in period 3 or later can appear with more than eight electrons in a valid Lewis diagram.

Part 1Valid with more than eight

You’ve already seen central atoms accepted with fewer than eight electrons.

The Lewis model also allows the opposite: some central atoms appear with more than eight.

Here is the accepted Lewis diagram of SF₆.

Count around S: six bonding pairs and no lone pairs — 12 electrons.

Each F still has an octet: three lone pairs plus its bonding pair.

There is a condition: only an atom in period 3 or a later period appears with more than eight electrons in a valid Lewis diagram.

S sits in period 3, so twelve electrons around S is accepted.

Atoms in period 2 — C, N, O, F — are never drawn with more than eight.

A central atom drawn with more than eight is said to have an expanded valence shell.

Treat the period condition as a rule of the model: the reason behind it is beyond this course.

What you are expected to be able to do

Accept and read Lewis diagrams in which a central atom from period 3 or later has more than eight electrons.

Part 2Check what you just learned

Check — numeric entry

How many electrons are counted around P in the PCl₅ Lewis diagram shown?

✓ ANSWER KEY: 10

Feedback when correct: Correct. P has five bonding pairs and no lone pairs: 5 × 2 = 10 electrons. P is in period 3, so the expanded count is accepted.

Anticipated wrong answers (feedback shown):

8	This answer stops at an octet instead of counting the diagram. P has five bonding pairs: 10 electrons — accepted, because P is in period 3.
5	This answer counts bonds, not electrons. Each bonding pair holds 2 electrons, so five bonds hold 10.
40	This answer counts every electron in the diagram. Count only the electrons around P: five bonding pairs, 10 electrons.

Check — multiple choice

Which of these atoms can appear as a central atom with more than eight electrons in a valid Lewis diagram?

Feedback if chosen: This option is wrong — you placed the expansion in period 2. N is a period-2 atom, and period-2 atoms are never drawn with more than eight electrons.

Feedback if chosen: This option is wrong — you placed the expansion in period 2. O is a period-2 atom, and period-2 atoms are never drawn with more than eight electrons.

CS ✓ KEY — this is the correct answer

Feedback if chosen: Correct. S is in period 3, so it can appear with more than eight electrons — twelve in SF₆, for example.

Feedback if chosen: This option is wrong — you placed the expansion in period 2. C is a period-2 atom, and period-2 atoms are never drawn with more than eight electrons.

Check — numeric entry

How many electrons are counted around Xe in the XeF₄ Lewis diagram shown?

✓ ANSWER KEY: 12

Feedback when correct: Correct. Four bonding pairs hold 8 electrons, and the two lone pairs on Xe hold 4 more: 12. Xe is in period 5, so the expanded count is accepted.

Anticipated wrong answers (feedback shown):

8	This answer counts only the bonding pairs. The two lone pairs on Xe count too: 8 + 4 = 12.
6	This answer counts pairs, not electrons. Six pairs around Xe — four bonding, two lone — hold 12 electrons.
36	This answer counts every electron in the diagram. Count only the electrons around Xe: four bonding pairs plus two lone pairs, 12 electrons.

U2-S09 — Lewis construction routine

10 lessons

U2-S09 · LESSON 76 · CSP-069

Counting valence electrons

Outcome: Calculate the total number of valence electrons in a neutral molecule.

Part 1Every diagram starts with a count

You’ve already seen that every valence electron of a molecule appears in its Lewis diagram: shared pairs as lines, unshared electrons as dots.

So the routine for drawing one starts with arithmetic: count how many valence electrons there are to place.

Each atom brings its valence electrons, and you’ve already seen where that number comes from: the atom’s group in the periodic table.

Valence electrons of some main-group atoms
Atom	Group	Valence electrons
H	1	1
C	14	4
N	15	5
O	16	6
F	17	7
P	15	5
S	16	6
Cl	17	7

The total for a neutral molecule is the sum of what its atoms bring — count every atom in the formula.

Part 2Worked examples

Q: How many valence electrons does CO₂ have in total?

Count the electrons each atom brings

C brings 4(C is in group 14)

each O brings 6(O is in group 16)

Total

4 + 6 + 6 = 16 electrons

CO₂ has 16 valence electrons

Q: How many valence electrons does NH₃ have in total?

Count the electrons each atom brings

N brings 5(N is in group 15)

each H brings 1

Total

5 + 1 + 1 + 1 = 8 electrons

NH₃ has 8 valence electrons

What you are expected to be able to do

Calculate the total number of valence electrons in a neutral molecule by adding the valence electrons its atoms bring.

Part 3Check what you just learned

Check — numeric entry

How many valence electrons does CF₄ have in total?

✓ ANSWER KEY: 32

Feedback when correct: Correct. C brings 4 and each of the four F atoms brings 7: 4 + 4 × 7 = 32.

Anticipated wrong answers (feedback shown):

11	This answer counts one F. The subscript means four F atoms, each bringing 7: 4 + 28 = 32.
42	This answer counts all of each atom’s electrons. Only valence electrons count: C brings 4, not 6, and each F brings 7, not 9 — the total is 32.
28	This answer leaves out the central C. Every atom in the formula contributes: 4 + 28 = 32.

Worked solution revealed after a second miss

Count the electrons each atom brings

C brings 4

each F brings 7

Total

4 + 7 + 7 + 7 + 7 = 32 electrons

Check — numeric entry

How many valence electrons does PCl₃ have in total?

✓ ANSWER KEY: 26

Feedback when correct: Correct. P brings 5 and each of the three Cl atoms brings 7: 5 + 3 × 7 = 26.

Anticipated wrong answers (feedback shown):

12	This answer counts one Cl. The subscript means three Cl atoms, each bringing 7: 5 + 21 = 26.
66	This answer counts all of each atom’s electrons. Only valence electrons count: P brings 5, not 15, and each Cl brings 7, not 17 — the total is 26.
21	This answer leaves out P. Every atom in the formula contributes: 5 + 21 = 26.

Worked solution revealed after a second miss

Count the electrons each atom brings

P brings 5

each Cl brings 7

Total

5 + 7 + 7 + 7 = 26 electrons

Check — numeric entry

How many valence electrons does C₂H₄ have in total?

✓ ANSWER KEY: 12

Feedback when correct: Correct. Each of the two C atoms brings 4 and each of the four H atoms brings 1: 2 × 4 + 4 × 1 = 12.

Anticipated wrong answers (feedback shown):

5	This answer ignores both subscripts. There are two C atoms and four H atoms: 8 + 4 = 12.
16	This answer counts all of each atom’s electrons. Only valence electrons count: each C brings 4, not 6 — the total is 12.
8	This answer leaves out the H atoms. Every atom in the formula contributes: 8 + 4 = 12.

Worked solution revealed after a second miss

Count the electrons each atom brings

each C brings 4

each H brings 1

Total

4 + 4 + 1 + 1 + 1 + 1 = 12 electrons

U2-S09 · LESSON 77 · CSP-070

Adjusting the count for an ion's charge

Outcome: Calculate the total number of valence electrons in a polyatomic ion.

Part 1The charge changes the count

You’ve already seen how to total a neutral molecule’s valence electrons: add up what every atom brings.

You’ve also already seen that an ion’s charge records gained or lost electrons: each negative charge is one electron gained, and each positive charge is one electron lost.

A polyatomic ion — a charged group of atoms — therefore holds a different total from what its atoms alone bring.

Add one electron to the total per negative charge; subtract one per positive charge.

Part 2Worked examples

Q: How many valence electrons does the nitrate ion, NO₃⁻, have in total?

Count the electrons each atom brings

N brings 5(N is in group 15)

each O brings 6(O is in group 16)

5 + 6 + 6 + 6 = 23 electrons

Adjust for the charge

the 1− charge means one electron gained(add 1)

23 + 1 = 24 electrons

Total

NO₃⁻ has 24 valence electrons

Q: How many valence electrons does the ammonium ion, NH₄⁺, have in total?

Count the electrons each atom brings

N brings 5(N is in group 15)

each H brings 1

5 + 1 + 1 + 1 + 1 = 9 electrons

Adjust for the charge

the 1+ charge means one electron lost(subtract 1)

9 − 1 = 8 electrons

Total

NH₄⁺ has 8 valence electrons

What you are expected to be able to do

Calculate a polyatomic ion’s total number of valence electrons by adding one electron per negative charge or subtracting one per positive charge.

Part 3Check what you just learned

Check — numeric entry

How many valence electrons does the sulfate ion, SO₄²⁻, have in total?

✓ ANSWER KEY: 32

Feedback when correct: Correct. S brings 6 and each O brings 6, giving 30; the 2− charge means two electrons gained: 30 + 2 = 32.

Anticipated wrong answers (feedback shown):

30	This answer forgets the charge. The 2− means the ion has gained two electrons: 30 + 2 = 32.
28	This answer subtracts for a negative charge. A negative charge means electrons gained, so add: 30 + 2 = 32.

Worked solution revealed after a second miss

Count the electrons each atom brings

S brings 6

each O brings 6

6 + 6 + 6 + 6 + 6 = 30

Adjust for the charge

the 2− charge means two electrons gained

30 + 2 = 32 electrons

Total

SO₄²⁻ has 32 valence electrons

Check — numeric entry

How many valence electrons does the hydronium ion, H₃O⁺, have in total?

✓ ANSWER KEY: 8

Feedback when correct: Correct. The three H atoms bring 3 and O brings 6, giving 9; the 1+ charge means one electron lost: 9 − 1 = 8.

Anticipated wrong answers (feedback shown):

9	This answer forgets the charge. The 1+ means the ion has lost one electron: 9 − 1 = 8.
10	This answer adds for a positive charge. A positive charge means an electron lost, so subtract: 9 − 1 = 8.

Worked solution revealed after a second miss

Count the electrons each atom brings

each H brings 1

O brings 6

1 + 1 + 1 + 6 = 9

Adjust for the charge

the 1+ charge means one electron lost

9 − 1 = 8 electrons

Total

H₃O⁺ has 8 valence electrons

Check — numeric entry

How many valence electrons does the cyanide ion, CN⁻, have in total?

✓ ANSWER KEY: 10

Feedback when correct: Correct. C brings 4 and N brings 5, giving 9; the 1− charge means one electron gained: 9 + 1 = 10.

Anticipated wrong answers (feedback shown):

9	This answer forgets the charge. The 1− means the ion has gained one electron: 9 + 1 = 10.
8	This answer subtracts for a negative charge. A negative charge means an electron gained, so add: 9 + 1 = 10.

Worked solution revealed after a second miss

Count the electrons each atom brings

C brings 4

N brings 5

4 + 5 = 9

Adjust for the charge

the 1− charge means one electron gained

9 + 1 = 10 electrons

Total

CN⁻ has 10 valence electrons

U2-S09 · LESSON 78 · CSP-071

Choosing the skeleton

Outcome: Identify the skeleton of a routine molecule or ion: use supplied connectivity, or place the less electronegative nonhydrogen atom in the middle.

Part 1Which atom sits in the middle

You’ve already seen the routine’s first step: for CO₂, the total is 16 valence electrons.

The next decision is the arrangement of the atoms — which one sits in the middle, and which attach around it. That arrangement is called the skeleton.

Here is the skeleton of CO₂: C in the middle, one O on each side. No bonds or dots yet — the skeleton is only the arrangement.

You’ve already seen electronegativity — how strongly a bonded atom attracts shared electrons — and its trend: it increases across a period toward the right, and up a group.

The rule for a routine skeleton: the less electronegative nonhydrogen atom goes in the middle.

C sits to the left of O in period 2, so C is less electronegative than O — C takes the middle of CO₂.

Hydrogen never takes the middle: bonded H is complete with one bonding pair, so it can hold only one bond — H always sits on the outside.

In H₂O, the only nonhydrogen atom is O, so O takes the middle.

Some questions supply the connectivity — a statement of which atom is joined to which. Supplied connectivity always wins: use it as given, and save the rule for when nothing is supplied.

What you are expected to be able to do

Identify the skeleton of a routine molecule or polyatomic ion: use supplied connectivity, or place the less electronegative nonhydrogen atom in the middle.

Part 2Check what you just learned

Check — multiple choice

Which statement chooses the central atom of SO₂ correctly?

AO, because it is more electronegative than S

Feedback if chosen: This option is wrong — you inverted the rule. The less electronegative atom goes in the middle, and S sits below O in group 16, so S is the less electronegative one.

BS, because it is the less electronegative atom ✓ KEY — this is the correct answer

Feedback if chosen: Correct. S sits below O in group 16, so S is less electronegative than O — S takes the middle.

CO, because the formula contains two O atoms

Feedback if chosen: This option is wrong — you chose the middle by headcount. How many times an atom appears is not the rule; the less electronegative atom, S, takes the middle.

DS, because sulfur is more electronegative than oxygen

Feedback if chosen: This option is wrong — you picked the right atom for a false reason. Oxygen’s electronegativity is 3.44 and sulfur’s is 2.58, so O is the more electronegative — S takes the middle because it is the less electronegative atom.

Check — multiple choice

Which atom is central in HCN?

AC — the less electronegative nonhydrogen atom ✓ KEY — this is the correct answer

Feedback if chosen: Correct. H never takes the middle, and C sits to the left of N in period 2, so C is less electronegative — C is central: H C N.

BH — it is written first in the formula

Feedback if chosen: This option is wrong — you used formula order, and you placed H in the middle. Bonded H holds only one bond, so H never takes the middle.

CN — the more electronegative of the two nonhydrogen atoms

Feedback if chosen: This option is wrong — you inverted the rule. The less electronegative nonhydrogen atom goes in the middle, and that is C, which sits to the left of N in period 2.

DEither C or N — with H on the outside, the middle is a free choice

Feedback if chosen: This option is wrong — you treated the middle as a free choice. The rule decides it: C is less electronegative than N, so C takes the middle.

Check — multiple choice

You are asked to draw HOCl, and the question supplies the connectivity: H is joined to O, and O is joined to Cl. Which atom is central?

ACl — it is the less electronegative nonhydrogen atom

Feedback if chosen: This option is wrong — you applied the middle-atom rule to a question that supplied the connectivity. Supplied connectivity always wins, and it puts O in the middle.

BH — it comes first in both the formula and the supplied connectivity

Feedback if chosen: This option is wrong — you placed H in the middle. Bonded H holds only one bond, so H never takes the middle; the supplied connectivity puts it on the outside, joined to O.

CO — the supplied connectivity places it between H and Cl ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The connectivity is supplied, and supplied connectivity always wins: H joined to O, O joined to Cl — O in the middle.

DYou cannot tell — the connectivity and the electronegativity rule point at different atoms

Feedback if chosen: This option is wrong — you treated the two as a contest. When the connectivity is supplied there is nothing to weigh: use it as given, with O in the middle.

U2-S09 · LESSON 79 · CSP-072

Joining the skeleton with single bonds

Outcome: Draw one single bond between each pair of connected atoms in a skeleton.

Part 1One line for every connected pair

You’ve already seen the routine bring CO₂ this far: a total of 16 valence electrons, and a skeleton with C in the middle.

The first marks to draw are the bonds: draw one single bond — one line — between each pair of connected atoms.

Here is CO₂ after this step: O–C–O.

You’ve already seen that each line holds one shared pair — 2 electrons — so the bonds spend part of the total: two lines use 4 of CO₂’s 16.

That leaves 12 electrons to place as dots in the coming steps.

Start every diagram with single bonds only — a double or triple bond comes later in the routine, and only where it is needed.

A second example: CF₄’s skeleton puts C in the middle with four F attached, so this step draws four single bonds — they use 8 of its 32 electrons.

What you are expected to be able to do

Draw one single bond between each pair of connected atoms in a Lewis skeleton.

Part 2Check what you just learned

Check — drawing task (pen and paper, self-marked)

The skeleton of PCl₃ puts P in the middle with three Cl atoms attached. Take a pen and paper and draw the skeleton joined with single bonds — bonds only, no dots yet. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows P in the middle with the three Cl atoms around it.
Your drawing shows exactly one line between P and each Cl — three lines in all.
Your drawing shows no line between any pair of Cl atoms — each Cl bonds to P only.
Your drawing shows no dots yet — this step draws bonds only.

Check — numeric entry

The skeleton of OF₂, joined with single bonds, is shown. OF₂ has 20 valence electrons in total. How many of them do the two bonds hold?

✓ ANSWER KEY: 4

Feedback when correct: Correct. Each line holds one shared pair — 2 electrons — so the two bonds hold 4.

Anticipated wrong answers (feedback shown):

2	This answer counts lines, not electrons. Each line holds a pair — 2 electrons — so two lines hold 4.
16	This answer gives the electrons left over (20 − 4), not the electrons in the bonds. The two lines themselves hold 4.
20	This answer gives the whole total. The two lines hold only 4 of the 20; the rest will be placed as dots.

Check — numeric entry

The skeleton of H₂S, joined with single bonds, is shown. H₂S has 8 valence electrons in total. How many electrons remain to be placed as dots?

✓ ANSWER KEY: 4

Feedback when correct: Correct. The two bonds hold 2 × 2 = 4 electrons, so 8 − 4 = 4 remain for dots.

Anticipated wrong answers (feedback shown):

6	This answer subtracts 1 per bond. Each line holds 2 electrons, so the bonds hold 4: 8 − 4 = 4.
8	This answer forgets that the bonds spend electrons. The two lines already hold 4 of the 8, so 4 remain.

U2-S09 · LESSON 80 · CSP-073

Completing the terminal atoms

Outcome: Complete the terminal atoms with lone pairs — octets, or duets for hydrogen — before filling the central atom.

Part 1Fill the outside atoms first

You’ve already seen the routine bring CF₄ this far: 32 valence electrons, C in the middle, and four single bonds holding 8 of them.

24 electrons are left to place as dots.

The atoms around the outside of a skeleton are called the terminal atoms — in CF₄, the four F atoms.

Place dots on the terminal atoms first: give each terminal atom lone pairs until it counts an octet, before anything goes on the central atom.

Each F already counts the 2 bonding electrons of its line, so it needs 6 more dots: three lone pairs.

Here is CF₄ after this step: three lone pairs on each F. The twelve lone pairs hold 24 electrons — the count is spent.

You’ve already seen that bonded hydrogen is complete with its duet of 2. A terminal H is already full — it never gets dots.

What you are expected to be able to do

Draw the remaining electrons on the terminal atoms to complete their octets — or duets for hydrogen — before filling the central atom.

Part 2Check what you just learned

Check — drawing task (pen and paper, self-marked)

NCl₃ has 26 valence electrons; its skeleton puts N in the middle, and its three single bonds hold 6 electrons, leaving 20. Take a pen and paper and draw NCl₃ as it stands after this step — skeleton, bonds, and completed terminal atoms. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows three lone pairs on each Cl — with its bond, each Cl now counts 8 electrons.
Your drawing shows no dots on N yet — the central atom is filled in the next step.
Your drawing still shows the three single bonds, unchanged.
Your drawing places 18 of the 20 remaining electrons; 2 are still unplaced.

Check — numeric entry

The skeleton of OF₂ (20 valence electrons in total) is shown, joined with single bonds. In this step, how many dots go on each terminal F?

✓ ANSWER KEY: 6

Feedback when correct: Correct. Each F already counts the 2 bonding electrons of its line, so 6 more dots — three lone pairs — complete its octet.

Anticipated wrong answers (feedback shown):

8	This answer forgets the bonding electrons. F already counts 2 from its line; 8 more dots would push it past an octet. It needs 6.
3	This answer counts lone pairs, not dots. Three lone pairs are 6 dots.
16	This answer gives all the remaining electrons. The 16 left over are shared out; each terminal F takes 6 of them.

Check — numeric entry

CH₄’s skeleton has four C–H single bonds. In this step, how many dots go on its terminal H atoms in total?

✓ ANSWER KEY: 0

Feedback when correct: Correct. Each bonded H already counts its duet of 2 from its line — terminal H never gets dots.

Anticipated wrong answers (feedback shown):

24	This answer gives each H an octet. Hydrogen is complete with 2 electrons, not 8, and each H already counts 2 from its bond: no dots.
8	This answer gives each H another pair. Bonded H already counts its duet of 2 from its line — it gets no dots.
4	This answer gives each H one more dot. Bonded H already counts its duet of 2 from its line — it gets no dots.

U2-S09 · LESSON 81 · CSP-074

Filling the central atom

Outcome: Place the electrons left after the terminal atoms are complete on the central atom.

Part 1Leftover electrons go to the middle

You’ve already seen the routine bring NH₃ this far: 8 valence electrons, N in the middle, three single bonds holding 6 — and the terminal H atoms complete with their duets.

8 − 6 = 2: two electrons are still unplaced, and every terminal atom is full.

Everything left after the terminal atoms are complete goes on the central atom, as lone pairs.

Here is NH₃ after this step: one lone pair on N.

Count around N: three bonds hold 6 and the lone pair holds 2 — an octet.

Sometimes nothing is left. CF₄ ended the last step with all 32 electrons placed, so C gets no dots — and its four bonds already count 8.

What you are expected to be able to do

Draw any electrons left after the terminal atoms are complete on the central atom.

Part 2Check what you just learned

Check — drawing task (pen and paper, self-marked)

H₂S has 8 valence electrons. Its skeleton, H S H, is joined by two single bonds, and the terminal H atoms are complete — 4 electrons are still unplaced. Take a pen and paper and draw H₂S as it stands after this step. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows two lone pairs — four dots — on S.
Your drawing shows no dots on the H atoms — each H is complete with its bonding pair.
Your drawing holds all 8 electrons: two bonds and two lone pairs.
Count around S: two bonds hold 4 and the two lone pairs hold 4 — an octet.

Check — numeric entry

PCl₃ has 26 valence electrons in total. The diagram shows it after the terminal Cl atoms were completed. How many electrons remain to be placed on P?

✓ ANSWER KEY: 2

Feedback when correct: Correct. The diagram holds 24 — three bonds (6) and nine lone pairs (18) — so 26 − 24 = 2: one lone pair for P.

Anticipated wrong answers (feedback shown):

0	This answer assumes nothing remains once every terminal atom has its octet. Count the diagram: it holds 24 of the 26, so 2 electrons are left for P.
20	This answer subtracts only the bonds (26 − 6). The 18 lone-pair electrons on the Cl atoms are placed too: 26 − 24 = 2.
8	This answer gives P a fresh octet of dots. P already counts 6 bonding electrons; only the 2 leftover electrons are added.

Check — multiple choice

The CF₄ diagram is shown after its terminal F atoms were completed; all 32 electrons are placed. What goes on the central C in this step?

AOne lone pair — the central atom always ends the routine with dots

Feedback if chosen: This option is wrong — you made central dots mandatory. The central atom gets only what is left over, and for CF₄ that is nothing; its four bonds already count 8.

BFour dots, so C matches the dots on each F

Feedback if chosen: This option is wrong — you placed dots for symmetry. Dots come from the leftover count, and CF₄ has no electrons left; C’s four bonds already count 8.

CTwo electrons taken back from one of the F atoms

Feedback if chosen: This option is wrong — you took electrons from a completed F. This step only places leftover electrons, and C’s four bonds already give it an octet — nothing needs to move.

DNothing — no electrons remain, and C’s four bonds already count 8 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. All 32 electrons are already placed, so C gets no dots — and its four bonds already count 8, an octet.

U2-S09 · LESSON 82 · CSP-075

Forming a double or triple bond when needed

Outcome: Draw an extra bond by converting a neighboring lone pair when the central atom lacks its expected octet.

Part 1A middle atom left short

You’ve already seen the routine bring CO₂ this far: 16 valence electrons, single bonds drawn, terminal O atoms complete — and all 16 electrons placed.

Count around C: two bonds and no dots — 4 electrons. C expects an octet, and no electrons are left to give it one.

The fix: convert a lone pair on an atom next to the central atom into an extra bond between them.

The converted pair stays in the diagram — it moves from dots on the neighbor into a second line — so the total does not change.

After a conversion, the neighbor still counts 8 — it traded a lone pair for a bonding pair — while the central atom counts 2 more.

C needs 4 more electrons, so convert twice: one lone pair from each O.

Here is CO₂ after this step: two double bonds. Count around C: 8. Count around each O: a double bond and two lone pairs — 8. The diagram still holds 16 electrons.

If the central atom already counts 8, convert nothing — the diagram is finished as it stands.

You’ve also already seen a few central atoms, like B, that are accepted with fewer than eight — leave those short too; do not force an extra bond.

What you are expected to be able to do

Draw an additional bond by converting a neighboring lone pair when the central atom lacks its expected octet and the total electron count permits it.

Part 2Check what you just learned

Check — drawing task (pen and paper, self-marked)

HCN has 10 valence electrons. The routine so far gives single bonds H–C and C–N, a complete terminal N with three lone pairs, and a complete H — all 10 electrons placed, but C counts only 4. Take a pen and paper and draw the finished diagram of HCN. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows a triple bond between C and N — two lone pairs on N were converted, one at a time, until C counted 8.
Your drawing shows one lone pair left on N — count around N: 6 + 2 = 8.
Your drawing shows the single H–C bond unchanged, with no dots on H.
Your drawing still holds all 10 electrons: 2 + 6 + 2.

Check — multiple choice

The finished OF₂ diagram is shown; the central O counts 8 electrons. A student converts a lone pair on one F into a second O–F bond anyway. What is the result?

AA wrong diagram — O would count 10 electrons, and a period-2 atom is never drawn with more than eight ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Conversion is a fix for a central atom left short. O already counted 8; the extra bond pushes it to 10, and a period-2 atom is never drawn with more than eight.

BA better diagram — more bonds always bring the central atom closer to its octet

Feedback if chosen: This option is wrong — you treated extra bonds as always helpful. O already counted 8; the conversion overshoots the octet instead of completing it.

CAn equally correct diagram — converting moves electrons without changing the total, so nothing is affected

Feedback if chosen: This option is wrong — the total is indeed unchanged, but the counts around the atoms move: O would count 10, more than a period-2 atom is ever drawn with.

DA wrong diagram — but only because the pair should have come from the other F

Feedback if chosen: This option is wrong — you blamed the choice of neighbor. Neither F should give up a pair: O already counted 8, so there was nothing to fix.

Check — numeric entry

Before its two conversions, the CO₂ diagram held all 16 valence electrons. The finished diagram is shown. How many electrons does it hold now?

✓ ANSWER KEY: 16

Feedback when correct: Correct. A conversion moves a pair; it never adds or removes one. Two double bonds hold 8 and four lone pairs hold 8: 16.

Anticipated wrong answers (feedback shown):

20	This answer adds 2 per new bond. A conversion moves an existing pair into the bond — nothing new arrives: the diagram still holds 16.
12	This answer removes the converted pairs. They are still in the diagram, as the second line of each double bond: 16.

U2-S09 · LESSON 83 · CSP-076

Brackets and charge for ions

Outcome: Draw brackets around a polyatomic ion's Lewis diagram with the overall charge outside.

Part 1The charge belongs to the whole ion

You’ve already seen that an ion’s charge changes its electron total — the hydroxide ion, OH⁻, counts 6 + 1 + 1 = 8 valence electrons.

Here is hydroxide after the routine’s drawing steps: one O–H bond and three lone pairs on O — all 8 electrons placed.

One thing is still missing: nothing in the drawing says this is an ion.

Finish a polyatomic ion’s diagram with square brackets around the whole drawing, and the ion’s charge outside the brackets at the top right.

Here is the finished hydroxide diagram.

The charge sits outside the brackets because it belongs to the whole ion — never write the ion's charge on one atom inside.

A second example: the ammonium ion, NH₄⁺, counts 9 − 1 = 8 electrons — four N–H bonds, brackets, and 1+ outside.

A neutral molecule gets no brackets — brackets and the outside charge are ion notation.

What you are expected to be able to do

Draw brackets around a polyatomic ion’s Lewis diagram and place the overall charge outside the brackets.

Part 2Check what you just learned

Check — drawing task (pen and paper, self-marked)

The cyanide ion, CN⁻, counts 10 valence electrons, and the routine’s drawing steps give a triple bond between C and N with one lone pair on each atom. Take a pen and paper and draw the finished diagram of CN⁻, complete with its ion notation. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows C and N joined by a triple bond, with one lone pair on each atom.
Your drawing shows square brackets around the whole diagram.
Your drawing shows 1− outside the brackets, at the top right.
Your drawing shows no charge on C or on N — the 1− belongs to the whole ion.

Check — multiple choice

A student finishes the Lewis diagram of the sulfate ion, SO₄²⁻. Where does the 2− go?

AAs a 2− on the O atom that carries the extra electrons

Feedback if chosen: This option is wrong — you localized the charge on one atom. The 2− is the charge of the whole ion, not of any single O, so it sits outside the brackets at the top right.

BSplit as 1− on two of the O atoms, so the charge adds up

Feedback if chosen: This option is wrong — you shared the charge out among atoms. The 2− belongs to the whole ion as one mark, outside the brackets at the top right.

COutside the brackets, at the top right ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Brackets enclose the whole diagram, and the overall charge sits outside them at the top right.

DInside the brackets, written once next to the diagram

Feedback if chosen: This option is wrong — you put the charge inside. Inside the brackets there are only atoms, bonds, and lone pairs; the 2− sits outside them, at the top right.

Check — multiple choice

Which of these species’ Lewis diagrams is finished with brackets, and why?

AH₂O — its O carries two lone pairs, and lone pairs must sit inside brackets

Feedback if chosen: This option is wrong — you tied brackets to lone pairs. Lone pairs appear in plenty of neutral molecules; brackets mark an overall charge, and H₂O has none.

BNO₃⁻ — it carries an overall charge, so its diagram ends with brackets and 1− outside ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Brackets and the outside charge are ion notation, and NO₃⁻ is the only ion here.

CH₂O₂ — the 2 in its formula is a charge, so it takes brackets

Feedback if chosen: This option is wrong — you read a subscript as a charge. The 2 in H₂O₂ counts atoms; the molecule is neutral, so no brackets.

DCO₂ — any molecule with double bonds takes brackets

Feedback if chosen: This option is wrong — you tied brackets to double bonds. Bond order has nothing to do with ion notation; CO₂ is neutral, so no brackets.

U2-S09 · LESSON 84 · CSP-077

Checking the electron count

Outcome: Evaluate whether a Lewis diagram holds exactly the calculated valence-electron total.

Part 1Count what you drew

You’ve already seen every drawing step of the routine. Each one spends electrons or moves them — so a correct finished diagram holds exactly the calculated total.

Here is the finished CO₂ diagram again; the calculated total is 16.

To count a diagram, add what every mark holds: 2 per single bond, 4 per double bond, 6 per triple bond, and 2 per lone pair.

CO₂’s marks: two double bonds hold 8, and four lone pairs hold 8 — 16.

Drawn 16, calculated 16: the diagram passes the check.

If the two numbers differ, the diagram is wrong — go back through the steps and find where the count slipped.

Here is a student’s diagram of H₂O. Count it: two bonds hold 4 and three lone pairs hold 6 — 10.

H₂O’s calculated total is 8, not 10 — the diagram fails the check: O carries one lone pair too many.

What you are expected to be able to do

Evaluate whether the electrons drawn as bonds and lone pairs equal the calculated total number of valence electrons.

Part 2Check what you just learned

Check — numeric entry

The finished diagram of the cyanide ion, CN⁻, is shown. How many electrons does it hold?

✓ ANSWER KEY: 10

Feedback when correct: Correct. The triple bond holds 6 and the two lone pairs hold 4: 10 — equal to CN⁻’s calculated total of 4 + 5 + 1 = 10, so the diagram passes the check.

Anticipated wrong answers (feedback shown):

6	This answer counts the triple bond as one pair. Three lines hold 6 electrons; with the two lone pairs, the diagram holds 10.
5	This answer counts pairs, not electrons. The five pairs — three bond lines and two lone pairs — hold 10 electrons.
11	This answer adds one for the 1− outside the brackets. That gained electron is already among the 10 drawn; the notation adds nothing to the count.

Check — numeric entry

A student’s diagram of NF₃ is shown. How many electrons does it hold?

✓ ANSWER KEY: 20

Feedback when correct: Correct. Three bonds hold 6, the six lone pairs on the F atoms hold 12, and the lone pair on N holds 2: 20.

Anticipated wrong answers (feedback shown):

26	This answer gives NF₃’s calculated total instead of counting the drawing. The check needs both numbers, counted separately — the drawing holds 20.
10	This answer counts pairs, not electrons. The ten pairs — three bonds and seven lone pairs — hold 20 electrons.
14	This answer counts only the lone pairs. The three bonds hold 6 more: 14 + 6 = 20.

Check — multiple choice

Count the electrons the NF₃ diagram shown holds, and compare your count with NF₃’s calculated total of 5 + 3 × 7 = 26. What do you conclude?

AThe calculation is wrong — a drawn diagram takes priority over the arithmetic

Feedback if chosen: This option is wrong — you trusted the drawing over the arithmetic. The total is fixed by what the atoms bring; a diagram that disagrees with it is a wrong diagram.

BThe diagram is wrong — six electrons were never placed, so lone pairs were left out along the way ✓ KEY — this is the correct answer

Feedback if chosen: Correct. 26 − 20 = 6: six electrons went missing — each F should carry three lone pairs, not two.

CThe diagram is fine — the count check does not apply when the central atom has a lone pair

Feedback if chosen: This option is wrong — you invented an exemption. Every correct diagram holds exactly the calculated total, whatever sits on the central atom.

DThe diagram is fine — the six missing electrons sit in the atoms’ inner shells

Feedback if chosen: This option is wrong — you sent the missing electrons to the inner shells. The total of 26 counts valence electrons only, and all 26 must appear in the diagram.

U2-S09 · LESSON 85 · CSP-078

Drawing a complete Lewis diagram

Outcome: Draw and check a complete Lewis diagram for a routine molecule or polyatomic ion.

Part 1The whole routine, start to finish

You’ve already seen every step of the Lewis routine. Here it is as one list.

1. Count the valence electrons each atom brings.

2. For an ion, adjust the total: add one electron per negative charge, subtract one per positive charge.

3. Choose the skeleton: use supplied connectivity, or put the less electronegative nonhydrogen atom in the middle — never H.

4. Join each connected pair with a single bond.

5. Complete the terminal atoms with lone pairs — octets, or duets for H.

6. Place whatever remains on the central atom.

7. If the central atom still lacks its expected octet, convert a neighboring lone pair into an extra bond.

8. For an ion, add brackets with the charge outside.

9. Check: the diagram must hold exactly the calculated total.

Part 2Worked through: the carbonate ion

Take the carbonate ion, CO₃²⁻. Count: C brings 4, each O brings 6, and the 2− charge adds 2 — 4 + 18 + 2 = 24.

Skeleton: C sits to the left of O in period 2, so C is the less electronegative atom — C takes the middle, with three O around it.

Single bonds: three lines hold 6 electrons, leaving 18.

Terminal atoms: three lone pairs on each O — 18 dots, and the count is spent.

Central atom: nothing is left for C, and C counts only 6 — it lacks its octet.

Convert one lone pair on one O into a second bond: C now counts 8.

Brackets with 2− outside finish the ion. Here is the result.

Check: the double bond holds 4, the two single bonds hold 4, and the eight lone pairs hold 16 — 24, equal to the calculated total.

This is one correct Lewis diagram of the carbonate ion.

What you are expected to be able to do

Draw and check a complete Lewis diagram for a routine molecule or polyatomic ion — electron count, skeleton, bonds, lone pairs, and ion notation.

Part 3Check what you just learned

Check — drawing task (pen and paper, self-marked)

Take a pen and paper and work through the whole routine to draw one correct complete Lewis diagram of sulfur dioxide, SO₂. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your diagram holds 18 electrons in total — S brings 6 and each O brings 6.
Your diagram shows S in the middle — S is the less electronegative atom.
Your diagram shows one double bond and one single bond, drawn as lines — the double bond may sit on either O.
Your diagram shows one lone pair on S, two on the double-bonded O, and three on the single-bonded O.
Count around every atom: S 8, each O 8.
Your diagram has no brackets — SO₂ is a neutral molecule.

Check — drawing task (pen and paper, self-marked)

Take a pen and paper and work through the whole routine to draw the complete Lewis diagram of the hypochlorite ion, ClO⁻. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your diagram holds 14 electrons — Cl brings 7, O brings 6, and the 1− charge adds 1.
Your diagram shows one single bond between Cl and O.
Your diagram shows three lone pairs on Cl and three on O — count around each atom: 8.
Your diagram shows square brackets around the whole drawing, with 1− outside at the top right.

Check — multiple choice

A student offers the diagram shown as the finished Lewis diagram of BH₄⁻ (calculated total: 3 + 4 × 1 + 1 = 8). Which check does it fail?

AThe electron count — the diagram holds fewer electrons than the calculated 8

Feedback if chosen: This option is wrong — count the diagram: four single bonds hold 8 electrons, exactly the calculated total.

BThe octet check — B counts fewer than eight electrons

Feedback if chosen: This option is wrong — count around B: four bonds hold 8 electrons, an octet.

CThe duet check — each H needs a lone pair to be complete

Feedback if chosen: This option is wrong — each H counts the 2 electrons of its bonding pair, a complete duet; H never takes lone pairs.

DThe ion notation — the brackets and the outside 1− are missing ✓ KEY — this is the correct answer

Feedback if chosen: Correct. BH₄⁻ is an ion, and nothing in the drawing says so — it needs square brackets with 1− outside at the top right.

U2-S10 — Formal charge and model selection

7 lessons

U2-S10 · LESSON 86 · CSP-079

What a formal charge is

Outcome: State what a formal charge is — a bookkeeping value from a counting rule, not a measured charge.

Part 1A number for each atom

Here is the Lewis diagram of the nitrate ion, NO₃⁻ — built the way you’ve already seen, with all 24 valence electrons placed as bonds and lone pairs.

The 1− outside the brackets is the ion’s overall charge — real and measurable.

Chemists also give each atom inside the diagram its own number, which tracks how the diagram splits the electrons between the atoms.

That number is the atom’s formal charge, and it is bookkeeping — an assigned value, not a measured one.

The counting rule that assigns it works from the diagram alone: start from the atom’s valence electrons, subtract its nonbonding electrons, and subtract half of its bonding electrons.

A formal charge of 0 means the diagram credits the atom with exactly as many electrons as a free atom of its element.

A positive formal charge means the diagram credits the atom with fewer electrons than the free atom; a negative one means more.

Applied to a diagram, the counting rule assigns each atom its own value — that atom's formal charge. You will soon learn to calculate these values.

Part 2Bookkeeping, not a real charge

A formal charge is not a real charge, and an atom carrying one has not become an ion. The real charge of an ion belongs to the whole ion; the formal charges on its atoms are the counting rule's way of splitting the bookkeeping between the atoms inside the diagram.

You’ve already seen partial charges, δ+ and δ−, which describe real, uneven sharing in a polar bond.

A formal charge is a different kind of number: it comes from the counting rule, not from any measurement.

What you are expected to be able to do

State that a formal charge is a bookkeeping value calculated from the Lewis diagram — valence electrons, minus nonbonding electrons, minus half of the bonding electrons — and not a measured charge.

Part 3Check what you just learned

Check — multiple choice

Which statement says what a formal charge is?

AA measured charge on the atom, found by experiment

Feedback if chosen: This option is wrong — you treated the formal charge as a measurement. No experiment finds it; it is assigned by the counting rule from the diagram alone.

BThe number of electrons an atom gained or lost when the bond formed

Feedback if chosen: This option is wrong — you read the formal charge as electrons gained or lost. The atoms in a covalent diagram share electrons; the counting rule only tracks how the diagram credits those shared and unshared electrons.

CThe overall charge of the ion, shared out equally among its atoms

Feedback if chosen: This option is wrong — you spread the ion’s overall charge across the atoms. The overall charge belongs to the whole ion; each atom’s formal charge comes from the counting rule, and the values usually differ from atom to atom.

DA bookkeeping value calculated from the Lewis diagram by a counting rule ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A formal charge is bookkeeping: valence electrons, minus nonbonding electrons, minus half of the bonding electrons, counted from the diagram.

Check — multiple choice

In the nitrate diagram shown, the counting rule gives N a formal charge of +1. What does the +1 mean?

AN has lost an electron and become an N⁺ ion

Feedback if chosen: This option is wrong — you read the formal charge as a real ionic charge. An atom carrying a formal charge has not become an ion; N is still a bonded atom inside the nitrate ion.

BA measurement would find a full +1 charge sitting on N

Feedback if chosen: This option is wrong — you treated the +1 as measurable. A formal charge comes from the counting rule, not from an experiment; no instrument would find a full +1 on N.

CThe diagram credits N with one electron fewer than a free N atom — bookkeeping, not a real charge ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The counting rule credits N with one electron fewer than a free N atom, so its formal charge is +1 — a bookkeeping value inside the diagram.

DThe +1 is the same as the δ+ partial charge on a polar bond, written as a whole number

Feedback if chosen: This option is wrong — you merged the formal charge with the partial charge. A δ+ marks real, uneven sharing in a polar bond; the +1 comes from the counting rule — bookkeeping on the diagram, not a description of the sharing.

Check — multiple choice

How is a formal charge different from the partial charges, δ+ and δ−, you’ve already seen on polar bonds?

APartial charges describe real, uneven sharing in a bond; a formal charge comes from a counting rule, not a measurement ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Partial charges mark real, uneven sharing in a polar bond. A formal charge is assigned by the counting rule from the diagram — bookkeeping, not measurement.

BThey are the same quantity, written with different symbols

Feedback if chosen: This option is wrong — you merged two different numbers. A polar bond’s partial charges reflect real electron distribution; a formal charge is assigned by the counting rule and can disagree with them.

CA formal charge is the measured one; partial charges are the bookkeeping

Feedback if chosen: This option is wrong — you swapped the two. Partial charges are the physically real ones; the formal charge is the bookkeeping value from the counting rule.

DPartial charges apply only to ions; formal charges apply only to neutral molecules

Feedback if chosen: This option is wrong — you split them by species type. Partial charges appear on polar bonds in molecules and ions alike, and formal charges are assigned to atoms in any Lewis diagram.

U2-S10 · LESSON 87 · CSP-080

Calculating formal charge

Outcome: Calculate the formal charge of an atom in a Lewis diagram.

Part 1The counting rule

You’ve already seen that formal charge is a bookkeeping value assigned to an atom in a Lewis diagram — not a measured charge.

