AP Chemistry · Handsworth Secondary 2026–27

Unit 2 · Compound Structure & Properties

Dr. Ras Mulinta
Handsworth Secondary
Notes Package

CED Unit 2, 7–9% of the AP exam, taught in 11 blocks. From BC Chemistry 11 you already meet ionic vs. covalent bonding and simple Lewis dots; here we go to AP depth: electronegativity-based bond type, potential-energy curves, ionic & metallic lattices, full Lewis diagrams, formal charge & resonance, VSEPR geometry & polarity, hybridization, and σ/π bonding. BC Chem 12 revisits bonding qualitatively, this unit is the quantitative, structure-first version.

Blocks:  1 → A  ·  2 → B  ·  3 → C–D  ·  4–5 → E  ·  6 → F  ·  7 → G  ·  8–9 → H  ·  10 → I  ·  11 → review

Name:Block:Date:

A · Types of Chemical Bonds CED 2.1 · LO 2.1.A

Three bonding models cover most matter: ionic (metal + nonmetal, electrons transferred), covalent (nonmetal + nonmetal, electrons shared), and metallic (metal atoms in a sea of delocalized electrons). Which one forms is read from electronegativity and the periodic table.

Electronegativity (EN) rises left→right across a period and falls top→bottom down a group (Coulomb's law + shell model). Bond type from ΔEN: roughly ΔEN ≈ 0 nonpolar covalent · 0 < ΔEN < ~1.7 polar covalent · ΔEN large (metal+nonmetal) ionic.
Properties tell the truth: ionic solids are hard, brittle, high-melting, conduct only when molten/aqueous; molecular covalent substances are soft, low-melting, poor conductors; metals are malleable, lustrous, conduct as solids (mobile electrons).
Worked example
Classify the bonding in NaCl, Cl₂, and Cu. (EN: Na 0.9, Cl 3.0, Cu 1.9)
NaCl: metal + nonmetal, ΔEN = 3.0 − 0.9 = 2.1 → ionic
Cl₂: two identical nonmetals, ΔEN = 0 → nonpolar covalent
Cu: metal lattice, electrons delocalized → metallic

Practice A

  1. Classify the bonding in MgO, HCl, and Fe. Justify each with EN / position.
  2. A solid is brittle, melts near 800 °C, and conducts electricity only when dissolved in water. What bond type? Explain.
  3. C–H bonds are treated as effectively nonpolar even though C (2.5) is slightly more electronegative than H (2.1). Why is that a reasonable approximation?

B · Intramolecular Force & Potential Energy CED 2.2 · LO 2.2.A

As two atoms approach, a graph of potential energy vs. internuclear distance tells the whole story of the bond: where it sits, how long it is, and how hard it is to break.

Reading the curve: the minimum is the equilibrium bond length (most stable separation); its depth below zero is the bond energy (energy to pull the atoms apart). Too close → repulsion shoots up; too far → no attraction.
Trends: higher bond order (single→double→triple) = shorter, stronger bond (deeper, leftward minimum). For ionic attraction, Coulomb's law F ∝ q₁q₂ / r² → larger ion charges and smaller ions give stronger, deeper wells.
Worked example
On one set of axes, compare the potential-energy wells of C–C, C=C, and C≡C.
Bond order increases 1 → 2 → 3, so bond length decreases (154 → 134 → 120 pm) and bond energy increases (~346 → 602 → 835 kJ/mol).
C≡C is the deepest well and sits farthest left (shortest); C–C is the shallowest and farthest right (longest).

Practice B

  1. Sketch potential-energy curves for H₂ vs. Cl₂ on the same axes. Which has the longer bond length, and which the deeper well? (Cl atoms are larger; H–H bond energy ≈ 436 kJ/mol, Cl–Cl ≈ 242.)
  2. Using Coulomb's law, explain why the MgO lattice is held together more strongly than the NaF lattice (both ~same ion sizes, but compare charges).
  3. On a potential-energy curve, what physically happens to the atoms at distances shorter than the equilibrium bond length?

C · Structure of Ionic Solids CED 2.3 · LO 2.3.A

An ionic compound isn't molecules, it's a giant 3-D lattice of alternating cations and anions, arranged to maximize attraction and minimize repulsion (Coulomb's law).

Why the properties follow: strong electrostatic attractions in every direction → high melting points & hardness. Brittle because shifting one layer lines up like charges, which repel and shatter the crystal. Conduct only when ions are free to move (molten or aqueous).
Worked example
Explain, with a particulate model, why NaCl shatters when struck but copper just bends.
NaCl: a blow slides a layer so Na⁺ meets Na⁺ and Cl⁻ meets Cl⁻; like charges repel → the lattice cleaves.
Cu: the electron sea is mobile, so displaced metal ions stay bonded by delocalized electrons → it deforms instead of cracking.

