AP Chemistry Unit 2 review: Compound Structure and Properties
Unit 2 explains how bonding and particle arrangement determine compound structure. It is 7–9% of multiple choice. Expect to count electrons, evaluate Lewis structures and use geometry, charge and bond energy to justify molecular or solid properties.
Unit 2 on the May 2027 exam
| 2027 exam | |
|---|---|
| Weight | 7–9% of the multiple-choice section |
| Exam date | Thursday, May 6, 2027, afternoon session |
| Multiple choice | 60 questions in 90 minutes, worth 50% of the exam score |
| Free response | 7 questions in 105 minutes, worth 50%: three long questions (10 points each) and four short questions (4 points each). |
| Calculator | Both sections allow scientific or approved graphing calculators. Four-function calculators are allowed but not recommended. Bluebook includes both Desmos scientific and graphing calculators. |
| Class time | About 12–13 class periods in College Board's pacing guide |
| 2026 results | 15% of students scored a 5 on the whole exam, and about 76% scored 3 or higher |
Sources: College Board's course and exam description, exam page, 2027 exam dates and score distributions.
What Unit 2 covers
College Board's course description splits the unit into 7 topics. The second column is what you should be able to do for each one.
| Topic | What you should be able to do |
|---|---|
| 2.1 Types of Chemical Bonds | Distinguish ionic, covalent and metallic bonding using electron distribution and electrostatic forces. |
| 2.2 Intramolecular Force and Potential Energy | Interpret potential energy versus internuclear distance and compare bond lengths and energies. |
| 2.3 Structure of Ionic Solids | Relate ion charge and separation to attraction within an ionic lattice. |
| 2.4 Structure of Metals and Alloys | Explain conductivity and deformation using delocalized electrons and alloy structure. |
| 2.5 Lewis Diagrams | Construct Lewis diagrams with the correct total valence-electron count. |
| 2.6 Resonance and Formal Charge | Calculate formal charges and use resonance contributors to describe delocalization. |
| 2.7 VSEPR and Hybridization | Predict molecular shape, bond angles, polarity and sp, sp² or sp³ hybridization. |
From a Lewis diagram to geometry and polarity
Count electron domains around the central atom; a multiple bond is one domain. Then account for lone pairs and the vector sum of bond dipoles.
| Domains and lone pairs | Geometry | Useful consequence |
|---|---|---|
| 2 domains, 0 lone pairs | Linear; 180°; sp | Equal opposite bond dipoles cancel, as in CO₂. |
| 3 domains, 0 lone pairs | Trigonal planar; 120°; sp² | A symmetric arrangement of identical bonds can be nonpolar. |
| 3 domains, 1 lone pair | Bent; angle below 120° | Bond dipoles need not cancel. |
| 4 domains, 0 lone pairs | Tetrahedral; about 109.5°; sp³ | Identical surrounding atoms can give a nonpolar molecule. |
| 4 domains, 1 lone pair | Trigonal pyramidal; angle below 109.5° | The lone pair changes shape and typically compresses angles. |
| 4 domains, 2 lone pairs | Bent; angle below 109.5° | Distinguish tetrahedral domain geometry from bent molecular shape. |
| Formal charge | Valence − nonbonding electrons − half bonding electrons | All atom formal charges must sum to the species charge. |
Common mistakes
Each of these loses points on questions you otherwise understand.
- Forgetting an ion’s charge when counting electrons. Add electrons for a negative charge and remove them for a positive charge. CED printed p. 41 identifies incorrect valence counts.
- Giving a second-period atom more than an octet. Recheck bonding and lone pairs before using a structure to predict shape. CED p. 41.
- Confusing electron-domain geometry with molecular shape. Lone pairs occupy domains but are not atoms in the named shape. CED p. 41 and topic 2.7.
- Treating resonance structures as molecules switching back and forth. They represent a single delocalized electron distribution. CED topic 2.6.
6 practice questions with answers
Written by Ashore for this page, in the style of the multiple-choice section. Try each one before you open the answer.
Question 1 · 2.5
How many total valence electrons belong in a Lewis diagram of NO₂⁻?
- A16
- B17
- C18
- D24
Show the answer
C. Nitrogen contributes 5, two oxygens contribute 12, and the negative charge adds 1: 5 + 12 + 1 = 18. Use this total to check all bonds and lone pairs.
Question 2 · 2.6
An oxygen atom has three lone pairs and one single bond. What is its formal charge?
- A−1
- B0
- C+1
- D−2
Show the answer
A. Formal charge = 6 − 6 − 2/2 = −1. Assign all lone-pair electrons and half the bonding electrons to oxygen for this bookkeeping calculation.
Question 3 · 2.7
A central atom has three single bonds and one lone pair. What is its molecular shape?
- ATrigonal planar
- BTetrahedral
- CSquare planar
- DTrigonal pyramidal
Show the answer
D. Four electron domains have tetrahedral domain geometry. With one occupied by a lone pair, the three bonded atoms form a trigonal pyramidal molecular shape.
Question 4 · 2.2
A diatomic molecule has a potential-energy minimum at −410 kJ mol⁻¹ relative to separated atoms at zero. How much energy is needed to separate one mole from that minimum?
- A−410 kJ
- B+410 kJ
- C+205 kJ
- D−205 kJ
Show the answer
B. Separation raises potential energy from −410 to 0 kJ mol⁻¹, an increase of 410 kJ mol⁻¹. Breaking the bond requires an energy input; the negative well depth describes the bound state.
Question 5 · 2.3
Two idealized ion pairs have the same separation. Pair A is +1 and −1; pair B is +2 and −2. What is the ratio of attractive-force magnitudes, B/A?
- A1/4
- B2
- C4
- D8
Show the answer
C. At equal separation, Coulombic force magnitude is proportional to |q₁q₂|. The charge products have magnitudes 1 and 4, so B/A = 4. Real lattice energies also depend on structure and distance.
Question 6 · 2.7
Which statement correctly compares CO₂ and H₂O polarity?
- ACO₂ is nonpolar; H₂O is polar
- BBoth are nonpolar because their bonds cancel
- CCO₂ is polar; H₂O is nonpolar
- DBoth are polar because each contains oxygen
Show the answer
A. CO₂ is linear with equal opposite C–O bond dipoles. H₂O is bent, so its O–H bond dipoles do not cancel. Bond polarity alone cannot establish molecular polarity.
How to study Unit 2
Use the same checklist for each Lewis diagram: electron total, connectivity, octets, formal charges and any equivalent resonance contributors. Keep the atom positions fixed when drawing resonance; only the electron placement changes.
After drawing a structure, count domains, identify lone pairs, name the shape and add dipole arrows. For solids, switch scale: identify the particles and explain which charges can move, which attractions hold the structure together and what happens when layers shift.
Ashore's Chemistry course has a short lesson for every topic in this unit, each ending with a few questions on what it taught.
- AP Chemistry: all 9 units, exam format and scores
- AP Chemistry Unit 1: Atomic Structure and Properties
- AP Chemistry Unit 3: Properties of Substances and Mixtures
- AP Chemistry Unit 4: Chemical Reactions
- AP Chemistry Unit 5: Kinetics
- AP Chemistry Unit 6: Thermochemistry
- AP Chemistry Unit 7: Equilibrium
- AP Chemistry Unit 8: Acids and Bases
- AP Chemistry Unit 9: Thermodynamics and Electrochemistry
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