To calculate an atom’s formal charge, count three numbers from the Lewis diagram, then combine them:

FC	formal charge
V	valence electrons of the free atom (from its group)
N	nonbonding electrons on the atom in the diagram (2 per lone pair)
B	bonding electrons shared by the atom (2 per bond, so 4 for a double bond, 6 for a triple bond)

N and B count electrons, not pairs and not bonds.

That is why B is halved: of each shared pair, the atom is credited with one electron.

What you are expected to be able to do

Calculate the formal charge of an atom in a supplied Lewis diagram using FC = V − N − B/2.

Part 2Worked examples

Here is the Lewis diagram of the nitrate ion, NO₃⁻.

Q: What is the formal charge on N in the nitrate ion Lewis diagram shown?

Write down the values from the diagram

V = 5(N is in group 15)

N = 0(no lone pairs on N)

B = 8(one double bond + two single bonds = 4 + 2 + 2 electrons)

Write down the equation

FC = V − N − B/2

Substitute in the values, and calculate

FC = 5 − 0 − 8/2

FC = +1

Q: What is the formal charge on the double-bonded O in the same nitrate diagram?

Write down the values from the diagram

V = 6(O is in group 16)

N = 4(two lone pairs)

B = 4(one double bond)

Write down the equation

FC = V − N − B/2

Substitute in the values, and calculate

FC = 6 − 4 − 4/2

FC = 0

A formal charge of zero is a normal, common result.

Write positive results with an explicit plus sign: +1, not 1.

Part 3Check what you just learned

Check — numeric entry

What is the formal charge on one of the single-bonded O atoms in the nitrate Lewis diagram shown?

✓ ANSWER KEY: -1

Feedback when correct: Correct. V = 6, N = 6 (three lone pairs), B = 2 (one single bond), so FC = 6 − 6 − 2/2 = −1.

Anticipated wrong answers (feedback shown):

-2	This answer comes from subtracting all of B. Only half of the bonding electrons count against the atom: FC = 6 − 6 − 2/2 = −1.
0	Check which O atom the question asks about. The single-bonded O has three lone pairs, so FC = 6 − 6 − 2/2 = −1; it is the double-bonded O that works out to 0.

Sign-error retry feedback:

1	You have the right magnitude but the wrong sign — write the formal charge with its sign.
+1	You have the right magnitude but the wrong sign — check whether you swapped N and B, and write the formal charge with its sign.

Worked solution revealed after a second miss

Write down the values from the diagram

V = 6

N = 6

B = 2

Write down the equation

FC = V − N − B/2

Substitute in the values, and calculate

FC = 6 − 6 − 2/2

FC = −1

Check — numeric entry

What is the formal charge on O in the hydronium ion Lewis diagram shown?

✓ ANSWER KEY: +1

Also accepted: 1

Feedback when correct: Correct. V = 6, N = 2 (one lone pair), B = 6 (three single bonds), so FC = 6 − 2 − 6/2 = +1.

Anticipated wrong answers (feedback shown):

-2	This answer comes from subtracting all 6 bonding electrons. Only half of B counts: FC = 6 − 2 − 6/2 = +1.
-1	This answer swaps N and B: N is the 2 lone-pair electrons, B is the 6 bonding electrons. FC = 6 − 2 − 6/2 = +1.
0	Zero is O's formal charge in water — two bonds and two lone pairs. In H₃O⁺ the O makes three bonds and keeps one lone pair: FC = 6 − 2 − 6/2 = +1.

Worked solution revealed after a second miss

Write down the values from the diagram

V = 6

N = 2

B = 6

Write down the equation

FC = V − N − B/2

Substitute in the values, and calculate

FC = 6 − 2 − 6/2

FC = +1

Check — numeric entry

What is the formal charge on C in the carbon monoxide Lewis diagram shown?

✓ ANSWER KEY: -1

Feedback when correct: Correct. V = 4, N = 2 (one lone pair), B = 6 (one triple bond), so FC = 4 − 2 − 6/2 = −1.

Anticipated wrong answers (feedback shown):

-4	This answer comes from subtracting all 6 bonding electrons. Only half of B counts: FC = 4 − 2 − 6/2 = −1.
0	This answer counts the triple bond as 4 bonding electrons, as if it were a double bond. A triple bond is 6 bonding electrons, so FC = 4 − 2 − 6/2 = −1.

Sign-error retry feedback:

1	You have the right magnitude but the wrong sign — write the formal charge with its sign.
+1	You have the right magnitude but the wrong sign — check how many bonding electrons the triple bond holds, and write the formal charge with its sign.

Worked solution revealed after a second miss

Write down the values from the diagram

V = 4

N = 2

B = 6

Write down the equation

FC = V − N − B/2

Substitute in the values, and calculate

FC = 4 − 2 − 6/2

FC = −1

U2-S10 · LESSON 88 · CSP-081

The sum of the formal charges

Outcome: Check that the formal charges in a Lewis diagram sum to zero for a neutral molecule or to the overall charge for an ion.

Part 1Add up the formal charges

You’ve already calculated the formal charges in this nitrate diagram: +1 on N, 0 on the double-bonded O, and −1 on each single-bonded O.

Add them: (+1) + 0 + (−1) + (−1) = −1.

That matches the 1− outside the brackets — and it must.

The formal charges split the diagram’s electron bookkeeping between its atoms, so together they always account for the species’ overall charge.

For a polyatomic ion, the formal charges sum to the overall charge; for a neutral molecule, they sum to zero.

You’ve already calculated −1 for C in this carbon monoxide diagram; the same counting gives O a formal charge of +1.

Add them: (−1) + (+1) = 0, and CO is a neutral molecule.

Part 2The sum as a quick check

The sum is a fast test of a finished diagram: when it does not match the overall charge, the drawing does not contain the right number of electrons.

A matching sum does not prove a diagram right, but a mismatched sum proves it wrong.

What you are expected to be able to do

Evaluate whether the formal charges in a supplied Lewis diagram sum to zero for a neutral molecule, or to the overall charge for a polyatomic ion.

Part 3Check what you just learned

Check — numeric entry

Use the hydroxide ion Lewis diagram shown. Calculate the formal charge of each atom, then add them. What is the sum of the formal charges?

✓ ANSWER KEY: -1

Feedback when correct: Correct. O: 6 − 6 − 2/2 = −1 and H: 1 − 0 − 2/2 = 0, so the sum is −1 — matching the 1− outside the brackets, as a valid diagram must.

Anticipated wrong answers (feedback shown):

-2	This answer comes from subtracting all of O’s bonding electrons. Only half of B counts: O is 6 − 6 − 2/2 = −1 and H is 0, so the sum is −1.
0	This answer assumes the formal charges of a finished diagram always cancel. They sum to the overall charge — 1− for hydroxide: O is −1 and H is 0, so the sum is −1.

Sign-error retry feedback:

1	You have the right magnitude but the wrong sign — write the sum with its sign.

Worked solution revealed after a second miss

Calculate the formal charge of each atom

O: FC = 6 − 6 − 2/2 = −1

H: FC = 1 − 0 − 2/2 = 0

Add the formal charges

(−1) + 0 = −1

Compare with the overall charge

The diagram shows 1− outside the brackets — the sum matches

Check — numeric entry

Use the carbonate ion Lewis diagram shown. What is the sum of the formal charges of all four atoms?

✓ ANSWER KEY: -2

Feedback when correct: Correct. C: 4 − 0 − 8/2 = 0; the double-bonded O: 6 − 4 − 4/2 = 0; each single-bonded O: 6 − 6 − 2/2 = −1. The sum is −2, matching the 2− outside the brackets.

Anticipated wrong answers (feedback shown):

-1	This answer counts only one of the two single-bonded O atoms. Both work out to −1, while C and the double-bonded O are both 0, so the sum is −2.
-3	This answer gives every O a −1. The double-bonded O works out to 6 − 4 − 4/2 = 0; only the two single-bonded O atoms are −1, so the sum is −2.
0	This answer treats carbonate like a neutral molecule. The formal charges of an ion sum to its overall charge: 0 + 0 + (−1) + (−1) = −2.

Sign-error retry feedback:

2	You have the right magnitude but the wrong sign — write the sum with its sign.

Worked solution revealed after a second miss

Calculate the formal charge of each atom

C: FC = 4 − 0 − 8/2 = 0

Double-bonded O: FC = 6 − 4 − 4/2 = 0

Each single-bonded O: FC = 6 − 6 − 2/2 = −1

Add the formal charges

0 + 0 + (−1) + (−1) = −2

Compare with the overall charge

The diagram shows 2− outside the brackets — the sum matches

Check — numeric entry

A student draws this diagram for H₂O, a neutral molecule. What do the drawing’s formal charges sum to?

✓ ANSWER KEY: -2

Feedback when correct: Correct. O here is 6 − 6 − 4/2 = −2 and each H is 0, so the sum is −2. A neutral molecule’s formal charges should sum to 0 — the mismatch shows the drawing holds 10 electrons where H₂O has 8.

Anticipated wrong answers (feedback shown):

0	This answer gives the sum a correct H₂O diagram would have. The question asks about this drawing: its O carries three lone pairs, so O is 6 − 6 − 4/2 = −2 and the sum is −2 — the mismatch is how the check exposes the mistake.
-1	This answer counts only one of O’s two bonds. B is 4 here (two single bonds), so O is 6 − 6 − 4/2 = −2, and the sum is −2.

Sign-error retry feedback:

2	You have the right magnitude but the wrong sign — write the sum with its sign.

Worked solution revealed after a second miss

Calculate the formal charge of each atom

O: FC = 6 − 6 − 4/2 = −2

Each H: FC = 1 − 0 − 2/2 = 0

Add the formal charges

(−2) + 0 + 0 = −2

Compare with the overall charge

H₂O is neutral, so the sum should be 0 — this drawing cannot be right

Check — multiple choice

A student’s Lewis diagram for a neutral molecule has formal charges that sum to +1. What does this tell you?

AThe molecule is actually a cation with charge 1+

Feedback if chosen: This option is wrong — you changed the species to fit the drawing. The molecule is given as neutral; a sum of +1 means the drawing is at fault, not the molecule.

BOne atom in the molecule really carries a permanent +1 charge

Feedback if chosen: This option is wrong — you read a formal charge as a real charge. Formal charges are bookkeeping; the mismatched sum means the drawing itself has an error.

CNothing — the sum does not need to match for neutral molecules

Feedback if chosen: This option is wrong — you exempted neutral molecules from the check. Their formal charges must sum to zero; +1 is a mismatch.

DThe drawing does not contain the right number of electrons, so it cannot be correct ✓ KEY — this is the correct answer

Feedback if chosen: Correct. For a neutral molecule the formal charges must sum to 0. A sum of +1 means the electrons drawn do not match the count the molecule actually has, so the diagram is wrong.

U2-S10 · LESSON 89 · CSP-082

Octets decide first

Outcome: Select, between candidate Lewis diagrams, the model that gives every atom its expected octet.

Part 1Two diagrams, both complete

You’ve already seen how to build a Lewis diagram and how to verify that the electrons drawn match the calculated total.

For many species, more than one diagram passes that test — all the valence electrons placed, every electron in a bond or a lone pair.

Here are two such candidate Lewis diagrams of CO₂. Model 1 joins C to each O by a double bond.

Model 2 joins C to one O by a double bond and to the other by a single bond, with an extra lone pair on the single-bonded O.

Both models place all 16 valence electrons, yet CO₂ is one molecule — the Lewis model needs a way to choose.

Part 2Octets decide first

Count around each atom in Model 1: C has 8 electrons, and each O has 8 — every atom has its expected octet.

Count around C in Model 2: one double bond and one single bond hold 6 electrons — no octet.

The first selection criterion: among otherwise plausible diagrams, prefer the model that gives every atom its expected octet — eight electrons, or two for hydrogen.

So Model 1 is the accepted diagram of CO₂.

The criterion assumes the species is not one of the accepted exceptions you’ve already seen: an odd electron total, an electron-deficient atom such as B or Be, or an expanded central atom from period 3 on.

For those species, the accepted diagram is allowed to break the octet pattern — and CO₂ is not one of them.

One family sits on the boundary: for species such as SO₂ and the sulfate ion, SO₄²⁻, an all-octet model and an expanded-octet model with smaller formal charges are both accepted on the AP exam. This course draws the all-octet version.

What you are expected to be able to do

State that, among otherwise plausible Lewis diagrams, the model satisfying expected octets is preferred, unless the species is an accepted exception.

Part 3Check what you just learned

Check — multiple choice

Two candidate Lewis diagrams of CS₂ are shown as Model 1 and Model 2. Both place all 16 valence electrons. Which model is preferred?

AModel 1, because every atom in it has an octet ✓ KEY — this is the correct answer

Feedback if chosen: Correct. In Model 1, C has two double bonds (8 electrons) and each S has a double bond plus two lone pairs (8). In Model 2, C is left with 6, and CS₂ is not one of the accepted exceptions.

BModel 2, because C is in period 2 and is accepted with fewer than eight electrons

Feedback if chosen: This option is wrong — you treated C as an accepted exception. The atoms accepted with fewer than eight are B and Be; C in Model 2 has only 6 electrons, which rules Model 2 out when an all-octet model exists.

CModel 1, because it has more bonds, and more bonds are always better

Feedback if chosen: This option is wrong — you picked the right model with a rule that does not exist. Bond count is not a criterion; Model 1 wins because every atom in it has its expected octet.

DThe two models are equally good, because both place all 16 valence electrons

Feedback if chosen: This option is wrong — you stopped at the electron count. Placing all 16 electrons makes both models plausible; the octet criterion then decides, and only Model 1 gives every atom an octet.

Check — multiple choice

Two candidate Lewis diagrams of formaldehyde, CH₂O, are shown as Model 1 and Model 2. Both place all 12 valence electrons. Which model is preferred?

AThe two models are equally acceptable, because both place all 12 valence electrons

Feedback if chosen: This option is wrong — you stopped at the electron count. Placing all 12 electrons makes a model plausible; the octet criterion then prefers Model 1, where C has 8 electrons rather than 6.

BModel 2, because O should keep as many lone pairs as possible

Feedback if chosen: This option is wrong — you invented a lone-pair rule. Lone-pair count is not a criterion, and in Model 2 the extra pair on O leaves C with only 6 electrons.

CModel 1, because every atom has its expected count — C 8, O 8, and each H its duet ✓ KEY — this is the correct answer

Feedback if chosen: Correct. In Model 1, C has two single bonds and a double bond (8 electrons), O has a double bond and two lone pairs (8), and each H has its duet. In Model 2, C is left with 6.

DModel 1, because double bonds are always preferred over single bonds

Feedback if chosen: This option is wrong — you picked the right model with a rule that does not exist. Plenty of accepted diagrams have only single bonds; Model 1 wins because every atom has its expected octet or duet.

Check — multiple choice

You’ve already seen that some central atoms are accepted without an octet. For which of these molecules does the octet criterion not decide between candidate diagrams, because the species is an accepted exception?

ACO₂

Feedback if chosen: This option is wrong — you treated CO₂ as an exception. It has an even electron count and no electron-deficient or expanded-shell atom, so the octet criterion applies and prefers its all-octet model.

BBF₃ ✓ KEY — this is the correct answer

Feedback if chosen: Correct. B is one of the electron-deficient atoms you’ve already seen: the accepted BF₃ diagram leaves B with 6 electrons, so a candidate cannot be rejected for that missing octet.

CHCN

Feedback if chosen: This option is wrong — you treated HCN as an exception. Its accepted diagram gives C and N full octets and H its duet, so the octet criterion applies in full.

DCH₂O

Feedback if chosen: This option is wrong — you treated CH₂O as an exception. Its accepted diagram gives C and O octets and each H a duet, so the octet criterion applies in full.

U2-S10 · LESSON 90 · CSP-083

Smaller and fewer formal charges

Outcome: Select, among Lewis diagrams with comparable octets, the model with smaller and fewer nonzero formal charges.

Part 1A tie the octet criterion cannot break

You’ve already seen that, among otherwise plausible Lewis diagrams, the model giving every atom its expected octet is preferred.

Here are two more candidate diagrams of CO₂. Model 1 is the double-bond model you’ve already seen.

Model 2 joins C to one O by a triple bond and to the other by a single bond.

Count around each atom in Model 2: the triple-bonded O has 8 electrons, C has 8, and the single-bonded O has 8 — every atom has an octet, just as in Model 1.

The octet criterion cannot choose here. The formal charges can.

Part 2Smaller and fewer

Work out Model 2’s formal charges with the counting rule you’ve already seen.

Triple-bonded O: FC = 6 − 2 − 6/2 = +1.

C: FC = 4 − 0 − 8/2 = 0.

Single-bonded O: FC = 6 − 6 − 2/2 = −1.

In Model 1, every atom works out to 0 — each O is 6 − 4 − 4/2 = 0, and C is 4 − 0 − 8/2 = 0.

Both sets sum to 0, so both models pass the sum check you’ve already seen — the sum tests validity, it does not pick between valid models.

The second selection criterion: among valid diagrams that tie on octets, prefer the model with smaller formal charges and fewer nonzero ones.

Model 1’s all-zero set beats Model 2’s +1 and −1, so Model 1 is the accepted diagram of CO₂.

What you are expected to be able to do

State that, among valid Lewis diagrams with comparable octets, the model with smaller and fewer nonzero formal charges is preferred.

Part 3Check what you just learned

Check — multiple choice

Two candidate Lewis diagrams of OCS are shown as Model 1 and Model 2. Every atom in both models has an octet. Which model is preferred?

AModel 2, because its formal charges sum to zero

Feedback if chosen: This option is wrong — you used the sum check to choose. Both models sum to zero, as any valid diagram of a neutral molecule must; the sum tests validity, not preference. Model 1’s charges are all 0, while Model 2 carries a −1 on O and a +1 on S.

BThe two models are equally good, because every atom in each has an octet

Feedback if chosen: This option is wrong — you stopped after the octet criterion. When octets tie, the formal charges decide: Model 1’s charges are all 0, while Model 2 carries a −1 on O and a +1 on S.

CModel 2, because a triple bond is stronger than a double bond

Feedback if chosen: This option is wrong — you chose by bond strength, which is not a selection criterion. The formal charges decide here: Model 1’s are all 0, while Model 2 carries a −1 and a +1.

DModel 1, because all of its formal charges work out to 0 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Model 1’s formal charges are all 0; Model 2’s work out to −1 on O, 0 on C, and +1 on S. Smaller and fewer nonzero formal charges are preferred, so Model 1 is the better model.

Check — multiple choice

Model 1 and Model 2 shown place the same 10 valence electrons with the atoms joined in a different order. Every atom has its expected octet or duet. Which model is preferred?

AModel 2, because its +1 and −1 cancel out

Feedback if chosen: This option is wrong — you credited Model 2 for a sum of zero. Both models sum to zero, as valid diagrams of a neutral molecule must; Model 2 still carries a +1 on N and a −1 on C, and Model 1 carries none.

BModel 1, because all of its formal charges are 0 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Model 1’s formal charges are 0 on H, C, and N; Model 2 puts +1 on N and −1 on C. The all-zero set is smaller and has fewer nonzero charges, so Model 1 is preferred.

CModel 1, because a diagram of a neutral molecule is not allowed to carry formal charges

Feedback if chosen: This option is wrong — you turned a preference into a ban. A valid diagram of a neutral molecule may carry nonzero formal charges — Model 2 does; Model 1 wins because its all-zero set is smaller and has fewer nonzero charges.

DThe two models are equally good, because both place all 10 electrons

Feedback if chosen: This option is wrong — you stopped at the electron count. Both models place all 10 electrons and tie on octets, so the formal charges decide: Model 1’s all-zero set is preferred.

Check — multiple choice

Formal charges of four candidate models
Model	Formal charges
Model 1	0, 0, +2, −2
Model 2	+1, −1, +1, −1
Model 3	0, 0, +1, −1
Model 4	+2, −1, −1, 0

Four valid Lewis diagrams of the same neutral molecule all give every atom an octet. The table shows each model’s formal charges. Which model is preferred?

AModel 1

Feedback if chosen: This option is wrong — you compared counts but not sizes. Model 1 has two nonzero charges, but they are size 2; Model 3 also has two nonzero charges, and neither is bigger than size 1.

BModel 2

Feedback if chosen: This option is wrong — you overlooked how many charges Model 2 carries. All four of its atoms are charged; Model 3 keeps the same size of charge on only two atoms.

CModel 3 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The octets are comparable across all four models, so compare the formal charges: Model 3 has the fewest nonzero charges (two) and the smallest (none bigger than size 1).

DModel 4

Feedback if chosen: This option is wrong — you accepted a +2 and three nonzero charges. Model 3 has only two nonzero charges and nothing bigger than size 1, so it is preferred on both grounds.

U2-S10 · LESSON 91 · CSP-084

Where the negative formal charge sits

Outcome: Select the model that places negative formal charge on the more electronegative atom.

Part 1One more tie to break

You’ve already seen two selection criteria: expected octets first, then smaller and fewer formal charges. Sometimes both leave a tie.

Here are two candidate diagrams of the cyanate ion, OCN⁻. Model 1 joins C to O by a single bond and to N by a triple bond.

Model 2 joins C to each neighbour by a double bond.

Every atom in both models has an octet, so the first criterion cannot choose.

Model 1, with the counting rule you’ve already seen: O is 6 − 6 − 2/2 = −1, while C and N both work out to 0.

Model 2: N is 5 − 4 − 4/2 = −1, while O and C both work out to 0.

Each model carries exactly one nonzero charge, and it is −1 in both — the patterns are comparable, so the second criterion cannot choose either.

Part 2Negative charge on the more electronegative atom

You’ve already seen electronegativity: how strongly a bonded atom attracts shared electrons. O sits further right than N in the same period, so O is the more electronegative atom.

The third selection criterion: when the formal-charge patterns are otherwise comparable, prefer the model that puts negative formal charge on the more electronegative atom.

The −1 marks the atom the diagram credits with one electron more than the free atom, and the atom that attracts electrons most strongly is the most plausible host for it.

Model 1 puts the −1 on O; Model 2 puts it on N. So Model 1 is the accepted diagram of cyanate.

What you are expected to be able to do

State that, when the formal-charge patterns are otherwise comparable, the model placing negative formal charge on the more electronegative atom is preferred.

Part 3Check what you just learned

Check — multiple choice

Two candidate diagrams of the thiocyanate ion, SCN⁻ (overall charge 1−), are shown. Every atom in both has an octet, and N is more electronegative than S. Which model is preferred, and why?

AModel 1, because it puts the −1 on N, the more electronegative atom ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Model 1 works out to −1 on N and 0 elsewhere; Model 2 to −1 on S and 0 elsewhere. The patterns are comparable, so the placement decides: negative formal charge belongs on the more electronegative atom, N.

BModel 2, because it puts the −1 on S, the larger atom

Feedback if chosen: This option is wrong — you chose by atomic size, which is not a selection criterion. With comparable patterns, the −1 belongs on the more electronegative atom — N — so Model 1 is preferred.

CModel 1, because a double bond on each side makes the diagram symmetric

Feedback if chosen: This option is wrong — you picked the right model for a reason that is not a criterion. Symmetry does not decide; Model 1 wins because its −1 sits on N, the more electronegative atom.

DModel 2, because its formal charges are smaller than Model 1’s

Feedback if chosen: This option is wrong — you compared magnitudes that are equal. Each model carries exactly one −1; when the patterns are comparable, the electronegativity criterion decides, and it prefers the −1 on N.

Check — multiple choice

When two valid models carry comparable formal-charge patterns, why is the model that puts the negative formal charge on the more electronegative atom preferred?

ABecause the more electronegative atom holds more valence electrons than its neighbours, so the −1 follows the bigger count

Feedback if chosen: This option is wrong — you swapped electronegativity for electron count. Electronegativity is how strongly a bonded atom attracts electrons, not how many valence electrons it has.

BBecause that atom attracts electrons most strongly, making it the most plausible host for the extra electron the −1 marks ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The −1 marks the atom the diagram credits with one electron more than the free atom. The atom that attracts electrons most strongly is the most plausible place for that extra electron, so that model describes the species better.

CBecause the negative formal charge turns that atom into an anion, and electronegative atoms form anions

Feedback if chosen: This option is wrong — you read the formal charge as a real charge. You’ve already seen that an atom carrying a formal charge has not become an ion; the criterion is about which bookkeeping is most plausible.

DBecause the more electronegative atom always takes the middle of the skeleton, and the middle atom carries the charge

Feedback if chosen: This option is wrong — you mixed in the skeleton rule, and backwards: the less electronegative atom usually sits in the middle. This criterion is about where the −1 sits, whatever the skeleton.

Check — multiple choice

Each model’s only nonzero formal charge
Model	Nonzero formal charge
Model 1	−1 on the S atom
Model 2	−1 on the Cl atom

Two valid Lewis diagrams of the same polyatomic ion give every atom an octet. Each model has exactly one nonzero formal charge, shown in the table. Which model is preferred?

AModel 1, because negative formal charge belongs on the less electronegative atom

Feedback if chosen: This option is wrong — you reversed the criterion. Negative formal charge is preferred on the more electronegative atom, and Cl sits further right than S in the same period.

BThe two models are equally good, because both charges have the same size

Feedback if chosen: This option is wrong — you stopped at the magnitude comparison. When the patterns are comparable, the placement still decides: Cl is more electronegative than S, so Model 2 is preferred.

CModel 2, because Cl is more electronegative than S ✓ KEY — this is the correct answer

Feedback if chosen: Correct. S and Cl sit in the same period, and electronegativity increases from left to right across a period, so Cl is the more electronegative atom — the −1 belongs there.

DModel 2, because Cl has more valence electrons than S

Feedback if chosen: This option is wrong — you picked the right model with the wrong quantity. Valence-electron count is not the criterion; Model 2 wins because Cl is more electronegative, not because it has seven valence electrons.

U2-S10 · LESSON 92 · CSP-085

Choosing the best Lewis diagram

Outcome: Evaluate candidate Lewis diagrams with all three selection criteria and select the best model.

Part 1Three criteria, in order

You’ve already seen the three selection criteria one at a time.

First: prefer models that give every atom its expected octet — unless the species is an accepted exception.

Second: among those, prefer smaller formal charges and fewer nonzero ones.

Third: when the patterns are still comparable, prefer negative formal charge on the more electronegative atom.

Apply them in order, and move to the next criterion only when the one before cannot decide.

Part 2A full evaluation of N₂O

Nitrous oxide, N₂O, is a real gas — an anesthetic — with 16 valence electrons and three atoms that can be joined in more than one order.

Model 1 joins the atoms N–N–O, with a triple bond then a single bond.

Model 2 keeps the same order with two double bonds.

Model 3 keeps the same order with a single bond then a triple bond.

Model 4 tries the other connectivity, N–O–N, with a double bond on each side.

All four models place all 16 valence electrons — each passes the electron-count check you’ve already seen.

First criterion: count around every atom. In all four models, every atom has an octet — no decision yet.

Second criterion: work out the formal charges. One sample with the counting rule — the O in Model 1: FC = 6 − 6 − 2/2 = −1.

Model 1, listed left to right: 0, +1, −1.

Model 2: −1, +1, 0.

Model 3: −2, +1, +1.

Model 4: −1, +2, −1.

All four sets sum to 0, as a neutral molecule’s must — every candidate passes the sum check.

Model 3 and Model 4 each contain a charge of size 2, so the second criterion removes them.

Model 1 and Model 2 each carry one +1 and one −1 — comparable patterns, so the third criterion decides.

Model 1 puts its −1 on O, the more electronegative atom; Model 2 puts its −1 on an N.

The best model of N₂O is Model 1: N≡N–O.

What you are expected to be able to do

Evaluate supplied Lewis diagrams with the three selection criteria, applied in order, and select the best model.

Part 3Check what you just learned

Check — multiple choice

Three candidate Lewis diagrams of CS₂ are shown, each placing all 16 valence electrons. Which model is preferred?

AModel 1

Feedback if chosen: This option is wrong — Model 1 carries a +1 on its triple-bonded S and a −1 on its single-bonded S. All three models tie on octets, so the smaller-and-fewer criterion picks Model 2, whose charges are all 0.

BModel 2 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Every atom in all three models has an octet, so the formal charges decide: Models 1 and 3 each carry a +1 and a −1, while Model 2’s charges are all 0. Model 2 — the double-bond model — is the best.

CModel 3

Feedback if chosen: This option is wrong — Model 3 carries a −1 on its single-bonded S and a +1 on its triple-bonded S. All three models tie on octets, so the smaller-and-fewer criterion picks Model 2, whose charges are all 0.

DModels 1, 2, and 3 are equally acceptable, because every atom in each has an octet

Feedback if chosen: This option is wrong — you stopped after the octet criterion. Octets only screen the candidates; when they tie, the formal charges decide, and only Model 2 has an all-zero set.

Check — multiple choice

Two candidate diagrams of carbon monoxide, CO, are shown; both place all 10 valence electrons. Which model is preferred?

AModel 1, because every formal charge in it is 0

Feedback if chosen: This option is wrong — you applied the formal-charge criterion first. The criteria run in order: C in Model 1 has only 6 electrons, and CO is not an accepted exception, so Model 2’s full octets win even though its charges are −1 and +1.

BModel 1, because a −1 formal charge on C would mean C had become an ion

Feedback if chosen: This option is wrong — you read a formal charge as a real charge. You’ve already seen that an atom carrying a formal charge has not become an ion; a −1 on C cannot disqualify a model.

CNeither model, because Model 2 puts −1 on C, the less electronegative atom

Feedback if chosen: This option is wrong — you used the electronegativity preference as a veto. It only compares models that are otherwise tied; Model 2 is the only model with full octets, so it is accepted, −1 on C and all.

DModel 2, because every atom in it has an octet — octets are checked before formal charges ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Model 2 gives C and O octets; Model 1 leaves C with 6 electrons, and CO is not an accepted exception. Octets are checked first, so Model 2 is preferred even though its formal charges are −1 and +1 rather than all zero.

Check — multiple choice

Four candidate diagrams for an ion with one C, one N, and one O atom and an overall charge of 1− are shown; all four place all 16 valence electrons. Which model is the best?

AModel 1

Feedback if chosen: This option is wrong — Model 1 survives the screening, but its single −1 sits on N. Model 3 has the same pattern with the −1 on O, the more electronegative atom, so Model 3 is preferred.

BModel 2

Feedback if chosen: This option is wrong — Model 2 joins the atoms in a different order, with N in the middle, and carries three nonzero formal charges: −1, +1, −1. Models 1 and 3 manage a single −1 each, so the smaller-and-fewer criterion removes Model 2 before electronegativity is even consulted.

CModel 3 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Every atom in all four models has an octet. Models 2 and 4 join the atoms in a different order, with N in the middle; Model 2 carries three nonzero charges and Model 4 contains a −2, so the smaller-and-fewer criterion removes both. That leaves Models 1 and 3; each carries a single −1, and Model 3 puts it on O, the more electronegative atom.

DModel 4

Feedback if chosen: This option is wrong — Model 4 also joins the atoms with N in the middle, and it contains a −2 on C. Models 1 and 3 keep every charge at size 1, so the smaller-and-fewer criterion removes Model 4.

U2-S11 — Resonance

7 lessons

U2-S11 · LESSON 93 · CSP-086

Resonance forms

Outcome: State that resonance forms are valid Lewis diagrams with the same atom connectivity but different electron placement.

Part 1Two valid diagrams for one ion

You’ve already seen how to construct a complete Lewis diagram for a polyatomic ion.

The nitrite ion, NO₂⁻, has 18 valence electrons: 5 from N, 6 from each of the two O atoms, and 1 more for the negative charge.

Here is a complete Lewis diagram of nitrite, with the N=O double bond on the left O.

Check it: three bonding pairs hold 6 electrons and six lone pairs hold 12 — all 18 placed, and every atom has an octet.

The double bond could sit on the right O instead — here is that diagram.

It checks out the same way: all 18 electrons placed, every atom with an octet.

Both diagrams follow every rule for a valid Lewis diagram; neither is wrong, and neither is preferred.

Part 2What resonance forms are

Chemists draw the two diagrams side by side, joined by a double-headed arrow, each in its own brackets with the 1− charge.

Diagrams related this way are called resonance forms: valid Lewis diagrams with the same atom connectivity but different electron placement.

Atom connectivity is which atoms are bonded to which — in both forms, each O is bonded to N, and every atom keeps its position.

Only the electron placement differs: the double bond and one lone pair traded places between the two O atoms.

If an atom had moved, the result would not be a resonance form — resonance forms keep every atom in place.

What you are expected to be able to do

State that resonance forms are valid Lewis diagrams with the same atom connectivity but different electron placement.

Part 3Check what you just learned

Check — multiple choice

The two resonance forms of ozone, O₃, are shown. What changes between the two forms?

AThe positions of the O atoms

Feedback if chosen: This option is wrong — you moved the atoms. Every O sits in the same position in both forms; only the double bond and a lone pair traded places.

BThe placement of the electrons only ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The atoms stay put and the total electron count is unchanged — the double bond and a lone pair traded places between the two outer O atoms.

CThe total number of valence electrons

Feedback if chosen: This option is wrong — you changed the electron count. Both forms place the same 18 valence electrons; what differs is where they sit.

DWhich atoms are bonded to which

Feedback if chosen: This option is wrong — you changed the connectivity. In both forms the central O is bonded to the same two outer O atoms.

Check — multiple choice

The two resonance forms of the formate ion, HCO₂⁻, are shown. Which feature is the same in both forms?

AWhich O atom carries the double bond

Feedback if chosen: This option is wrong — that is exactly what differs: the double bond is on the upper O in one form and on the lower O in the other.

BThe number of lone pairs on each O atom

Feedback if chosen: This option is wrong — the lone pairs moved with the double bond: the double-bonded O has two lone pairs and the single-bonded O has three, and those roles swap between the forms.

CThe placement of every valence electron in it

Feedback if chosen: This option is wrong — electron placement is the one thing that differs between resonance forms; connectivity and the total electron count are what stay the same.

DHow the atoms are connected to each other ✓ KEY — this is the correct answer

Feedback if chosen: Correct. In both forms C is bonded to the same H and the same two O atoms — resonance forms keep the atom connectivity unchanged.

Check — numeric entry

The two diagrams shown are resonance forms of dinitrogen monoxide, N₂O. How many valence electrons does each form show?

✓ ANSWER KEY: 16

Feedback when correct: Correct. Each form shows 4 bonding pairs and 4 lone pairs — 8 pairs, 16 electrons — and resonance forms always place the same total.

Anticipated wrong answers (feedback shown):

8	This answer counts pairs, not electrons. Each form has 8 pairs — 4 bonding and 4 lone — and each pair holds 2 electrons: 16.
24	This answer gives each atom its own octet of 8. Shared pairs serve both bonded atoms, so the total drawn is 4 bonding pairs plus 4 lone pairs: 16 electrons.

U2-S11 · LESSON 94 · CSP-087

The resonance hybrid

Outcome: State that the real electron distribution is one resonance hybrid, not a structure flipping between the forms.

Part 1One blend, not a flip

You’ve already seen the two resonance forms of the nitrite ion: same atom connectivity, different electron placement.

Each form is a valid diagram — but measurements of the real ion match neither form alone.

The real electron distribution is a single blend of the two drawn placements.

That single blended structure is called the resonance hybrid, and the drawn forms together represent it.

The nitrite ion is the hybrid at all times: its two N–O bonds are identical at every moment.

The ion does not flip back and forth between the forms.

A sample is not a mixture either: every nitrite ion is the same one hybrid.

Part 2What the double-headed arrow means

One Lewis diagram cannot show electrons in two placements at once, so the model uses the whole set of forms to stand for the hybrid.

The double-headed arrow means exactly that: these diagrams together represent one hybrid.

It does not mean that one form changes into the other — nothing converts, because there is only ever the one blended structure.

What you are expected to be able to do

State that the real electron distribution is one resonance hybrid represented by the set of forms — not a structure flipping between them.

Part 3Check what you just learned

Check — multiple choice

A student looks at the two resonance forms of ozone shown and says the molecule flips rapidly back and forth between them. Which statement is correct?

AThe student is right — the double-headed arrow shows the flipping.

Feedback if chosen: This option is wrong — you read the double-headed arrow as a change. It joins diagrams that together represent one hybrid; nothing converts.

BThe student is right, but the flipping is too fast to detect.

Feedback if chosen: This option is wrong — you kept the flipping and only hid it. There is no flipping at any speed: the molecule is the one blended structure at all times.

CThe student is wrong — half of the ozone molecules are one form and half are the other.

Feedback if chosen: This option is wrong — you turned the two drawings into two kinds of molecule. Every ozone molecule is the same single hybrid; the sample is not a mixture of forms.

DThe student is wrong — every ozone molecule is one blended structure at all times. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The two drawn forms together represent one resonance hybrid — a single blended structure with two identical O–O bonds.

Check — multiple choice

In the nitrite resonance forms shown, what does the double-headed arrow between the two diagrams mean?

AThe two diagrams together represent one resonance hybrid. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The arrow joins the forms into one set, and the set stands for the single blended structure of the real ion.

BThe left diagram changes into the right diagram over time.

Feedback if chosen: This option is wrong — you read the arrow as a change over time. Nothing converts: the ion is the one blended structure at every moment.

CThe two diagrams show two different ions with the same formula.

Feedback if chosen: This option is wrong — you split one species into two. Both diagrams describe the same nitrite ion; they differ only in drawn electron placement.

DEither diagram on its own is a complete picture of the ion.

Feedback if chosen: This option is wrong — you let one form stand alone. Measurements match neither form by itself; only the blend the full set represents describes the real ion.

Check — multiple choice

The two resonance forms of the formate ion, HCO₂⁻, are shown. Which statement describes the two C–O bonds in the real ion?

AOne is a double bond and the other is a single bond, as one of the drawn forms shows.

Feedback if chosen: This option is wrong — you took one drawn form at face value. The real ion matches neither form alone; it is the blend of both.

BEach switches between single and double as the electrons move around.

Feedback if chosen: This option is wrong — you made the ion flip between the forms. The electron distribution does not move around; it is the one blend at all times.

CThey are identical at all times — each a blend of the drawn single and double bond. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The hybrid gives the two C–O bonds one identical, blended character at every moment.

DBoth are double bonds, because the hybrid takes the double bond from each drawn form.

Feedback if chosen: This option is wrong — you upgraded both bonds to doubles. Blending sits between the drawn single and double; it does not turn both bonds into doubles.

U2-S11 · LESSON 95 · CSP-088

Recognizing equivalent resonance forms

Outcome: Identify whether supplied Lewis diagrams are equivalent resonance forms.

Part 1When two forms count as equivalent

You’ve already seen the nitrite resonance forms: the atoms keep their positions, and only the electron placement differs.

Compare what the two forms show: each has one double bond, one single bond, and the matching lone pairs — only the roles of the two O atoms are swapped.

The two O positions are identical: each O is bonded only to N.

Forms like these are called equivalent resonance forms: apart from which identical position holds the double bond, the diagrams look exactly the same.

Part 2Resonance forms that are not equivalent

Here are two valid Lewis diagrams of dinitrogen monoxide, N₂O — each places all 16 valence electrons.

They pass the resonance-form checks: every atom keeps its position, and only the electron placement differs.

But the forms do not look the same: one has a triple bond and a single bond; the other has two double bonds.

The two bonds trading electrons are different positions — one joins N to N, the other joins N to O — so no swap of identical atoms can make the forms match.

These are resonance forms, but not equivalent ones.

Part 3Not resonance forms at all

Here is a pair a student wrote as resonance forms of N₂O — but in the second diagram, the O sits in the middle.

An atom moved, so the atom connectivity changed: these are two different arrangements of the atoms, not two electron placements of one arrangement.

They are not resonance forms at all.

So run the checks in order: atoms unmoved and connected the same way; only the electron placement differing; and forms that match apart from which identical position holds the double bond.

What you are expected to be able to do

Identify whether supplied Lewis diagrams are equivalent resonance forms: unchanged atom connectivity, only electron placement differing, and forms that match apart from which identical position holds the double bond.

Part 4Check what you just learned

Check — multiple choice

A student writes the two diagrams of ozone, O₃, shown. How should the pair be classified?

AEquivalent resonance forms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The atoms stay put, only the electron placement differs, and the two outer O positions are identical — each bonded only to the central O — so the forms match apart from which one holds the double bond.

BResonance forms, but not equivalent

Feedback if chosen: This option is wrong — you stopped short at the matching check. The two positions trading the double bond are identical outer O atoms, and both forms show the same one-double, one-single pattern, so the forms are equivalent.

CNot resonance forms — the atom connectivity differs

Feedback if chosen: This option is wrong — you moved an atom that did not move. In both diagrams the central O is bonded to the same two outer O atoms.

DNot valid Lewis diagrams — the electron counts are wrong

Feedback if chosen: This option is wrong — you rejected two valid diagrams. Count either form: 3 bonding pairs and 6 lone pairs place all 18 valence electrons, and every atom has an octet.

Check — multiple choice

A student writes the two diagrams of the thiocyanate ion, SCN⁻, shown. How should the pair be classified?

AEquivalent resonance forms

Feedback if chosen: This option is wrong — you skipped the matching check. One form has a single bond and a triple bond; the other has two double bonds — the patterns differ, because the S–C and C–N positions are not identical.

BResonance forms, but not equivalent ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The atoms stay in place and only the electron placement differs, so they are resonance forms — but the bonds trading electrons join different pairs of elements (S–C and C–N), so the forms do not match and are not equivalent.

CNot resonance forms — the atom connectivity differs

Feedback if chosen: This option is wrong — you moved an atom that did not move. Both diagrams keep the order S, C, N with the same neighbors.

DNot valid Lewis diagrams — the electron counts are wrong

Feedback if chosen: This option is wrong — you rejected two valid diagrams. Each form places all 16 valence electrons and gives every atom an octet.

Check — multiple choice

A student writes the two diagrams shown as resonance forms of hydrogen cyanide, HCN. How should the pair be classified?

AEquivalent resonance forms

Feedback if chosen: This option is wrong — you accepted a moved atom. In the first diagram H is bonded to C; in the second it is bonded to N. Resonance forms never change which atoms are bonded to which.

BResonance forms, but not equivalent

Feedback if chosen: This option is wrong — the pair fails an earlier check. Before equivalence can even be asked, the atom connectivity must be unchanged, and here the H changed partners from C to N.

CNot resonance forms — the atom connectivity differs ✓ KEY — this is the correct answer

Feedback if chosen: Correct. H is bonded to C in one diagram and to N in the other — the atom connectivity changed, so these are two different arrangements, not resonance forms of one species.