Not tested: knowledge of specific crystal structures (e.g. naming rock-salt vs. cesium-chloride lattices) is excluded, you only need the general alternating-array model.

Practice C

  1. Use a particulate diagram to explain why molten NaCl conducts electricity but solid NaCl does not.
  2. Rank the lattice energies of NaCl, MgCl₂, and MgO using Coulomb's law (ion charges and sizes).
  3. Why are ionic solids generally hard and high-melting compared with molecular solids like ice?

D · Structure of Metals & Alloys CED 2.4 · LO 2.4.A

A metal is a lattice of positive metal ions immersed in a "sea" of delocalized valence electrons. That single picture explains conductivity, luster, and malleability.

Sea of electrons: mobile electrons carry charge (electrical conduction) and heat, reflect light (luster), and let ion layers slide without breaking bonds (malleable, ductile). Two alloy types: interstitial (small atoms fill gaps between large ones, e.g. C in Fe → steel) and substitutional (similar-radius atoms swap in, e.g. Zn for Cu → brass).
Worked example
Carbon (r ≈ 70 pm) is added to iron (r ≈ 126 pm) to make steel. Interstitial or substitutional? Why is steel harder than pure iron?
Radii differ greatly (70 vs. 126 pm) → carbon fits in the gaps → interstitial.
The wedged carbon atoms block layers of iron ions from sliding past each other, so the metal resists deformation → harder.

Practice D

  1. Brass is copper (r ≈ 128 pm) with zinc (r ≈ 134 pm). Predict interstitial vs. substitutional and justify with radii.
  2. Use the sea-of-electrons model to explain why metals conduct electricity in the solid state but ionic solids do not.
  3. Why can a copper wire be drawn out thin (ductile) without snapping, unlike a salt crystal?

E · Lewis Diagrams CED 2.5 · LO 2.5.A

A Lewis diagram is the working drawing of a molecule: every valence electron placed as a bond or a lone pair. Master the recipe and everything downstream (geometry, polarity, hybridization) follows.

Recipe: (1) total valence electrons (add 1 per negative charge, subtract 1 per positive). (2) least electronegative atom in the centre (never H). (3) single bonds out. (4) fill outer atoms to an octet. (5) leftovers as lone pairs on the centre. (6) short on octets → make double/triple bonds.
Worked example
Draw the Lewis diagram of CO₂.
Valence: C 4 + 2×O 6 = 16 e⁻. C is central: O–C–O.
Single bonds use 4 e⁻; 12 left fill the oxygens, but C has only 4 e⁻ around it. Short on the octet → form two C=O double bonds.
O=C=O, with two lone pairs on each O. Check: C has 8 e⁻ (two double bonds), each O has 8 e⁻. Total used = 16 ✓

Practice E

  1. Draw the Lewis diagram of water (H₂O) and of ammonia (NH₃). How many lone pairs on the central atom in each?
  2. Draw the Lewis diagram of the ammonium ion, NH₄⁺. (Remember to subtract one electron for the + charge.)
  3. Draw N₂ and show that it contains a triple bond. How many bonding and lone pairs total?

F · Resonance & Formal Charge CED 2.6 · LO 2.6.A

Sometimes one Lewis diagram isn't enough or isn't unique. Resonance averages several equivalent structures; formal charge picks the best among non-equivalent ones.

Formal charge FC = (valence e⁻) − (lone-pair e⁻) − ½(bonding e⁻). The best Lewis structure keeps formal charges closest to zero, with any negative charge on the most electronegative atom.
Resonance: when two or more equivalent structures differ only in where double bonds / lone pairs sit, the real molecule is the average (delocalized). All such bonds are identical and intermediate in length.
Worked example
Ozone, O₃ (18 valence e⁻), can be drawn O=O–O or O–O=O. Use formal charge to describe it.
Central O (one double, one single bond, one lone pair): FC = 6 − 2 − ½(6) = +1.
Double-bonded terminal O (2 lone pairs): FC = 6 − 4 − ½(4) = 0. Single-bonded terminal O (3 lone pairs): FC = 6 − 6 − ½(2) = −1.
Neither single structure is "right", the two are equivalent, so O₃ is a resonance hybrid with two identical bonds of order 1.5 (between single and double).

Not tested in depth: the Lewis model has limits, odd-electron (radical) species like NO can't satisfy every octet. You should recognize this limitation, not master radical bookkeeping.

Practice F

  1. Draw the three resonance structures of the nitrate ion, NO₃⁻ (24 valence e⁻). What is the bond order of each N–O bond?
  2. For CO₂ (O=C=O), compute the formal charge on carbon and on each oxygen. Is this the best structure?
  3. Two Lewis structures of OCN⁻ are possible. Briefly state how formal charge helps you pick the better one.