DNot valid Lewis diagrams — the electron counts are wrong

Feedback if chosen: This option is wrong — the counts are fine: both diagrams place 10 valence electrons. What disqualifies the pair is the changed connectivity.

U2-S11 · LESSON 96 · CSP-089

Drawing a complete set of resonance forms

Outcome: Draw the complete set of equivalent resonance forms for a molecule or polyatomic ion.

Part 1A routine for the complete set

You’ve already seen how to construct a complete Lewis diagram for a polyatomic ion.

The nitrate ion, NO₃⁻, has 24 valence electrons: 5 from N, 6 from each of the three O atoms, and 1 for the charge.

Step one: construct one complete, valid diagram — here the double bond sits on the top O.

Step two: find every position identical to the one holding the double bond. All three O atoms are bonded only to N, so all three positions are identical.

Step three: redraw the diagram with the double bond on each identical position in turn — moving only electrons, never atoms.

Here is the complete set: three equivalent resonance forms, joined by double-headed arrows.

Check every form: all 24 electrons placed, every atom with an octet, and each form in its own brackets with the 1− charge.

The set is complete when each identical position has had its turn — no form missing, and no form drawn twice.

The routine is the same for a neutral molecule, except the forms get no brackets.

What you are expected to be able to do

Draw the complete set of equivalent resonance forms for a molecule or polyatomic ion, moving only electrons — never atoms.

Part 2Check what you just learned

Check — numeric entry

The carbonate ion is CO₃²⁻. How many valence electrons must each of its resonance forms show?

✓ ANSWER KEY: 24

Feedback when correct: Correct. C brings 4, each O brings 6 (18 in all), and the 2− charge adds 2 more electrons: 4 + 18 + 2 = 24.

Anticipated wrong answers (feedback shown):

22	This answer leaves out the charge. A 2− charge means the ion holds 2 extra electrons: 4 + 18 + 2 = 24.
20	This answer subtracts the charge. A negative charge means gained electrons, not lost ones: 4 + 18 + 2 = 24.
32	This answer gives every atom its own octet of 8. Shared pairs serve both bonded atoms, so the total to place is the valence electrons supplied: 4 + 18 + 2 = 24.

Check — drawing task (pen and paper, self-marked)

Take a pen and paper and draw the complete set of equivalent resonance forms for the carbonate ion, CO₃²⁻ — a central C bonded to three O atoms. Include brackets and the charge on every form. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows three resonance forms — one for each O position holding the double bond.
The atoms sit in the same positions in every form — only electrons moved.
The forms differ only in electron placement: the double bond is on a different O in each form, that O has two lone pairs, and each single-bonded O has three.
Each form is wrapped in its own brackets with the 2− charge.
Each form shows all 24 valence electrons.

Check — drawing task (pen and paper, self-marked)

Ozone, O₃, is a neutral molecule with a central O bonded to two outer O atoms. Take a pen and paper and draw the complete set of resonance forms for ozone. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows two resonance forms — one for each outer O holding the double bond.
The atoms sit in the same positions in both forms — only the double bond and a lone pair traded places.
In each form the double-bonded O has two lone pairs, the single-bonded O has three, and the central O has one.
Each form shows all 18 valence electrons.
No brackets and no charge — ozone is a neutral molecule, not an ion.

U2-S11 · LESSON 97 · CSP-090

Finding the flaw in a resonance set

Outcome: Evaluate a resonance set and identify the specific flaw if it has one.

Part 1Three ways a set can fail

You’ve already seen the routine for drawing a complete set of equivalent resonance forms.

A written set is checked the same way it is built — three checks, run in order.

Check one — connectivity: every atom keeps its position in every form, bonded to the same neighbors.

You’ve already seen this pair fail: the O moved to the middle, so the second diagram is not a resonance form at all.

Check two — electron count: every form places the same total number of valence electrons.

This claimed nitrite pair fails check two: the left form places 18 electrons, but the right form places 20 — an extra lone pair was added to the double-bonded O.

A form with the wrong count is not a valid diagram of the species, so it cannot belong to the set.

Check three — completeness: every identical position takes its turn with the double bond, so no equivalent form is missing.

This claimed carbonate set fails check three: the double bond visits only two of the three identical O positions.

When a set fails, name the specific flaw: which check failed, and where.

What you are expected to be able to do

Evaluate a supplied resonance set against atom connectivity, electron count, and completeness — and identify the specific flaw when a check fails.

Part 2Check what you just learned

Check — multiple choice

A student writes the three diagrams shown as the resonance set for the nitrate ion, NO₃⁻. Which statement identifies the specific flaw, if any?

ANo flaw — the set is valid and complete.

Feedback if chosen: This option is wrong — you passed a set that fails the electron-count check. The middle form places 26 electrons; the other two place 24.

BAn atom changed position between the forms.

Feedback if chosen: This option is wrong — no atom moved. All three forms keep N bonded to the same three O atoms; the flaw is in the electron count.

CThe forms do not all place the same number of electrons. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The first and third forms place 24 valence electrons, but the middle form’s double-bonded O keeps three lone pairs, giving 26 — every form must place the same electrons.

DAn equivalent form is missing, so the set is incomplete.

Feedback if chosen: This option is wrong — no form is missing: the double bond visits all three O positions. The flaw is that the middle form’s electron count does not match.

Check — multiple choice

A student writes the two diagrams shown as resonance forms of the cyanate ion, OCN⁻. Which statement identifies the specific flaw, if any?

ANo flaw — the set is valid and complete.

Feedback if chosen: This option is wrong — you passed a pair that fails the connectivity check: the C and N atoms traded places between the diagrams.

BAn atom changed position between the forms. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. In the first diagram the central atom is C; in the second it is N — atoms moved, so the second diagram is not a resonance form of the first.

CThe forms do not all place the same number of electrons.

Feedback if chosen: This option is wrong — both diagrams place 16 valence electrons. The flaw is the changed atom positions, not the count.

DAn equivalent form is missing, so the set is incomplete.

Feedback if chosen: This option is wrong — the problem is not a missing form. Before completeness can be judged, the diagrams must show one connectivity, and these show two.

Check — multiple choice

A student writes the single diagram shown as the complete resonance representation of the formate ion, HCO₂⁻. Which statement identifies the specific flaw, if any?

ANo flaw — the set is valid and complete.

Feedback if chosen: This option is wrong — the diagram itself is valid, but the representation is not complete: the two O positions are identical, and only one of them holds the double bond here.

BAn atom changed position between the forms.

Feedback if chosen: This option is wrong — with only one diagram, no atom has moved. What is missing is the equivalent form with the double bond on the other O.

CThe forms do not all place the same number of electrons.

Feedback if chosen: This option is wrong — there is no second count to disagree with. The flaw is the missing equivalent form, not the electrons in this one.

DAn equivalent form is missing, so the set is incomplete. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The lower O position is identical to the upper one, so the form with the double bond on the lower O must be drawn too — the set is missing an equivalent form.

Check — multiple choice

A student writes the two diagrams shown as the resonance set for ozone, O₃. Which statement identifies the specific flaw, if any?

ANo flaw — the set is valid and complete. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The atoms stay put, both forms place all 18 valence electrons, and each identical outer O position takes its turn with the double bond — the set passes every check.

BAn atom changed position between the forms.

Feedback if chosen: This option is wrong — the atoms sit in the same positions in both forms; only the double bond and a lone pair traded places.

CThe forms do not all place the same number of electrons.

Feedback if chosen: This option is wrong — count them: each form has 3 bonding pairs and 6 lone pairs, 18 electrons.

DAn equivalent form is missing, so the set is incomplete.

Feedback if chosen: This option is wrong — only the two outer O positions are identical, and both take a turn holding the double bond. Two forms complete the set.

U2-S11 · LESSON 98 · CSP-091

Bond lengths in a resonance hybrid

Outcome: Predict that bonds exchanged among equivalent resonance forms have equal measured lengths.

Part 1One form against the measurement

You’ve already seen that the real nitrite ion is one resonance hybrid: its two N–O bonds are identical at all times.

You’ve already seen that, between the same two elements, as bond order increases, bond length decreases.

Take either form at face value: it shows one N=O double bond and one N–O single bond.

If the ion really matched one form, the double bond would be shorter than the single bond.

Here are the measured lengths.

Measured N–O bond lengths in the nitrite ion (pm), rounded.

Both N–O bonds measure 124 pm — exactly equal.

The hybrid explains the measurement: both bonds are identical blends, so their measured lengths must be equal.

This is the general prediction: bonds that trade places among equivalent resonance forms have equal measured lengths.

Part 2The same test on carbonate

Here are the measured C–O bond lengths in the carbonate ion.

Measured C–O bond lengths in the carbonate ion (pm), rounded.

All three C–O bonds trade places among the three equivalent forms, and all three measure 129 pm.

A bond that does not trade places is not covered by the prediction — only the exchanged bonds must match.

What you are expected to be able to do

Predict that bonds exchanged among equivalent resonance forms have equal measured lengths.

Part 3Check what you just learned

Check — multiple choice

The three equivalent resonance forms of the nitrate ion, NO₃⁻, are shown. Which prediction about the measured N–O bond lengths is correct?

AAll three N–O bonds have the same measured length. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The three N–O bonds trade places among equivalent forms, so the hybrid gives all three one identical blended character — and equal measured lengths.

BOne N–O bond is shorter than the other two.

Feedback if chosen: This option is wrong — you read one form at face value. No single form matches the real ion; the hybrid makes the three bonds identical, so no bond is the short one.

CThe three lengths keep changing as the double bond moves from O to O.

Feedback if chosen: This option is wrong — you set the ion flipping. The double bond does not move around in the real ion: the ion is one blended structure at all times, so the lengths are steady and equal.

DThe lengths cannot be predicted from the resonance forms.

Feedback if chosen: This option is wrong — you gave up on a prediction the model makes. Bonds exchanged among equivalent forms are predicted equal, and measurement confirms it.

Check — multiple choice

The two equivalent resonance forms of the formate ion, HCO₂⁻, are shown. Which prediction about the measured bond lengths is correct?

AAll three bonds — both C–O bonds and the C–H bond — are predicted to have equal measured lengths.

Feedback if chosen: This option is wrong — you included a bond that never trades places. The C–H bond is identical in both forms; the prediction covers only the exchanged C–O bonds.

BThe two C–O bonds have different lengths: one matches a single bond and one matches a double bond.

Feedback if chosen: This option is wrong — you read one drawn form at face value. The real ion is the hybrid: both C–O bonds are identical blends with equal lengths.

CThe two C–O bond lengths trade values back and forth over time.

Feedback if chosen: This option is wrong — you set the lengths flipping. The ion is one blended structure at all times, so both C–O bonds hold one equal length.

DThe two C–O bonds are predicted equal; the C–H bond is not part of the prediction. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The two C–O bonds trade places between the equivalent forms, so they are predicted equal; the C–H bond is not exchanged, so it keeps its own length.

Check — numeric entry

Measured bond lengths in the formate ion, HCO₂⁻
Bond	Measured length (pm)
C–O (bond a)	125
C–O (bond b)	?
C–H	110

Formate’s two equivalent resonance forms exchange its two C–O bonds. Using the measured data shown, what is the length of C–O bond b, in pm?

✓ ANSWER KEY: 125 pm

Feedback when correct: Correct. Bonds exchanged among equivalent resonance forms have equal measured lengths, so bond b matches bond a: 125 pm.

Anticipated wrong answers (feedback shown):

110	That is the C–H bond’s length. The C–H bond is not exchanged between the forms; the bond equal to bond a is the other C–O bond, so bond b measures 125 pm.

U2-S11 · LESSON 99 · CSP-092

Bond order in a resonance hybrid

Outcome: Predict the average bond order of bonds exchanged among equivalent resonance forms.

Part 1Averaging the drawn bond orders

You’ve already seen bond order: 1 for a single bond, 2 for a double bond, and 3 for a triple bond.

You’ve already seen that the real nitrite ion is one blend of its two resonance forms.

Pick one N–O position and read its drawn bond order in each form: 2 in the form where it holds the double bond, and 1 in the other.

The hybrid blends the equivalent forms equally, so the real bond order at that position is the average of the drawn orders.

Q: What is the average bond order of each N–O bond in the nitrite hybrid?

Write down the drawn orders at one N–O position, one per form

order = 2(the form where that O holds the double bond)

order = 1(the other form)

Write down the equation

average order = sum of drawn orders ÷ number of forms

Substitute in the values, and calculate

average order = (2 + 1) ÷ 2

average order = 1.5

Both N–O positions give the same average: each real bond has order 1.5 at all times.

1.5 sits between the whole-number orders drawn in the forms — between the 1 of the single bond and the 2 of the double bond.

And because order 1.5 sits between 1 and 2, each bond’s measured length sits between a single bond’s and a double bond’s — matching the equal lengths you’ve already seen.

Part 2Three forms instead of two

Here is the complete set of carbonate resonance forms.

Q: What is the average bond order of each C–O bond in the carbonate hybrid?

Write down the drawn orders at one C–O position, one per form

order = 2(the form where that O holds the double bond)

order = 1(each of the other two forms)

Write down the equation

average order = sum of drawn orders ÷ number of forms

Substitute in the values, and calculate

average order = (2 + 1 + 1) ÷ 3

average order = 4/3 ≈ 1.33

4/3 is again between 1 and 2 — closer to 1, because the position holds the double bond in only one of the three forms.

The general prediction: bonds exchanged among equivalent resonance forms have an average bond order between the whole-number bond orders shown in the individual forms.

What you are expected to be able to do

Predict the average bond order of bonds exchanged among equivalent resonance forms — between the whole-number bond orders shown in the individual forms.

Part 3Check what you just learned

Check — multiple choice

The two equivalent resonance forms of the formate ion are shown. What is the average bond order of each C–O bond in the hybrid?

A1.0

Feedback if chosen: This option is wrong — you read only the form where that C–O bond is single. Average the position across both forms: (2 + 1) ÷ 2 = 1.5.

B1.5 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each C–O position is drawn with order 2 in one form and order 1 in the other: (2 + 1) ÷ 2 = 1.5.

C2.0

Feedback if chosen: This option is wrong — you read only the form where that C–O bond is double. Average across both forms: (2 + 1) ÷ 2 = 1.5.

D3.0

Feedback if chosen: This option is wrong — you added the two drawn orders without dividing. The average is the sum divided by the number of forms: (2 + 1) ÷ 2 = 1.5.

Check — multiple choice

The three equivalent resonance forms of the nitrate ion are shown. What is the average bond order of each N–O bond in the hybrid?

Feedback if chosen: This option is wrong — you read only the forms where that N–O bond is single. Across the three forms the position is drawn 2, 1, 1: (2 + 1 + 1) ÷ 3 = 4/3.

B3/2

Feedback if chosen: This option is wrong — you averaged one double and one single, but nitrate has three forms. The position is drawn 2, 1, 1: (2 + 1 + 1) ÷ 3 = 4/3.

C4/3 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each N–O position is drawn with order 2 in one form and order 1 in the other two: (2 + 1 + 1) ÷ 3 = 4/3.

Feedback if chosen: This option is wrong — you read only the form where that N–O bond is double. Average across all three forms: (2 + 1 + 1) ÷ 3 = 4/3.

Check — multiple choice

The average N–O bond order in the nitrite ion is 1.5. What does this value mean?

AEach N–O bond has order 1.5 at all times. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The hybrid blends the drawn forms, so every N–O bond carries the in-between order 1.5 at every moment.

BEach N–O bond switches between order 1 and order 2.

Feedback if chosen: This option is wrong — you set the bonds flipping between the drawn orders. The drawn 1 and 2 belong to the forms; the real bond holds the blend, 1.5, at all times.

COne N–O bond has order 1 while the other has order 2.

Feedback if chosen: This option is wrong — you took one drawn form at face value. The real ion matches neither form alone: both bonds are identical, each with order 1.5.

DHalf of the nitrite ions have order-1 bonds and half have order-2 bonds.

Feedback if chosen: This option is wrong — you split the sample into two kinds of ion. Every nitrite ion is the same hybrid, and both of its bonds have order 1.5.

AFTER U2-S11 · CHECKPOINT · U2-G3 · 8 items

Checkpoint 3 — Lewis diagrams, formal charge, and resonance

Check — multiple choice

Four candidate Lewis diagrams of sulfur dichloride, SCl₂, are shown (calculated total: 6 + 2 × 7 = 20 electrons). Which model passes every check of the construction routine — total electron count, octets and duets, formal charges kept small, and a plausible skeleton?

AModel 1 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Model 1 places all 20 electrons — two S–Cl bonds, two lone pairs on S, three on each Cl — S is bonded to both Cl atoms, and every atom counts an octet.

BModel 2

Feedback if chosen: This option is wrong — count Model 2's electrons: two bonds plus six lone pairs hold only 16 of the calculated 20. The two lone pairs on S were never placed, and S is left short of an octet.

CModel 3

Feedback if chosen: This option is wrong — count Model 3's electrons: it places 18 of the calculated 20, so it fails the electron-count check.

DModel 4

Feedback if chosen: This option is wrong — Model 4 places 20 electrons, but its atoms are strung S–Cl–Cl. In SCl₂ the single S atom is the central atom, bonded to each Cl.

Check — multiple choice

Two candidate Lewis diagrams of dinitrogen monoxide, N₂O, are shown; both place all 16 valence electrons. Which model is preferred?

AModel 2, because its formal charges are smaller in magnitude

Feedback if chosen: This option is wrong — compute the formal charges: Model 1 gives 0, +1, −1 and Model 2 gives −1, +1, 0. The magnitudes match; what differs is which atom carries the −1.

BModel 1, because diagrams with a triple bond are preferred over diagrams with two double bonds

Feedback if chosen: This option is wrong — bond multiplicity is not a selection criterion. Both models pass the earlier checks; formal charge and electronegativity decide between them.

CModel 1, because its −1 formal charge sits on O, the more electronegative atom ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both models carry charges of 0, +1, and −1, so the tiebreaker is placement: Model 1 puts the −1 on O, the more electronegative atom, while Model 2 puts it on N.

DModels 1 and 2 are equally acceptable, because both place 16 electrons and give every atom an octet

Feedback if chosen: This option is wrong — passing the electron-count and octet checks only makes both models valid candidates. Formal charge then selects the better model, and the two differ in where the −1 sits.

Check — multiple choice

The acetate ion, CH₃CO₂⁻, has a central C bonded to the CH₃ group and to two O atoms, and each of its resonance forms must place 24 valence electrons. You are asked to draw the complete set of equivalent resonance forms. Which description gives the correct set?

AOne form, with C=O to one O atom and C–O to the other — no second form is needed

Feedback if chosen: This option is wrong — you stopped after one valid diagram. The double bond can sit on either O equally well, so the complete set must show both equivalent forms.

BTwo forms: one with C=O to the first O and one with C=O to the second O; the atoms stay in place, and each form places all 24 electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The double bond has exactly two equivalent positions — one per O atom — so the complete set is two forms that differ only in where the electrons are drawn.

CTwo forms in which the two O atoms trade places while the double bond stays where it is

Feedback if chosen: This option is wrong — you moved atoms. Resonance forms redistribute electrons only; swapping the positions of the two O atoms is not a resonance form.

DThree forms: the two with C=O to each O atom, plus one with the double bond moved onto the C–CH₃ bond

Feedback if chosen: This option is wrong — you moved the double bond onto the C–CH₃ bond. The CH₃ carbon already shares four pairs, so a double bond there would give it ten electrons; no such form exists.

Check — numeric entry

How many valence electrons does the phosphate ion, PO₄³⁻, have in total?

✓ ANSWER KEY: 32

Feedback when correct: Correct. P contributes 5, the four O atoms contribute 4 × 6 = 24, and the 3− charge adds 3 more electrons: 5 + 24 + 3 = 32.

Anticipated wrong answers (feedback shown):

29	This count leaves the charge out: 5 + 24 = 29 counts only the atoms. A 3− charge means the ion holds 3 extra electrons: 29 + 3 = 32.
26	This count subtracts 3 for the charge. A negative charge means extra electrons, so add one per negative charge: 5 + 24 + 3 = 32.
30	This count adds only 1 electron for the charge. Add one electron per unit of negative charge — 3 of them here: 5 + 24 + 3 = 32.

Worked solution revealed after a second miss

Write down the values from the formula

P: 5 valence electrons (group 15)

O: 6 each (group 16), and 4 × 6 = 24

charge 3−: add 3 electrons

Write down the equation

total = P + O + charge electrons

Substitute in the values, and calculate

total = 5 + 24 + 3

total = 32 electrons

Check — multiple choice

A student's Lewis diagram of HOBr is shown. Its calculated valence-electron total is 1 + 6 + 7 = 14. What do you conclude?

AThe diagram is fine — the two missing electrons are inner-shell electrons, which Lewis diagrams leave out

Feedback if chosen: This option is wrong — you excused the gap with inner-shell electrons, but the calculated total of 14 already counts valence electrons only. Every one of them must appear in the diagram.

BThe diagram is wrong — it holds 12 electrons, two short of the calculated total, so a lone pair is missing ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Two bonds (4 electrons) plus two lone pairs on O (4) plus two lone pairs on Br (4) hold 12 of the 14 calculated electrons — Br needs a third lone pair.

CThe diagram is fine — H is drawn without a lone pair, and that accounts for the difference

Feedback if chosen: This option is wrong — H is complete with just its shared pair and never carries a lone pair, so H cannot account for missing electrons. The gap is a real error on Br.

DThe diagram is wrong — it holds 16 electrons, two too many

Feedback if chosen: This option is wrong — recount: two bonds hold 4 electrons and the four lone pairs hold 8, a total of 12. The diagram holds too few electrons, not too many.

Check — numeric entry

What is the formal charge on S in the sulfite ion Lewis diagram shown?

✓ ANSWER KEY: +1

Also accepted: 1

Feedback when correct: Correct. V = 6, N = 2 (one lone pair), B = 6 (three single bonds), so FC = 6 − 2 − 6/2 = +1.

Anticipated wrong answers (feedback shown):

-2	This answer comes from subtracting all 6 bonding electrons. Only half of B counts against the atom: FC = 6 − 2 − 6/2 = +1.
-1	This answer swaps N and B in the equation. N is the 2 lone-pair electrons and B is the 6 bonding electrons: FC = 6 − 2 − 6/2 = +1.
+3	This answer misses the lone pair on S. S has one lone pair, so N = 2: FC = 6 − 2 − 6/2 = +1.

Worked solution revealed after a second miss

Write down the values from the diagram

V = 6

N = 2

B = 6

Write down the equation

FC = V − N − B/2

Substitute in the values, and calculate

FC = 6 − 2 − 6/2

FC = +1

Check — multiple choice

A student writes the two diagrams shown as the complete resonance set for the carbonate ion, CO₃²⁻. Which statement identifies the specific flaw, if any?

ANo flaw — the set is valid and complete.

Feedback if chosen: This option is wrong — each form is valid on its own, but the double bond has a third equivalent position that the set never shows.

BAn atom changed position between the forms.

Feedback if chosen: This option is wrong — check the positions: C and the three O atoms sit in the same places in both forms; only the electrons move.

CThe forms do not all place the same number of electrons.

Feedback if chosen: This option is wrong — count again: both forms place all 24 electrons (four bonds' worth plus eight lone pairs).

DAn equivalent form is missing, so the set is incomplete. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The double bond can sit on any of the three equivalent O atoms, so the complete set has three forms — the form with C=O to the lower-right O is missing.

Check — multiple choice

The two equivalent resonance forms of ozone, O₃, are shown. What is the average bond order of each O–O bond in the hybrid?

A1.0

Feedback if chosen: This option is wrong — you took the single bond from one form. Average the orders each bond takes across the set: (1 + 2)/2 = 1.5.

B1.5 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Across the two forms, each O–O bond is a single bond once and a double bond once, so its average order is (1 + 2)/2 = 1.5.

C2.0

Feedback if chosen: This option is wrong — you took the double bond from one form. Each bond is double in only one of the two forms: (1 + 2)/2 = 1.5.

D3.0

Feedback if chosen: This option is wrong — you added the two orders instead of averaging them: (1 + 2)/2 = 1.5.

U2-S12 — VSEPR and molecular geometry

15 lessons

U2-S12 · LESSON 100 · CSP-093

The VSEPR model

Outcome: State how VSEPR places the regions of valence electron density around a central atom.

Part 1A flat diagram, a three-dimensional molecule

You’ve already seen the Lewis diagram of CH₄: it shows which atoms are joined, drawn flat on the page.

A real CH₄ molecule is not flat — its atoms sit in three dimensions.

To predict the arrangement, chemists use a model called VSEPR — short for valence shell electron pair repulsion.

The model starts from something you’ve already seen: electrons repel one another.

Part 2As far apart as possible

Around a central atom, the valence electrons sit in regions of valence electron density — the electrons of each bond form one region, clustered between the two atoms.

Because the regions all carry negative charge, they repel one another.

VSEPR places the regions around the central atom as far apart as possible, because that arrangement reduces the repulsion between them.

Here is what that gives for the four bonding regions of CH₄: they point away from one another into three dimensions, not into a flat cross.

In the drawing, a solid wedge is a bond coming toward you and a dashed bond points away from you — that is how a flat page shows three dimensions.

With only two regions, the prediction is simpler: the two regions end up on opposite sides of the central atom, as far apart as they can get.

What you are expected to be able to do

State that VSEPR predicts the arrangement around a central atom by placing the regions of valence electron density as far apart as possible, which reduces the repulsion between them.

Part 3Check what you just learned

Check — multiple choice

How does VSEPR place the regions of valence electron density around a central atom?

APointing toward the outer atoms’ nuclei, because electrons and nuclei attract

Feedback if chosen: This option is wrong — you built the prediction on electron–nucleus attraction. VSEPR is built on the repulsion between the regions themselves: they spread as far apart as possible.

BAs close together as possible, because the regions attract one another

Feedback if chosen: This option is wrong — you gave the regions an attraction they do not have. Regions of electron density all carry negative charge, so they repel, and the model spreads them as far apart as possible.

CFlat in one plane, because Lewis diagrams are drawn flat

Feedback if chosen: This option is wrong — you read the flat drawing as the real arrangement. The Lewis diagram only shows which atoms are joined; VSEPR spreads the regions into three dimensions, as far apart as possible.

DAs far apart as possible, because the regions repel one another ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each region carries negative charge, so the regions repel — and spreading them as far apart as possible reduces that repulsion.

Check — multiple choice

A central atom has exactly two regions of valence electron density around it. Where does VSEPR place them?

AWherever the diagram is easiest to draw — the model makes no prediction

Feedback if chosen: This option is wrong — you treated the arrangement as a drawing choice. VSEPR does make a prediction: the two regions repel, so they end up as far apart as possible, on opposite sides.

BBoth on the same side of the central atom, so they can share space

Feedback if chosen: This option is wrong — you placed the regions where their repulsion is largest. Two negatively charged regions push apart, so they end up on opposite sides of the central atom.

COn opposite sides of the central atom, as far apart as possible ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Opposite sides is as far apart as two regions can get, and that arrangement reduces the repulsion between them.

DBoth pulled toward whichever outer atom attracts electrons more strongly

Feedback if chosen: This option is wrong — you let attraction to an outer atom set the arrangement. VSEPR places regions by the repulsion between the regions themselves: as far apart as possible.

U2-S12 · LESSON 101 · CSP-094

Electron domains

Outcome: State what one electron domain around a selected atom is.

Part 1One region gets a name

You’ve already seen that VSEPR works with regions of valence electron density around a central atom.

Counting those regions comes up so often that each one gets a name: one region of valence electron density around a selected atom is called an electron domain.

Here is the Lewis diagram of CO₂.

Look at the C=O bond on the left: its bonding electrons sit clustered between C and the left O — one region of electron density around C, so one electron domain around C.

The bond on the right is a second region — a second electron domain around C.

A domain always belongs to a selected atom: the same bond that is one domain around C is also one domain around the O it reaches.

Bonds are not the only regions: a lone pair on the selected atom is also one region of electron density, so it is also an electron domain.

What you are expected to be able to do

State that an electron domain is one region of valence electron density around a selected atom.

Part 2Check what you just learned

Check — multiple choice

Use the H₂O Lewis diagram shown. Which of these is one electron domain around O?

AOne electron taken from an O–H bond

Feedback if chosen: This option is wrong — you split a region into single electrons. A domain is a whole region of electron density, such as all the bonding electrons of one bond, not one electron from it.

BThe bonding electrons of one O–H bond ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The bonding electrons of one O–H bond sit clustered between O and that H — one region of valence electron density around O, so one electron domain.

CThe whole H₂O molecule

Feedback if chosen: This option is wrong — you took the whole molecule as the region. A domain is one region of electron density around one selected atom, not the molecule itself.

DThe two H atoms taken together

Feedback if chosen: This option is wrong — you counted atoms. A domain is a region of electron density, not a set of atoms — around O, each bond’s electrons form one domain.

Check — multiple choice

Which statement says what an electron domain is?

AOne shared electron within a bond

Feedback if chosen: This option is wrong — you made the domain a single electron. A domain is a whole region of electron density — a bond’s bonding electrons together form one domain.

BThe space a molecule takes up

Feedback if chosen: This option is wrong — you stretched the region over the whole molecule. A domain is one region of electron density around one selected atom.

CAny atom directly bonded to the central atom

Feedback if chosen: This option is wrong — you counted bonded atoms instead of electron regions. A domain is a region of valence electron density; lone pairs can be domains even though no atom sits there.

DOne region of valence electron density around a selected atom ✓ KEY — this is the correct answer

Feedback if chosen: Correct. An electron domain is one region of valence electron density around a selected atom — the electrons of a bond, for example.

U2-S12 · LESSON 102 · CSP-095

One bond, one domain

Outcome: State that a single, double, or triple bond counts as one electron domain around a selected atom.

Part 1Bond order does not add domains

You’ve already seen that an electron domain is one region of valence electron density around a selected atom.

Here is the Lewis diagram of CO₂ again.

Each C=O double bond holds two shared pairs — but all four of those electrons sit clustered in the same place, between C and that O.

One cluster is one region, so the double bond counts as one electron domain around C — not two.

The rule covers every bond order: a single, double, or triple bond to a selected atom counts as one electron domain around that atom.

So around C in CO₂ there are exactly two domains: one per C=O bond.

What you are expected to be able to do

Count any single, double, or triple bond to a selected atom as one electron domain around that atom.

Part 2Check what you just learned

Check — multiple choice

The diagram shows a correct Lewis diagram of SO₂. How does the S=O double bond count toward the electron domains around S?

AAs two domains, one per shared pair

Feedback if chosen: This option is wrong — you counted shared pairs instead of regions. Both pairs of the double bond sit clustered between S and that O — one region, one domain.

BAs four domains, one per bonding electron

Feedback if chosen: This option is wrong — you counted electrons instead of regions. All four bonding electrons sit in the same cluster between S and that O — one region, one domain.

CAs one electron domain ✓ KEY — this is the correct answer

Feedback if chosen: Correct. All the bonding electrons of the S=O bond sit clustered between S and that O — one region of electron density, so one domain around S.

DIt does not count — only lone pairs form electron domains

Feedback if chosen: This option is wrong — you dropped bonds from the count. A bond’s electrons are a region of valence electron density, so every bond is one domain around the selected atom.

Check — multiple choice

An atom is joined to a neighbor by a triple bond. How does that triple bond count toward the electron domains around the atom?

AAs one electron domain ✓ KEY — this is the correct answer

Feedback if chosen: Correct. All six bonding electrons of a triple bond sit clustered between the same two atoms — one region of electron density, so one domain.

BAs three domains, one per shared pair

Feedback if chosen: This option is wrong — you counted shared pairs instead of regions. The three pairs of a triple bond share one cluster between the two atoms — one region, one domain.

CAs six domains, one per bonding electron

Feedback if chosen: This option is wrong — you counted electrons instead of regions. The six bonding electrons sit together in one cluster — one region, one domain.

DIt depends on which two elements the bond joins

Feedback if chosen: This option is wrong — you added a condition the rule does not have. The count comes from the region, not the elements: any bond, between any two atoms, is one domain.

U2-S12 · LESSON 103 · CSP-096

Each lone pair is its own domain

Outcome: State that each lone pair on a selected atom counts as one electron domain around that atom.

Part 1Lone pairs join the count

You’ve already seen that every bond to a selected atom counts as one electron domain.

Lone pairs count too. Here is the Lewis diagram of H₂O.

Each lone pair on O is a pair of valence electrons sitting in its own cluster on O — its own region of electron density.

So each lone pair on a selected atom counts as one electron domain around that atom.

The two lone pairs on O are two separate regions, so they count as two domains — they never merge into one.

A lone pair carries negative charge just as a bond’s electrons do, so it repels the other domains and takes up a direction of its own.

What you are expected to be able to do

Count each lone pair on a selected atom as one electron domain around that atom.

Part 2Check what you just learned

Check — multiple choice

Use the NH₃ Lewis diagram shown. How does the lone pair on N count toward the electron domains around N?

AAs two domains, one per electron

Feedback if chosen: This option is wrong — you counted electrons instead of regions. The two electrons of the lone pair sit together in one cluster — one region, one domain.

BAs half a domain, because it is not shared with another atom

Feedback if chosen: This option is wrong — you discounted the lone pair for being unshared. Sharing is not the test; the lone pair is a full region of electron density, so it is a full domain.

CIt does not count, because it joins N to no other atom

Feedback if chosen: This option is wrong — you required a domain to join two atoms. A domain is a region of electron density, and a lone pair is exactly that — it counts even though no atom sits at its end.

DAs one electron domain, just as each N–H bond does ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The lone pair is its own cluster of valence electrons on N — one region of electron density, so one domain, on equal footing with each bond.

Check — multiple choice

A central atom carries three lone pairs. How many electron domains do those lone pairs contribute around it?

AOne — lone pairs on the same atom merge into a single region

Feedback if chosen: This option is wrong — you merged separate regions. Each lone pair sits in its own cluster, so three lone pairs are three separate domains.

BThree — each lone pair is one domain ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each lone pair is its own region of electron density, so three lone pairs contribute three domains.

CSix — one per electron

Feedback if chosen: This option is wrong — you counted electrons instead of regions. Each pair’s two electrons share one cluster: three pairs, three domains.

DNone — lone pairs sit too close to the atom to repel anything

Feedback if chosen: This option is wrong — you dropped lone pairs from the count. A lone pair carries negative charge in its own region, so it repels the other domains and counts fully.

U2-S12 · LESSON 104 · CSP-097

Counting electron domains

Outcome: Count the electron domains around a selected atom from its Lewis diagram.

Part 1The counting routine

You’ve already seen the two counting rules: every bond to a selected atom is one electron domain, and every lone pair on it is one more.

So the count is a two-step routine: count the bonds to the atom (a double or triple bond still counts once), then add the lone pairs on the atom.

Here is the Lewis diagram of H₂O.

Around O: two O–H bonds, so two bonding domains.

Plus two lone pairs on O: two more domains.

Total: 2 + 2 = 4 electron domains around O.

Now BF₃, where the central atom has no lone pairs.

Around B: three B–F bonds and no lone pairs — 3 + 0 = 3 electron domains.

Count only what sits on the selected atom: the lone pairs on the F atoms belong to the F atoms’ own counts, not to B’s.

What you are expected to be able to do

Count the electron domains around a selected atom from a supplied Lewis diagram: bonds to the atom plus lone pairs on the atom.

Part 2Check what you just learned

Check — numeric entry

Use the NH₃ Lewis diagram shown. How many electron domains are around N?

✓ ANSWER KEY: 4

Feedback when correct: Correct. Three N–H bonds plus one lone pair on N: 3 + 1 = 4 domains.

Anticipated wrong answers (feedback shown):

3	This answer counts only the bonds. The lone pair on N is a region of electron density too: 3 bonds + 1 lone pair = 4 domains.
1	This answer counts only the lone pair. Each N–H bond is one domain as well: 3 bonds + 1 lone pair = 4 domains.
8	This answer counts the electrons around N. Count regions instead: three bonds + one lone pair = 4 domains.

Check — numeric entry

Use the SF₄ Lewis diagram shown. How many electron domains are around S?

✓ ANSWER KEY: 5

Feedback when correct: Correct. Four S–F bonds plus one lone pair on S: 4 + 1 = 5 domains.

Anticipated wrong answers (feedback shown):

4	This answer counts only the bonds. The lone pair on S counts too: 4 bonds + 1 lone pair = 5 domains.
10	This answer counts the electrons around S. Count regions instead: four bonds + one lone pair = 5 domains.
17	This answer sweeps in the lone pairs on the F atoms. Count only what sits on the selected atom: four bonds + one lone pair on S = 5 domains.

Check — numeric entry

Use the CO₂ Lewis diagram shown. How many electron domains are around C?

✓ ANSWER KEY: 2

Feedback when correct: Correct. Two C=O bonds and no lone pairs on C: each double bond counts once, so 2 domains.

Anticipated wrong answers (feedback shown):

4	This answer counts each double bond twice. A double bond is one region of electron density, so it is one domain: two bonds, 2 domains.
8	This answer counts the bonding electrons around C. Count regions instead: two double bonds = 2 domains.
6	This answer adds the lone pairs on the O atoms. Those belong to the O atoms’ own counts — around C there are just the two bonds: 2 domains.

U2-S12 · LESSON 105 · CSP-098

Electron-domain geometry and molecular geometry

Outcome: Classify a described arrangement as electron-domain geometry or molecular geometry.

Part 1Two arrangements from one molecule

You’ve already seen that O in H₂O has four electron domains: two bonds and two lone pairs.

Here is H₂O with all four domains drawn — the two O–H bonds plus the two lone pairs, each pointing in its own direction.

There are two different questions you can ask about this picture, and each has its own name.

The arrangement of ALL the domains — bonds and lone pairs alike — is called the electron-domain geometry.

The arrangement of only the bonded atoms — the shape you would see if you could photograph the atoms — is called the molecular geometry.

In H₂O the two arrangements differ: all four domains spread into three dimensions, but the visible molecule is just O with two H atoms attached.

Lone pairs count in the electron-domain geometry, but they are invisible in the molecular geometry — no atom sits at their ends.

When the central atom has no lone pairs, every domain ends in a bonded atom, so the two arrangements coincide.

What you are expected to be able to do

Classify the electron-domain geometry as the arrangement of all domains and the molecular geometry as the arrangement of only the bonded atoms.

Part 2Check what you just learned

Check — multiple choice

Which of these does molecular geometry describe?

AThe arrangement of only the bonded atoms around the central atom ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Molecular geometry is the arrangement of the bonded atoms only — the shape the atoms themselves make.

BThe arrangement of all electron domains, lone pairs included

Feedback if chosen: This option is wrong — you named the other geometry. All domains together give the electron-domain geometry; molecular geometry keeps only the bonded atoms.

CThe arrangement of only the lone pairs around the central atom

Feedback if chosen: This option is wrong — you kept exactly the wrong part. Lone pairs are the domains molecular geometry leaves out; it describes the bonded atoms.

DThe arrangement of the electrons within one bond

Feedback if chosen: This option is wrong — you zoomed inside a single bond. Both geometries describe arrangements around the central atom, and molecular geometry is the arrangement of its bonded atoms.

Check — multiple choice

A central atom carries one lone pair. Which geometry counts that lone pair as part of its arrangement?

AThe molecular geometry only

Feedback if chosen: This option is wrong — you put the lone pair in the atoms-only picture. Molecular geometry keeps only the bonded atoms; the lone pair belongs to the electron-domain geometry.

BNeither geometry — lone pairs are never part of an arrangement

Feedback if chosen: This option is wrong — you dropped the lone pair from both pictures. A lone pair is a full electron domain, so the electron-domain geometry counts it.

CThe electron-domain geometry only ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Electron-domain geometry arranges all domains, lone pairs included; molecular geometry keeps only the bonded atoms.

DBoth geometries

Feedback if chosen: This option is wrong — you counted the lone pair in the atoms-only picture too. No atom sits at the lone pair’s end, so the molecular geometry leaves it out.

U2-S12 · LESSON 106 · CSP-099

Electron-domain geometry for two to four domains

Outcome: Predict the electron-domain geometry from a count of two, three, or four domains.

Part 1The count fixes the arrangement

You’ve already seen that VSEPR spreads the electron domains around a central atom as far apart as possible.

For a given number of domains there is exactly one such arrangement — so the domain count alone fixes the electron-domain geometry.

Two domains spread to opposite sides of the central atom. That arrangement is called linear. C in CO₂ has two domains, so its electron-domain geometry is linear.

Three domains spread into a flat triangle around the central atom. That arrangement is called trigonal planar. B in BF₃ has three domains.

Four domains spread into three dimensions, each pointing to a corner of a tetrahedron — a pyramid with a triangular base. That arrangement is called tetrahedral. C in CH₄ has four domains.

Part 2Lone pairs are domains too

The count includes every domain, lone pairs included.

N in NH₃ has four domains — three bonds and one lone pair — so its electron-domain geometry is also tetrahedral: all four domains are directed at the corners of a tetrahedron, and the lone pair simply occupies one of them.

What you are expected to be able to do

Predict linear, trigonal planar, or tetrahedral electron-domain geometry from a count of two, three, or four electron domains.

Part 3Check what you just learned

Check — multiple choice

Use the H₂O Lewis diagram shown. Which electron-domain geometry does VSEPR predict around O?

ALinear

Feedback if chosen: This option is wrong — you counted only the two bonded H atoms. The two lone pairs on O are domains too: 2 + 2 = 4 domains, so the electron-domain geometry is tetrahedral.

BTetrahedral ✓ KEY — this is the correct answer

Feedback if chosen: Correct. O has four domains — two bonds plus two lone pairs — and four domains spread to the corners of a tetrahedron.

CTrigonal planar

Feedback if chosen: This option is wrong — you merged the two lone pairs into one domain. Each lone pair is its own domain: 2 bonds + 2 lone pairs = 4 domains, so tetrahedral.

DVSEPR makes no prediction when lone pairs are present

Feedback if chosen: This option is wrong — you dropped lone pairs from the model. A lone pair is a full domain, so it joins the count: four domains, tetrahedral.

Check — multiple choice

The diagram shows a correct Lewis diagram of SO₂. Which electron-domain geometry does VSEPR predict around S?

ATetrahedral

Feedback if chosen: This option is wrong — you counted the double bond as two domains. Any bond counts once: two bonds + one lone pair = 3 domains, so trigonal planar.

BLinear

Feedback if chosen: This option is wrong — you counted only the two bonded O atoms. The lone pair on S is a domain too: 2 + 1 = 3 domains, so trigonal planar.

CVSEPR makes no prediction when the bonds differ

Feedback if chosen: This option is wrong — you added a condition the model does not have. The count is all that matters: two bonds + one lone pair = 3 domains, so trigonal planar.

DTrigonal planar ✓ KEY — this is the correct answer

Feedback if chosen: Correct. S has three domains — two bonds (the double bond counts once) plus one lone pair — and three domains spread into a flat triangle: trigonal planar.

U2-S12 · LESSON 107 · CSP-100

Five domains: trigonal bipyramidal

Outcome: Predict trigonal-bipyramidal electron-domain geometry from a count of five domains.

Part 1The five-domain arrangement

You’ve already seen that the domain count fixes the electron-domain geometry: two domains give linear, three trigonal planar, four tetrahedral.

Five domains spread into an arrangement called trigonal bipyramidal: three domains form a flat triangle around the middle of the central atom, and the other two point straight up and straight down from it.

P in PCl₅ has five domains — five bonds, no lone pairs.

So its electron-domain geometry is trigonal bipyramidal.

Lone pairs join the count here just as before: S in SF₄ has four bonds plus one lone pair — five domains — so its electron-domain geometry is also trigonal bipyramidal.

What you are expected to be able to do

Predict trigonal-bipyramidal electron-domain geometry from a count of five electron domains.

Part 2Check what you just learned

Check — multiple choice

Use the XeF₂ Lewis diagram shown. Which electron-domain geometry does VSEPR predict around Xe?

ALinear

Feedback if chosen: This option is wrong — you counted only the two bonded F atoms. The three lone pairs on Xe are domains too: 2 + 3 = 5 domains, so trigonal bipyramidal.

BTetrahedral

Feedback if chosen: This option is wrong — you dropped one lone pair from the count. Xe carries three lone pairs: 2 bonds + 3 lone pairs = 5 domains, so trigonal bipyramidal.

CTrigonal bipyramidal ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Xe has five domains — two bonds plus three lone pairs — and five domains take the trigonal bipyramidal arrangement.

DTrigonal planar

Feedback if chosen: This option is wrong — you merged the three lone pairs into one domain. Each lone pair is its own domain: 2 + 3 = 5 domains, so trigonal bipyramidal.

Check — numeric entry

A central atom’s electron-domain geometry is trigonal bipyramidal. How many electron domains does it have?

✓ ANSWER KEY: 5

Feedback when correct: Correct. Trigonal bipyramidal is the five-domain arrangement: three domains in a triangle around the middle plus two pointing straight up and down.

Anticipated wrong answers (feedback shown):

3	This answer counts only the triangle the ‘trigonal’ part names. Add the two domains pointing straight up and down: 3 + 2 = 5.
2	This answer counts only the up-and-down pair. Add the three domains in the triangle around the middle: 3 + 2 = 5.
4	This answer belongs to tetrahedral, the four-domain arrangement. Trigonal bipyramidal is the five-domain one: 3 + 2 = 5.
6	This answer overshoots by one. Count the positions: three in the triangle plus one up and one down — 5.

U2-S12 · LESSON 108 · CSP-101

Six domains: octahedral

Outcome: Predict octahedral electron-domain geometry from a count of six domains.

Part 1The six-domain arrangement

You’ve already seen the arrangements for two to five domains.

Six domains spread into an arrangement called octahedral: four domains form a flat square around the middle of the central atom, and the other two point straight up and straight down from it.

All six positions in an octahedral arrangement are alike — each domain has the same set of neighbors as every other.

S in SF₆ has six domains — six bonds, no lone pairs — so its electron-domain geometry is octahedral.

Lone pairs join the count as always: Xe in XeF₄ has four bonds plus two lone pairs — six domains — so its electron-domain geometry is also octahedral.

What you are expected to be able to do

Predict octahedral electron-domain geometry from a count of six electron domains.

Part 2Check what you just learned

Check — multiple choice

Around the central atom
Species	Bonds to the central atom	Lone pairs on the central atom
BrF₅	5	1
SF₄	4	1
ClF₃	3	2
CH₄	4	0

Use the table. Around which central atom is the electron-domain geometry octahedral?

AThe Br in BrF₅ ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Br has 5 bonds + 1 lone pair = 6 domains, and six domains take the octahedral arrangement.

BThe S in SF₄

Feedback if chosen: This option is wrong — count S's domains: 4 bonds + 1 lone pair = 5, which gives trigonal bipyramidal, not octahedral.

CThe Cl in ClF₃

Feedback if chosen: This option is wrong — you counted six domains for ClF₃. Cl has 3 bonds + 2 lone pairs = 5 domains — trigonal bipyramidal, not octahedral.

DThe C in CH₄

Feedback if chosen: This option is wrong — you picked a four-domain atom. C has 4 bonds + 0 lone pairs = 4 domains — tetrahedral, not octahedral.

Check — numeric entry

A central atom’s electron-domain geometry is octahedral. How many electron domains does it have?

✓ ANSWER KEY: 6

Feedback when correct: Correct. Octahedral is the six-domain arrangement: four domains in a square around the middle plus two pointing straight up and down.

Anticipated wrong answers (feedback shown):

8	This answer reads ‘octa’ as the domain count. The eight in the name counts the eight faces of an octahedron; the domains sit at its six corners.
5	This answer belongs to trigonal bipyramidal, the five-domain arrangement. Octahedral is the six-domain one: four around the middle plus two up and down.
4	This answer counts only the square around the middle. Add the two domains pointing straight up and down: 4 + 2 = 6.

U2-S12 · LESSON 109 · CSP-102

Molecular geometry with up to four domains

Outcome: Predict the molecular geometry for a central atom with two, three, or four electron domains.

Part 1Name only what you can see

You’ve already seen the two-step picture: all the domains spread out (the electron-domain geometry), and the bonded atoms sit at some of the positions (the molecular geometry).

To predict a molecular geometry: place all the domains first, then name the arrangement of only the bonded atoms.

With no lone pairs, every position holds an atom, so the molecular geometry keeps the electron-domain name: CO₂ is linear, BF₃ is trigonal planar, CH₄ is tetrahedral.

Part 2Four domains with lone pairs

N in NH₃ has four domains — three bonds and one lone pair — so all four spread to tetrahedral corners. The three bonded H atoms alone form a three-sided pyramid with N at its top: that molecular geometry is called trigonal pyramidal.

O in H₂O has four domains — two bonds and two lone pairs. The two bonded H atoms alone make a V shape with O at the corner: that molecular geometry is called bent.

Part 3Three domains with a lone pair

Three domains with one lone pair leave two bonded atoms, again in a V shape — also called bent. The two bent cases differ only in where the V comes from: one lone pair out of three domains, or two out of four.

Here are the five names in one table.

Molecular geometry, two to four domains
Electron domains	Lone pairs	Molecular geometry
2	0	linear
3	0	trigonal planar
3	1	bent
4	0	tetrahedral
4	1	trigonal pyramidal
4	2	bent

What you are expected to be able to do

Predict linear, trigonal planar, tetrahedral, trigonal pyramidal, or bent molecular geometry for a central atom with two, three, or four electron domains.

Part 4Check what you just learned

Check — multiple choice

The diagram shows a correct Lewis diagram of SO₂. Which molecular geometry does VSEPR predict for it?

ALinear

Feedback if chosen: This option is wrong — you placed only the two bonded O atoms and let them spread straight. The lone pair on S takes a domain position too, so the two O atoms sit in a V: bent.

BTrigonal planar

Feedback if chosen: This option is wrong — you named the electron-domain geometry. All three domains are trigonal planar, but the molecular geometry keeps only the two bonded O atoms: bent.

CTrigonal pyramidal

Feedback if chosen: This option is wrong — you counted the double bond as two domains, reaching four with a lone pair. Any bond counts once: three domains, one of them a lone pair, so bent.

DBent ✓ KEY — this is the correct answer

Feedback if chosen: Correct. S has three domains — two bonds and one lone pair. The two bonded O atoms alone make a V shape: bent.

Check — multiple choice

A central atom has four electron domains, one of them a lone pair. Which molecular geometry does VSEPR predict?

ATetrahedral

Feedback if chosen: This option is wrong — you named the electron-domain geometry. All four domains are tetrahedral, but the molecular geometry keeps only the three bonded atoms: trigonal pyramidal.

BTrigonal pyramidal ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The three bonded atoms sit at three tetrahedral corners, forming a three-sided pyramid with the central atom at its top: trigonal pyramidal.

CBent

Feedback if chosen: This option is wrong — you took two lone pairs instead of one. One lone pair leaves three bonded atoms — trigonal pyramidal; bent needs only two bonded atoms.

DTrigonal planar

Feedback if chosen: This option is wrong — you flattened the three bonded atoms as if the lone pair took no position. The lone pair holds the fourth tetrahedral corner, so the three atoms form a pyramid, not a flat triangle.

Check — multiple choice

A central atom has four electron domains, two of them lone pairs. Which molecular geometry does VSEPR predict?

ABent ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The two bonded atoms sit at two of the four tetrahedral corners, making a V shape: bent.

BTetrahedral

Feedback if chosen: This option is wrong — you named the electron-domain geometry. All four domains are tetrahedral, but only the two bonded atoms count for the molecular geometry: bent.

CLinear

Feedback if chosen: This option is wrong — you spread the two bonded atoms to opposite sides as if there were only two domains. All four domains are placed first, so the two atoms sit at tetrahedral corners: bent.

DTrigonal pyramidal

Feedback if chosen: This option is wrong — you took one lone pair instead of two. Two lone pairs leave two bonded atoms — bent; trigonal pyramidal needs three bonded atoms.

U2-S12 · LESSON 110 · CSP-103

Axial and equatorial positions

Outcome: Classify a position in trigonal-bipyramidal electron-domain geometry as axial or equatorial.

Part 1Five positions, two kinds

You’ve already seen the trigonal bipyramidal arrangement: three domains in a flat triangle around the middle of the central atom, and two pointing straight up and straight down.

In the arrangements for two, three, and four domains, every position is like every other. Trigonal bipyramidal is different: its five positions come in two kinds.

The three positions in the flat triangle circle the middle of the central atom, like the equator circles the middle of the Earth — they are called equatorial positions.

The two positions perpendicular to that triangle — one straight up, one straight down, along the axis of the arrangement — are called axial positions.

In PCl₅, the Cl straight above P and the Cl straight below P sit at the axial positions; the other three Cl atoms sit at the equatorial positions.

What you are expected to be able to do

Classify a position in trigonal-bipyramidal electron-domain geometry as axial or equatorial.

Part 2Check what you just learned

Check — multiple choice

In a trigonal bipyramidal arrangement, which positions are the axial ones?

AThe three positions in the flat triangle around the middle

Feedback if chosen: This option is wrong — you swapped the two labels. The triangle positions are the equatorial ones; axial names the up-and-down pair on the axis.

BAll five positions — they are interchangeable

Feedback if chosen: This option is wrong — you treated the five positions as all alike. That holds for two to four domains, but trigonal bipyramidal has two kinds: three equatorial and two axial.

CThe two positions perpendicular to the triangle, straight up and straight down ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The axial positions lie on the axis of the arrangement — perpendicular to the equatorial triangle, one above and one below.

DWhichever two positions the largest atoms happen to occupy

Feedback if chosen: This option is wrong — you let the occupants define the position. Axial and equatorial name places in the arrangement, fixed by the geometry, not by which atoms sit there.

Check — numeric entry

How many equatorial positions does a trigonal bipyramidal arrangement have?

✓ ANSWER KEY: 3

Feedback when correct: Correct. The equatorial positions are the three in the flat triangle around the middle; the other two positions are axial.

Anticipated wrong answers (feedback shown):

2	This answer counts the axial pair. The equatorial positions are the three in the triangle around the middle.
5	This answer counts every position. Only the three in the triangle are equatorial; the up-and-down pair is axial.
4	This answer mixes the two kinds. Trigonal bipyramidal splits five positions into three equatorial plus two axial.

U2-S12 · LESSON 111 · CSP-104

Where a lone pair sits in a trigonal bipyramid

Outcome: Predict that a lone pair takes an equatorial position in trigonal-bipyramidal electron-domain geometry.

Part 1The two kinds of position get different crowding

You’ve already seen that trigonal bipyramidal positions come in two kinds: three equatorial, two axial.

The two kinds are crowded differently. A domain in an axial position has three close neighbors — the three equatorial domains.

A domain in an equatorial position has only two close neighbors — the two axial domains. The other two equatorial domains sit farther around the triangle.

A lone pair’s electron density spreads out more than a bond’s, so a lone pair repels its neighbors more strongly.

Putting the strongest repeller where it has the fewest close neighbors keeps the total repulsion lowest — so a lone pair takes an equatorial position before an axial one.

S in SF₄ has five domains: four bonds and one lone pair. The lone pair sits equatorial, in the triangle around the middle.

What you are expected to be able to do

Predict that a lone pair occupies an equatorial position before an axial position in trigonal-bipyramidal electron-domain geometry.

Part 2Check what you just learned

Check — multiple choice

Why does the lone pair in SF₄ take an equatorial position rather than an axial one?

AThe lone pair is pulled toward the axial F atoms

Feedback if chosen: This option is wrong — you placed the lone pair by attraction. Domains repel one another; the lone pair sits where its strong repulsion does the least: the less crowded equatorial position.

BAn equatorial domain has two close neighbors instead of three, so the strongly repelling lone pair causes less total repulsion there ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Equatorial means two close neighbors (the axial pair); axial means three (the equatorial triangle). The strongest repeller goes where it has the fewest close neighbors.

CA lone pair repels other domains more weakly than a bond does, so it takes whatever position is left over

Feedback if chosen: This option is wrong — you inverted the lone pair’s strength. A lone pair spreads out more and repels more strongly than a bond, which is exactly why it claims the roomier equatorial position.

DThe axial positions are reserved for the largest atoms

Feedback if chosen: This option is wrong — you assigned positions by atom size. Positions go by repulsion between domains, and the lone pair’s strong repulsion sends it equatorial.

Check — multiple choice

Cl in ClF₃ has five electron domains, two of them lone pairs. Which positions do the two lone pairs take?

ABoth axial, at the two ends of the axis

Feedback if chosen: This option is wrong — you sent the lone pairs to the more crowded kind of position. Each lone pair takes an equatorial position before an axial one, and that holds for both of them.

BOne axial and one equatorial, to balance the arrangement

Feedback if chosen: This option is wrong — you split the pair as a compromise. The rule applies to each lone pair in turn: equatorial before axial, so both go equatorial.

CEither — the five positions are all equivalent

Feedback if chosen: This option is wrong — you treated the five positions as all alike. Trigonal bipyramidal has two kinds of position, and the strongly repelling lone pairs take the less crowded equatorial kind.

DBoth equatorial, in the triangle around the middle ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each lone pair takes an equatorial position before an axial one, so both sit in the triangle around the middle.

U2-S12 · LESSON 112 · CSP-105

Molecular geometry with five domains

Outcome: Predict the molecular geometry for a central atom with five electron domains.

Part 1Five domains, four shapes

You’ve already seen the routine: place all the domains, put lone pairs equatorial, then name the arrangement of only the bonded atoms.

With no lone pairs, all five positions hold atoms and the molecular geometry keeps the electron-domain name: PCl₅ is trigonal bipyramidal.

One lone pair (equatorial) leaves four bonded atoms — two axial and two equatorial. That shape is called seesaw: tip it on its side and the two axial atoms are the plank, the two equatorial ones the legs. SF₄ is seesaw.

Two lone pairs (both equatorial) leave three bonded atoms — two axial and one equatorial. The three atoms trace a letter T: that shape is called T-shaped. ClF₃ is T-shaped.

Three lone pairs fill all three equatorial positions, leaving two bonded atoms at the two axial positions — on exactly opposite sides. That molecular geometry is linear. XeF₂ is linear.

Here are the four names in one table.

Molecular geometry, five domains
Lone pairs	Bonded atoms	Molecular geometry
0	5	trigonal bipyramidal
1	4	seesaw
2	3	T-shaped
3	2	linear

What you are expected to be able to do

Predict trigonal bipyramidal, seesaw, T-shaped, or linear molecular geometry for a central atom with five electron domains.

Part 2Check what you just learned

Check — multiple choice

A central atom has five electron domains, one of them a lone pair. Which molecular geometry does VSEPR predict?

ASeesaw ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The lone pair takes an equatorial position, leaving two axial and two equatorial bonded atoms — the seesaw shape.

BTrigonal bipyramidal

Feedback if chosen: This option is wrong — you named the electron-domain geometry. All five domains are trigonal bipyramidal, but the molecular geometry keeps only the four bonded atoms: seesaw.

CT-shaped

Feedback if chosen: This option is wrong — you took two lone pairs instead of one. One lone pair leaves four bonded atoms — seesaw; T-shaped needs three bonded atoms.

DTrigonal pyramidal

Feedback if chosen: This option is wrong — you reached for the four-domain family. Trigonal pyramidal comes from four domains with one lone pair; five domains with one lone pair give seesaw.

Check — multiple choice

A central atom has five electron domains, two of them lone pairs. Which molecular geometry does VSEPR predict?

ASeesaw

Feedback if chosen: This option is wrong — you took one lone pair instead of two. Two lone pairs leave three bonded atoms — T-shaped; seesaw needs four bonded atoms.

BTrigonal planar

Feedback if chosen: This option is wrong — you spread the three bonded atoms into a flat triangle of their own. Both lone pairs hold equatorial positions, so the atoms sit at two axial spots plus one equatorial — a T.

CT-shaped ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The two lone pairs go equatorial, leaving bonded atoms at both axial positions and one equatorial position — the three atoms trace a T.

DLinear

Feedback if chosen: This option is wrong — you took three lone pairs instead of two. Three lone pairs leave two bonded atoms — linear; with two lone pairs, three atoms remain: T-shaped.

Check — multiple choice

A central atom has five electron domains, three of them lone pairs. Which molecular geometry does VSEPR predict?

AT-shaped

Feedback if chosen: This option is wrong — you took two lone pairs instead of three. Three lone pairs leave only two bonded atoms — linear; T-shaped needs three bonded atoms.

BLinear ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The three lone pairs fill the equatorial triangle, so the two bonded atoms sit at the axial positions — exactly opposite each other: linear.

CBent

Feedback if chosen: This option is wrong — you bent the two bonded atoms as if a lone pair sat between them. All three lone pairs are equatorial, so the two atoms hold the two axial positions: a straight line.

DTrigonal bipyramidal

Feedback if chosen: This option is wrong — you named the electron-domain geometry. All five domains are trigonal bipyramidal, but only the two bonded atoms count for the molecular geometry: linear.

U2-S12 · LESSON 113 · CSP-106

Where two lone pairs sit in an octahedron

Outcome: Predict that two lone pairs take opposite positions in octahedral electron-domain geometry.

Part 1Two strong repellers, kept apart

You’ve already seen that in an octahedral arrangement all six positions are alike — so a single lone pair can sit anywhere.

With two lone pairs, the positions relative to each other start to matter.

You’ve already seen that a lone pair repels other domains more strongly than a bond does — and two lone pairs repel each other most strongly of all.

Two octahedral positions can be neighbors, or they can be exactly opposite each other, on either side of the central atom — as far apart as two positions get.

Placing the two lone pairs opposite each other keeps the strongest repulsion — lone pair against lone pair — as small as possible.

Xe in XeF₄ has six domains: four bonds and two lone pairs. The two lone pairs sit opposite each other, one above and one below, with the four F atoms around the middle.

What you are expected to be able to do

Predict that two lone pairs occupy opposite positions in octahedral electron-domain geometry, which keeps their mutual repulsion as small as possible.

Part 2Check what you just learned

Check — multiple choice

A central atom has octahedral electron-domain geometry with two lone pairs. Where does VSEPR place the two lone pairs?

ANext to each other, so the bonded atoms can spread out

Feedback if chosen: This option is wrong — you put the two strongest repellers side by side. Lone pair against lone pair is the strongest repulsion, so the pairs take positions as far apart as possible: opposite.

BOne in the square around the middle and one above it — the exact places never matter

Feedback if chosen: This option is wrong — you treated the pairs’ relative placement as free. Any single position is like any other, but two lone pairs must end up opposite each other to keep their mutual repulsion smallest.

CAs close to the central atom’s nucleus as possible, tucked between the bonds

Feedback if chosen: This option is wrong — you tucked the lone pairs between positions. Each lone pair is a full domain and holds a full octahedral position — and the two of them take opposite ones.

DAt two positions exactly opposite each other ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Opposite positions are as far apart as two octahedral positions get, which keeps the strong lone-pair-to-lone-pair repulsion as small as possible.

Check — multiple choice

Why do the two lone pairs in XeF₄ sit opposite each other?

ATwo lone pairs repel each other strongly, and opposite positions keep them as far apart as possible ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Lone pair against lone pair is the strongest repulsion in the arrangement, and opposite positions put the largest possible separation between them.

BThe two lone pairs attract each other, and opposite positions balance that pull

Feedback if chosen: This option is wrong — you gave the lone pairs an attraction they do not have. Both carry negative charge, so they repel — and land as far apart as possible.

CLone pairs must take equatorial positions, and the equatorial positions sit opposite one another

Feedback if chosen: This option is wrong — you carried the trigonal bipyramidal rule into the octahedron. An octahedral arrangement has no equatorial kind of position — all six are alike — so the pairs simply go opposite each other.

DOpposite positions bring the lone pairs closer to the F atoms

Feedback if chosen: This option is wrong — you placed the lone pairs by attraction to the F atoms. Placement follows repulsion between domains, and opposite positions minimize the lone-pair-to-lone-pair repulsion.

U2-S12 · LESSON 114 · CSP-107

Molecular geometry with six domains

Outcome: Predict the molecular geometry for a central atom with six electron domains.

Part 1Six domains, three shapes

You’ve already seen the routine, and for six domains the steps are the same: place all six domains, put two lone pairs opposite each other, then name the arrangement of only the bonded atoms.

With no lone pairs, all six positions hold atoms and the molecular geometry keeps the electron-domain name: SF₆ is octahedral.

One lone pair leaves five bonded atoms: four in a square with one more atom above its center — a pyramid on a square base. That shape is called square pyramidal. BrF₅ is square pyramidal.

Two lone pairs sit opposite each other, leaving four bonded atoms in a flat square around the central atom. That shape is called square planar. XeF₄ is square planar.

Here are the three names in one table.

Molecular geometry, six domains
Lone pairs	Bonded atoms	Molecular geometry
0	6	octahedral
1	5	square pyramidal
2	4	square planar

What you are expected to be able to do

Predict octahedral, square pyramidal, or square planar molecular geometry for a central atom with six electron domains.

Part 2Check what you just learned

Check — multiple choice

A central atom has six electron domains, one of them a lone pair. Which molecular geometry does VSEPR predict?

AOctahedral

Feedback if chosen: This option is wrong — you named the electron-domain geometry. All six domains are octahedral, but the molecular geometry keeps only the five bonded atoms: square pyramidal.

BSquare planar

Feedback if chosen: This option is wrong — you took two lone pairs instead of one. One lone pair leaves five bonded atoms — square pyramidal; square planar needs four.

CSquare pyramidal ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Five bonded atoms remain: four in a square plus one above its center — a pyramid on a square base.

DTrigonal bipyramidal

Feedback if chosen: This option is wrong — you moved to the five-domain family. Trigonal bipyramidal is the five-domain arrangement; six domains with one lone pair give square pyramidal.

Check — multiple choice

A central atom has six electron domains, two of them lone pairs. Which molecular geometry does VSEPR predict?

ASquare pyramidal

Feedback if chosen: This option is wrong — you took one lone pair instead of two. Two lone pairs leave four bonded atoms — square planar; square pyramidal needs five.

BSquare planar ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The two lone pairs sit opposite each other, so the four bonded atoms form a flat square around the central atom.

COctahedral

Feedback if chosen: This option is wrong — you named the electron-domain geometry. All six domains are octahedral, but the molecular geometry keeps only the four bonded atoms: square planar.

DSeesaw

Feedback if chosen: This option is wrong — you reached for the five-domain family. Seesaw comes from five domains with one lone pair; six domains with two lone pairs give square planar.

Check — numeric entry

A molecule’s central atom has six electron domains, and its molecular geometry is square pyramidal. How many of the six domains are lone pairs?

✓ ANSWER KEY: 1

Feedback when correct: Correct. Square pyramidal means five bonded atoms out of six positions, so exactly one domain is a lone pair.

Anticipated wrong answers (feedback shown):

0	This answer keeps all six positions bonded, which is octahedral. Square pyramidal has five bonded atoms, so one domain is a lone pair.
2	This answer leaves only four bonded atoms, which is square planar. Square pyramidal has five bonded atoms, so one lone pair.
5	This answer counts the bonded atoms. Five of the six domains hold atoms, so the lone pairs number 6 − 5 = 1.

U2-S13 — Bond angles and VSEPR synthesis

11 lessons

U2-S13 · LESSON 115 · CSP-108

What a bond angle is

Outcome: Identify the bond angle between two bonds that share a central atom, and name it by its three atoms.

Part 1The angle between two bonds

You’ve already seen that molecular geometry describes the arrangement of the bonded atoms around a central atom.

Here is the geometry of H₂O; the marked arc is the angle between its two O–H bonds.

An angle like this — between two bonds that share a central atom — is called a bond angle.

A bond angle is named by three atoms: the two outer atoms, with the shared central atom written in the middle.

The marked angle is the H–O–H bond angle: two H atoms on the outside, the shared O in the middle.

Every pair of bonds that meets at the same central atom defines one bond angle — NH₃, with three N–H bonds, has three H–N–H bond angles.

A lone pair is not a bond, so no bond angle is measured to a lone pair.

What you are expected to be able to do

Identify and name the bond angle between two bonds that share a central atom in a supplied molecule.

Part 2Check what you just learned

Check — multiple choice

What is a bond angle?

AThe angle between a bond and a lone pair on the central atom

Feedback if chosen: This option is wrong — you measured to a lone pair. A lone pair is not a bond; a bond angle is measured between two bonds.

BThe distance between two atoms joined by a bond

Feedback if chosen: This option is wrong — you described a length, not an angle. The distance between two bonded atoms is the bond length; a bond angle is measured between two bonds that meet at the same central atom.

CThe angle between any two bonds anywhere in a molecule

Feedback if chosen: This option is wrong — you dropped the shared central atom. Two bonds define a bond angle only when they meet at the same central atom.

DThe angle between two bonds that share a central atom ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Two bonds meeting at the same central atom define one bond angle, named with that central atom in the middle.

Check — multiple choice

Use the geometry of NH₃ shown. Which of these is a bond angle in NH₃?

AThe N–H bond length

Feedback if chosen: This option is wrong — you named a length, not an angle. The N–H bond length is a distance between two atoms; a bond angle is measured between two bonds that meet at N.

BThe angle between the lone pair and an N–H bond

Feedback if chosen: This option is wrong — you measured to the lone pair. The lone pair is not a bond; the bond angles in NH₃ are measured between two N–H bonds.

CThe H–N–H angle ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Two N–H bonds share the central N, so the angle between them is the H–N–H bond angle.

DThe N–H–H angle

Feedback if chosen: This option is wrong — you named the atoms in the wrong order. The shared central atom goes in the middle of the name; in NH₃ every bond meets at N, so the angle is H–N–H, not N–H–H.

Check — multiple choice

In the Lewis diagram of CH₃Cl shown, between which two bonds is an H–C–Cl bond angle measured?

ABetween two C–H bonds

Feedback if chosen: This option is wrong — you chose the bonds of an H–C–H angle. The name H–C–Cl lists the outer atoms H and Cl, so the two bonds are one C–H bond and the C–Cl bond.

BBetween an H atom and the Cl atom, measured straight across

Feedback if chosen: This option is wrong — you measured between two atoms instead of two bonds. Straight across from H to Cl is a distance; the bond angle sits at C, between the two bonds that share that C.

CBetween the C–Cl bond and a lone pair on Cl

Feedback if chosen: This option is wrong — you measured to a lone pair. The lone pairs on Cl are not bonds; the angle is measured between the C–H and C–Cl bonds.

DBetween a C–H bond and the C–Cl bond ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The name lists the outer atoms H and Cl around the shared central C, so the angle sits at C, between a C–H bond and the C–Cl bond.

U2-S13 · LESSON 116 · CSP-109

The ideal angle for linear geometry

Outcome: State that linear electron-domain geometry has an ideal bond angle of 180°, and assign it in a molecule.

Part 1Two domains, opposite directions

You’ve already seen the Lewis diagram of CO₂.

Each double bond counts as one electron domain around C, so C has two electron domains.

You’ve already seen that two electron domains give linear electron-domain geometry.

Here is that geometry with its bond angle marked.

The two domains repel each other and move as far apart as they can, so they end up pointing in opposite directions — a straight line through C.

Opposite directions along a straight line are 180° apart, so the O–C–O bond angle is 180°.

The angle a geometry sets by itself is called the ideal bond angle: for linear electron-domain geometry, the ideal bond angle is 180°.

What you are expected to be able to do

State that linear electron-domain geometry has an ideal bond angle of 180°, and assign 180° to the bond angle in a molecule with two electron domains.

Part 2Check what you just learned

Check — numeric entry

Use the Lewis diagram of BeCl₂ shown. What is the ideal Cl–Be–Cl bond angle, in degrees?

✓ ANSWER KEY: 180 degrees

Feedback when correct: Correct. Be has two electron domains, which point in opposite directions along a straight line — and opposite directions are 180° apart.

Anticipated wrong answers (feedback shown):

90	This answer takes a right angle. Two electron domains move as far apart as they can, which is opposite directions along a straight line: 180°.
120	This is the ideal angle for three electron domains sharing a flat circle. Be has two domains, and two domains sit opposite each other: 180°.
360	A full circle is 360°. The two bonds point in opposite directions — half a circle apart — so the angle between them is 180°.

Check — multiple choice

Use the Lewis diagram of CS₂ shown. What is the ideal S–C–S bond angle?

A90°

Feedback if chosen: This option is wrong — you assumed the two domains sit at right angles. Repulsion pushes them as far apart as possible: opposite directions, 180°.

B109.5°

Feedback if chosen: This option is wrong — you counted each double bond as two domains. A bond of any order counts once, so C has two electron domains: linear, 180°.

C120°

Feedback if chosen: This option is wrong — you counted the molecule’s three atoms instead of the central atom’s domains. C has two electron domains: linear, 180°.

D180° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each S=C double bond counts as one electron domain, so C has two domains pointing in opposite directions: 180°.

Check — multiple choice

Why is the ideal bond angle for two electron domains 180°?

ATwo domains repel each other, and opposite directions put them as far apart as possible ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Electron domains repel one another, and for two domains the farthest-apart arrangement is opposite directions — 180°.

BTwo domains attract each other until they line up on the same side of the atom

Feedback if chosen: This option is wrong — you replaced repulsion with attraction. Domains are regions of negative charge and repel one another; repulsion pushes them to opposite sides of the central atom.

CThe two terminal atoms pull the bonds until they lie in a straight line

Feedback if chosen: This option is wrong — you gave the terminal atoms the deciding role. The arrangement is set by the electron domains repelling one another around the central atom.

DElectron domains always start at right angles and then swing apart slightly

Feedback if chosen: This option is wrong — you started the domains at right angles. Nothing holds domains at 90°; they move as far apart as repulsion allows, and for two domains that is 180°.

U2-S13 · LESSON 117 · CSP-110

The ideal angles for trigonal-planar geometry

Outcome: State that trigonal-planar electron-domain geometry has ideal bond angles of 120°, and assign them in a molecule.

Part 1Three domains share a flat circle

You’ve already seen the Lewis diagram of BF₃.

B has three bonding domains and no lone pairs: three electron domains.

You’ve already seen that three electron domains give trigonal-planar electron-domain geometry.

Here is that geometry with a bond angle marked.

The three domains repel one another equally and spread out in one flat plane, sharing the full circle around B.

A full circle is 360°, split three ways: 360° ÷ 3 = 120°.

So trigonal-planar electron-domain geometry has ideal bond angles of 120° — each F–B–F angle in BF₃ is 120°.

What you are expected to be able to do

State that trigonal-planar electron-domain geometry has ideal bond angles of 120°, and assign 120° to a bond angle in a molecule with three electron domains.

Part 2Check what you just learned

Check — multiple choice

Use the Lewis diagram of BCl₃ shown. What is the ideal Cl–B–Cl bond angle?

A90°

Feedback if chosen: This option is wrong — you assumed right angles. Three domains share the full flat circle equally: 360° ÷ 3 = 120°.

B109.5°

Feedback if chosen: This option is wrong — that is the ideal angle for four electron domains. B has three bonding domains and no lone pairs: three domains, 120°.

C120° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. B has three electron domains — trigonal planar — and three domains split the 360° circle equally: 120°.

D180°

Feedback if chosen: This option is wrong — that is the ideal angle for two electron domains. B has three, and three domains sharing the circle sit 120° apart.

Check — numeric entry

Every ideal bond angle in BF₃ is 120°. How many electron domains surround the central B atom?

✓ ANSWER KEY: 3

Feedback when correct: Correct. Ideal 120° angles mean the domains split the full 360° circle equally: 360° ÷ 120° = 3 electron domains.

Anticipated wrong answers (feedback shown):

2	This answer counts one fewer domain than atoms attached, as if the bonds formed a chain. Every one of the three F atoms bonds straight to B, and each bond is its own domain: 3 domains.
4	This answer counts every atom in the molecule, including the central B. Domains are counted around the central atom, not including it: three B–F bonds, 3 domains.
6	This answer counts the bonding electrons — each of the three bonds holds 2 electrons. A domain is a whole region, not a single electron: three bonds, 3 domains.

Check — multiple choice

Why is the ideal bond angle for three electron domains 120°?

AThree bonds drawn to the corners of a triangle make 60° angles at each corner

Feedback if chosen: This option is wrong — you measured the corners of the triangle instead of the angles at the central atom. The three bonds radiate from the center, splitting the 360° circle into 120° each.

B120° is halfway between the two-domain angle and a right angle

Feedback if chosen: This option is wrong — you averaged other geometries’ angles (and halfway between 180° and 90° is 135°, not 120°). The 120° comes from splitting the full circle three ways.

CThe three domains attract one another until they settle into a plane

Feedback if chosen: This option is wrong — you replaced repulsion with attraction. Domains repel one another, and that repulsion is what spreads them 120° apart.

DThe three domains spread into one flat plane and share the full 360° circle equally ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Three mutually repelling domains get farthest apart in one flat plane, splitting the full circle three ways: 360° ÷ 3 = 120°.

U2-S13 · LESSON 118 · CSP-111

The ideal angles for tetrahedral geometry

Outcome: State that tetrahedral electron-domain geometry has ideal bond angles of 109.5°, and assign them in a molecule.

Part 1Four domains leave the flat page

You’ve already seen the Lewis diagram of CH₄: four bonding domains around C and no lone pairs — four electron domains.

On the page the four bonds are drawn 90° apart, but the flat drawing records bonds and electrons — not the molecule’s shape.

You’ve already seen that four electron domains give tetrahedral electron-domain geometry.

Here is that geometry with a bond angle marked — the solid wedge comes toward you, and the dashed wedge points back behind the page.

In three dimensions the four domains spread to the corners of a tetrahedron — farther apart than any flat arrangement allows.

The angle between any two of the four bonds is then 109.5°, not 90°.

So tetrahedral electron-domain geometry has ideal bond angles of 109.5° — each H–C–H angle in CH₄ is 109.5°.

109.5° is not a round number; it is simply the angle the corners of a tetrahedron make at its center.

What you are expected to be able to do

State that tetrahedral electron-domain geometry has ideal bond angles of 109.5°, and assign 109.5° to a bond angle in a molecule with four electron domains.

Part 2Check what you just learned

Check — multiple choice

Use the Lewis diagram of SiCl₄ shown. What is the ideal Cl–Si–Cl bond angle?

A90°

Feedback if chosen: This option is wrong — you read the angle off the flat drawing. The molecule is three-dimensional: four domains spread to the corners of a tetrahedron, 109.5° apart.

B109.5° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Si has four electron domains — tetrahedral — so the ideal angle between any two Si–Cl bonds is 109.5°.

C120°

Feedback if chosen: This option is wrong — that is the ideal angle for three electron domains. Si has four domains: tetrahedral, 109.5°.

D180°

Feedback if chosen: This option is wrong — that is the ideal angle for two electron domains. Si has four domains: tetrahedral, 109.5°.

Check — multiple choice

Use the Lewis diagram of the ammonium ion, NH₄⁺, shown. A student claims the ideal H–N–H angle is 90°, because the drawing shows the four bonds at right angles. Which evaluation of the claim is correct?

AThe claim is correct — the Lewis diagram shows the true angles between the bonds

Feedback if chosen: This option is wrong — you read the flat drawing as the real shape. A Lewis diagram shows connections, not 3-D geometry; the four domains spread into three dimensions, 109.5° apart.

BThe claim is wrong — the ion is three-dimensional, and its four domains spread to tetrahedron corners, 109.5° apart ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The drawing is flat, but the ion is not: its four electron domains spread to the corners of a tetrahedron, so the ideal H–N–H angle is 109.5°.

CThe claim is wrong — the ideal angle is 120°, because the three N–H bonds share the flat circle equally

Feedback if chosen: This option is wrong — you miscounted the domains. N here has four bonding domains, not three; 120° is the ideal angle for three domains, and four give 109.5°.

DThe claim is wrong — the ideal angle is 180°, because domains always spread as far apart as they can get

Feedback if chosen: This option is wrong — you applied the two-domain result to four domains. Only two domains can sit fully opposite at 180°; four domains settle at 109.5°.

Check — multiple choice

The Lewis diagram of CH₄ draws its four bonds 90° apart, yet the ideal H–C–H angle is 109.5°. Why?

AThe four domains spread into three dimensions, which puts them farther apart than any flat arrangement ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Leaving the flat page lets the four domains reach the corners of a tetrahedron, 109.5° apart — farther apart than the 90° of a flat arrangement.

BThe flat drawing is accurate, and the molecule really has 90° angles

Feedback if chosen: This option is wrong — you trusted the drawing’s angles. A Lewis diagram records bonds and electrons, not shape; the molecule is three-dimensional.

CThe H atoms attract one another and pull the angles open beyond 90°

Feedback if chosen: This option is wrong — you replaced repulsion between electron domains with attraction between atoms. The domains repel, and spreading into three dimensions is how they get farthest apart.

DLone pairs on the H atoms push the four bonds apart

Feedback if chosen: This option is wrong — you gave H lone pairs. Bonded H has none; the spreading comes from the four bonding domains repelling one another.

U2-S13 · LESSON 119 · CSP-112

The ideal angles for five domains

Outcome: State the ideal angles in trigonal-bipyramidal geometry — 90°, 120°, and 180° — and assign the right one to a pair of bonds.

Part 1Five domains, two kinds of position

You’ve already seen the Lewis diagram of PCl₅: five bonding domains around P and no lone pairs — five electron domains.

You’ve already seen that five electron domains give trigonal-bipyramidal electron-domain geometry.

You’ve also seen its two kinds of position: three equatorial positions around the middle, and two axial positions, one above and one below.

Here is that geometry with two of its bond angles marked.

The three equatorial bonds share a flat circle around P, so two equatorial bonds are 120° apart.

Each axial bond stands perpendicular to that circle, so an axial bond and an equatorial bond are 90° apart.

The two axial bonds point in opposite directions, so they are 180° apart.

So trigonal-bipyramidal electron-domain geometry contains ideal angles of 90°, 120°, and 180° — which value applies depends on which pair of bonds you pick.

What you are expected to be able to do

State the ideal angles in trigonal-bipyramidal electron-domain geometry — 90°, 120°, and 180° — and assign the right one to a chosen pair of bonds.

Part 2Check what you just learned

Check — multiple choice

Use the geometry of PF₅ shown. What is the ideal angle between an axial P–F bond and an equatorial P–F bond?

A90° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. An axial bond stands perpendicular to the flat circle of equatorial bonds, so the axial–equatorial angle is 90°.

B109.5°

Feedback if chosen: This option is wrong — that is the ideal angle for four electron domains. PF₅ has five, and an axial bond stands perpendicular to the equatorial circle: 90°.

C120°

Feedback if chosen: This option is wrong — 120° is the angle between two equatorial bonds. An axial bond stands perpendicular to the equatorial circle: 90°.

D180°

Feedback if chosen: This option is wrong — 180° is the angle between the two axial bonds. An axial and an equatorial bond are 90° apart.

Check — numeric entry

In the PF₅ geometry shown, what is the ideal angle between the two axial P–F bonds, in degrees?

✓ ANSWER KEY: 180 degrees

Feedback when correct: Correct. The two axial bonds point in opposite directions — one straight up, one straight down — and opposite directions are 180° apart.

Anticipated wrong answers (feedback shown):

90	That is the angle between an axial and an equatorial bond. The two axial bonds point in opposite directions: 180°.
120	That is the angle between two equatorial bonds. The two axial bonds point in opposite directions: 180°.
360	A full circle is 360°. The two axial bonds point in opposite directions — half a circle apart — so the angle between them is 180°.

Check — numeric entry

AsF₅, shown, also has five electron domains. What is the ideal angle between two equatorial As–F bonds, in degrees?

✓ ANSWER KEY: 120 degrees

Feedback when correct: Correct. The three equatorial bonds share a flat circle around As: 360° ÷ 3 = 120°.

Anticipated wrong answers (feedback shown):

72	This answer spreads all five domains around one flat circle (360° ÷ 5). Only the three equatorial bonds share the circle; the other two point straight up and down. 360° ÷ 3 = 120°.
90	That is the angle between an axial and an equatorial bond. Two equatorial bonds share the flat circle: 120°.
180	That is the angle between the two axial bonds. Two equatorial bonds share the flat circle: 120°.

U2-S13 · LESSON 120 · CSP-113

The ideal angles for six domains

Outcome: State the ideal angles in octahedral geometry — 90° and 180° — and assign the right one to a pair of bonds.

Part 1Six domains, all positions alike

You’ve already seen the Lewis diagram of SF₆: six bonding domains around S and no lone pairs — six electron domains.

You’ve already seen that six electron domains give octahedral electron-domain geometry.

Here is that geometry with one bond angle marked.

Four bonds form a square around the middle, one bond points straight up, and one points straight down — and all six positions are equivalent.