G · VSEPR: Electron & Molecular Geometry CED 2.7 · LO 2.7.A

Electron pairs (bonds + lone pairs) around a central atom repel and spread as far apart as possible (Coulomb's law). Count the groups → get the shape and the bond angles.

By electron-domain count (lone pairs push harder, so they bend angles a little): 2 → linear (180°) · 3 → trigonal planar (120°) · 4 → tetrahedral (109.5°). With lone pairs on a 4-domain centre: 1 LP → trigonal pyramidal (~107°), 2 LP → bent (~104.5°).
Worked example
Give the electron geometry, molecular geometry, and approximate bond angle of H₂O.
O has 2 bonding pairs + 2 lone pairs = 4 electron domains → electron geometry tetrahedral.
Two of those are lone pairs, so the atoms form a bent shape.
Bent; angle ≈ 104.5° (less than 109.5° because the two lone pairs compress the H–O–H angle).

Practice G

  1. Give electron geometry, molecular geometry, and bond angle for CH₄, NH₃, and CO₂.
  2. Why is the bond angle in NH₃ (~107°) smaller than in CH₄ (109.5°)?
  3. SO₃ has three bonding domains and no lone pairs on S. Predict its shape and bond angle.

H · Bond Polarity & Molecular Dipoles CED 2.7 · EK 2.7.A.2.v

A polar bond doesn't always make a polar molecule. Bond dipoles are vectors, add them up using the molecule's geometry, and symmetry can cancel them out.

Rule: a molecule is polar if it has polar bonds and their dipole vectors don't cancel. Symmetric shapes (linear AX₂, trigonal planar AX₃, tetrahedral AX₄) with identical outer atoms → dipoles cancel → nonpolar. Lone pairs or unlike atoms break the symmetry → polar.
Worked example
CO₂ and H₂O both have polar bonds. Which molecule is polar overall?
CO₂ is linear (O=C=O); the two C=O dipoles point opposite and equal → they cancel → nonpolar.
H₂O is bent; the two O–H dipoles add to a net downward dipole and don't cancel.
H₂O is polar; CO₂ is nonpolar. Shape, not bond polarity alone, decides.

Practice H

  1. Is NH₃ polar or nonpolar? Justify using its shape and the lone pair.
  2. CCl₄ has four polar C–Cl bonds yet is nonpolar. Explain with geometry.
  3. Compare CO₂ and SO₂: both are AX₂-type O compounds, but one is polar. Which, and why?

I · Hybridization & σ / π Bonds CED 2.7 · EK 2.7.A.2.vi

Hybridization names the electron arrangement around a central atom, and bonds come in two flavours: σ (head-on overlap, in every bond) and π (side-on overlap, only in the 2nd and 3rd lines of a multiple bond).

Count electron domains on the central atom: 2 → sp (180°) · 3 → sp² (120°) · 4 → sp³ (109.5°). Bonds: a single bond = 1 σ; a double = 1 σ + 1 π; a triple = 1 σ + 2 π. A π bond locks rotation (geometric isomers) and is weaker than a σ bond.
Bond order ↔ length ↔ energy: higher bond order = more shared pairs = shorter and stronger bond. So C≡C < C=C < C–C in length, and the reverse in energy.
Worked example
For ethyne, H–C≡C–H, give the hybridization of each carbon and count all σ and π bonds.
Each C has 2 electron domains (one H, one C) → sp hybridized, 180° linear.
Bonds: 2 C–H singles = 2 σ. The C≡C triple = 1 σ + 2 π.
Hybridization: both C are sp. Total = 3 σ bonds and 2 π bonds.

Not tested: deriving or depicting the hybrid orbitals themselves, hybridization using d orbitals (5+ domains, give the shape only), and molecular-orbital diagrams (bonding/antibonding filling). Know the sp/sp²/sp³ names and σ vs. π, not the orbital math.

Practice I

  1. For ethene, H₂C=CH₂, give each carbon's hybridization and count all σ and π bonds.
  2. In the carbonate ion, CO₃²⁻, what is the hybridization of carbon, and how many σ and π bonds does it contain?
  3. Rank the C–C, C=C, and C≡C bonds by length and by energy, and explain the trend using bond order.
AP Chemistry · Unit 2, Compound Structure & Properties · Dr. Ras Mulinta · Handsworth Secondary 2026–27. Pegged to the College Board AP Chemistry CED (topics 2.1–2.7) and BC Chemistry 11 (chemical bonding) / BC Chemistry 12. Atomic data from the IUPAC periodic table; bond lengths/energies from standard tables.