Neighboring bonds are 90° apart: around the square, and from the square to the bond above or below.

Each bond also has one bond directly opposite it, and opposite bonds are 180° apart — in SF₆, each S–F bond sits 180° from the S–F bond across from it.

So octahedral electron-domain geometry contains ideal angles of 90° between neighboring bonds and 180° between opposite bonds.

There is no 120° anywhere in this geometry: no three of its bonds are spread evenly around a flat circle.

What you are expected to be able to do

State the ideal angles in octahedral electron-domain geometry — 90° and 180° — and assign the right one to a chosen pair of bonds.

Part 2Check what you just learned

Check — numeric entry

SeF₆, shown, has six electron domains around Se. What is the ideal angle between two neighboring Se–F bonds, in degrees?

✓ ANSWER KEY: 90 degrees

Feedback when correct: Correct. Four bonds form a square and two point straight up and down; every pair of neighboring bonds is 90° apart.

Anticipated wrong answers (feedback shown):

60	This answer spreads all six domains around one flat circle (360° ÷ 6). They do not share one circle — four form a square and two point up and down — so neighbors are 90° apart.
120	That angle belongs to trigonal-bipyramidal geometry’s equatorial bonds. Octahedral geometry contains only 90° and 180°; neighbors are 90° apart.
109	That is the ideal angle for four electron domains (109.5°). Six domains give octahedral geometry, where neighboring bonds are 90° apart.
180	That is the angle between opposite bonds. Neighboring bonds are 90° apart.

Check — numeric entry

In the SeF₆ geometry shown, what is the ideal angle between two opposite Se–F bonds, in degrees?

✓ ANSWER KEY: 180 degrees

Feedback when correct: Correct. Opposite bonds point in opposite directions through Se, and opposite directions are 180° apart.

Anticipated wrong answers (feedback shown):

90	That is the angle between neighboring bonds. Opposite bonds point in opposite directions through Se: 180°.
120	Octahedral geometry contains no 120° angles. Opposite bonds point in opposite directions: 180°.
360	A full circle is 360°. Opposite bonds are half a circle apart: 180°.

Check — multiple choice

Which set lists every ideal bond angle found in octahedral electron-domain geometry?

A90° only

Feedback if chosen: This option is wrong — you dropped the opposite pairs. Each bond has a partner directly opposite it, 180° away, so 180° occurs too.

B90° and 180° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Neighboring bonds are 90° apart and opposite bonds are 180° apart — and no other angle occurs.

C90°, 120°, and 180°

Feedback if chosen: This option is wrong — you carried 120° over from trigonal-bipyramidal geometry. No three octahedral bonds are spread evenly around a flat circle, so 120° never occurs.

D109.5° and 90°

Feedback if chosen: This option is wrong — 109.5° is the four-domain (tetrahedral) ideal angle, not an octahedral one. Octahedral geometry has 90° and 180°.

U2-S13 · LESSON 121 · CSP-114

Ranking electron-domain repulsions

Outcome: Rank electron-domain repulsions from strongest to weakest, and identify each type in a molecule.

Part 1Not all domains push equally

You’ve already seen that the electron domains around a central atom repel one another.

You’ve also seen the two kinds of electron pair: a bonding pair shared between two atoms, and a lone pair that stays with one atom.

Here is H₂O: two bonding pairs and two lone pairs around O.

A bonding pair is pulled out between two nuclei, so its charge sits partly away from the central atom.

A lone pair is held by the central atom alone, so its charge stays close to the central atom and spreads wider.

The closer and wider a domain’s charge, the harder it pushes on its neighbors.

That gives a ranking, strongest to weakest: lone pair–lone pair, then lone pair–bonding pair, then bonding pair–bonding pair.

In H₂O, the strongest repulsion is between the two lone pairs, and the weakest is between the two bonding pairs.

What you are expected to be able to do

Rank electron-domain repulsions from strongest to weakest — lone pair–lone pair, lone pair–bonding pair, bonding pair–bonding pair — and identify each type in a molecule.

Part 2Check what you just learned

Check — multiple choice

Which ranking of electron-domain repulsions is correct, from strongest to weakest?

Abonding pair–bonding pair > lone pair–bonding pair > lone pair–lone pair

Feedback if chosen: This option is wrong — you reversed the ranking. A lone pair’s charge stays close to the central atom and spreads wider, so it pushes hardest: lone pair–lone pair is the strongest pairing.

Blone pair–lone pair > bonding pair–bonding pair > lone pair–bonding pair

Feedback if chosen: This option is wrong — you swapped the middle and the bottom. A pairing with one lone pair pushes harder than a pairing with none, so lone pair–bonding pair outranks bonding pair–bonding pair.

Clone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The more lone pairs in the pairing, the stronger the repulsion: lone pair–lone pair, then lone pair–bonding pair, then bonding pair–bonding pair.

DAll three types of repulsion are equally strong

Feedback if chosen: This option is wrong — you treated every domain as identical. A lone pair’s charge sits closer to the central atom and spreads wider than a bonding pair’s, so pairings with lone pairs push harder.

Check — multiple choice

Use the geometry of NH₃ shown: three bonding pairs and one lone pair around N. Which repulsion between two of these domains is the strongest one present?

AA lone pair–lone pair repulsion

Feedback if chosen: This option is wrong — you picked a pairing NH₃ does not have. With only one lone pair, no lone pair–lone pair repulsion exists here; the strongest present is lone pair–bonding pair.

BA lone pair–bonding pair repulsion ✓ KEY — this is the correct answer

Feedback if chosen: Correct. NH₃ has one lone pair and three bonding pairs, so the strongest repulsion present is between the lone pair and a bonding pair.

CA bonding pair–bonding pair repulsion

Feedback if chosen: This option is wrong — bonding pair–bonding pair is the weakest type in the ranking. The lone pair–bonding pair repulsions here are stronger.

DNone is strongest — all the repulsions in NH₃ are equal

Feedback if chosen: This option is wrong — you treated every domain as identical. The lone pair’s charge sits closer to N and spreads wider, so it pushes the bonding pairs harder than they push one another.

Check — multiple choice

Bonds to the central atom of four molecules
Molecule	Bonds to the central atom
CH₄	4
BF₃	3
NH₃	3
H₂O	2

Use the table shown. Work out how many lone pairs each central atom holds. In which molecule does a lone pair–lone pair repulsion occur around the central atom?

ACH₄

Feedback if chosen: This option is wrong — CH₄’s central atom holds no lone pairs, so no pairing around it involves a lone pair at all.

BBF₃

Feedback if chosen: This option is wrong — BF₃’s central atom holds no lone pairs, so no pairing around it involves a lone pair at all.

CNH₃

Feedback if chosen: This option is wrong — one lone pair cannot pair with itself. NH₃’s strongest repulsion is lone pair–bonding pair; a lone pair–lone pair repulsion needs two lone pairs.

DH₂O ✓ KEY — this is the correct answer

Feedback if chosen: Correct. H₂O’s central O holds two lone pairs, so a lone pair–lone pair repulsion occurs between them.

U2-S13 · LESSON 122 · CSP-115

How lone pairs squeeze bond angles

Outcome: Predict that lone-pair repulsion generally compresses neighboring bond angles below the ideal angle.

Part 1Stronger pushes close the angle

You’ve already seen the repulsion ranking, strongest to weakest: lone pair–lone pair, lone pair–bonding pair, bonding pair–bonding pair.

Here is NH₃: four electron domains around N, one of them a lone pair.

With four electron domains, the ideal bond angle is 109.5°.

The lone pair repels each N–H bonding pair more strongly than the bonding pairs repel one another.

So the three N–H bonds are pushed slightly closer together, and each H–N–H angle closes to slightly less than 109.5°.

The measured H–N–H angle is about 107°.

The model predicts the direction, not the number: from the diagram alone, predict the angle as less than 109.5° — do not invent an exact value.

Here is H₂O: four electron domains around O, two of them lone pairs.

Its two lone pairs push the two O–H bonds together, and the measured H–O–H angle is about 104.5° — again, the prediction is simply less than 109.5°.

The squeeze is local: a lone pair generally compresses the bond angles next to it, not every angle in the molecule.

What you are expected to be able to do

Predict that lone-pair repulsion generally compresses neighboring bond angles to slightly less than the ideal angle.

Part 2Check what you just learned

Check — multiple choice

Use the Lewis diagram of NF₃ shown. What is the best prediction for the F–N–F bond angle?

AExactly 109.5°

Feedback if chosen: This option is wrong — you ignored the lone pair’s stronger repulsion. It pushes the three N–F bonding pairs together, closing the angle below 109.5°.

BLess than 109.5° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Four electron domains set the ideal at 109.5°, and the lone pair repels the bonding pairs more strongly, squeezing the N–F bonds together: below 109.5°.

CMore than 109.5°

Feedback if chosen: This option is wrong — you pushed the angle the wrong way. The lone pair repels the bonding pairs more strongly, closing the angle, not opening it.

DExactly 120°

Feedback if chosen: This option is wrong — you counted only the bonded atoms. The lone pair counts as a domain too: four domains, ideal 109.5°, squeezed below it.

Check — multiple choice

Why is the H–O–H angle in H₂O smaller than the 109.5° ideal?

AThe lone pairs repel the O–H bonding pairs more strongly than the bonding pairs repel each other, pushing the two bonds together ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The lone pairs push harder on the bonding pairs than the bonding pairs push on each other, so the two O–H bonds close together — slightly below 109.5°.

BThe lone pairs attract the O–H bonding pairs, drawing the two bonds toward themselves

Feedback if chosen: This option is wrong — you replaced repulsion with attraction. Electron pairs are regions of negative charge and repel one another; the stronger push from the lone pairs is what closes the angle.

CThe two H atoms attract each other, drawing the two bonds together

Feedback if chosen: This option is wrong — you moved the cause to the atoms. The angle is set by the electron domains repelling one another, not by attraction between the H atoms.

DO–H bonds are shorter than most bonds, and shorter bonds make smaller angles

Feedback if chosen: This option is wrong — you tied the angle to bond length. The squeeze comes from the lone pairs repelling the bonding pairs more strongly, not from how long the bonds are.

Check — multiple choice

NH₃ and H₂O each have four electron domains around the central atom: N in NH₃ holds one lone pair, and O in H₂O holds two. Which species has the smaller bond angle?

ANH₃, because its three bonding pairs crowd one another more than H₂O’s two

Feedback if chosen: This option is wrong — you blamed the bonding pairs. Bonding pair–bonding pair repulsion is the weakest kind; the squeeze comes from lone pairs, and H₂O has more of them.

BH₂O, because its two lone pairs repel the bonding pairs more strongly than NH₃’s one lone pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Lone pairs repel bonding pairs more strongly than bonding pairs repel each other, so two lone pairs squeeze the bonds harder than one: H₂O’s angle (about 104.5°) closes further below 109.5° than NH₃’s (about 107°).

CH₂O, because O is a smaller atom than N, and a smaller central atom makes smaller angles

Feedback if chosen: This option is wrong — you picked the right species for the wrong reason. The extra squeeze comes from H₂O’s second lone pair, not from the size of the O atom.

DNeither — both have four electron domains, so both bond angles are exactly 109.5°

Feedback if chosen: This option is wrong — you stopped at the ideal angle. Lone pairs repel bonding pairs more strongly than bonding pairs repel each other, so both angles close below 109.5° — and H₂O’s two lone pairs close it further.

U2-S13 · LESSON 123 · CSP-116

Double and triple bonds push harder

Outcome: State that a multiple-bond domain repels neighboring domains more strongly than a single-bond domain.

Part 1One domain, more electrons

You’ve already seen that a single, double, or triple bond to an atom counts as one electron domain.

Counting once does not mean pushing equally.

Here is the Lewis diagram of formaldehyde, H₂CO.

Around C there are three electron domains: the C=O double bond and two C–H single bonds.

A single bond holds two shared electrons; a double bond holds four.

More shared electrons means more negative charge in the domain, so the C=O domain repels its neighboring domains more strongly than either C–H domain does.

A triple bond holds six shared electrons and pushes harder still.

What you are expected to be able to do

State that a double- or triple-bond domain repels neighboring domains more strongly than a single-bond domain, and identify the most repulsive domain around an atom.

Part 2Check what you just learned

Check — multiple choice

Use the Lewis diagram of COCl₂ shown. Which electron domain around C repels its neighboring domains most strongly?

AThe C=O double-bond domain ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The C=O domain holds four shared electrons — more negative charge than a two-electron single bond — so it pushes hardest on its neighbors.

BA C–Cl single-bond domain

Feedback if chosen: This option is wrong — a C–Cl single bond holds only two shared electrons. The C=O domain holds four, so it carries more charge and pushes harder.

CThe lone pairs on the O atom

Feedback if chosen: This option is wrong — those lone pairs sit on O, not around C. They are not electron domains around the C atom.

DNone — all three domains around C repel equally

Feedback if chosen: This option is wrong — you treated ‘counts as one domain’ as ‘pushes equally’. Counting sets the geometry; the strength of the push follows the electrons, and the double bond holds four.

Check — multiple choice

Why does a double-bond domain repel its neighbors more strongly than a single-bond domain?

ABecause a double bond counts as two electron domains around the atom

Feedback if chosen: This option is wrong — a double bond still counts as one domain. What changes is how much charge that one domain holds: four shared electrons instead of two.

BBecause a double bond is longer than a single bond, so it sweeps out more space

Feedback if chosen: This option is wrong — you tied the push to length, and in the wrong direction anyway: double bonds are shorter than single bonds between the same atoms. The push comes from the extra shared electrons.

CBecause it holds four shared electrons instead of two, giving it more negative charge ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Four shared electrons carry more negative charge than two, and more charge pushes neighboring domains harder.

DBecause the atom at the far end of a double bond is always larger than one at a single bond

Feedback if chosen: This option is wrong — the terminal atom’s size is not the cause. The domain’s own electron count is what sets the strength of its push.

Check — multiple choice

Use the Lewis diagram of HCN shown. Which domain around C repels its neighbor more strongly?

AThe C–H domain, because H is the smaller atom

Feedback if chosen: This option is wrong — you tied the push to the terminal atom’s size. Repulsion follows the electrons in the domain, and the C≡N domain holds six shared electrons to the C–H domain’s two.

BThe C≡N domain, because it holds six shared electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A triple bond holds six shared electrons — more negative charge than the two in the C–H single bond — so the C≡N domain pushes harder.

CNeither — the two domains repel equally, because each counts as one domain

Feedback if chosen: This option is wrong — you treated ‘counts as one domain’ as ‘pushes equally’. Counting sets the geometry; the strength of the push follows how many electrons the domain holds.

DThe C≡N domain, because a triple bond counts as three domains

Feedback if chosen: This option is wrong — right domain, wrong reason: a triple bond still counts as one domain. It pushes harder because it holds six shared electrons.

U2-S13 · LESSON 124 · CSP-117

How multiple bonds reshape bond angles

Outcome: Predict that a multiple-bond domain can open its neighboring bond angles slightly above the ideal, closing the remaining angle slightly below it.

Part 1The push shows up in the angles

You’ve already seen that the C=O domain in formaldehyde, H₂CO, repels its neighboring domains more strongly than the C–H domains repel each other.

Three electron domains around C set the ideal bond angle at 120°.

The C=O domain pushes both C–H bonds away from itself, so each O–C–H angle can open to slightly more than 120°.

The three angles around C lie in one plane and must still total 360°, so the H–C–H angle between the single bonds closes to slightly less than 120°.

Measured: the O–C–H angles are about 122°, and the H–C–H angle is about 116°.

The prediction stays qualitative: next to a multiple bond, expect slightly more than the ideal; between the single bonds it pushes together, expect slightly less.

What you are expected to be able to do

Predict that a multiple-bond domain can open its neighboring bond angles slightly above the ideal angle, closing the remaining angle slightly below it.

Part 2Check what you just learned

Check — multiple choice

Use the Lewis diagram of COCl₂ shown. What is the best prediction for the Cl–C–Cl bond angle?

ALess than 120° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The C=O domain pushes both C–Cl bonds away from itself, so the Cl–C–Cl angle between the single bonds closes below 120°.

BExactly 120°

Feedback if chosen: This option is wrong — you treated the three domains as pushing equally. The four-electron C=O domain pushes harder, reshaping the angles around C.

CMore than 120°

Feedback if chosen: This option is wrong — you opened the wrong angle. The angles next to the C=O open; the Cl–C–Cl angle between the two single bonds is the one that closes.

DExactly 109.5°

Feedback if chosen: This option is wrong — you counted four domains. The double bond counts once: three domains, ideal 120°, adjusted by the stronger C=O push.

Check — multiple choice

In ethene, H₂C=CH₂, shown, each C has three electron domains. What is the best prediction for each H–C–H bond angle?

AExactly 120°

Feedback if chosen: This option is wrong — you treated the three domains as pushing equally. The four-electron C=C domain pushes the two C–H bonds harder than they push each other.

BSlightly more than 120°

Feedback if chosen: This option is wrong — you opened the wrong angle. The H–C=C angles next to the double bond open; the H–C–H angle between the single bonds closes.

CSlightly less than 120° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The C=C domain pushes both C–H bonds away from itself, closing the H–C–H angle between them to slightly less than 120°.

DSlightly less than 109.5°

Feedback if chosen: This option is wrong — you counted four domains. The C=C double bond counts once, so each C has three domains and an ideal of 120°, squeezed slightly below.

Check — multiple choice

In formaldehyde, H₂CO, why are the O–C–H angles slightly larger than 120°?

AThe O atom is larger than the H atoms, so it needs more room

Feedback if chosen: This option is wrong — you tied the angle to the terminal atom’s size. The push comes from the four shared electrons in the C=O domain, not from how big O is.

BThe C=O domain repels the two C–H domains more strongly than they repel each other, pushing both away from itself ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The four-electron C=O domain pushes both C–H bonds away from itself, opening each O–C–H angle slightly beyond 120°.

CThe C=O domain counts as two domains, so it claims two shares of the circle around C

Feedback if chosen: This option is wrong — a double bond counts as one domain. It claims one direction around C; what changes is how hard that one domain pushes.

DThe two C–H bonds attract each other, which opens the angles next to the double bond

Feedback if chosen: This option is wrong — you replaced repulsion with attraction. Bonding pairs repel one another; the reshaping comes from the C=O domain’s stronger push.

U2-S13 · LESSON 125 · CSP-118

From Lewis diagram to geometry and angle

Outcome: Predict electron-domain geometry, molecular geometry, and the approximate bond angle from a Lewis diagram.

Part 1The whole routine

You’ve already seen every step this lesson uses; what is new is chaining them into one routine.

Step 1 — count the electron domains around the central atom: a bond of any order counts once, and each lone pair counts once.

Step 2 — the domain count gives the electron-domain geometry.

Step 3 — the arrangement of the bonded atoms alone gives the molecular geometry.

Step 4 — the electron-domain geometry sets the ideal bond angle.

Step 5 — adjust qualitatively: a lone pair generally squeezes its neighboring angles slightly below the ideal, and a multiple bond can open its neighboring angles slightly above it.

Part 2The routine on NH₃

Here is the Lewis diagram of NH₃.

Count the domains around N: three N–H bonds and one lone pair — four electron domains.

Four domains: tetrahedral electron-domain geometry.

Here is the resulting shape — the bonded atoms alone form a pyramid: trigonal-pyramidal molecular geometry.

Four domains set the ideal angle at 109.5°.

The lone pair squeezes the three N–H bonds together: predict each H–N–H angle as slightly less than 109.5°.

The full answer: tetrahedral electron-domain geometry, trigonal-pyramidal molecular geometry, and H–N–H angles slightly less than 109.5°.

What you are expected to be able to do

Predict electron-domain geometry, molecular geometry, and the approximate bond angle from a supplied Lewis diagram, justifying any deviation from the ideal angle.

Part 3Check what you just learned

Check — multiple choice

Use the Lewis diagram of PCl₃ shown. Which prediction is correct?

ATrigonal-planar electron-domain geometry; trigonal-planar molecular geometry; Cl–P–Cl exactly 120°

Feedback if chosen: This option is wrong — you did not count the lone pair on P. Three bonds plus one lone pair is four domains, not three, so the electron-domain geometry is tetrahedral.

BTetrahedral electron-domain geometry; tetrahedral molecular geometry; Cl–P–Cl exactly 109.5°

Feedback if chosen: This option is wrong — you treated the lone pair as if it were a bonded atom. Molecular geometry describes the bonded atoms only (trigonal pyramidal), and the lone pair’s stronger repulsion closes the angle below 109.5°.

CTetrahedral electron-domain geometry; trigonal-pyramidal molecular geometry; Cl–P–Cl less than 109.5° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Three bonds plus one lone pair is four domains — tetrahedral; the three bonded Cl atoms form a pyramid — trigonal pyramidal; and the lone pair squeezes Cl–P–Cl below 109.5°.

DTrigonal-pyramidal electron-domain geometry; trigonal-pyramidal molecular geometry; Cl–P–Cl less than 109.5°

Feedback if chosen: This option is wrong — you named the electron-domain geometry after the molecular one. The domain count (four) names it tetrahedral; trigonal pyramidal describes only the bonded atoms.

Check — multiple choice

Use the Lewis diagram of H₂S shown. Which prediction is correct?

ATetrahedral electron-domain geometry; bent molecular geometry; H–S–H less than 109.5° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Two bonds plus two lone pairs is four domains — tetrahedral; the two bonded H atoms make it bent; and the two lone pairs squeeze H–S–H below 109.5°.

BLinear electron-domain geometry; linear molecular geometry; H–S–H exactly 180°

Feedback if chosen: This option is wrong — you counted only the bonded atoms as domains. The two lone pairs on S count too: four domains, tetrahedral, not linear.

CTetrahedral electron-domain geometry; bent molecular geometry; H–S–H exactly 109.5°

Feedback if chosen: This option is wrong — geometry right, angle unadjusted. The two lone pairs repel the bonding pairs more strongly and close the angle below 109.5°.

DTrigonal-planar electron-domain geometry; bent molecular geometry; H–S–H less than 120°

Feedback if chosen: This option is wrong — you counted only one of S’s two lone pairs. Two bonds plus two lone pairs is four domains, so the ideal is 109.5°, not 120°.

Check — multiple choice

Use the Lewis diagram of HCN shown. Which prediction is correct for the geometry around C and the H–C–N angle?

ALinear electron-domain geometry; linear molecular geometry; H–C–N slightly less than 180°

Feedback if chosen: This option is wrong — you applied a squeeze with nothing to do the squeezing. C holds no lone pair, and with only two domains there is no third neighbor to trade space with: the domains sit exactly opposite, 180°.

BTetrahedral electron-domain geometry; tetrahedral molecular geometry; H–C–N about 109.5°

Feedback if chosen: This option is wrong — you counted the triple bond as three domains. A bond of any order counts once, so C has two domains: linear.

CTrigonal-planar electron-domain geometry; bent molecular geometry; H–C–N slightly less than 120°

Feedback if chosen: This option is wrong — you counted the lone pair on N as a domain around C. Domains are counted at the central atom, and that lone pair belongs to N: C has two domains.

DLinear electron-domain geometry; linear molecular geometry; H–C–N exactly 180° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. C–H and C≡N are two domains — linear electron-domain and molecular geometry — and with no lone pair on C and no third domain, nothing shifts the angle: exactly 180°.

Check — multiple choice

Use the Lewis diagram of COF₂ shown. Which prediction is correct?

ATrigonal-planar electron-domain geometry; trigonal-planar molecular geometry; F–C–F exactly 120°

Feedback if chosen: This option is wrong — geometry right, angle unadjusted. The C=O domain holds four shared electrons and pushes both C–F bonds away from itself, closing F–C–F below 120°.

BTrigonal-planar electron-domain geometry; trigonal-planar molecular geometry; F–C–F less than 120° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The C=O counts once, so C has three domains and no lone pairs — trigonal planar for both — and the C=O domain’s stronger push closes F–C–F below 120°.

CTetrahedral electron-domain geometry; tetrahedral molecular geometry; F–C–F about 109.5°

Feedback if chosen: This option is wrong — you counted the double bond as two domains. A bond of any order counts once: three domains, not four.

DTrigonal-planar electron-domain geometry; bent molecular geometry; F–C–F less than 120° with the O left out of the shape

Feedback if chosen: This option is wrong — you dropped a bonded atom. Bent would need a lone pair on C; C has none, and O is a bonded atom, so the molecular geometry matches the electron-domain geometry: trigonal planar.

AFTER U2-S13 · CHECKPOINT · U2-G4 · 6 items

Checkpoint 4 — Electron domains and molecular geometry

Check — multiple choice

Use the Lewis diagram of the nitrite ion, NO₂⁻, shown. Which prediction is correct?

ATrigonal-planar electron-domain geometry; trigonal-planar molecular geometry; bond angle 120°

Feedback if chosen: This option is wrong — you described the positions of all three domains. Molecular geometry names atom positions only, so with one domain a lone pair the molecule is bent, and the lone pair squeezes the angle below 120°.

BTetrahedral electron-domain geometry; bent molecular geometry; slightly less than 109.5°

Feedback if chosen: This option is wrong — you counted the double bond as two domains. One double bond is one domain, so N has three domains, not four, and the ideal angle to adjust from is 120°.

CTrigonal-planar electron-domain geometry; bent molecular geometry; slightly less than 120° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. N has three electron domains — one single bond, one double bond, one lone pair — so the domains are trigonal planar; the two bonded O atoms make the molecule bent, and the lone pair compresses the O–N–O angle to slightly less than 120°.

DLinear electron-domain geometry; linear molecular geometry; bond angle 180°

Feedback if chosen: This option is wrong — you counted only the bonded atoms. The lone pair on N is a domain too, so N has three domains, and the molecule cannot be linear.

Check — numeric entry

Use the XeF₂ Lewis diagram shown. How many electron domains are around Xe?

✓ ANSWER KEY: 5

Feedback when correct: Correct. Xe has two bonding domains — one per Xe–F single bond — and three lone pairs: 2 + 3 = 5 electron domains.

Anticipated wrong answers (feedback shown):

2	This counts only the bonds. Each lone pair is a domain too, so add Xe's three lone pairs: 2 + 3 = 5.
3	This counts only the lone pairs. Each bond is a domain as well, so add the two Xe–F bonds: 2 + 3 = 5.

Check — multiple choice

The diagram shows a correct Lewis diagram of formaldehyde, H₂CO. Which molecular geometry does VSEPR predict for it?

ABent

Feedback if chosen: This option is wrong — you treated one of C's domains as a lone pair. All three domains around C are bonds (two C–H and one C=O), so no domain is invisible in the molecular geometry.

BTrigonal planar ✓ KEY — this is the correct answer

Feedback if chosen: Correct. C has three electron domains — two single bonds and one double bond — and all three hold bonded atoms, so the molecular geometry is trigonal planar.

CTetrahedral

Feedback if chosen: This option is wrong — you counted the double bond as two domains. A double bond is one domain, so C has three domains, not four.

DTrigonal pyramidal

Feedback if chosen: This option is wrong — trigonal pyramidal needs four domains with one lone pair. C has three domains and no lone pair.

Check — multiple choice

A central atom has five electron domains, none of them lone pairs. Which molecular geometry does VSEPR predict?

ASquare pyramidal

Feedback if chosen: This option is wrong — square pyramidal belongs to the six-domain family (six domains, one lone pair). Five bonding domains take the five-domain parent shape.

BSeesaw

Feedback if chosen: This option is wrong — seesaw is the five-domain geometry when one domain is a lone pair. Here every domain holds a bonded atom.

CTrigonal bipyramidal ✓ KEY — this is the correct answer

Feedback if chosen: Correct. With five domains and no lone pairs, the atoms sit at all five positions: trigonal bipyramidal.

DTrigonal pyramidal

Feedback if chosen: This option is wrong — trigonal pyramidal is a four-domain geometry (four domains, one lone pair), not a five-domain one.

Check — multiple choice

Use the SF₆ Lewis diagram shown. Which molecular geometry does VSEPR predict for it?

AOctahedral ✓ KEY — this is the correct answer

Feedback if chosen: Correct. S has six electron domains — six S–F bonds and no lone pairs — so the atoms occupy all six positions: octahedral.

BTrigonal bipyramidal

Feedback if chosen: This option is wrong — trigonal bipyramidal is the five-domain parent shape. Count the bonds in the diagram: S holds six.

CSquare pyramidal

Feedback if chosen: This option is wrong — square pyramidal is the six-domain geometry when one domain is a lone pair. S has no lone pair here; all six domains hold F atoms.

DSquare planar

Feedback if chosen: This option is wrong — square planar needs six domains with two lone pairs. All six of S's domains are bonds.

Check — multiple choice

Use the Lewis diagram of NCl₃ shown. What is the best prediction for the Cl–N–Cl bond angle?

AExactly 109.5°

Feedback if chosen: This option is wrong — 109.5° is the ideal for four domains with no lone pair. N's lone pair repels the bonding pairs more strongly and squeezes the angle below the ideal.

BExactly 120°

Feedback if chosen: This option is wrong — 120° belongs to three electron domains. N has four domains (three bonds and a lone pair), so start from 109.5° and adjust.

CSlightly less than 109.5° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. N has four electron domains, one a lone pair; lone-pair repulsion compresses the adjacent Cl–N–Cl angles to slightly less than the 109.5° ideal.

DSlightly more than 109.5°

Feedback if chosen: This option is wrong — the lone pair pushes the bonding pairs together, not apart, so the angle closes below 109.5° rather than opening above it.

U2-S14 — Hybridization

5 lessons

U2-S14 · LESSON 126 · CSP-119

The hybridization model

Outcome: State what hybridization is: a model that combines an atom's valence atomic orbitals into hybrid orbitals that set the directions of its bonds and lone pairs.

Part 1A model for where bonds and lone pairs point

Here is the Lewis diagram of methane, CH₄.

You've already seen how to count electron domains: C has four bonds and no lone pairs, so four electron domains.

And you've already seen the arrangement four electron domains take: tetrahedral.

You've also seen, in Unit 1, that an atom's valence electrons occupy valence atomic orbitals — an s orbital and p orbitals.

Hybridization is a model that combines an atom's valence atomic orbitals into a new set of orbitals.

The new orbitals are called hybrid atomic orbitals — hybrid orbitals, for short.

The hybrid orbitals set the directions of the atom's bonds and lone pairs: one hybrid orbital points along each electron domain.

For the C in CH₄, the model combines C's valence atomic orbitals into four hybrid orbitals, one pointing along each of the four bonds.

Lone pairs get hybrid orbitals too. A lone pair is an electron domain, so it occupies a hybrid orbital pointing along its own direction.

Hybridization is a description chemists apply to an atom — a model — not an event that happens to the atom.

Different domain counts call for different sets of hybrid orbitals. The sets have short names — sp, sp², and sp³ — and the next three lessons pair each name with its domain count.

What you are expected to be able to do

State that hybridization is a model that combines an atom's valence atomic orbitals into hybrid orbitals, and that the hybrid orbitals set the directions of the atom's bonds and lone pairs.

Part 2Check what you just learned

Check — multiple choice

Which statement says what the hybridization model does?

AIt combines an atom's valence atomic orbitals into hybrid orbitals that set the directions of the atom's bonds and lone pairs ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Hybridization works on one atom's own valence atomic orbitals, and the hybrid orbitals it produces set the directions of that atom's bonds and lone pairs.

BIt combines the valence atomic orbitals of two bonded atoms into orbitals shared between the atoms

Feedback if chosen: This option is wrong — you combined orbitals across two atoms. Hybridization describes one atom: it combines that atom's own valence atomic orbitals.

CIt combines all of an atom's orbitals, core and valence, into hybrid orbitals that set the directions of the atom's bonds and lone pairs

Feedback if chosen: This option is wrong — you pulled the core orbitals into the combination. Only the atom's valence atomic orbitals are combined; the core orbitals are left alone.

DIt combines an atom's valence atomic orbitals into hybrid orbitals that set the strengths of the atom's bonds

Feedback if chosen: This option is wrong — you gave the hybrid orbitals the wrong job. Hybrid orbitals set the directions of the atom's bonds and lone pairs, not their strengths.

Check — multiple choice

Use the Lewis diagram of ammonia, NH₃. According to the hybridization model, how are the electron domains around the N atom described?

AThree hybrid orbitals — one for each bond, with the lone pair left in an unchanged valence atomic orbital

Feedback if chosen: This option is wrong — you left the lone pair out. Hybrid orbitals set the directions of lone pairs as well as bonds, so the lone pair occupies a hybrid orbital too: four in all.

BFive hybrid orbitals — one for each valence electron of N

Feedback if chosen: This option is wrong — you counted valence electrons instead of electron domains. N's four electron domains — three bonds and one lone pair — are described with four hybrid orbitals.

CFour hybrid orbitals — made by combining N's valence atomic orbitals with the H atoms' orbitals

Feedback if chosen: This option is wrong — you have the right count but combined orbitals across atoms. The four hybrid orbitals come from combining N's own valence atomic orbitals.

DFour hybrid orbitals — one for each electron domain: three bonds and one lone pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. One hybrid orbital points along each of N's four electron domains — the three bonds and the lone pair.

Check — multiple choice

Use the Lewis diagram of water, H₂O. According to the hybridization model, how many hybrid orbitals set the directions of the O atom's bonds and lone pairs?

ATwo — one for each bond

Feedback if chosen: This option is wrong — you counted only the bonds. The two lone pairs occupy hybrid orbitals too, so four hybrid orbitals set the directions of O's domains.

BFour — one for each electron domain: two bonds and two lone pairs ✓ KEY — this is the correct answer

Feedback if chosen: Correct. O has four electron domains — two bonds and two lone pairs — and one hybrid orbital points along each of them.

CSix — one for each valence electron of O

Feedback if chosen: This option is wrong — you counted O's valence electrons instead of its electron domains. The four domains — two bonds and two lone pairs — take four hybrid orbitals.

DEight — one for each of the eight electrons drawn around O in the diagram

Feedback if chosen: This option is wrong — you counted every electron around O instead of its electron domains. The four domains — two bonds and two lone pairs — take four hybrid orbitals.

Part 3Why chemists find this model useful (enrichment — not assessed)

This last part is for interest only — nothing in it will be tested, in this lesson or anywhere in the course.

Chemists keep the hybridization model because of a mismatch: the four bonds of CH₄ are identical and point at the corners of a tetrahedron, but an isolated C atom's valence atomic orbitals — one s and three p — are not four identical orbitals pointing that way.

Combining those orbitals into four identical hybrid orbitals, one along each bond, gives a picture of the atom that matches the bonds and bond angles the molecule is observed to have.

U2-S14 · LESSON 127 · CSP-120

sp hybridization

Outcome: Recognize that an atom with two electron domains is sp hybridized.

Part 1Two electron domains

Here is the Lewis diagram of carbon dioxide, CO₂.

You've already seen how to count electron domains: each double bond is one domain, so C has two electron domains.

And you've already seen the arrangement two electron domains take: linear, 180° apart.

The hybridization model describes two electron domains with two hybrid orbitals, one pointing along each domain.

A set of two hybrid orbitals is called sp. An atom with two electron domains is sp hybridized.

The name records what was combined: one s orbital and one p orbital give the two sp hybrid orbitals.

An atom has three p orbitals, and the mixing here uses only one of them — an sp atom keeps two unhybridized p orbitals, left out of the mixing.

The two domains can be any mix — two bonds, or a bond and a lone pair. The count is what fixes the name.

So when you meet a two-domain atom — like the C in CO₂ — the label to attach is sp.

What you are expected to be able to do

State that two electron domains around an atom correspond to sp hybridization, and recognize that sp describes a two-domain atom such as the C in CO₂.

Part 2Check what you just learned

Check — multiple choice

Use the Lewis diagram of hydrogen cyanide, HCN. Which hybridization describes the C atom?

Asp² — C has two electron domains

Feedback if chosen: This option is wrong — you have the right count but the wrong name. Two electron domains correspond to sp hybridization; sp² is the name for three.

Bsp³ — C has four electron domains, counting each line of the triple bond as its own domain

Feedback if chosen: This option is wrong — you counted each line of the triple bond separately. A triple bond is one electron domain — one region of electron density — so C has two domains and is sp hybridized.

Csp — C has two electron domains: the single bond to H and the triple bond to N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each bond, single or triple, is one electron domain. C has two domains, and two domains correspond to sp hybridization.

Dsp² — C has three electron domains: two bonds and N's lone pair

Feedback if chosen: This option is wrong — you counted N's lone pair as one of C's domains. Only the domains around C itself count: the two bonds. Two domains correspond to sp hybridization.

Check — multiple choice

Use the Lewis diagram of carbon monoxide, CO. Which hybridization describes the C atom?

Asp — C has two electron domains: the triple bond and the lone pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The triple bond is one electron domain and the lone pair is another. Two domains correspond to sp hybridization.

Bsp — C has one electron domain, because lone pairs do not count

Feedback if chosen: This option is wrong — you reached the right name with the wrong count. A lone pair is an electron domain, so C has two domains — and it is the count of two that corresponds to sp.

Csp³ — C has four electron domains, counting each line of the triple bond and the lone pair

Feedback if chosen: This option is wrong — you counted each line of the triple bond separately. The triple bond is one electron domain; with the lone pair, C has two domains, so it is sp hybridized.

Dsp² — C has three electron domains, counting each line of the triple bond but not the lone pair

Feedback if chosen: This option is wrong — you counted each line of the triple bond separately and left the lone pair out. The triple bond is one domain and the lone pair is one more: two domains, sp.

U2-S14 · LESSON 128 · CSP-121

sp² hybridization

Outcome: Recognize that an atom with three electron domains is sp² hybridized.

Part 1Three electron domains

Here is the Lewis diagram of boron trifluoride, BF₃.

You've already seen how to count electron domains: B has three bonds and no lone pairs, so three electron domains.

And you've already seen the arrangement three electron domains take: trigonal planar, 120° apart.

The hybridization model describes three electron domains with three hybrid orbitals, one pointing along each domain.

A set of three hybrid orbitals is called sp². An atom with three electron domains is sp² hybridized.

The name records what was combined: one s orbital and two p orbitals give the three sp² hybrid orbitals — the superscript counts the p orbitals.

The mixing uses two of the atom's three p orbitals — an sp² atom keeps one unhybridized p orbital, left out of the mixing.

The three domains can be any mix of bonds and lone pairs, and a double bond still counts as one domain.

So when you meet a three-domain atom — like the B in BF₃ — the label to attach is sp².

What you are expected to be able to do

State that three electron domains around an atom correspond to sp² hybridization, and recognize that sp² describes a three-domain atom such as the B in BF₃.

Part 2Check what you just learned

Check — multiple choice

Use the Lewis diagram of formaldehyde, H₂CO. Which hybridization describes the C atom?

Asp³ — C has four electron domains, counting each line of the double bond as its own domain

Feedback if chosen: This option is wrong — you counted each line of the double bond separately. The double bond is one electron domain, so C has three domains — sp².

Bsp² — C has three electron domains: two bonds to H and one double bond to O ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The double bond is one electron domain, so C has three domains in all, and three domains correspond to sp² hybridization.

Csp — C has three electron domains

Feedback if chosen: This option is wrong — you have the right count but the wrong name. Three electron domains correspond to sp² hybridization; sp is the name for two.

Dsp³ — C has five electron domains: three bonds and O's two lone pairs

Feedback if chosen: This option is wrong — you counted O's lone pairs as C's domains. Only the domains around C itself count: three bonds, so three domains — sp².

Check — multiple choice

Use the Lewis diagram of nitrosyl chloride, NOCl. Which hybridization describes the N atom?

Asp — N has two electron domains: the two bonded atoms

Feedback if chosen: This option is wrong — you counted only the bonded atoms and left the lone pair out. The lone pair is an electron domain too, so N has three domains — sp².

Bsp³ — N has four electron domains, counting each line of the double bond and the lone pair

Feedback if chosen: This option is wrong — you counted each line of the double bond separately. The double bond is one electron domain; with the single bond and the lone pair, N has three domains — sp².

Csp³ — N has three electron domains

Feedback if chosen: This option is wrong — you have the right count but the wrong name. Three electron domains correspond to sp² hybridization; sp³ is the name for four.

Dsp² — N has three electron domains: two bonds and one lone pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The double bond is one domain, the single bond is another, and the lone pair is the third. Three domains correspond to sp² hybridization.

U2-S14 · LESSON 129 · CSP-122

sp³ hybridization

Outcome: Recognize that an atom with four electron domains is sp³ hybridized.

Part 1Four electron domains

Here is the Lewis diagram of water, H₂O.

You've already seen how to count electron domains: O has two bonds and two lone pairs, so four electron domains.

And you've already seen the arrangement four electron domains take: tetrahedral — here, two of the four domains are lone pairs.

The hybridization model describes four electron domains with four hybrid orbitals, one pointing along each domain.

A set of four hybrid orbitals is called sp³. An atom with four electron domains is sp³ hybridized.

The name records what was combined: one s orbital and three p orbitals give the four sp³ hybrid orbitals.

You've already seen the C in CH₄ described with four hybrid orbitals, one along each of its four bonds — so the C in CH₄ is sp³ hybridized too.

Two bonds and two lone pairs, or four bonds — any mix of four domains gets the same label.

What you are expected to be able to do

State that four electron domains around an atom correspond to sp³ hybridization, and recognize that sp³ describes a four-domain atom such as the O in H₂O.

Part 2Check what you just learned

Check — multiple choice

Use the Lewis diagram of ammonia, NH₃. Which hybridization describes the N atom?

Asp² — N has three electron domains, because the lone pair does not count

Feedback if chosen: This option is wrong — you left the lone pair out. A lone pair is an electron domain, so N has four domains, and four domains correspond to sp³.

Bsp² — N has four electron domains

Feedback if chosen: This option is wrong — you have the right count but the wrong name. Four electron domains correspond to sp³ hybridization; sp² is the name for three.

Csp³ — N has four electron domains: three bonds and one lone pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Three bonds plus one lone pair make four electron domains, and four domains correspond to sp³ hybridization.

Dsp³ — N has three electron domains: one for each H atom

Feedback if chosen: This option is wrong — you reached the right name with the wrong count. N's lone pair counts along with the three bonds: four domains, and it is four that corresponds to sp³.

Check — multiple choice

Use the Lewis diagram of methanol, CH₃OH. Which hybridization describes the O atom?

Asp³ — O has four electron domains: two bonds and two lone pairs ✓ KEY — this is the correct answer

Feedback if chosen: Correct. O's two bonds and two lone pairs make four electron domains, and four domains correspond to sp³ hybridization.

Bsp — O has two electron domains: the bond to C and the bond to H

Feedback if chosen: This option is wrong — you counted only the bonds. O's two lone pairs are electron domains too, so O has four domains — sp³.

Csp² — O has three electron domains: two bonds and one lone pair

Feedback if chosen: This option is wrong — you missed one lone pair. The diagram shows two lone pairs on O; with the two bonds that is four domains — sp³.

Dsp³ — O has four electron domains, because C has four bonds

Feedback if chosen: This option is wrong — you reached the right name by counting the wrong atom's domains. The question asks about O, and it is O's own four domains — two bonds and two lone pairs — that make it sp³.

U2-S14 · LESSON 130 · CSP-123

Predicting an atom's hybridization

Outcome: Predict whether an atom is sp, sp², or sp³ hybridized from its electron-domain count, never from the molecular geometry alone.

Part 1Count domains, then name the hybridization

You've already seen the three correspondences one at a time. Here they are together.

Electron domains and hybridization
Electron domains around the atom	Hybridization
2	sp
3	sp²
4	sp³

To assign an atom's hybridization: read the Lewis diagram, count that atom's electron domains — bonds and lone pairs — and match the count to the name.

Count the domains of the selected atom only. Different atoms in the same molecule can have different hybridizations.

The table stops at four: you will not be asked to assign a hybridization to an atom with five or six electron domains.

Part 2The trap: molecular geometry alone

Here is a water molecule drawn from its molecular geometry alone: bent, with two bonded H atoms.

From this picture you might count two and assign sp.

But molecular geometry shows only the bonded atoms. O's Lewis diagram shows two lone pairs as well — four electron domains in all.

Four domains — sp³. The O in bent H₂O is sp³ hybridized.

That is the rule: assign hybridization from the atom's electron-domain count, never from the molecular geometry alone.

What you are expected to be able to do

Predict an atom's sp, sp², or sp³ hybridization by counting its electron domains from the Lewis diagram — never from the molecular geometry alone.

Part 3Check what you just learned

Check — multiple choice

Sulfur difluoride, SF₂, is a bent molecule. Use its Lewis diagram. Which hybridization describes the S atom?

Asp — the molecule is bent with two bonded atoms

Feedback if chosen: This option is wrong — you assigned from the molecular geometry alone. The bent geometry shows only the two bonded atoms; the Lewis diagram shows two lone pairs on S as well, so S has four electron domains — sp³.

Bsp³ — S has four electron domains: two bonds and two lone pairs ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Two bonds and two lone pairs make four electron domains, so S is sp³ — even though the molecule is bent.

Csp² — S has three electron domains: two bonds and one lone pair

Feedback if chosen: This option is wrong — you missed one lone pair. The diagram shows two lone pairs on S; with the two bonds that is four domains — sp³.

Dsp — S has two electron domains, because lone pairs do not count

Feedback if chosen: This option is wrong — you left the lone pairs out of the count. A lone pair is an electron domain, so S has four domains, and four domains correspond to sp³.

Check — multiple choice

Phosphorus trifluoride, PF₃, has trigonal pyramidal molecular geometry. Use its Lewis diagram. Which hybridization describes the P atom?

Asp² — P has three electron domains: one for each bonded F atom

Feedback if chosen: This option is wrong — you counted only the bonded atoms. P's lone pair is an electron domain too, so P has four domains — sp³.

BIt cannot be assigned from the Lewis diagram — the bond angles must be measured first

Feedback if chosen: This option is wrong — you asked for more information than the assignment needs. The electron-domain count from the Lewis diagram settles it: four domains around P, so sp³.

Csp³ — P has four electron domains: three bonds and one lone pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Three bonds and one lone pair make four electron domains, so P is sp³ — the trigonal pyramidal molecular geometry does not change the count.

Dsp² — trigonal pyramidal molecular geometry comes from three electron domains

Feedback if chosen: This option is wrong — you assigned from the molecular geometry alone. Trigonal pyramidal is what four domains look like when one of them is a lone pair: the geometry shows three bonded atoms, but the domain count is four — sp³.

Check — multiple choice

Use the Lewis diagram of hydrogen cyanide, HCN. Which hybridization describes the N atom?

Asp² — N has three electron domains, counting each line of the triple bond but not the lone pair

Bsp³ — N has four electron domains, counting each line of the triple bond and the lone pair

Feedback if chosen: This option is wrong — you counted each line of the triple bond separately. The triple bond is one electron domain; with the lone pair, N has two domains — sp.

Csp² — N has three electron domains, one for each of the molecule’s three atoms

Feedback if chosen: This option is wrong — you counted the atoms in the whole molecule instead of the domains around N. Domains belong to one atom: around N, the triple bond is one domain and the lone pair is the second — two domains, sp.

Dsp — N has two electron domains: the triple bond and the lone pair ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The triple bond is one electron domain and the lone pair is the second. Two domains correspond to sp hybridization.

Check — multiple choice

A molecule is bent: a central atom X bonded to two Y atoms. No Lewis diagram is supplied. Can X's hybridization be assigned from the molecular geometry alone?

ANo — a bent atom can have three electron domains or four, so the count must come from the Lewis diagram ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Bent molecular geometry happens with three domains (two bonds and one lone pair) and with four (two bonds and two lone pairs), so the geometry alone cannot decide between sp² and sp³. Count X's domains from the Lewis diagram.

BYes — X is bonded to two atoms, so X has two electron domains and is sp hybridized

Feedback if chosen: This option is wrong — you counted bonded atoms instead of electron domains. A bent central atom carries one or two lone pairs, and those are domains too.

CYes — bent molecular geometry comes from four electron domains around the central atom, so X is sp³

Feedback if chosen: This option is wrong — you matched the geometry to a single count. Bent also happens with three domains — two bonds and one lone pair — so the geometry alone cannot decide between sp² and sp³.

DNo — hybridization is never assigned to central atoms that carry lone pairs, whatever the geometry

Feedback if chosen: This option is wrong — you gave the right verdict with the wrong reason. Atoms with lone pairs are assigned hybridizations all the time — the O in H₂O is sp³. The geometry alone fails because it cannot tell three domains from four.

U2-S15 — Orbital overlap and geometric isomerism

14 lessons

U2-S15 · LESSON 131 · CSP-124

The internuclear axis

Outcome: State that the internuclear axis is the straight line joining the nuclei of two bonded atoms.

Part 1A line through two nuclei

Here are two bonded atoms drawn as overlapping orbitals. The dot at the center of each orbital is that atom’s nucleus.

You’ve already seen internuclear distance: how far apart the nuclei of two bonded atoms sit.

The straight line drawn through the two nuclei is called the internuclear axis.

Internuclear distance is a length measured along that line; the internuclear axis is the line itself.

In a C–C bond, the internuclear axis is the straight line joining the two C nuclei.

What you are expected to be able to do

State that the internuclear axis is the straight line joining the nuclei of two bonded atoms.

Part 2Check what you just learned

Check — multiple choice

Two H atoms are bonded in H₂. What is the internuclear axis of this bond?

AThe distance between the two H nuclei

Feedback if chosen: This option is wrong — you named internuclear distance. The distance is how far apart the nuclei sit; the internuclear axis is the straight line joining them.

BThe region where the two orbitals overlap

Feedback if chosen: This option is wrong — you named the overlap region. The internuclear axis is not a region: it is the straight line joining the two nuclei.

CThe straight line joining the two H nuclei ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The internuclear axis is the straight line joining the nuclei of the two bonded atoms.

DA line through the midpoint of the bond, perpendicular to it

Feedback if chosen: This option is wrong — you turned the line 90°. The internuclear axis extends from one nucleus to the other, not across the bond.

Check — multiple choice

A C atom and an O atom are bonded in a molecule. The internuclear axis of the C–O bond joins which two points?

AThe C nucleus and the O nucleus ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The internuclear axis joins the nuclei of the two bonded atoms — here, the C nucleus and the O nucleus.

BThe outer edge of the C orbital and the outer edge of the O orbital

Feedback if chosen: This option is wrong — you used the orbitals’ edges. The axis is defined by the nuclei: it joins the C nucleus to the O nucleus.

CThe two shared electrons

Feedback if chosen: This option is wrong — you used the electrons. Electrons move throughout the orbitals; the internuclear axis is fixed by the two nuclei.

DThe C nucleus and the midpoint of the bond

Feedback if chosen: This option is wrong — you stopped halfway. The internuclear axis extends the whole way, from the C nucleus to the O nucleus.

U2-S15 · LESSON 132 · CSP-125

Sigma bonds

Outcome: State that a sigma bond forms from end-to-end orbital overlap, with electron density along the internuclear axis.

Part 1End-to-end overlap

Here are the two H atoms of H₂, drawn as orbitals. You’ve already seen that each H atom holds its electron in a spherical s orbital.

The two s orbitals overlap end-to-end on the internuclear axis, and the shaded region is that overlap.

The shared pair spends its time in that overlap: the bond’s electron density lies along the internuclear axis, between the two nuclei.

A bond formed from end-to-end orbital overlap, with electron density along the internuclear axis, is called a sigma (σ) bond.

The two labels describe the same overlap.

p orbitals can do it too: two p orbitals pointing at each other along the internuclear axis overlap end-to-end and form a sigma bond.

What makes a bond sigma is the overlap, not which orbitals meet: end-to-end overlap along the internuclear axis gives a sigma bond every time.

The H–H single bond is a sigma bond, and so is every other single bond.

What you are expected to be able to do

State that a sigma bond forms from end-to-end orbital overlap, with its electron density along the internuclear axis.

Part 2Check what you just learned

Check — multiple choice

Which description fits a sigma bond?

AIt forms from side-by-side orbital overlap, with electron density on opposite sides of the internuclear axis.

Feedback if chosen: This option is wrong — you described the opposite overlap. A sigma bond’s orbitals overlap end-to-end, and its electron density lies along the internuclear axis.

BIt forms from end-to-end orbital overlap, with electron density along the internuclear axis. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. End-to-end overlap puts the shared pair’s electron density along the internuclear axis, between the two nuclei.

CIt forms from end-to-end orbital overlap, with electron density on opposite sides of the internuclear axis.

Feedback if chosen: This option is wrong — you paired the right overlap with the wrong density. End-to-end overlap sits on the internuclear axis, so the electron density lies along the axis, not on opposite sides of it.

DIt forms only when two s orbitals overlap, whatever the direction of the overlap.

Feedback if chosen: This option is wrong — you classified by orbital type. Two p orbitals pointing at each other along the internuclear axis also form a sigma bond; what matters is the end-to-end overlap.

Check — multiple choice

In HCl, the single bond forms when an s orbital on H overlaps end-to-end with a p orbital on Cl. Where does this bond’s electron density lie?

AOn opposite sides of the internuclear axis, above and below it

Feedback if chosen: This option is wrong — you placed the density where side-by-side overlap would put it. This overlap is end-to-end, so the density lies along the internuclear axis.

BAt the two nuclei themselves

Feedback if chosen: This option is wrong — you put the shared pair on the nuclei. The nuclei sit at the ends; the shared pair’s density fills the overlap between them.

CSpread evenly over every atom in the molecule

Feedback if chosen: This option is wrong — you spread the pair over the molecule. A sigma bond’s density is concentrated in the overlap region, along the internuclear axis.

DAlong the internuclear axis, between the two nuclei ✓ KEY — this is the correct answer

Feedback if chosen: Correct. End-to-end overlap — s with p included — puts the electron density along the internuclear axis, between the two nuclei.

U2-S15 · LESSON 133 · CSP-126

Pi bonds

Outcome: State that a pi bond forms from side-by-side overlap of unhybridized p orbitals, with electron density on opposite sides of the internuclear axis.

Part 1Side-by-side overlap

Here is the Lewis diagram of ethene, C₂H₄. The two C atoms are joined by a double bond.

You’ve already seen sp² hybridization: each C in ethene has three electron domains, so it is sp² hybridized and keeps one unhybridized p orbital, left out of the mixing.

The two unhybridized p orbitals sit side-by-side, one on each C, with lobes above and below the internuclear axis.

The two p orbitals overlap side-by-side: their upper lobes overlap above the internuclear axis, and their lower lobes overlap below it.

So this bond’s electron density lies in two regions, on opposite sides of the internuclear axis.

A bond formed from side-by-side overlap of unhybridized p orbitals, with electron density on opposite sides of the internuclear axis, is called a pi (π) bond.

The two regions together hold one shared pair and count as one pi bond — not two.

In ethene’s C=C double bond, the additional bond — the second line — is a pi bond.

What you are expected to be able to do

State that a pi bond forms from side-by-side overlap of unhybridized p orbitals, with its electron density on opposite sides of the internuclear axis.

Part 2Check what you just learned

Check — multiple choice

Which description fits a pi bond?

AIt forms from side-by-side overlap of unhybridized p orbitals, with electron density on opposite sides of the internuclear axis. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Side-by-side overlap of the unhybridized p orbitals puts the electron density in two regions, on opposite sides of the internuclear axis.

BIt forms from end-to-end orbital overlap, with its electron density lying along the internuclear axis, between the two nuclei.

Feedback if chosen: This option is wrong — you described a sigma bond. A pi bond’s orbitals overlap side-by-side, and its density sits on opposite sides of the axis.

CIt forms from side-by-side overlap of unhybridized p orbitals, with electron density along the internuclear axis.

Feedback if chosen: This option is wrong — you paired the right overlap with the wrong density. Side-by-side overlap happens above and below the axis, so the density sits on opposite sides of it.

DIt forms whenever two p orbitals overlap, whatever the direction of the overlap.

Feedback if chosen: This option is wrong — you classified by orbital type. Two p orbitals pointing at each other end-to-end form a sigma bond; a pi bond needs side-by-side overlap.

Check — multiple choice

A pi bond’s electron density sits in two regions, one above and one below the internuclear axis. How many bonds, holding how many shared pairs, is this?

ATwo separate pi bonds, each region holding its own shared pair

Feedback if chosen: This option is wrong — you counted each region as a separate bond. The two regions are the two halves of one side-by-side overlap: one pi bond, one shared pair.

BOne sigma bond above the axis and one pi bond below it

Feedback if chosen: This option is wrong — you split the regions into different bond types. Both regions belong to the same side-by-side overlap, and a sigma bond’s density would lie on the axis, not above it.

COne pi bond, with one shared pair spread over the two regions ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The two regions together make one side-by-side overlap: one pi bond holding one shared pair.

DOne pi bond holding two shared pairs, one per region

Feedback if chosen: This option is wrong — you gave the bond two pairs. One bond holds one shared pair; that pair’s density is split between the regions above and below the axis.

Check — multiple choice

In the ethene Lewis diagram shown, the C=C double bond is drawn as two lines. One of them is the sigma bond. What is the other?

AA second sigma bond, identical to the first

Feedback if chosen: This option is wrong — you doubled the sigma bond. The additional bond forms from the unhybridized p orbitals overlapping side-by-side: a pi bond.

BA pi bond, formed from side-by-side overlap of the two unhybridized p orbitals ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The additional bond of C=C is a pi bond, formed from side-by-side overlap of the unhybridized p orbital on each C.

CA pi bond, formed from the end-to-end overlap of the two sp² hybrid orbitals

Feedback if chosen: This option is wrong — right name, wrong overlap. A pi bond forms from side-by-side overlap of unhybridized p orbitals, not end-to-end overlap of hybrids.

DHalf of the sigma bond, drawn as its own line

Feedback if chosen: This option is wrong — you split one bond across two lines. Each line is a full bond holding its own shared pair.

U2-S15 · LESSON 134 · CSP-127

Classifying overlap: sigma or pi

Outcome: Classify a supplied orbital-overlap diagram as sigma or pi.

Part 1One question to ask

Here are the two kinds of overlap side by side.

To classify an overlap diagram, ask one question: where does the overlap sit relative to the internuclear axis?

Overlap on the axis — end-to-end overlap — is a sigma bond.

Overlap above and below the axis — orbitals sitting side-by-side — is a pi bond.

Both diagrams here show p orbitals, yet one bond is sigma and one is pi. The orbitals’ type does not decide; the position of the overlap does.

What you are expected to be able to do

Classify a supplied orbital-overlap diagram as sigma or pi from where its overlap sits relative to the internuclear axis.

Part 2Check what you just learned

Check — multiple choice

Classify the bond shown in the orbital-overlap diagram.

AA sigma bond

Feedback if chosen: This option is wrong — you placed the overlap on the axis. The shaded overlap sits above and below the internuclear axis — side-by-side overlap — so this is a pi bond.

BA pi bond ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The overlap sits above and below the internuclear axis — side-by-side overlap — so this is a pi bond.

CBoth a sigma bond and a pi bond

Feedback if chosen: This option is wrong — you counted the two shaded regions as two bonds. They are the two halves of one side-by-side overlap: one pi bond, and nothing overlaps on the axis itself.

DNeither — the orbitals shown do not overlap

Feedback if chosen: This option is wrong — the shaded regions are overlap. The two p orbitals overlap side-by-side, above and below the internuclear axis: a pi bond.

Check — multiple choice

Two p orbitals overlap as shown. Classify the bond.

AA pi bond

Feedback if chosen: This option is wrong — you classified by orbital type. p orbitals can form either bond; here the lobes overlap end-to-end and the overlap sits on the internuclear axis, so it is a sigma bond.

BNeither — the orbitals shown do not overlap

Feedback if chosen: This option is wrong — the shaded region between the nuclei is overlap. The two lobes overlap end-to-end on the internuclear axis.

CA sigma bond ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The overlap sits on the internuclear axis — end-to-end overlap — so this is a sigma bond, even though the orbitals are p orbitals.

DBoth a sigma bond and a pi bond

Feedback if chosen: This option is wrong — there is only one overlap region, and it sits on the axis. That is one sigma bond; a pi bond would need overlap above and below the axis.

Check — multiple choice

Two s orbitals overlap as shown. Classify the bond.

AA sigma bond ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The overlap sits on the internuclear axis between the nuclei — end-to-end overlap — so this is a sigma bond.

BA pi bond

Feedback if chosen: This option is wrong — you moved the overlap off the axis. A pi bond’s overlap sits above and below the internuclear axis; this overlap sits on it.

CBoth a sigma bond and a pi bond

Feedback if chosen: This option is wrong — there is one overlap region, on the axis: one sigma bond.

DNeither — s orbitals cannot form bonds

Feedback if chosen: This option is wrong — s orbitals form bonds whenever they overlap; the H–H bond in H₂ forms exactly this way.

U2-S15 · LESSON 135 · CSP-128

What a double bond is made of

Outcome: Identify a double bond as one sigma bond and one pi bond.

Part 1Two lines, two kinds of bond

Here is ethene again. The C=C double bond is drawn as two lines.

The two lines are not two copies of the same bond.

The first bond between the two C atoms forms end-to-end on the internuclear axis: it is a sigma bond.

The second forms from the unhybridized p orbitals overlapping side-by-side: it is a pi bond.

So a double bond is one sigma bond plus one pi bond.

This holds for every double bond, whichever atoms it joins.

What you are expected to be able to do

Identify a double bond as one sigma bond and one pi bond.

Part 2Check what you just learned

Check — multiple choice

In the O₂ Lewis diagram shown, the two O atoms are joined by a double bond. What is that double bond made of?

ATwo sigma bonds

Feedback if chosen: This option is wrong — you made both bonds sigma. The first bond forms end-to-end, and the additional bond forms from side-by-side p-orbital overlap: a pi bond.

BTwo pi bonds

Feedback if chosen: This option is wrong — you made both bonds pi. The first bond between the atoms forms end-to-end on the internuclear axis: a sigma bond.

COne sigma bond and one pi bond ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A double bond is one sigma bond (end-to-end) plus one pi bond (side-by-side).

DOne sigma bond and two pi bonds

Feedback if chosen: This option is wrong — you added a pi bond the diagram does not show. Two lines mean two bonds: the sigma bond plus one pi bond.

Check — multiple choice

In the formaldehyde (H₂CO) Lewis diagram shown, what is the C=O double bond made of?

ATwo sigma bonds

Feedback if chosen: This option is wrong — you made both bonds sigma. Only the first bond forms end-to-end; the additional bond is a pi bond from side-by-side overlap.

BOne pi bond and nothing else

Feedback if chosen: This option is wrong — you dropped the sigma bond. Every bonded pair of atoms has a sigma bond first; the pi bond is the addition on top of it.

CTwo pi bonds

Feedback if chosen: This option is wrong — you made both bonds pi. The first bond between C and O forms end-to-end on the internuclear axis: a sigma bond.

DOne sigma bond and one pi bond ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Like every double bond, C=O is one sigma bond plus one pi bond.

U2-S15 · LESSON 136 · CSP-129

What a triple bond is made of

Outcome: Identify a triple bond as one sigma bond and two pi bonds.

Part 1Three lines between two atoms

Here is ethyne, C₂H₂. The two C atoms are joined by a triple bond, drawn as three lines.

The first bond between the two C atoms forms end-to-end on the internuclear axis: one sigma bond.

Each of the other two forms from side-by-side overlap: two pi bonds.

So a triple bond is one sigma bond plus two pi bonds.

However many lines join two atoms, exactly one is the sigma bond; every additional line is a pi bond.

What you are expected to be able to do

Identify a triple bond as one sigma bond and two pi bonds.

Part 2Check what you just learned

Check — multiple choice

In the N₂ Lewis diagram shown, the two N atoms are joined by a triple bond. What is it made of?

AThree sigma bonds

Feedback if chosen: This option is wrong — you counted every line as end-to-end. Only the first bond forms on the internuclear axis; the other two form side-by-side.

BOne sigma bond and two pi bonds ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A triple bond is one sigma bond (end-to-end) plus two pi bonds (side-by-side).

CTwo sigma bonds and one pi bond

Feedback if chosen: This option is wrong — you gave the pair a second sigma bond. Between two atoms, exactly one bond forms end-to-end; both additional bonds are pi bonds.

DThree pi bonds

Feedback if chosen: This option is wrong — you left out the sigma bond. The first bond forms end-to-end on the internuclear axis.

Check — numeric entry

The C and N atoms in hydrogen cyanide, HCN, are joined by a triple bond. How many pi bonds does that triple bond contain?

✓ ANSWER KEY: 2

Feedback when correct: Correct. One of the three bonds is the sigma bond; the other two are pi bonds.

Anticipated wrong answers (feedback shown):

3	This answer counts every line of the triple bond as a pi bond. The first bond is the sigma bond; the pi bonds are the other two.
1	This answer gives the triple bond a double bond’s share. A double bond has one pi bond; a triple bond has two.
0	A triple bond is more than its sigma bond: the two additional bonds are pi bonds.

U2-S15 · LESSON 137 · CSP-130

Counting sigma and pi bonds

Outcome: Calculate the total number of sigma and pi bonds in a supplied Lewis diagram.

Part 1The counting rule

You’ve already seen what each bond is made of: a single bond is one sigma bond; a double bond is one sigma plus one pi; a triple bond is one sigma plus two pi.

To count sigma and pi bonds in a Lewis diagram, work through the pairs of bonded atoms one at a time.

Each pair of bonded atoms contributes exactly one sigma bond, whatever its bond order.

Every line beyond the first between a pair — the second line of a double bond, the second and third of a triple bond — is one pi bond.

Count ethene: it has four C–H single bonds and one C=C double bond.

Sigma bonds: one per C–H pair is four, plus one from the C=C pair — 5 sigma bonds.

Pi bonds: the second line of C=C — 1 pi bond.

What you are expected to be able to do

Calculate the total number of sigma and pi bonds in a supplied Lewis diagram.

Part 2Check what you just learned

Check — numeric entry

How many sigma bonds does the ethyne Lewis diagram shown contain?

✓ ANSWER KEY: 3

Feedback when correct: Correct. Each pair of bonded atoms contributes one sigma bond: H–C, C≡C, and C–H — 3 sigma bonds.

Anticipated wrong answers (feedback shown):

5	This answer counts every line as a sigma bond. A pair of bonded atoms contributes exactly one sigma bond, so the triple bond contributes one — not three.
2	This answer skips the triple bond. The C≡C pair also contributes exactly one sigma bond: 2 + 1 = 3.
4	This answer gives the triple bond two sigma bonds. Each pair of bonded atoms contributes exactly one: two C–H pairs and one C–C pair make 3.

Check — numeric entry

How many pi bonds does the CO₂ Lewis diagram shown contain?

✓ ANSWER KEY: 2

Feedback when correct: Correct. Each C=O double bond is one sigma plus one pi; two double bonds give 2 pi bonds.

Anticipated wrong answers (feedback shown):

4	This answer counts every line of the two double bonds as a pi bond. The first line of each double bond is its sigma bond, so each C=O contributes one pi bond.
1	This answer takes one pi bond for the whole molecule. Each double bond contributes its own pi bond, and CO₂ has two double bonds.
0	CO₂ has no single bonds, but each C=O double bond carries one pi bond on top of its sigma bond: 2 in all.

Check — numeric entry

How many sigma bonds does the HCN Lewis diagram shown contain?

✓ ANSWER KEY: 2

Feedback when correct: Correct. Two pairs of bonded atoms — H–C and C≡N — contribute one sigma bond each.

Anticipated wrong answers (feedback shown):

4	This answer counts every line as a sigma bond. The C≡N pair contributes exactly one sigma bond; its other two lines are pi bonds.
1	This answer skips the triple bond. The C≡N pair also contributes exactly one sigma bond: 1 + 1 = 2.
3	This answer counts the triple bond’s lines and skips the H–C bond. Count pairs instead: H–C and C≡N give one sigma bond each — 2.

Check — numeric entry

How many pi bonds does the same HCN Lewis diagram contain?

✓ ANSWER KEY: 2

Feedback when correct: Correct. The H–C single bond has no pi bond; the C≡N triple bond carries two.

Anticipated wrong answers (feedback shown):

3	This answer counts every line of the triple bond as a pi bond. One of the three is the sigma bond; the triple bond carries two pi bonds.
1	This answer gives C≡N a double bond’s share. A triple bond carries two pi bonds.
4	This answer counts every line in the diagram as a pi bond. The H–C line and the first line of C≡N are sigma bonds; only the remaining two are pi bonds.

U2-S15 · LESSON 138 · CSP-131

Comparing sigma and pi bond strength

Outcome: State that a sigma bond generally has greater bond energy than a corresponding pi bond between the same two atoms.

Part 1Comparing the two components

You’ve already seen bond energy: the energy required to pull a bonded atom pair completely apart.

The sigma and pi components of a double bond each contribute part of its bond energy, so the two components can be compared.

Between the same two atoms, the sigma bond generally has greater bond energy than the pi bond.

Generally, not always: for some pairings of atoms — C and O is one — the pi bond turns out stronger.

Here are average bond energies between two C atoms.

Average bond energies between two C atoms
Bond	Components	Bond energy (kJ/mol)
C–C	one sigma bond	347
C=C	one sigma bond + one pi bond	614

The C=C double bond adds a pi bond to the sigma bond, and the addition raises the bond energy from 347 to 614 kJ/mol. The pi bond’s share is 614 − 347 = 267 kJ/mol.

267 kJ/mol is less than the sigma bond’s 347 kJ/mol: between two C atoms, the sigma bond is the stronger component.

What you are expected to be able to do

State that a sigma bond generally has greater bond energy than a corresponding pi bond between the same two atoms.

Part 2Check what you just learned

Check — multiple choice

In the C=C double bond of ethene, which component generally has the greater bond energy?

AThe sigma bond ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Between the same two atoms, the sigma bond generally has greater bond energy than the pi bond — for C=C, 347 against 267 kJ/mol.

BThe pi bond

Feedback if chosen: This option is wrong — you picked the weaker component. Between the same two C atoms, the pi bond contributes less bond energy than the sigma bond.

CThe two components have equal bond energy

Feedback if chosen: This option is wrong — you treated the two lines as identical. They form differently, and their shares of the bond energy differ: 347 against 267 kJ/mol for C=C.

DNo comparison is possible — only the whole double bond has a bond energy

Feedback if chosen: This option is wrong — the components’ shares can be separated: the pi bond’s share is the rise from C–C to C=C, and it is smaller than the sigma bond’s.

Check — multiple choice

The rule you’ve just seen compares bond components between the same two atoms. Which comparison does it cover?

AA sigma bond between two C atoms against a pi bond between a C atom and an O atom

Feedback if chosen: This option is wrong — you crossed atom pairs. The rule compares sigma against pi between the same two atoms; C-and-C against C-and-O is outside it.

BA sigma bond in H–H against a pi bond in C=C

Feedback if chosen: This option is wrong — you crossed atom pairs (H-and-H against C-and-C). The rule only compares components between the same two atoms.

CA sigma bond between two C atoms against another sigma bond between two N atoms

Feedback if chosen: This option is wrong — you compared sigma against sigma. The rule compares a sigma bond against a corresponding pi bond.

DThe sigma component and the pi component of one C=C double bond ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both components join the same two C atoms, so the rule applies: the sigma component is generally the stronger.

Check — multiple choice

Average bond energies between two C atoms
Bond	Components	Bond energy (kJ/mol)
C–C	one sigma bond	347
C=C	one sigma bond + one pi bond	614

Use the supplied bond energies. Which statement do the data support?

AThe pi component of C=C contributes 267 kJ/mol — less than the sigma component’s 347 kJ/mol. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The pi share is the rise from C–C to C=C: 614 − 347 = 267 kJ/mol, which is less than 347 kJ/mol.

BThe pi component of C=C contributes more bond energy than the sigma component, because 614 is greater than 347.

Feedback if chosen: This option is wrong — you compared the whole double bond (614) with the sigma bond (347). The pi component’s share is the difference, 614 − 347 = 267 kJ/mol.

CThe sigma and pi components of C=C contribute 307 kJ/mol each.

Feedback if chosen: This option is wrong — you split 614 evenly. The components are not equal: the sigma bond contributes 347 kJ/mol, and the pi bond the remaining 267.

DThe data show the C=C double bond is exactly twice as strong as the C–C single bond.

Feedback if chosen: This option is wrong — twice 347 is 694, not 614. The double bond is stronger than the single bond but less than twice as strong, because the added pi bond is the weaker component.

U2-S15 · LESSON 139 · CSP-132

Why sigma bonds are generally stronger

Outcome: Explain why sigma bonding is generally stronger than corresponding pi bonding.

Part 1Comparing the two overlaps

Here are the two overlaps again, drawn with the same pair of p orbitals.

On the left, the two p orbitals point directly at each other along the internuclear axis. Because each lobe approaches the other tip-first, a large part of each lobe reaches the other: the end-to-end overlap is extensive.

On the right, the two p orbitals run parallel. Only the edges of their lobes reach each other, so the side-by-side overlap is less extensive.

The more extensive the overlap, the more shared electron density sits between the two nuclei, attracting both — and the stronger the bond.

That is why, between the same two atoms, the sigma bond is generally stronger than the pi bond: end-to-end overlap is more extensive than side-by-side overlap.

What you are expected to be able to do

Explain that sigma bonding is generally stronger than corresponding pi bonding because end-to-end overlap is more extensive than side-by-side overlap.

Part 2Check what you just learned

Check — multiple choice

Why does the sigma component of a C=C double bond generally have greater bond energy than the pi component?

AThe sigma bond holds more electrons than the pi bond.

Feedback if chosen: This option is wrong — you gave the sigma bond extra electrons. Each component holds one shared pair; the difference is in how extensively the orbitals overlap.

BSigma bonds form from s orbitals, which are held more tightly than p orbitals.

Feedback if chosen: This option is wrong — you classified by orbital type. A sigma bond can form from p or hybrid orbitals too; the strength difference comes from the overlap, not the orbital type.

CThe end-to-end overlap of the sigma bond is more extensive than the side-by-side overlap of the pi bond. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Because the orbitals point directly at each other, the end-to-end overlap is more extensive, putting more shared electron density between the nuclei.

DThe sigma bond forms first, and the first bond formed between two atoms is always the strongest one.

Feedback if chosen: This option is wrong — you invented a rule about forming first. The comparison rests on overlap: end-to-end overlap is more extensive than side-by-side overlap.

Check — multiple choice

Use the diagram. Which feature of the two overlaps explains why the sigma bond is generally the stronger?

AThe pi overlap has two shaded regions, and two regions make a more extensive overlap.

Feedback if chosen: This option is wrong — you counted regions as extent. The two pi regions are edge-to-edge brushes of parallel lobes; together they are still less extensive than the direct end-to-end overlap.

BThe sigma overlap sits on the internuclear axis, where the orbitals point directly at each other, so it is more extensive. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Orbitals pointing directly at each other along the axis overlap more extensively than parallel orbitals overlapping edge-to-edge — and more extensive overlap means a stronger bond.

CThe two overlaps are equally extensive, so the difference must come from electron count.

Feedback if chosen: This option is wrong — the overlaps differ. Orbitals pointing directly at each other overlap more extensively than parallel orbitals meeting at their edges.

DThe sigma overlap sits closer to the two nuclei, and distance from the nuclei alone sets bond strength.

Feedback if chosen: This option is wrong — you swapped in a distance rule. What sets the difference is how extensive the overlap is, and end-to-end overlap is the more extensive.

U2-S15 · LESSON 140 · CSP-133

Rotation about a sigma bond

Outcome: Explain why rotation about a sigma bond can preserve its end-to-end overlap.

Part 1Spinning without breaking

Look at a sigma bond’s end-to-end overlap. The overlap sits on the internuclear axis itself.

Because the overlap is symmetric around the axis, rotating either atom about the axis leaves the overlap exactly as it was.

So rotation about a sigma bond can happen without breaking the bond.

Here is ethane, CH₃–CH₃. The bond between the two C atoms is a single bond: one sigma bond.

The two CH₃ groups can rotate about the C–C bond, and the end-to-end overlap is preserved through every angle of the turn.

What you are expected to be able to do

Explain that rotation about a sigma bond can preserve its end-to-end overlap because the overlap is symmetric around the internuclear axis.

Part 2Check what you just learned

Check — multiple choice

Why can the two CH₃ groups of ethane rotate about the C–C bond without breaking it?

AThe C–C bond is strong enough to survive the strain of rotation.

Feedback if chosen: This option is wrong — you answered with strength. Rotation is possible because the overlap is unchanged by turning about the axis, not because the bond withstands damage.

BThe H atoms are small enough to swing past one another.

Feedback if chosen: This option is wrong — you answered with atom size. The reason lies in the bond itself: its overlap is symmetric around the axis, so rotation does not disturb it.

CThe shared pair releases the atoms during the rotation and re-forms the bond once the turn is complete.

Feedback if chosen: This option is wrong — you broke the bond. Nothing lets go: the end-to-end overlap looks the same at every angle of the turn, so the bond persists throughout.

DThe sigma bond’s end-to-end overlap is symmetric around the internuclear axis, so rotation leaves it unchanged. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The overlap sits on the axis and is symmetric around it, so turning either group leaves the overlap — and the bond — intact.

Check — multiple choice

Methanol, CH₃–OH, contains a single C–O bond. Can the CH₃ group rotate about that bond without breaking it?

AYes — the C–O sigma bond’s overlap is symmetric around the internuclear axis, so rotation leaves it unchanged. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A single bond is a sigma bond, and its end-to-end overlap is symmetric around the axis — whichever atoms it joins.

BNo — C and O are different elements, so the overlap between their orbitals is not symmetric around the axis.

Feedback if chosen: This option is wrong — you tied the symmetry to the atoms’ identities. The symmetry that matters is around the axis: end-to-end overlap looks the same at every angle, whichever atoms it joins.

CYes — but only because O carries lone pairs that steady the bond during the turn.

Feedback if chosen: This option is wrong — you credited the lone pairs. They play no part; the single bond is a sigma bond, and its overlap is symmetric around the axis.

DNo — rotation about any covalent bond breaks it.

Feedback if chosen: This option is wrong — you made all rotation bond-breaking. Rotation about a sigma bond preserves its end-to-end overlap.

U2-S15 · LESSON 141 · CSP-134

Why a double bond resists rotation

Outcome: Explain why the pi component of a double bond restricts rotation.

Part 1A bond that locks

Look at a pi bond’s side-by-side overlap. It exists only while the two p orbitals stay parallel.

Now imagine rotating one CH₂ group of ethene by 90° about the C=C bond. The group’s p orbital turns with it.

With one p orbital turned 90°, the two are no longer parallel: the side-by-side overlap is destroyed, and the pi bond is broken.

Breaking the pi bond costs energy, so the rotation does not happen on its own: the pi component restricts rotation about the double bond.

The sigma component alone would allow the turn — its overlap is symmetric around the axis. It is the pi component that locks the bond.

What you are expected to be able to do

Explain that rotation about a double bond would disrupt the side-by-side p-orbital overlap, so the pi component restricts rotation.

Part 2Check what you just learned

Check — multiple choice

Why is rotation about the C=C double bond of ethene restricted?

ARotating one end would break the sigma component, whose overlap cannot survive any turn about the axis.

Feedback if chosen: This option is wrong — you blamed the sigma component. Its end-to-end overlap is symmetric around the axis and would survive the turn; it is the pi bond’s side-by-side overlap that cannot.

BRotating one end would turn its p orbital out of parallel and destroy the pi bond’s side-by-side overlap. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The turn would leave the two p orbitals out of parallel, destroying the side-by-side overlap — and breaking the pi bond costs energy.

CThe four H atoms collide with one another when the ends turn.

Feedback if chosen: This option is wrong — you answered with collisions. The restriction comes from the bond itself: the turn would destroy the pi bond’s side-by-side overlap.

DA double bond holds two shared pairs, and four electrons are too many to move.

Feedback if chosen: This option is wrong — you counted electrons. The number of pairs is not the point; the pi bond’s side-by-side overlap needs parallel p orbitals, and rotation removes that.

Check — multiple choice

The ends of ethane (CH₃–CH₃) rotate freely about the C–C bond, but the ends of ethene (CH₂=CH₂) do not. Why?

AEthene is a flat molecule, and flat molecules cannot rotate.

Feedback if chosen: This option is wrong — you argued from the molecule’s shape. The restriction comes from the pi bond: rotating one end would destroy its side-by-side overlap.

BEthane’s C–C bond is weaker than ethene’s C=C, and weaker bonds leave the ends free to rotate.

Feedback if chosen: This option is wrong — you argued from strength. Rotation about ethane’s C–C is free because sigma overlap is symmetric around the axis, not because the bond is weak.

CEthane’s C–C is a sigma bond only; ethene’s C=C adds a pi bond that the turn would destroy. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Sigma overlap survives the turn; the pi bond’s side-by-side overlap does not — so the double bond is locked while the single bond is free.

DEthane’s H atoms are farther from the C–C bond, leaving more room to turn.

Feedback if chosen: This option is wrong — you argued from crowding. The difference is the pi bond: ethene has one, ethane does not.

U2-S15 · LESSON 142 · CSP-135

Geometric isomers

Outcome: State that geometric isomers have the same connectivity but different fixed spatial arrangements around a rotation-restricted bond.

Part 1Same connections, different shapes

Here is a molecule called cis-2-butene. Each C of the C=C double bond is bonded to one CH₃ group and one H, and the two CH₃ groups sit on the same side of the double bond.

Here is trans-2-butene. Each double-bond C is again bonded to one CH₃ group and one H — but the two CH₃ groups sit on opposite sides.

Compare the connectivity — which atom is bonded to which. It is identical in the two molecules.

You’ve already seen that the pi component restricts rotation about a double bond. Neither molecule can turn one end to become the other, so each arrangement is fixed.

Molecules with the same connectivity but different fixed arrangements around a rotation-restricted bond are called geometric isomers.

The name cis marks the form with the matching groups on the same side; trans marks the form with them on opposite sides.

What you are expected to be able to do

State that geometric isomers have the same connectivity but different fixed spatial arrangements around a rotation-restricted bond.

Part 2Check what you just learned

Check — multiple choice

What makes cis-2-butene and trans-2-butene geometric isomers?

AThey have different connectivity: the atoms in one are bonded to different partners than in the other.

Feedback if chosen: This option is wrong — you changed the connectivity. In both molecules each double-bond C is bonded to one CH₃ group and one H; what differs is the fixed arrangement.

BThey have different molecular formulas.

Feedback if chosen: This option is wrong — both are C₄H₈. Geometric isomers share a formula and connectivity; the difference is the arrangement around the double bond.

CThey are the same molecule, drawn at different angles on the page.

Feedback if chosen: This option is wrong — you flattened the difference into drawing style. No turn of the page puts same-side CH₃ groups on opposite sides; the two arrangements are genuinely different molecules.

DThey have the same connectivity but different fixed arrangements around the rotation-restricted double bond. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Same connectivity, different fixed arrangements around the rotation-restricted double bond — the definition of geometric isomers.

Check — multiple choice

The two structures shown have a CH₃ group and an H on one double-bond C, and a C₂H₅ group and an H on the other. Are they geometric isomers?

AYes — same connectivity, different fixed arrangements around the rotation-restricted double bond. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each atom keeps the same partners in both structures; what differs is which side of the double bond the CH₃ and C₂H₅ groups occupy, and the pi bond keeps each arrangement fixed.

BNo — they are the same molecule drawn twice.

Feedback if chosen: This option is wrong — you merged the two arrangements. In the left structure the CH₃ and C₂H₅ groups share a side; in the right they do not — and the double bond cannot rotate one into the other.

CNo — geometric isomers must have different connectivity.

Feedback if chosen: This option is wrong — you inverted the definition. Geometric isomers have the same connectivity; it is the fixed arrangement around the double bond that differs.

DYes — any two molecules that share the same molecular formula count as geometric isomers.

Feedback if chosen: This option is wrong — a shared formula is not enough. Geometric isomers must also share connectivity and differ only in the fixed arrangement around a rotation-restricted bond.

U2-S15 · LESSON 143 · CSP-136

When a double bond gives cis and trans forms

Outcome: Predict whether a carbon-carbon double bond produces a cis-trans pair from its substituents.

Part 1The substituent test

Each double-bond C holds two attached atoms or groups. Call them its substituents.

Here are the two forms of CHCl=CHCl. On each double-bond C, the substituents are one H and one Cl — two different substituents.

Because each C carries two different substituents, the same-side and opposite-side arrangements are genuinely different molecules: a cis-trans pair exists.

Now look at CH₂=CHCl. The right-hand C carries H and Cl. The left-hand C carries H and H — two identical substituents.

Swap the two H atoms on the left-hand C and nothing changes: there is only one arrangement, so no cis-trans pair exists.

The rule: a C=C double bond produces a cis-trans pair only when each double-bond C carries two different substituents.

One C with two identical substituents is enough to rule the pair out.

A cis-trans pair is a pair of geometric isomers — the term you've already seen, and the term the AP exam uses.

This substituent test goes beyond what the AP exam asks. The exam expects only that the pi bond restricts rotation, and that restricted rotation leads to geometric isomers. Keep the test anyway — it tells you when a pair actually exists before you name one.

What you are expected to be able to do

Predict whether a C=C double bond produces a cis-trans pair by checking the substituents on each double-bond C.

Part 2Check what you just learned

Check — multiple choice

Use the structure shown. Does the molecule have a cis-trans pair?

AYes — the right-hand C carries two different substituents.

Feedback if chosen: This option is wrong — you checked only one C. The test must pass on each double-bond C, and the left-hand C carries two identical H atoms.

BNo — the left-hand C carries two identical H substituents. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Swapping the two H atoms on the left-hand C changes nothing, so there is only one arrangement — no cis-trans pair.

CYes — every C=C double bond produces a cis-trans pair.

Feedback if chosen: This option is wrong — the double bond alone is not enough. Each double-bond C must carry two different substituents, and the left-hand C here does not.

DNo — a cis-trans pair needs the same group on both double-bond C atoms.

Feedback if chosen: This option is wrong — you invented a matching rule. The substituents need not match across the two C atoms; each C just needs its own two to differ.

Check — multiple choice

Does the structure shown have a cis-trans pair?

AYes — the molecule contains three CH₃ groups, and CH₃ groups create cis-trans pairs.

Feedback if chosen: This option is wrong — you argued from the groups themselves. What matters is whether each double-bond C carries two different substituents, and the left-hand C carries two identical CH₃ groups.

BYes — the right-hand C carries two different substituents.

Feedback if chosen: This option is wrong — you checked only one C. The other double-bond C carries two identical CH₃ groups, and one such C rules the pair out.

CNo — the left-hand C carries two identical CH₃ substituents. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Swapping the left-hand C’s two CH₃ groups changes nothing, so the two would-be forms are one molecule.

DYes — every carbon-carbon double bond gives a cis and a trans form.

Feedback if chosen: This option is wrong — the double bond alone is not enough. Each double-bond C must also carry two different substituents, and the left-hand C here carries two identical CH₃ groups.

Check — multiple choice

Predict: does the molecule shown have a cis-trans pair?

ANo — the two C atoms carry different substituent sets from each other.

Feedback if chosen: This option is wrong — you compared across the two C atoms. The test is within each C: one carries H and F, the other H and Cl — both pass, so the pair exists.

BYes — the molecule contains halogen atoms, which force fixed arrangements.

Feedback if chosen: This option is wrong — you credited the halogens. The fixed arrangements come from the double bond’s restricted rotation; the pair exists because each C carries two different substituents.

CNo — a cis-trans pair needs a CH₃ group on each double-bond C.

Feedback if chosen: This option is wrong — you overfitted to 2-butene. Any two different substituents will do; CH₃ groups are not required.

DYes — each double-bond C carries two different substituents. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. H and F differ on one C, and H and Cl differ on the other — both C atoms pass the test, so same-side and opposite-side forms are different molecules.

U2-S15 · LESSON 144 · CSP-137

Why cis and trans forms stay distinct

Outcome: Explain how pi overlap restricts double-bond rotation and preserves distinct geometric isomers.

Part 1The whole chain

Here are cis-2-butene and trans-2-butene again: same connectivity, different fixed arrangements.

First, the substituent condition: each double-bond C carries one CH₃ group and one H — two different substituents. So two genuinely different arrangements exist.

Turning one form into the other would require rotating one end of the molecule about the C=C bond.

Partway through that turn, the rotating C’s p orbital is no longer parallel to its partner: the side-by-side overlap is destroyed, and the pi bond breaks.

Breaking the pi bond costs energy, so under ordinary conditions the turn does not happen — each molecule keeps its arrangement.

That is the whole chain: side-by-side pi overlap restricts rotation; restricted rotation fixes each arrangement; and because each double-bond C carries two different substituents, the two fixed arrangements are distinct geometric isomers.

What you are expected to be able to do

Explain how side-by-side pi overlap restricts double-bond rotation and preserves distinct geometric isomers when each double-bond C carries two different substituents.

Part 2Check what you just learned

Check — multiple choice

Why can cis-2-butene not turn into trans-2-butene under ordinary conditions?

ARotating one end would destroy the pi bond’s side-by-side overlap, and breaking that bond costs energy. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The turn would leave the two p orbitals out of parallel, destroying the side-by-side overlap — and that energy cost keeps each form fixed.

BRotating one end would break the sigma bond that joins the two C atoms along the internuclear axis.

Feedback if chosen: This option is wrong — you blamed the sigma bond. Its overlap is symmetric around the axis and would survive the turn; the pi bond’s side-by-side overlap would not.

CThe two molecules have different connectivity, so no rotation could turn one into the other.

Feedback if chosen: This option is wrong — the connectivity is the same in both. The barrier is the pi bond: the turn would destroy its side-by-side overlap.

DThe CH₃ groups are too heavy for the molecule to swing around.

Feedback if chosen: This option is wrong — you answered with mass. The barrier is the energy cost of breaking the pi bond’s side-by-side overlap.

Check — multiple choice

cis-CHCl=CHCl and trans-CHCl=CHCl are distinct, stable compounds. Which pair of facts explains this?

AThe sigma bond restricts rotation, and each double-bond C carries two different substituents.

Feedback if chosen: This option is wrong — you gave the sigma bond the locking role. Its overlap is symmetric around the axis and permits rotation; the pi bond is what restricts it.

BThe pi bond restricts rotation, and the two C atoms carry identical substituents.

Feedback if chosen: This option is wrong — half right. The pi bond does restrict rotation, but each C here carries H and Cl — different substituents, which is exactly why the two arrangements are distinct compounds.

CEach double-bond C carries two different substituents, and the sigma bond’s symmetric overlap holds them in place.

Feedback if chosen: This option is wrong — you swapped the bonds’ roles. Symmetric sigma overlap allows rotation; the fixing comes from the pi bond’s side-by-side overlap.

DThe pi bond restricts rotation, and each double-bond C carries two different substituents. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Restricted rotation from the pi bond, plus two different substituents (H and Cl) on each C, keeps the same-side and opposite-side forms distinct.

Check — multiple choice

In the molecule shown, the two CH₂Cl groups are joined by a single bond. Same-side and opposite-side drawings of its Cl atoms do not count as distinct compounds. Why not?

AThe two Cl atoms repel each other, forcing the molecule into a single fixed arrangement.

Feedback if chosen: This option is wrong — you fixed the arrangement instead of freeing it. With only a sigma bond, the groups rotate freely, so no arrangement stays fixed.

BThe C–C single bond is too weak to hold either arrangement of the Cl atoms in place.

Feedback if chosen: This option is wrong — you argued from strength. The bond survives rotation because sigma overlap is symmetric around the axis; that free rotation is what merges the drawings.

CRotation about the C–C sigma bond is free, so the two drawings interconvert and are one compound. ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Without a pi bond there is no restriction: rotation about the sigma bond turns one drawing into the other, so they are the same compound.

DDistinct forms would need a third, different substituent attached to each of the two C atoms.

Feedback if chosen: This option is wrong — you changed the substituent rule. The reason is the missing pi bond: nothing restricts rotation about a single bond.

AFTER U2-S15 · CHECKPOINT · U2-G5 · 6 items

Checkpoint 5 — Hybridization, orbital overlap, and cis-trans isomers

Check — multiple choice

Use the Lewis diagram of CO₂ shown. Which hybridization describes the C atom?

Asp³ — C shares four electron pairs, and each shared pair is its own domain

Feedback if chosen: This option is wrong — you counted shared pairs instead of domains. A multiple bond counts as one domain no matter how many pairs it holds, so C has two domains: sp.

BIt cannot be assigned from the Lewis diagram — the measured bond angle is needed first

Feedback if chosen: This option is wrong — hybridization is assigned from the electron-domain count, which the Lewis diagram supplies directly: two domains around C give sp.

Csp² — the two double bonds and the two O atoms give C three regions to point toward

Feedback if chosen: This option is wrong — you invented a third region. Each double bond is one electron domain, so C in CO₂ has two domains, and two domains correspond to sp.

Dsp — C has two electron domains: each double bond is one domain ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each double bond is one electron domain, and C has no lone pairs, so C has two domains — sp hybridization.

Check — multiple choice

1,2-difluoroethene, CHF=CHF, exists as separate cis and trans compounds that do not interconvert under ordinary conditions. Why?

ATwisting one end would break the C–C sigma bond, which is too strong for rotation

Feedback if chosen: This option is wrong — the sigma bond's end-to-end overlap is symmetric about the internuclear axis and survives rotation; it is the pi overlap that twisting would destroy.

BTwisting one end would pull the pi bond's side-by-side overlap apart, and each double-bond C carries two different substituents (H and F) ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Rotation would disrupt the side-by-side p-orbital overlap of the pi bond, which costs too much energy — and because each double-bond C carries two different substituents, the two frozen arrangements really are distinct compounds.

CThe two F atoms attract each other strongly enough to hold each molecule in one fixed arrangement

Feedback if chosen: This option is wrong — no attraction between the F atoms locks the shape. The restriction comes from the pi bond's overlap geometry, not from the substituents pulling on each other.

DThe cis and trans forms have different atom connectivity, so no rotation could turn one into the other

Feedback if chosen: This option is wrong — the two forms have identical connectivity. They differ only in the arrangement of the groups around the rotation-restricted double bond.

Check — multiple choice

A student classifies the bond in the orbital-overlap diagram shown as a pi bond, because it is built from p orbitals. Evaluate the claim.

AThe student is right — bonds formed from p orbitals are pi bonds

Feedback if chosen: This option is wrong — the orbital type does not decide the classification. p orbitals that point at each other overlap end-to-end on the internuclear axis, which makes a sigma bond.

BThe student is right — the overlap region lies between the two nuclei, which is what pi means

Feedback if chosen: This option is wrong — overlap between the nuclei on the axis is the sigma arrangement. Pi overlap sits side-by-side, above and below the axis.

CThe student is wrong — this is a sigma bond, because the orbitals overlap end-to-end and the overlap sits on the internuclear axis ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The classification comes from where the overlap sits: end-to-end on the internuclear axis is sigma, whatever orbitals it is built from.

DThe student is wrong — this is both a sigma and a pi bond, because each p orbital has two lobes

Feedback if chosen: This option is wrong — one overlap makes one bond. The diagram shows a single overlap region on the axis, so this is one sigma bond, not a sigma plus a pi.

Check — numeric entry

How many sigma bonds does the propene Lewis diagram shown contain?

✓ ANSWER KEY: 8

Feedback when correct: Correct. The six C–H bonds and the C–C single bond are one sigma each, and the C=C double bond holds one sigma plus one pi: 7 + 1 = 8 sigma bonds.

Anticipated wrong answers (feedback shown):

9	This counts both lines of the double bond as sigma bonds. A double bond is one sigma and one pi, so the total is 8.
7	This leaves the double bond out of the sigma count. A double bond contains one sigma bond (plus one pi), so add it: 8.
2	This counts only the carbon–carbon bonds. Every bond contains one sigma bond, including each C–H bond: 8.

Check — multiple choice

Average bond energies between two C atoms
Bond	Bond energy (kJ/mol)
C–C (one sigma bond)	347
C≡C (one sigma bond + two pi bonds)	839

In ethyne, HC≡CH, the C≡C triple bond contains one sigma bond and two pi bonds. Use the supplied bond energies to estimate each component's share. Which statement do the data support?

AEach pi bond contributes about 246 kJ/mol, so the sigma bond is the strongest component ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The C–C single bond is the sigma bond alone: 347 kJ/mol. The two pi bonds add 839 − 347 = 492 kJ/mol, so each pi bond contributes 492/2 = 246 kJ/mol — less than the sigma bond's 347.

BEach of the three components contributes about 280 kJ/mol, so the sigma and pi bonds are equally strong

Feedback if chosen: This option is wrong — you split the 839 kJ/mol evenly across the three components. The C–C single bond shows the sigma component alone: 347 kJ/mol, more than an even third.

CEach pi bond contributes about 492 kJ/mol, so each pi bond is stronger than the sigma bond

Feedback if chosen: This option is wrong — you gave the whole addition to one pi bond. The triple bond holds two pi bonds, so the extra 839 − 347 = 492 kJ/mol is shared between them: 246 kJ/mol each.

DThe data cannot separate the sigma contribution from the pi contributions

Feedback if chosen: This option is wrong — the single-bond row isolates the sigma component at 347 kJ/mol. Subtracting it from the triple bond's 839 kJ/mol leaves the two pi bonds' share.

Check — multiple choice

Use the Lewis diagram shown. Could this molecule exist as a cis-trans pair?

AYes — the right-hand C carries two different substituents (H and Cl)

Feedback if chosen: This option is wrong — the condition must hold at both ends of the double bond. The right-hand C passes, but the left-hand C carries two Cl atoms, so swapping them changes nothing.

BNo — the left-hand C carries two identical Cl substituents ✓ KEY — this is the correct answer

Feedback if chosen: Correct. With two identical Cl substituents on the left-hand C, exchanging them gives back the same molecule — no distinct cis and trans arrangements exist.

CYes — a double bond flanked by three halogen atoms is locked in place

Feedback if chosen: This option is wrong — halogens do not lock a molecule by themselves. The test is whether each double-bond C carries two different substituents, and the left-hand C fails it.

DNo — a cis-trans pair requires a CH₃ group on each double-bond carbon

Feedback if chosen: This option is wrong — the substituents can be any groups. What matters is that the two substituents on each double-bond C differ from each other.

U2-S16 — Molecular polarity

7 lessons

U2-S16 · LESSON 145 · CSP-138

The molecular dipole moment

Outcome: State that a molecular dipole moment is the vector sum of all the bond dipoles in the molecule's three-dimensional geometry.

Part 1One molecule, one dipole

Hold a charged rod near a thin stream of water: the stream bends toward the rod.

One explanation: each water molecule as a whole has an electron-rich end and an electron-poor end — a dipole belonging to the whole molecule, not just to one bond.

Here is H₂O with its two bond dipoles drawn. You’ve already seen that each polar bond carries a bond dipole, drawn as an arrow pointing toward the more electronegative atom — here, toward O.

You’ve already seen the geometry too: four electron domains with two lone pairs give H₂O its bent molecular geometry, so the two arrows sit at an angle.

Part 2Adding the arrows

To find the molecule’s overall dipole, add the arrows — with direction counted, not just size.

Two equal arrows pointing in exactly opposite directions add to zero: the pulls undo each other.

Two arrows at an angle do not undo each other: they add to a single arrow pointing between them.

Adding arrows with their directions counted is called a vector sum.

In H₂O, both arrows point partly toward O, so their vector sum is a single arrow pointing at O — drawn beneath the molecule and labeled net dipole.

This vector sum of all the bond dipoles is the molecule’s overall dipole: the molecular dipole moment.

The sum must be taken in the molecule’s three-dimensional geometry, because the geometry sets the direction of every arrow.

The molecular dipole moment gives each water molecule an electron-rich end and an electron-poor end for the charged rod to pull on. Support, not proof: a charged rod can bend other liquid streams too, so the bend suggests the molecular dipole rather than proving it.

What you are expected to be able to do

State that a molecular dipole moment is the vector sum of all the bond dipoles in the molecule's three-dimensional geometry.

Part 3Check what you just learned

Check — multiple choice

Which of these is a molecule’s dipole moment?

AThe bond dipole of the single most polar bond in the molecule

Feedback if chosen: This option is wrong — you kept only one bond. Every bond dipole contributes to the sum; the molecular dipole moment combines all of them.

BThe total number of polar bonds the molecule contains

Feedback if chosen: This option is wrong — you counted the polar bonds instead of adding their dipoles. A count cannot cancel or reinforce; the dipole moment is a vector sum, where directions decide the result.

CThe sum of the bond-dipole sizes, with the directions of the arrows ignored

Feedback if chosen: This option is wrong — you dropped the directions. Direction is what lets arrows cancel or reinforce; ignoring it would give CO₂ and H₂O the same kind of result, and they are opposites.

DThe vector sum of all the bond dipoles in the molecule’s three-dimensional geometry ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Each bond dipole is an arrow, and the molecule’s dipole moment is the vector sum of all of them, taken in the three-dimensional geometry that sets each arrow’s direction.

Check — multiple choice

The diagram shows bent SO₂ with its two bond-dipole arrows drawn, each pointing toward an O atom. What is the vector sum of the two arrows?

AZero — the two arrows point in different directions, so they cancel

Feedback if chosen: This option is wrong — you treated any two directions as opposites. Only equal arrows pointing in exactly opposite directions add to zero; these two point partly the same way — both away from S.

BA single arrow pointing from S toward the gap between the two O atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both arrows share a pull away from S toward the O side, so they add to one arrow pointing between the two O atoms.

CTwo separate arrows — bond dipoles cannot be combined into one

Feedback if chosen: This option is wrong — you stopped before adding. The arrows do combine: their vector sum is the molecule’s one overall dipole.

DZero — SO₂ is a neutral molecule, so its arrows must add to zero

Feedback if chosen: This option is wrong — you mixed up charge and dipole. A molecule can be neutral overall and still have an electron-rich end and an electron-poor end — that is exactly what a dipole is.

Check — multiple choice

Why does finding a molecule’s dipole moment require the three-dimensional geometry, not just a list of its bonds?

AThe geometry sets the direction of each arrow, and the directions decide how the arrows add ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A vector sum needs directions, and the directions of the bond-dipole arrows come from where the geometry points each bond.

BThe geometry changes each bond’s electronegativity difference

Feedback if chosen: This option is wrong — you moved the electronegativities. Each bond’s electronegativity difference depends only on its two atoms; the geometry changes where the arrows point, not how strong they are.

CThe geometry determines how many bonds the molecule contains

Feedback if chosen: This option is wrong — you gave the geometry a counting job. The bonds are fixed by the molecule; the geometry’s job is to point them in three dimensions.

DThe geometry makes every bond more polar than the list suggests

Feedback if chosen: This option is wrong — you let the geometry change the bonds themselves. Each bond dipole keeps its size; the geometry only directs the arrows, which is what decides the sum.

U2-S16 · LESSON 146 · CSP-139

A polar bond is not enough

Outcome: State that a nonzero dipole moment requires at least one polar bond, and that polar bonds alone do not guarantee one.

Part 1No polar bonds, nothing to add

You’ve already seen that a molecule’s dipole moment is the vector sum of its bond dipoles, drawn as arrows in its three-dimensional geometry.

Here is methane, CH₄. You’ve already seen that C–H bonds are effectively nonpolar, so methane has no bond dipoles at all.

With no bond dipoles, there is nothing to add: the vector sum is zero, whatever the geometry.

So a molecule must contain at least one polar bond to have a nonzero dipole moment.

Part 2Polar bonds do not guarantee it

Now carbon dioxide, CO₂ — linear, with two polar C=O bonds. O is clearly more electronegative than C, so each bond carries a bond dipole pointing toward its O.

The two arrows are equal in size and point in exactly opposite directions, so their vector sum is zero.

CO₂ has polar bonds, yet its dipole moment is zero.

So a polar bond is necessary for a nonzero dipole moment, but not sufficient — the geometry decides whether the bond dipoles add up or cancel.

Throughout this course, molecular polarity is modeled exactly this way: as the vector sum of the bond dipoles.

What you are expected to be able to do

State that a molecule must contain at least one polar bond to have a nonzero dipole moment, and that polar bonds alone do not guarantee one — the geometry decides.

Part 3Check what you just learned

Check — multiple choice

All the bonds in ethane, C₂H₆, are C–C bonds or C–H bonds — none of them polar. What do you conclude about ethane’s dipole moment?

AIt is nonzero if the geometry happens to gather the bonds toward one side

Feedback if chosen: This option is wrong — you gathered arrows that do not exist. Effectively nonpolar bonds carry no bond dipoles, so there is nothing for the geometry to gather.

BIt is zero — with no bond dipoles, there is nothing to add ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A nonzero dipole moment needs at least one polar bond; with no bond dipoles at all, the vector sum is zero no matter the shape.

CIt is nonzero — every molecule with more than one bond has a dipole moment

Feedback if chosen: This option is wrong — you counted bonds instead of bond dipoles. Bonds without polarity contribute nothing to the sum.

DIt cannot be decided without the full three-dimensional geometry

Feedback if chosen: This option is wrong — you reached for the geometry too early. The geometry directs the arrows, but here there are no arrows to direct: the sum is zero regardless.

Check — multiple choice

A molecule contains two clearly polar bonds. What can you conclude about its dipole moment?

AIt is nonzero — polar bonds always produce a nonzero dipole moment

Feedback if chosen: This option is wrong — you treated a necessary condition as a sufficient one. CO₂ has two polar bonds and a dipole moment of zero; the geometry decides.

BIt is zero — two equal bond dipoles always cancel each other

Feedback if chosen: This option is wrong — you generalized CO₂ to every molecule. Equal arrows cancel only when the geometry points them in exactly opposite directions; at an angle, they add.

CNothing yet — the vector sum depends on how the geometry directs the two arrows ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Polar bonds supply the arrows, but whether the arrows cancel or add is set by the three-dimensional geometry — so the question is not yet decided.

DIt equals the two bond-dipole sizes added together

Feedback if chosen: This option is wrong — you added sizes without directions. A vector sum can be anything from zero up to that total, depending on where the geometry points the arrows.

Check — multiple choice

A classmate reasons: “CCl₄ contains four polar C–Cl bonds, so it must have a nonzero dipole moment.” How do you evaluate the reasoning?

AValid — a molecule with polar bonds always has a nonzero dipole moment

Feedback if chosen: This option is wrong — you accepted the exact leap the classmate made. Polar bonds only supply the arrows; CO₂ shows that supplied arrows can still cancel.

BValid — four bond dipoles are too many to cancel one another

Feedback if chosen: This option is wrong — you put a limit on cancellation that does not exist. Any number of equal arrows can cancel if the geometry spreads them evenly.

CNot valid — polar bonds never produce a nonzero dipole moment

Feedback if chosen: This option is wrong — you flipped the error. Polar bonds often do produce a nonzero dipole moment — in H₂O they do; the point is that they do not guarantee it.

DNot valid — polar bonds are necessary but not sufficient; the geometry decides whether the four arrows cancel ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The first half of the claim is fine — polar bonds are required — but the conclusion does not follow: whether the four arrows cancel is decided by the geometry, not by the bonds alone.

U2-S16 · LESSON 147 · CSP-140

When bond dipoles cancel

Outcome: Predict that equal bond dipoles cancel when the geometry arranges them symmetrically, so the net dipole is zero.

Part 1Equal and opposite

Here is CO₂ again, with its two bond dipoles drawn. You’ve already seen that their vector sum is zero.

The cancellation needs two things at once: the arrows are equal in size, and the linear geometry points them in exactly opposite directions.

The arrows are equal because the bonds are identical — the same element, O, at both positions.

Part 2Cancelling without an opposite partner

Here is BF₃: trigonal planar, with three identical polar B–F bonds and three equal arrows pointing outward toward the F atoms.

No single arrow has a direct opposite.

But together the three pull outward evenly, 120° apart — no direction wins, and the vector sum is zero.

The same happens in three dimensions. CCl₄ is tetrahedral: four identical polar C–Cl bonds, four equal arrows pointing outward evenly into space — vector sum zero.

That is the pattern: when a geometry arranges equal bond dipoles symmetrically — pointing outward evenly, no direction favored — their vector sum is zero, and the molecule has no net dipole.

Both conditions matter: equal arrows, arranged evenly. Identical bonds at every position supply the equal arrows.

What you are expected to be able to do

Predict a net dipole of zero when a molecular geometry arranges equal bond dipoles symmetrically — as in linear CO₂, trigonal-planar BF₃, and tetrahedral CCl₄.

Part 3Check what you just learned

Check — multiple choice

BCl₃ is trigonal planar with three identical polar B–Cl bonds; the diagram shows its three bond dipoles. Predict its net dipole.

AZero — the three equal arrows point outward evenly, 120° apart, and cancel ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Equal arrows spread evenly cancel: no direction wins, so the vector sum — the net dipole — is zero.

BNonzero — a molecule with three polar bonds must have a net dipole

Feedback if chosen: This option is wrong — you went from polar bonds straight to a net dipole. Polar bonds are necessary but not sufficient; this symmetric arrangement cancels them.

CNonzero — three arrows cannot cancel, because none of them has a direct opposite

Feedback if chosen: This option is wrong — you required an opposite partner for each arrow. Cancellation only needs the arrows to balance as a group, and three equal arrows 120° apart do exactly that.

DZero — trigonal planar molecules are never polar, whatever bonds they contain

Feedback if chosen: This option is wrong — you let the shape decide alone. A trigonal planar molecule with three different bonds would have unequal arrows and a net dipole; the zero here comes from three equal arrows spread evenly.

Check — multiple choice

CF₄ is tetrahedral with four identical polar C–F bonds, shown with their bond dipoles. Predict its net dipole.

ANonzero — arrows pointing into three dimensions cannot cancel

Feedback if chosen: This option is wrong — you kept cancellation flat. A tetrahedron spreads its four arrows evenly in three dimensions, and evenly spread equal arrows cancel just as they do in a plane.

BNonzero — C–F bonds are strongly polar, and strong bond dipoles do not cancel

Feedback if chosen: This option is wrong — you let the arrow size decide. Strong arrows cancel exactly like weak ones when the arrangement balances them; what matters is equal sizes and an even spread.

CZero — the four equal arrows point outward evenly into space and cancel ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Four identical bonds give four equal arrows, and the tetrahedral geometry points them outward evenly — the vector sum is zero.

DZero — each C–F arrow is cancelled by a direct opposite across the molecule

Feedback if chosen: This option is wrong — you gave each arrow a direct opposite. No two bonds of a tetrahedron point exactly opposite each other; the four equal arrows cancel as a group because they spread evenly.

Check — multiple choice

Which combination guarantees that a molecule’s bond dipoles cancel?

AAn even number of polar bonds, so that the arrows can pair off

Feedback if chosen: This option is wrong — you made cancellation a matter of counting. BF₃ cancels with three arrows and no pairs; the even spread does the work, not the pair-off.

BEqual bond dipoles, arranged by the geometry to point outward evenly ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Cancellation needs both conditions at once: arrows of equal size, and an arrangement that spreads them evenly so no direction is favored.

CIdentical bonds at every position, whatever the geometry

Feedback if chosen: This option is wrong — you dropped the arrangement. H₂O’s two O–H bonds are identical, yet its bent geometry points the arrows partly the same way and leaves a net dipole.

DBond dipoles that all lie along one straight line

Feedback if chosen: This option is wrong — you trusted the line to cancel. Two arrows on one line pointing the same way reinforce; and BF₃ cancels with no line at all. It is the even spread of equal arrows that guarantees zero.

U2-S16 · LESSON 148 · CSP-141

When bond dipoles do not cancel

Outcome: Predict a nonzero net dipole when the arrangement or the sizes of the bond dipoles leave a nonzero vector sum.

Part 1Arrows gathered to one side

Here is H₂O again. You’ve already seen that its two arrows point partly the same way — both toward O — so they add to a net dipole instead of cancelling.

Compare CO₂: equal arrows there too, but its linear geometry points them in exactly opposite directions.

The only difference is the geometry — bent instead of linear — and that alone leaves H₂O with a net dipole.

NH₃ is trigonal pyramidal: all three N–H arrows point partly upward, toward N.

Nothing on the far side of N opposes them, so the three arrows add to a net dipole pointing toward N.

Compare BF₃: three equal arrows again, but its flat trigonal-planar geometry spreads them evenly. A pyramid gathers the arrows to one side; a plane balances them.

Part 2Unequal arrows

Geometry is not the only way to break the balance. Here is CHCl₃ — tetrahedral like CCl₄, but with one Cl replaced by H.

You’ve already seen that C–H is effectively nonpolar: the C–H position contributes no bond dipole.

The three C–Cl arrows pull toward their Cl atoms, and nothing at the H position opposes them — the vector sum is a net dipole pointing toward the Cl side.

So bond dipoles fail to cancel in two ways: the geometry gathers them to one side, or the arrows at the positions are not all equal.

What you are expected to be able to do

Predict a nonzero net dipole when the three-dimensional arrangement or the sizes of the bond dipoles leave a nonzero vector sum — as in bent H₂O, trigonal-pyramidal NH₃, and CHCl₃.

Part 3Check what you just learned

Check — multiple choice

In OF₂, F is more electronegative than O; the diagram shows bent OF₂ with its two bond dipoles pointing toward the F atoms. Predict its net dipole.

AZero — two equal arrows always cancel, whatever the geometry

Feedback if chosen: This option is wrong — you let equal sizes settle it alone. Equal arrows cancel only when they point in exactly opposite directions; the bent geometry points these two partly the same way.

BNonzero, pointing from the F atoms toward O — a net dipole points at the central atom

Feedback if chosen: This option is wrong — you turned H₂O’s direction into a rule. Arrows point toward the more electronegative atoms — here the F atoms — so the sum points away from O, not at it.

CNonzero, pointing toward one of the two F atoms

Feedback if chosen: This option is wrong — you followed a single arrow. The net dipole is the sum of both, and the sum of two equal arrows at an angle points between them, not along either one.

DNonzero, pointing from O toward the gap between the two F atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both arrows point partly away from O, so they add to a net dipole pointing from O toward the F side of the molecule.

Check — multiple choice

PCl₃ is trigonal pyramidal, and Cl is more electronegative than P; the diagram shows its three bond dipoles. Predict its net dipole.

ANonzero — the pyramid gathers all three arrows toward the Cl side, so they add instead of cancelling ✓ KEY — this is the correct answer

Feedback if chosen: Correct. In a pyramid the three arrows all aim at the same side of P, so nothing opposes them and their sum is a net dipole toward the Cl side.

BZero — three equal bond dipoles cancel, as they do in BF₃

Feedback if chosen: This option is wrong — you read the pyramid as a plane. BF₃’s arrows lie flat and spread evenly; a pyramid tilts all three arrows to one side, where they add.

CZero — the three Cl atoms pull equally hard, so their pulls balance out

Feedback if chosen: This option is wrong — you balanced sizes and ignored directions. Equal pulls cancel only when their directions balance too; the pyramid tilts all three arrows to the same side, where they add.

DNonzero, pointing toward P — the odd atom out carries the net dipole

Feedback if chosen: This option is wrong — you aimed the sum at the lone atom instead of following the arrows. The arrows point toward the Cl atoms, so their sum points toward the Cl side, away from P.

Check — multiple choice

CHF₃ is tetrahedral, as shown; F is clearly more electronegative than C, while C–H is effectively nonpolar. Predict its net dipole.

AZero — tetrahedral geometry cancels bond dipoles, as it does in CF₄

Feedback if chosen: This option is wrong — you let the shape decide alone. CF₄ cancels because its four arrows are equal; replace one F with H and one arrow disappears, so the remaining three no longer balance.

BZero — the C–H bond dipole balances the three C–F bond dipoles

Feedback if chosen: This option is wrong — you gave C–H an arrow. C–H is effectively nonpolar, so it contributes no bond dipole and cannot balance anything.

CNonzero, toward the F side — three C–F arrows add, and the C–H position supplies no arrow to oppose them ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Only the three C–F bonds carry arrows; they all pull toward the F side, and with nothing at the H position to oppose them, the vector sum is nonzero.

DNonzero, toward H — the net dipole points at the position with the fewest atoms

Feedback if chosen: This option is wrong — you pointed the sum at the odd position instead of following the arrows. The arrows point toward the F atoms, so their sum points toward the F side, away from H.

U2-S16 · LESSON 149 · CSP-142

Drawing a dipole diagram

Outcome: Draw bond-dipole arrows on a molecular geometry, and add a net-dipole arrow when the bond dipoles do not cancel.

Part 1The four steps

You’ve already seen finished dipole diagrams — now you learn to draw one. Here is the finished diagram for H₂O.

Step one: draw the molecular geometry — the bend, the pyramid, the tetrahedron — not just a flat sketch of the bonds.

Step two: on each polar bond, draw one arrow parallel to the bond, pointing toward the more electronegative atom.

Effectively nonpolar bonds get no arrow — a C–H bond contributes no bond dipole.

Step three: decide whether the arrows cancel. Equal arrows spread evenly cancel; arrows gathered to one side, or unequal arrows, do not.

Step four: if the arrows do not cancel, draw one labeled net-dipole arrow pointing in the direction the arrows add up to. If they cancel, draw no net arrow — write that the bond dipoles cancel instead.

Part 2The steps on H₂O and CO₂

Take H₂O. Step one gives the bent geometry.

Step two: both O–H bonds are polar, so each gets an arrow pointing toward O.

Steps three and four: the two arrows point partly the same way, so they do not cancel — draw the net dipole pointing toward O, and label it.

Now CO₂: linear geometry, an arrow toward each O — and the equal, opposite arrows cancel, so no net arrow. Write the cancellation instead.

What you are expected to be able to do

Draw one bond-dipole arrow per polar bond on the molecular geometry, each pointing toward the more electronegative atom, and add a labeled net-dipole arrow only when the bond dipoles do not cancel.

Part 3Check what you just learned

Check — drawing task (pen and paper, self-marked)

Ammonia, NH₃, is trigonal pyramidal, and N is more electronegative than H. Take a pen and paper and draw the dipole diagram of NH₃: the geometry, its bond-dipole arrows, and the net dipole if the arrows do not cancel. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows trigonal-pyramidal NH₃ with one arrow on each of the three N–H bonds — three arrows in total.
Your drawing shows every arrow pointing toward N, the more electronegative atom.
Your drawing shows a net-dipole arrow — the three arrows all point partly toward N, so they do not cancel.
Your drawing shows the net dipole labeled and pointing toward N, the side all three arrows favor.

Check — drawing task (pen and paper, self-marked)

Boron trichloride, BCl₃, is trigonal planar, and Cl is more electronegative than B. Take a pen and paper and draw the dipole diagram of BCl₃: the geometry, its bond-dipole arrows, and the net dipole only if the arrows do not cancel. When you are done, select Check to compare your drawing with the model answer.

Model answer shown to the student

Self-mark checklist — the drawing should show:

Your drawing shows trigonal-planar BCl₃ with one arrow on each of the three B–Cl bonds.
Your drawing shows every arrow pointing outward, toward a Cl atom — Cl is the more electronegative atom.
Your drawing shows no net-dipole arrow: the three equal arrows point outward evenly, 120° apart, and cancel.
Your drawing notes the cancellation in words — bond dipoles cancel, no net dipole.

Check — multiple choice

You are drawing the dipole diagram of CH₃Cl, shown; C–Cl is polar and C–H is effectively nonpolar. Which bonds get bond-dipole arrows?

AAll four bonds — every bond in a dipole diagram gets an arrow

Feedback if chosen: This option is wrong — you gave arrows to bonds without bond dipoles. An arrow stands for a bond dipole, and the effectively nonpolar C–H bonds have none.

BOnly the C–Cl bond — the three C–H bonds contribute no bond dipole ✓ KEY — this is the correct answer

Feedback if chosen: Correct. One arrow per polar bond: the C–Cl bond gets an arrow toward Cl, and the three effectively nonpolar C–H bonds get none.

CNo bonds — CH₃Cl is tetrahedral, and tetrahedral molecules have no bond dipoles

Feedback if chosen: This option is wrong — you confused cancelling with not existing. Geometry decides whether arrows cancel; it does not remove them, and the polar C–Cl bond carries an arrow either way.

DThe C–Cl bond and one C–H bond, so that the two arrows can balance

Feedback if chosen: This option is wrong — you invented an arrow to force a balance. Arrows follow the bonds as they are: one polar bond here, so one arrow, and the diagram ends with a net dipole along it.

U2-S16 · LESSON 150 · CSP-143

Checking a dipole diagram

Outcome: Evaluate a supplied dipole diagram's arrow directions and cancellation, naming the specific flaw when there is one.

Part 1Three checks

You’ve already seen the drawing steps — evaluating a supplied dipole diagram applies the same steps you used in the checks. Here is a correct diagram of H₂O to check against.

Check one: the right bonds carry arrows — one per polar bond, none on effectively nonpolar bonds.

Check two: every arrow points toward the more electronegative atom.

Check three: the cancellation is handled honestly — a net-dipole arrow only when the arrows do not cancel, pointing where they add; a cancellation note when they do.

When a diagram fails, name the one specific check it fails.

Part 2A worked evaluation

Here is a classmate’s diagram of BF₃. Run the checks.

Check one passes: all three polar B–F bonds carry arrows.

Check two fails: the arrows point inward, toward B — but F is the more electronegative atom, so every arrow should point outward, toward its F.

That is the flaw to report: the arrow directions, and exactly which way the arrows should point.

What you are expected to be able to do

Evaluate whether a supplied dipole diagram has correct bond-dipole directions and correct cancellation, and identify the one specific flaw when it does not.

Part 3Check what you just learned

Check — multiple choice

A classmate drew this dipole diagram of NH₃. N is more electronegative than H. What is your evaluation?

ACorrect as drawn — bond-dipole arrows point away from the central atom, as they should

Feedback if chosen: This option is wrong — you invented an away-from-the-center rule. Arrows point toward the more electronegative atom, and here that atom is the central one, N.

BFlawed — the three equal bond dipoles cancel, so there should be no net-dipole arrow at all

Feedback if chosen: This option is wrong — you cancelled arrows a pyramid cannot cancel. The pyramid gathers all three arrows to one side, so a net arrow belongs; the flaw is which way everything points.

CFlawed — the arrows point toward the H atoms; N is the more electronegative atom, so the bond dipoles and the net dipole must point toward N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The diagram fails the direction check: every arrow points at H, the less electronegative atom. Reversed, the three arrows add to a net dipole toward N.

DFlawed — the geometry is wrong: NH₃ is trigonal planar

Feedback if chosen: This option is wrong — you flagged a correct feature. NH₃ has four electron domains with one lone pair, so trigonal pyramidal is exactly right.

Check — multiple choice

A classmate drew this dipole diagram of bent SO₂ and marked its bond dipoles as cancelling. O is more electronegative than S. What is your evaluation?

AFlawed — the arrows point the wrong way; they should point inward, toward S

Feedback if chosen: This option is wrong — you flagged a correct feature. O is the more electronegative atom, so arrows toward the O atoms pass the direction check; the failure is elsewhere.

BCorrect — two equal bond dipoles always cancel

Feedback if chosen: This option is wrong — you let equal sizes settle it alone. Equal arrows cancel only when they point in exactly opposite directions, and a bend never points them that way.

CFlawed — SO₂ should be drawn linear, and then the cancellation note would be right

Feedback if chosen: This option is wrong — you fixed the diagram by changing the molecule. SO₂ really is bent — three electron domains with one lone pair — so the diagram must handle the bend honestly.

DFlawed — the cancellation claim: the bent geometry points the two arrows partly the same way, so they do not cancel, and a net-dipole arrow toward the O side is missing ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The diagram fails the cancellation check: two equal arrows at an angle add instead of cancelling, so the diagram needs a net-dipole arrow pointing from S toward the O side.

Check — multiple choice

A classmate drew this dipole diagram of linear CO₂ and marked its bond dipoles as cancelling. O is more electronegative than C. What is your evaluation?

ACorrect — the arrows point toward the O atoms, and the equal, opposite arrows cancel, so no net arrow belongs ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The diagram passes all three checks: arrows on both polar bonds, both pointing toward the more electronegative O, and an honest cancellation note in place of a net arrow.

BFlawed — CO₂’s bonds are polar, so a net-dipole arrow is missing

Feedback if chosen: This option is wrong — you went from polar bonds straight to a net dipole. Polar bonds are necessary but not sufficient, and here the linear geometry cancels the two equal arrows exactly.

CFlawed — the arrows should point inward, toward C

Feedback if chosen: This option is wrong — you reversed the direction rule. Arrows point toward the more electronegative atom, and O outpulls C, so outward is correct.

DFlawed — a finished dipole diagram must always end with a net-dipole arrow

Feedback if chosen: This option is wrong — you made the net arrow compulsory. A net arrow belongs only when the bond dipoles fail to cancel; when they cancel, the correct ending is the cancellation note.

U2-S16 · LESSON 151 · CSP-144

Polar or nonpolar?

Outcome: Predict whether a molecule is polar or nonpolar from its bond polarity and three-dimensional geometry, and the direction of its net dipole when it is polar.

Part 1The full prediction

Every piece is now in place. This lesson puts them together into one prediction: does a molecule have a net dipole, and which way does it point?

A molecule with a nonzero net dipole is a polar molecule. A molecule whose bond dipoles leave no net dipole is a nonpolar molecule.

Step one: classify each bond from the electronegativities — polar or effectively nonpolar.

Step two: get the molecular geometry from VSEPR.

Step three: picture one arrow per polar bond, each pointing toward the more electronegative atom.

Step four: take the vector sum. Arrows that cancel: nonpolar. Arrows that do not cancel: polar, with the net dipole pointing where the arrows add.

Part 2The prediction on CCl₄ and CHCl₃

Run it on CCl₄: four identical polar C–Cl bonds, tetrahedral geometry, four equal arrows pointing outward evenly — they cancel. CCl₄ is nonpolar.

Now CHCl₃: the same tetrahedral geometry, but one position holds H, and C–H is effectively nonpolar — only the three C–Cl bonds get arrows.

The three arrows pull toward the Cl side and nothing at the H position opposes them: they add. CHCl₃ is polar, with its net dipole pointing from the H side toward the Cl side.

One geometry, two verdicts — the prediction always needs both the bonds and the geometry, never the shape alone.

What you are expected to be able to do

Predict whether a molecule is polar (nonzero net dipole, with its direction) or nonpolar (bond dipoles cancel) from its bond polarity and three-dimensional geometry.

Part 3Check what you just learned

Check — multiple choice

Electronegativities
Element	Electronegativity
H	2.20
C	2.55
N	3.04

Hydrogen cyanide, H–C≡N, is linear. Using the table shown, classify the molecule, giving the direction of its net dipole if it has one.

ANonpolar — linear geometries cancel bond dipoles, as in CO₂

Feedback if chosen: This option is wrong — you let the shape decide alone. CO₂ cancels because it has two equal arrows pointing in exactly opposite directions; in HCN nothing opposes the C≡N arrow, which points toward N.

BPolar, with the net dipole pointing toward N ✓ KEY — this is the correct answer

Feedback if chosen: Correct. C≡N is polar toward N (3.04 against C's 2.55). And even if the smaller H–C difference (2.20 against 2.55) is counted too, that arrow also points away from H, toward the N end — the arrows add, and the net dipole points toward N.

CPolar, with the net dipole pointing toward H

Feedback if chosen: This option is wrong — you flipped the arrow. A bond dipole points toward the more electronegative atom, and the table puts N above C, so the arrow points toward N.

DNonpolar — both bonds in HCN are effectively nonpolar

Feedback if chosen: This option is wrong — you classified C≡N along with C–H. The table puts N well above C (3.04 against 2.55), so C≡N is polar — and nothing opposes its arrow.

Check — multiple choice

Xenon tetrafluoride, XeF₄, has the square-planar molecular geometry shown, and its four Xe–F bonds are identical and polar, toward F. Classify the molecule.

APolar — any molecule with four polar bonds is polar

Feedback if chosen: This option is wrong — you went from polar bonds straight to a polar molecule. Polar bonds are necessary but not sufficient; this symmetric square arranges the four equal arrows so they cancel.

BPolar — with six electron domains, the arrows point into three dimensions and cannot cancel

Feedback if chosen: This option is wrong — you counted domains instead of following the arrows. The four bonded atoms sit in one plane, in two equal, opposite pairs — an arrangement that cancels exactly.

CNonpolar — the Xe–F bond dipoles are too small to add up to anything

Feedback if chosen: This option is wrong — you made the result a matter of size. Xe–F bonds are clearly polar; the zero comes from the symmetric arrangement, not from weak dipoles.

DNonpolar — the four equal arrows point outward evenly in the square plane, each opposed by its partner across Xe, so they cancel ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The square puts each Xe–F arrow directly opposite an equal partner, so the vector sum is zero and XeF₄ is nonpolar.

Check — multiple choice

Water, H₂O, is bent, as shown, and O is clearly more electronegative than H. Classify the molecule, giving the direction of its net dipole if it has one.

APolar, with the net dipole pointing toward O ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both O–H bonds are polar toward O, and the bent geometry points the two arrows partly the same way, so they add to a net dipole toward O.

BNonpolar — the two identical O–H bond dipoles cancel

CPolar, with the net dipole pointing toward one of the two H atoms

Feedback if chosen: This option is wrong — you followed one arrow and flipped it. Each arrow points toward O, the more electronegative atom, so their sum points toward O as well.

DNonpolar — each O–H bond is effectively nonpolar

Feedback if chosen: This option is wrong — you misclassified the bonds. O and H are clearly unequal in electronegativity, so each O–H bond is polar and carries an arrow.

Check — multiple choice

Which one of these four molecules is polar?

ACO₂ (linear)

Feedback if chosen: This option is wrong — you may have gone from polar bonds to a polar molecule. CO₂’s two equal arrows point in exactly opposite directions and cancel: nonpolar.

BBF₃ (trigonal planar)

Feedback if chosen: This option is wrong — you picked a symmetric arrangement. BF₃’s three equal arrows point outward evenly, 120° apart, and cancel: nonpolar.

CCCl₄ (tetrahedral)

Feedback if chosen: This option is wrong — you picked a symmetric arrangement. CCl₄’s four equal arrows point outward evenly into space and cancel: nonpolar.

DNH₃ (trigonal pyramidal) ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The pyramid gathers all three N–H arrows toward N, nothing opposes them, and the net dipole points toward N: NH₃ is polar.

AFTER U2-S16 · CHECKPOINT · U2-G6 · 9 items

Checkpoint 6 — Unit 2 synthesis

Check — multiple choice

Phosphorus pentafluoride, PF₅, has the trigonal-bipyramidal molecular geometry shown, and its five P–F bonds are identical and polar, toward F. Classify the molecule.

APolar — a molecule with five polar bonds must be polar

Feedback if chosen: This option is wrong — polar bonds alone do not make a polar molecule. The geometry decides whether the bond dipoles cancel, and here they do.

BNonpolar — the five equal bond dipoles point outward symmetrically and sum to zero ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The two axial dipoles are opposite and cancel, and the three equal equatorial dipoles at 120° also sum to zero — the vector sum over all five is zero, so PF₅ has no net dipole.

CPolar — the two axial bond dipoles cancel, but the three equatorial ones cannot

Feedback if chosen: This option is wrong — three equal dipoles spaced 120° apart in a plane sum to zero, just as the two axial ones do, so every group cancels.

DNonzero — the two axial arrows point along one line but the three equatorial arrows do not, so the two sets cannot balance

Feedback if chosen: This option is wrong — you treated the two sets as if they had to cancel against each other. Each set cancels within itself: the two axial arrows are direct opposites, and the three equatorial arrows are arranged symmetrically at 120°, so their vector sum is zero too.

Check — multiple choice

Measured properties of substance T
Property	Measurement
Melting point	113 °C
Conducts as a solid?	no
Conducts when molten?	no
Struck with a hammer	shatters

Substance T was tested, and the measurements are shown in the table. Weighing all of the evidence, how should T be classified?

AIonic

Feedback if chosen: This option is wrong — you leaned on the brittleness and the nonconducting solid. An ionic substance conducts once molten; T does not, so its particles are not ions.

BMetallic

Feedback if chosen: This option is wrong — a metal conducts in the solid state and deforms rather than shattering; T does neither.

CMolecular ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A low melting point and no conduction in either state together identify discrete neutral molecules; the brittleness of the solid cannot outweigh that evidence.

DThe data cannot decide among the three

Feedback if chosen: This option is wrong — the evidence set is decisive: nonconduction when molten rules out ionic, nonconduction as a solid rules out metallic, and the low melting point fits molecular.

Check — numeric entry

Use the potential-energy curve shown for two iodine atoms. What is the bond energy of I–I, in kJ/mol?

✓ ANSWER KEY: 151 kJ/mol

Feedback when correct: Correct. The minimum sits at −151 kJ/mol below the zero line, so separating the bonded pair requires 151 kJ/mol.

Anticipated wrong answers (feedback shown):

267	This is the equilibrium bond length read from the horizontal axis, in pm. Bond energy is the vertical gap between the minimum and the zero line: 151 kJ/mol.
0	The zero line is where the separated atoms sit, not the bond energy. Read how far the minimum sits below it: 151 kJ/mol.

Sign-error retry feedback:

-151	Right gap, wrong sign. Bond energy is the energy required to pull the atoms apart, so it is positive: 151 kJ/mol. The minus sign on the axis label marks where the minimum sits relative to zero.

Check — multiple choice

Ion charges and distances between the ion centers
Pair	Ion charges	Distance between the ion centers (pm)
NaCl	1+ and 1−	283
SrS	2+ and 2−	302

Use the data shown. Which solid has the stronger attraction between its ions: NaCl or SrS?

ANaCl, because its distance between the ion centers is smaller

Feedback if chosen: This option is wrong — you let the 19 pm distance difference decide alone. Weigh both factors: SrS's charge product is four times NaCl's, and that dominates a distance increase of a few percent.

BNeither — SrS's larger charges and NaCl's smaller distance offset each other exactly

Feedback if chosen: This option is wrong — you assumed the effects cancel without weighing them. Quadrupling the charge product far outweighs a distance about 7% larger.

CSrS, because Sr²⁺ and S²⁻ are larger ions, and larger ions attract more strongly

Feedback if chosen: This option is wrong — right solid, wrong reason. Larger ions actually sit farther apart, which weakens attraction; SrS wins because of its 2+ and 2− charges, despite the larger distance.

DSrS, because its charge product is four times larger, while its distance is only about 7% larger ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Charges: 2+ with 2− gives a charge product four times that of 1+ with 1−. Distance: 302 pm is only slightly larger than 283 pm. The fourfold charge factor dominates.

Check — multiple choice

A student offers the diagram shown as the finished Lewis diagram of the hydroxide ion, OH⁻ (calculated total: 6 + 1 + 1 = 8). Which check does it fail?

AThe electron count — the diagram holds fewer electrons than the calculated total

Feedback if chosen: This option is wrong — count them: one bond (2 electrons) plus three lone pairs on O (6) holds all 8 calculated electrons.

BThe octet check — O counts fewer than eight electrons

Feedback if chosen: This option is wrong — O counts its three lone pairs plus the shared pair: 6 + 2 = 8, a full octet.

CThe ion notation — the brackets and the outside 1− are missing ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The electrons are all placed correctly, but an ion's diagram must sit in brackets with the 1− charge written outside — without them the diagram shows a neutral fragment, not the hydroxide ion.

DThe duet check — H needs a lone pair to be complete

Feedback if chosen: This option is wrong — H is complete with its one shared pair; H never carries a lone pair in a correct diagram.

Check — multiple choice

Two candidate Lewis diagrams are shown for a molecule with one H, one C, and one N atom; both place all 10 valence electrons. Which model is preferred?

AModel 1, because every formal charge in it is 0 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Model 1 (H–C≡N) works out to formal charges of 0 on every atom, while Model 2 (H–N≡C) carries +1 on N and −1 on C — and that −1 sits on C, the less electronegative of the two.

BModel 2, because the more electronegative atom belongs at the center

Feedback if chosen: This option is wrong — you reversed the skeleton rule: the less electronegative nonhydrogen atom goes in the middle, so electronegativity argues for C at the center, not N. Model 2's charges (+1 on N, −1 on C) are also worse, not better.

CNeither model is acceptable, because a valid diagram never places a nonzero formal charge on any atom

Feedback if chosen: This option is wrong — valid diagrams can carry nonzero formal charges; the charges are a comparison tool, not a validity test. And Model 1 in fact reaches 0 on every atom, so it is acceptable and preferred.

DModels 1 and 2 are equally acceptable, because both place 10 electrons and give C and N octets

Feedback if chosen: This option is wrong — passing the count and octet checks makes both models candidates; formal charge then decides, and only Model 1 reaches all zeros.

Check — multiple choice

Use the Lewis diagram of the amide ion, NH₂⁻, shown. Which prediction is correct?

ALinear electron-domain geometry; linear molecular geometry; bond angle 180°

Feedback if chosen: This option is wrong — you counted only the two bonded atoms. N's two lone pairs are domains too, so N has four domains, not two.

BTetrahedral electron-domain geometry; bent molecular geometry; less than 109.5° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. N has four electron domains — two N–H bonds and two lone pairs — so the domains are tetrahedral; the two bonded H atoms make the molecule bent, and the lone pairs compress the H–N–H angle below 109.5°.

CTrigonal-planar electron-domain geometry; bent molecular geometry; less than 120°

Feedback if chosen: This option is wrong — you counted only one of the two lone pairs. N has four domains (two bonds, two lone pairs), so the starting ideal is 109.5°, not 120°.

DTetrahedral electron-domain geometry; bent molecular geometry; more than 109.5°

Feedback if chosen: This option is wrong — right geometry, wrong deviation. Lone pairs repel the bonding pairs more strongly and squeeze the angle below the ideal, not above it.

Check — multiple choice

Use the Lewis diagram of the ammonium ion, NH₄⁺, shown. Which hybridization describes the N atom?

Asp — N has no lone pairs, and atoms without lone pairs are sp

Feedback if chosen: This option is wrong — lone pairs are not what sets the label. Count all domains: N's four bonds are four domains, so sp³.

Bsp² — one of the four N–H bonds comes from the 1+ charge, so it does not count as a domain

Feedback if chosen: This option is wrong — every bond in the diagram is a domain, whatever its bookkeeping origin. N holds four bonds, so it has four domains: sp³.

CIt cannot be assigned — hybridization applies only to neutral molecules

Feedback if chosen: This option is wrong — the routine is the same for ions: count the electron domains around the atom in the Lewis diagram. Four domains give sp³.

Dsp³ — N has four electron domains: four bonding pairs ✓ KEY — this is the correct answer

Feedback if chosen: Correct. N has four electron domains — the four N–H bonding pairs — so its hybridization is sp³.

Check — numeric entry

How many pi bonds does the but-2-yne Lewis diagram shown contain?

✓ ANSWER KEY: 2

Feedback when correct: Correct. Only the C≡C triple bond holds pi bonds — one sigma and two pi. The single bonds are sigma only, so the molecule has 2 pi bonds.

Anticipated wrong answers (feedback shown):

1	This treats the triple bond as one sigma and one pi. A triple bond is one sigma and two pi bonds: 2.
3	This counts all three lines of the triple bond as pi bonds. One of the three is the sigma bond, so the pi count is 2.

AFTER U2-S16 · UNIT TEST · U2-T1 · 20 items

Unit 2 test

Check — multiple choice

Electronegativity values
Element	Electronegativity
Br	2.96
I	2.66
F	3.98

Interhalogen compounds contain a single covalent bond between two different halogen atoms. Using the electronegativity values in the table, identify the bond with the larger bond dipole: the I–F bond in the molecule IF, or the Br–F bond in the molecule BrF.

AThe Br–F bond, because Br is more electronegative than I

Feedback if chosen: This option is wrong — you compared the two atoms' own electronegativities instead of the difference across each bond. A bond dipole grows with the electronegativity DIFFERENCE, and 3.98 − 2.66 = 1.32 for I–F is larger than 3.98 − 2.96 = 1.02 for Br–F.

BThe two bond dipoles are equal, because each bond has an F atom on one end

Feedback if chosen: This option is wrong — you treated the dipole as set by the F atom alone. The dipole measures how unevenly the pair shares electrons, so the other atom's electronegativity matters: I holds electrons more loosely than Br, making I–F the more lopsided bond.

CThe I–F bond, because the electronegativity difference between I and F is larger ✓ KEY — this is the correct answer

Feedback if chosen: Correct. For comparable single bonds, the larger electronegativity difference produces the larger bond dipole: I–F (3.98 − 2.66 = 1.32) beats Br–F (3.98 − 2.96 = 1.02).

DThe Br–F bond, because the shorter bond holds the partial charges closer together

Feedback if chosen: This option is wrong — you used bond length as the deciding factor. For comparable single bonds it is the electronegativity difference that sets the dipole magnitude, and the I–F difference is larger.

Check — multiple choice

A student draws the particulate model of solid RbCl shown in the figure. Which statement correctly evaluates the model?

AThe model is flawed: two like-charged ions sit side by side where oppositely charged neighbors belong ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A valid ionic-solid model maximizes opposite-charge attractions and minimizes like-charge repulsions, so every ion should sit among oppositely charged neighbors — two like-charged ions side by side add repulsion the real lattice avoids.

BThe model is flawed: the ions should be grouped into pairs, each Rb⁺ attached to one particular Cl⁻

Feedback if chosen: This option is wrong — you described an ionic solid as a collection of separate ion pairs. In the lattice each ion is attracted to ALL of its oppositely charged neighbors; no Rb⁺ belongs to one particular Cl⁻.

CThe model is correct: it shows a repeating three-dimensional pattern of alternating cations and anions

Feedback if chosen: This option is wrong — you checked the overall pattern without checking every neighbor. Two like-charged ions sit directly beside each other, which a lattice that minimizes repulsion would not allow.

DThe model is flawed: the Rb⁺ and Cl⁻ circles should be drawn the same size

Feedback if chosen: This option is wrong — you flagged the size difference, but Rb⁺ and Cl⁻ genuinely have different radii, so drawing them as different-sized circles is correct. The actual flaw is the pair of like-charged neighbors.

Check — multiple choice

Atomic radii
Atom	Atomic radius (pm)
Zr	155
H	25

Zirconium metal absorbs hydrogen to form an alloy. Using the atomic radii in the table, predict the type of alloy that forms.

AA substitutional alloy, because the H atoms take the places of some Zr atoms within the metallic lattice

Feedback if chosen: This option is wrong — you applied the substitutional model to atoms of very different sizes. Substitution works when the atoms have comparable radii; a 25 pm H atom cannot stand in for a 155 pm Zr atom.

BAn interstitial alloy, because the much smaller H atoms fit into the spaces between Zr atoms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. At 25 pm, an H atom is far smaller than a 155 pm Zr atom, so the H atoms occupy the spaces between the Zr atoms — an interstitial alloy.

CNo alloy can form, because hydrogen is not a metal

Feedback if chosen: This option is wrong — you required every component of an alloy to be a metal. An alloy is a metallic material that can contain nonmetal atoms mixed in, the way steel contains C in Fe.

DAn ionic solid, because a metal is combining with a nonmetal

Feedback if chosen: This option is wrong — you applied the metal-plus-nonmetal rule for ionic bonding, but the stem describes hydrogen absorbing into the metal: the Zr keeps its metallic structure, and the small H atoms sit in its spaces.

Check — multiple choice

Measured properties of substance J
Property	Observation
Melting point	993 °C
Conducts electricity as a solid?	no
Conducts electricity when molten?	yes
Behavior when struck	shatters

A student measures the properties of an unknown solid, substance J, shown in the table. Which classification is most consistent with this evidence?

AMetallic, because only strong bonding throughout an extended solid produces such a high melting point

Feedback if chosen: This option is wrong — you weighed only the melting point. A metal conducts electricity in the solid state and dents rather than shatters; substance J does neither.

BMolecular covalent, because substance J does not conduct as a solid

Feedback if chosen: This option is wrong — you used only the solid-state conductivity test. A substance made of neutral molecules would stay nonconducting when molten and would melt at a relatively low temperature, but J conducts when molten and melts at 993 °C.

CMetallic, but as an alloy of two different metals

Feedback if chosen: This option is wrong — an alloy is still metallic: it conducts as a solid and deforms rather than shatters. Substance J does not conduct as a solid and shatters.

DIonic, because the ions are fixed in place in the solid but carry current once molten ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Every observation fits an ionic solid: strong Coulombic attractions give the high melting point, the ions cannot move in the solid but carry current once molten, and shifting the lattice brings like charges together, so it shatters.

Check — multiple choice

A student draws the particulate model of solid potassium shown in the figure. Which statement correctly evaluates the model?

AThe model is consistent with metallic bonding — the alternating charges balance, so the solid is neutral overall

Feedback if chosen: This option is wrong — balancing the charges with K⁻ ions describes an ionic solid. A metal is neutral because the positive ions sit within the delocalized electrons the atoms released.

BThe model is flawed — the ions of a real metal are scattered randomly, not arranged in a repeating array

Feedback if chosen: This option is wrong — the positive ions of a metallic solid do sit in a regular repeating array. The flaw is the negative ions, which a metal does not contain.

CThe model is flawed — the charge labels should be removed, because the particles of a metal are drawn neutral

Feedback if chosen: This option is wrong — the circles in a metallic model are positive ions, and the model must say so: symbol and charge on every ion. The flaw is the negative ions.

DThe model is flawed — a metal contains no negative ions; its only negative charge is the delocalized electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. A metal contains no negative ions: each atom releases its valence electrons, and the resulting positive ions sit within one shared region of delocalized electrons — which this drawing is missing.

Check — multiple choice

A student sketches the potential-energy curve for an I atom and a Cl atom as they form the I–Cl bond in ICl, shown in the figure. Which evaluation of the sketch is correct?

AThe sketch is wrong: the minimum of the curve should sit exactly on the zero-energy line

Feedback if chosen: This option is wrong — you put the bonded pair at zero energy. The zero level belongs to the separated atoms; the bonded pair at the minimum sits BELOW zero, which is why energy is required to pull the bond apart.

BThe sketch is correct: it rises steeply at short distance, has a single minimum, and levels off to a flat region at large internuclear distance

Feedback if chosen: This option is wrong — you checked that the curve flattens but not WHERE it flattens. The flat region at large distance must sit at zero, the energy of the separated, noninteracting atoms; this sketch levels off below zero.

CThe sketch is wrong: the curve should flatten to zero at large distance, where the far-apart atoms no longer interact ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Far-apart atoms no longer interact, and that state defines zero potential energy — so the curve must flatten to the zero line at large distance. This sketch levels off below zero, which is its error.

DThe sketch is wrong: the potential energy should keep falling as the atoms move farther apart

Feedback if chosen: This option is wrong — you have the attraction backwards. Pulling the atoms apart works against their attraction, so the potential energy RISES toward zero as distance grows beyond the minimum; it does not keep falling.

Check — multiple choice

Silicon forms a single covalent bond with each halogen. Predict which bond is longer: the Si–Br bond or the Si–Cl bond.

AThe Si–Cl bond, because the larger Si–Cl electronegativity difference pulls the atoms closer together

Feedback if chosen: This option is wrong — you used the electronegativity difference to set the length. That difference sets the bond DIPOLE; at the same bond order, it is the size of the bonded atoms that sets the length.

BThe Si–Br bond, because Br atoms are larger than Cl atoms, so the bonded nuclei sit farther apart ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both bonds are single bonds, so atomic size decides: Br sits one period below Cl, its atoms are larger, and the Si and Br nuclei sit farther apart than Si and Cl.

CThe two bonds have equal lengths, because both are single bonds

Feedback if chosen: This option is wrong — you stopped at bond order. Equal order makes the bonds comparable, but the atoms still differ in size: the larger Br atom makes Si–Br the longer bond.

DThe lengths cannot be compared without measured bond-order data

Feedback if chosen: This option is wrong — the bond orders are already given: the stem states that both are single bonds. With order fixed, the periodic size trend (Br larger than Cl) is enough to compare the lengths.

Check — multiple choice

Two ion pairs
Ion pair	Ion charges	Distance between ion centers (pm)
Mg²⁺ and F⁻	2+ and 1−	205
Rb⁺ and Br⁻	1+ and 1−	348

The table gives the ion charges and the distance between the ion centers for two ion pairs. Which pair has the stronger Coulombic attraction?

ARb⁺ and Br⁻, because the larger ions contain many more electrons available to attract the other ion

Feedback if chosen: This option is wrong — you counted electrons instead of applying Coulombic reasoning. The attraction depends on the ions' charges and the distance between their centers, not on how many electrons each ion contains.

BThe attractions are equal, because each pair contains one cation and one anion

Feedback if chosen: This option is wrong — you treated every cation-anion pair as identical. The charge MAGNITUDES differ (2+ versus 1+) and so do the distances (205 pm versus 348 pm); both differences matter.

CRb⁺ and Br⁻, because heavier ions attract each other more strongly

Feedback if chosen: This option is wrong — you used mass, which plays no role in Coulombic attraction. Charge magnitude and the distance between the ion centers are what decide the strength.

DMg²⁺ and F⁻, because the cation's charge is larger and the ion centers are closer together ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Both factors favor Mg²⁺ and F⁻: the 2+ charge is twice the 1+ charge, and the centers sit much closer (205 pm versus 348 pm), so the Coulombic attraction is far stronger.

Check — multiple choice

A student draws the Lewis diagram of KrF₂ shown in the figure, with ten electrons around the central Kr atom. Is this diagram acceptable?

AYes — a central atom in period 3 or later, like period-4 Kr, can appear with more than eight electrons ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The octet rule binds period-2 atoms, but a central atom from period 3 or later, like period-4 Kr, can appear with more than eight electrons in an AP-level Lewis diagram — here, two bonds plus three lone pairs.

BNo — every atom in a valid Lewis diagram must end up with exactly eight electrons, with no exceptions allowed

Feedback if chosen: This option is wrong — you applied the octet rule with no exceptions. Central atoms in period 3 or later can hold more than eight electrons, so ten around period-4 Kr is acceptable.

CNo — the two extra electrons should be relocated as an additional lone pair on one of the F atoms

Feedback if chosen: This option is wrong — you tried to relocate the extra electrons, but each F atom already has a complete octet, and F is in period 2, so it can never hold more than eight.

DNo — noble-gas atoms already hold complete octets, so they cannot share electrons with other atoms

Feedback if chosen: This option is wrong — you ruled out noble-gas bonding entirely, but heavier noble gases do form compounds: Xe appears in accepted Lewis diagrams with expanded valence shells, and KrF₂ is a real compound.

Check — multiple choice

Calculate the total number of valence electrons in the Lewis diagram of the bromate ion, BrO₃⁻.

A24

Feedback if chosen: This option is wrong — you subtracted an electron for the negative charge. A NEGATIVE charge means the ion has gained an electron, so one electron is added: 7 + 3(6) + 1 = 26.

B25

Feedback if chosen: This option is wrong — you left out the charge. The 1− charge adds one electron to the count from the atoms: 7 + 3(6) + 1 = 26.

C26 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Br contributes 7, each of the three O atoms contributes 6, and the 1− charge adds one more: 7 + 18 + 1 = 26.

D32

Feedback if chosen: This option is wrong — you gave each O atom eight valence electrons. O is in group 16, so each O contributes six: 7 + 3(6) + 1 = 26.

Check — multiple choice

Three candidate Lewis diagrams of carbonyl sulfide, OCS, are shown in the figure. Each places all 16 valence electrons and gives every atom an octet. Which model is preferred?

AModel 1 ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Every atom in all three models has an octet, so the formal charges decide. Model 1 (O=C=S) works out to 0 on every atom; Model 2 carries +1 on O and −1 on S, and Model 3 carries −1 on O and +1 on S. All zeros is the smaller pattern, so Model 1 is preferred.

BModel 2

Feedback if chosen: This option is wrong — Model 2 (O≡C–S) works out to +1 on O and −1 on S. Model 1 reaches 0 on every atom, so the smaller-and-fewer criterion prefers Model 1.

CModel 3

Feedback if chosen: This option is wrong — Model 3 (O–C≡S) works out to −1 on O and +1 on S. Its −1 does sit on the more electronegative atom, but that tie-breaker is never reached: Model 1 reaches 0 on every atom and wins first.

DModels 1, 2, and 3 are equally acceptable, because every atom in each has an octet

Feedback if chosen: This option is wrong — you stopped after the octet criterion. Octets only screen the candidates; the formal charges then decide, and only Model 1 has an all-zero set.

Check — multiple choice

The figure shows the three equivalent resonance forms of sulfur trioxide, SO₃. Which statement describes the actual bonding in an SO₃ molecule?

AOne third of all SO₃ molecules match each of the three resonance forms

Feedback if chosen: This option is wrong — you read the forms as a population census of different molecules. Every SO₃ molecule has the same, single electron distribution: the hybrid represented jointly by the three forms.

BEvery S–O bond is identical, with a bond order between 1 and 2: the molecule is a single hybrid of the three forms ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The actual electron distribution is one resonance hybrid: the double-bond character is shared over all three S–O bonds, so each bond is identical, with an order between 1 and 2.

CThe double bond moves rapidly from one O atom to another, so each molecule alternates in time among the three forms

Feedback if chosen: This option is wrong — you pictured the electrons hopping between arrangements over time. The molecule never occupies any single form; it always has the one intermediate distribution the forms jointly represent.

DIn each molecule, one S–O bond stays permanently double while the other two stay permanently single

Feedback if chosen: This option is wrong — you took one diagram literally. Measured SO₃ has three IDENTICAL S–O bonds, which no single form shows; only the hybrid of all three forms matches the molecule.

Check — multiple choice

The Lewis diagram of TeCl₄ is shown in the figure. Identify the molecular geometry of TeCl₄.

ASeesaw ✓ KEY — this is the correct answer

Feedback if chosen: Correct. Te has five electron domains (four bonding, one lone pair), giving trigonal-bipyramidal electron-domain geometry; the lone pair takes an equatorial position, and the four bonded Cl atoms form a seesaw.

BTrigonal bipyramidal

Feedback if chosen: This option is wrong — you named the arrangement of all five domains. Molecular geometry describes only the bonded atoms, and one of the five domains is a lone pair, so the four Cl atoms make a seesaw.

CTetrahedral

Feedback if chosen: This option is wrong — you ignored the lone pair on Te. It counts as a fifth electron domain, so the domains adopt the trigonal-bipyramidal arrangement, not the tetrahedral one.

DSquare planar

Feedback if chosen: This option is wrong — you used the six-domain pattern with two lone pairs on opposite sides. Te has five domains with a single lone pair, which sits equatorial and leaves the atoms in a seesaw.

Check — multiple choice

The Lewis diagram of IF₅ is shown in the figure. Identify the molecular geometry of IF₅.

AOctahedral

Feedback if chosen: This option is wrong — you named the arrangement of all six domains. One domain is a lone pair, so the five bonded F atoms form a square pyramid, not an octahedron.

BTrigonal bipyramidal

Feedback if chosen: This option is wrong — you counted only the five bonded atoms and used the five-domain geometry. The lone pair on I is a sixth domain, so the domains adopt the octahedral arrangement and the atoms form a square pyramid.

CSquare pyramidal ✓ KEY — this is the correct answer

Feedback if chosen: Correct. I has six electron domains (five bonding, one lone pair) in an octahedral arrangement; with the lone pair in one position, the five F atoms form a square pyramid.

DSquare planar

Feedback if chosen: This option is wrong — square planar requires TWO lone pairs on opposite sides of the six-domain arrangement. IF₅ has only one lone pair, so the atoms form a square pyramid.

Check — multiple choice

The figure shows SbCl₅. Which set lists the ideal bond angles that occur between Sb–Cl bonds in this molecule?

A109.5° only

Feedback if chosen: This option is wrong — you used the tetrahedral ideal, which belongs to four electron domains. SbCl₅ has five domains, split between axial and equatorial positions with three different angles.

B90° and 180° only

Feedback if chosen: This option is wrong — you used the octahedral set, which belongs to six electron domains. Five domains add the 120° angles between the equatorial bonds.

C120° only

Feedback if chosen: This option is wrong — you counted only the equatorial-equatorial angles. The two axial bonds make 90° with each equatorial bond and 180° with each other.

D90°, 120°, 180° ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The three equatorial bonds sit 120° apart, each axial bond makes 90° with the equatorial bonds, and the two axial bonds are 180° from each other.

Check — multiple choice

The Lewis diagram of acetonitrile, CH₃CN, is shown in the figure. Predict the hybridization of the C atom that is bonded to the N atom.

Asp²

Feedback if chosen: This option is wrong — you counted the triple bond as two electron domains. All the electrons shared between the same two atoms form ONE domain, so this C has two domains, which corresponds to sp.

Bsp ✓ KEY — this is the correct answer

Feedback if chosen: Correct. This C has two electron domains — the single bond to the other C and the triple bond to N — and two domains correspond to sp hybridization.

Csp³

Feedback if chosen: This option is wrong — you counted each line of the triple bond separately, reaching four domains. A multiple bond counts once, so this C has two domains and is sp hybridized.

DThe hybridization cannot be found without first knowing the molecular geometry

Feedback if chosen: This option is wrong — you made geometry a prerequisite. Hybridization follows directly from the electron-domain count in the Lewis diagram: two domains around this C means sp.

Check — multiple choice

Using the Lewis diagram of acrylonitrile, CH₂=CH−C≡N, in the figure, calculate the number of sigma bonds and pi bonds in one molecule.

A6 sigma and 3 pi ✓ KEY — this is the correct answer

Feedback if chosen: Correct. The three C–H bonds and the C–C single bond are four sigma bonds; the C=C adds 1 sigma + 1 pi; the C≡N adds 1 sigma + 2 pi. Total: 6 sigma and 3 pi.

B9 sigma and 0 pi

Feedback if chosen: This option is wrong — you counted every bond line as a sigma bond. Within a double or triple bond, only one component is sigma; every additional shared pair is a pi bond.

C4 sigma and 5 pi

Feedback if chosen: This option is wrong — you counted the double and triple bonds as entirely pi. Each multiple bond contains exactly one sigma component, so the C=C and C≡N contribute two of the six sigma bonds.

D6 sigma and 2 pi

Feedback if chosen: This option is wrong — you gave the triple bond only one pi component. A triple bond is one sigma plus TWO pi bonds, so the molecule has three pi bonds in total.

Check — multiple choice

Each molecule below is listed with its molecular geometry. Which molecule has a dipole moment of zero?

ASeO₂ (bent)

Feedback if chosen: This option is wrong — you let the two identical Se–O bonds cancel, but the bent geometry points them partly the same way, so their vector sum is not zero and SeO₂ is polar.

BSbCl₃ (trigonal pyramidal)

Feedback if chosen: This option is wrong — you treated three identical bonds as always canceling. In a trigonal pyramid all three Sb–Cl dipoles tilt toward the same side of the molecule, leaving a nonzero sum.

CCHBr₃ (tetrahedral)

Feedback if chosen: This option is wrong — you used the tetrahedral shape alone. Cancellation needs four IDENTICAL bond dipoles; the C–H bond dipole differs from the three C–Br dipoles, so a net dipole remains.

DSiCl₄ (tetrahedral) ✓ KEY — this is the correct answer

Feedback if chosen: Correct. SiCl₄ has four identical Si–Cl bond dipoles arranged symmetrically in a tetrahedron, so the four vectors sum to zero and the molecule has no dipole moment.

Free-response question

Source note (not shown as part of the question): authored to the Lewis/VSEPR + bonding-energetics long archetype (modeled on 2023 FRQ Q2(c)-(d), 2024 FRQ Q6(c), and 2026 FRQ Q7(c)); no unit-2-pure real long FRQ exists in 2021-2026

A student investigates the bonding in several bromine-containing substances. Phosphorus tribromide, PBr₃, melts at −41 °C, and neither solid nor liquid PBr₃ conducts electricity. In PBr₃, the P atom is bonded to each of the three Br atoms. The figure shows the potential-energy curve for an H atom and a Br atom as they form the H–Br bond in HBr.

(a) A student claims that the melting point and conductivity of PBr₃ show that it is composed of discrete neutral molecules. Do you agree with the student's claim? Justify your answer using both properties. [1 point]

(b) Draw the complete Lewis diagram of PBr₃, showing all bonding and nonbonding valence electrons. [1 point]

(c)(i) Identify the electron-domain geometry around the P atom in PBr₃. Justify your answer in terms of the number of electron domains around P. [1 point]

(c)(ii) Identify the molecular geometry of PBr₃. [1 point]

(c)(iii) Is the Br–P–Br bond angle in PBr₃ equal to, less than, or greater than 109.5°? Justify your answer. [1 point]

(d) Does PBr₃ have a nonzero dipole moment? Justify your answer in terms of the P–Br bond dipoles and the molecular geometry. (Electronegativities: P = 2.19, Br = 2.96.) [1 point]

(e)(i) Using the graph, identify the equilibrium bond length of the H–Br bond, in picometers. [1 point]

(e)(ii) Using the graph, identify the bond energy of the H–Br bond, in kJ/mol. [1 point]

(f) BaS(s) melts at 2226 °C, while CsI(s) melts at 626 °C. In BaS the centers of neighboring Ba²⁺ and S²⁻ ions are 319 pm apart; in CsI the centers of neighboring Cs⁺ and I⁻ ions are 395 pm apart. Using Coulomb's law, explain why the melting point of BaS is so much higher than that of CsI. [2 points]

✓ Model response (the key)

(a)

Agree. A very low melting point (−41 °C) fits a solid made of discrete neutral molecules, because the attractions between neutral molecules are weak compared with the forces inside ionic or metallic solids.

The absence of conductivity in both the solid and the liquid also fits neutral molecules: there are no mobile charged particles to carry current in either state.

(b)

Total valence electrons: 5 (P) + 3 × 7 (Br) = 26.

The drawing shows P singly bonded to each of the three Br atoms, one lone pair on P, and three lone pairs on each Br.

Electron check: 3 bonds (6 electrons) + 1 lone pair on P (2 electrons) + 9 lone pairs on the Br atoms (18 electrons) = 26 electrons.

(c)(i)

Tetrahedral electron-domain geometry.

P has four electron domains: three bonding domains (one to each Br) plus one lone pair.

(c)(ii)

Trigonal pyramidal — molecular geometry describes only the bonded atoms, and the three Br atoms form a pyramid with P at its apex while the fourth domain is a lone pair.

(c)(iii)

Less than 109.5°.

The lone pair–bonding pair repulsion is stronger than the bonding pair–bonding pair repulsion, so the lone pair pushes the three P–Br bonds closer together than the tetrahedral ideal.

(d)

Yes — PBr₃ has a nonzero dipole moment.

Each P–Br bond is polar (Br, at 2.96, is more electronegative than P, at 2.19), and in the trigonal-pyramidal geometry the three bond dipoles all tilt toward the same side of the molecule, so their vector sum is not zero.

(e)(i)

141 pm — the internuclear distance at the minimum of the curve.

(e)(ii)

366 kJ/mol.

The minimum sits at −366 kJ/mol, and pulling the bonded pair completely apart raises it to the zero level of the separated atoms, so +366 kJ/mol must be added.

(f)

By Coulomb's law, the attraction between opposite charges grows with larger charge magnitudes and with a smaller distance between the charge centers. In BaS the ions carry 2+ and 2− charges with centers 319 pm apart; in CsI the ions carry only 1+ and 1− charges with centers farther apart at 395 pm. Both factors make the Coulombic attractions between ions in BaS much stronger.

Melting requires enough energy for the ions to move past one another against these attractions, so BaS must be heated to a much higher temperature than CsI before it melts.

Scoring rubric — 10 points available

(a) One point is earned for agreeing with a justification that uses both properties: the low melting point fits the relatively weak attractions between discrete neutral molecules, and the lack of conductivity in both states fits the absence of mobile charged particles. (Using only one of the two properties does not earn the point.) [1 pt]
(b) One point is earned for a diagram with a single bond from P to each of the three Br atoms, exactly one lone pair on P, and three lone pairs on each Br — 26 electrons in total. (A diagram whose electron total is not 26 does not earn the point.) [1 pt]
(c)(i) One point is earned for tetrahedral together with the domain count: three bonding domains plus one lone pair make four electron domains around P. (Naming the geometry without the domain count does not earn the point.) [1 pt]
(c)(ii) One point is earned for trigonal pyramidal, consistent with your diagram in part (b). [1 pt]
(c)(iii) One point is earned for less than 109.5° with a valid justification: the lone pair repels the bonding pairs more strongly than the bonding pairs repel one another, compressing the Br–P–Br angles below the tetrahedral ideal. (The answer without the repulsion comparison does not earn the point.) [1 pt]
(d) One point is earned for yes with a justification naming both requirements: the P–Br bonds are polar because Br is more electronegative than P, and the trigonal-pyramidal geometry does not cancel the three bond dipoles, leaving a nonzero vector sum. (Stating that the molecule is asymmetrical, with no reference to the bond dipoles, does not earn the point.) [1 pt]
(e)(i) One point is earned for 141 pm, read from the distance-axis guide at the curve's minimum. [1 pt]
(e)(ii) One point is earned for 366 kJ/mol, reported as a positive quantity: the energy required to raise the bonded pair from the minimum at −366 kJ/mol to the zero level of the separated atoms. (An answer of −366 kJ/mol does not earn the point — bond energy is energy that must be added.) [1 pt]
(f) One point is earned for comparing the two solids with Coulomb's law using the given values: the Ba²⁺ and S²⁻ ions carry twice the charge of Cs⁺ and I⁻ AND their centers sit closer together (319 pm versus 395 pm), so the Coulombic attractions between ions are much stronger in BaS. (Naming Coulomb's law without using the charges or the distances does not earn the point.) [1 pt]
(f) One point is earned for connecting attraction strength to melting: melting requires the ions to move past one another, so overcoming the stronger attractions in BaS takes more energy, which corresponds to a higher melting temperature. [1 pt]

Free-response question

Source note (not shown as part of the question): authored to the Lewis/resonance short archetype (modeled on 2024 FRQ Q6(c) and 2022 FRQ Q7(a)); no unit-2-pure real short FRQ exists in 2021-2026

N–O bond lengths
Bond	Length (pm)
N–O bond 1 in CH₃NO₂ (measured)	122
N–O bond 2 in CH₃NO₂ (measured)	122
typical N–O single bond	140
typical N=O double bond	117

Nitromethane, CH₃NO₂, is used as a fuel in drag racing. In the nitromethane molecule, each of the three H atoms is bonded to the C atom, the C atom is bonded to the N atom, and the N atom is bonded to each of the two O atoms. The table gives the measured N–O bond lengths in nitromethane, together with typical lengths for clearly single N–O bonds and clearly double N=O bonds in other molecules.

(a) Calculate the total number of valence electrons in the Lewis diagram of nitromethane. [1 point]

(b) Nitromethane has two equivalent resonance forms. Draw the complete Lewis diagram of both forms, keeping every atom in the same position in each form. [1 point]

(c) Using the data in the table, explain why the two measured N–O bond lengths are equal to each other, and why each is between the typical single-bond and double-bond lengths. [1 point]

(d) Predict the approximate O–N–O bond angle in nitromethane. Justify your prediction using the VSEPR model. [1 point]

✓ Model response (the key)

(a)

4 (C) + 3 × 1 (H) + 5 (N) + 2 × 6 (O) = 24 valence electrons

(b)

Both drawings show the three H atoms bonded to C, C bonded to N, and N bonded to both O atoms. In one form the N=O double bond goes to one O atom (which carries two lone pairs) and the N–O single bond goes to the other O atom (which carries three lone pairs); in the second form the double bond and the single bond are exchanged between the same two O atoms.

Atom positions are identical in the two forms — only the electron placement differs — and each drawing contains 24 electrons.

(c)

The actual molecule is one resonance hybrid of the two forms, not a pair of structures it switches between: the double-bond character is shared equally by the two N–O bonds, so each has the same bond order, between 1 and 2.

Because the two bonds are identical, their lengths are equal, and an intermediate bond order gives a length (122 pm) between the typical single-bond length (140 pm) and the typical double-bond length (117 pm).

(d)

Approximately 120°.

In either resonance form the N atom has three electron domains — the bond to C and a bond to each O — and no lone pair, so the domains spread into a trigonal-planar arrangement with ideal 120° angles.

Scoring rubric — 4 points available

(a) One point is earned for the correct total with the contributions shown: 4 + 3(1) + 5 + 2(6) = 24 valence electrons. [1 pt]
(b) One point is earned for two complete 24-electron diagrams with unchanged atom positions, in which the N=O double bond goes to a different O atom in each form, the singly bonded O carries three lone pairs, and the doubly bonded O carries two lone pairs. (Diagrams that move an atom, or whose electron totals differ from 24, do not earn the point.) [1 pt]
(c) One point is earned for an explanation that treats the molecule as one resonance hybrid: each N–O bond has the same intermediate bond order (between 1 and 2), so the two bonds have equal lengths, and that length falls between the 140 pm single-bond value and the 117 pm double-bond value. (Saying that the molecule switches back and forth between the two forms does not earn the point.) [1 pt]
(d) One point is earned for approximately 120° with a valid justification: N has three electron domains and no lone pair, giving a trigonal-planar arrangement of domains around N. [1 pt]

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