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It’s important to review AP Chemistry Unit 2 so you can understand bonding, Lewis structures, VSEPR, hybridization, and molecular polarity. These are all important ideas that come up throughout the whole AP Chemistry course. If you know Unit 2 well, you’ll do better on Units 3–9.
This guide goes over all of the sub-topics (2.1–2.7) and includes important formulas, VSEPR tables, practice questions with answers, common mistakes, and a focused study plan that is completely in line with the College Board CED.
| AP Resource | What’s Included | Download |
| AP Chemistry Unit 2 Review Notes | VSEPR, hybridization, polarity, and bonding | Download Unit 2 Notes PDF |
| AP Chemistry Unit 2 Practice Questions | MCQs and conceptual problems at the exam level | Download Practice Questions |
| AP Chemistry Unit 2 MCQ PDF | Fast revision with multiple-choice questions | Download MCQ PDF |
| AP Chemistry Unit 2 FRQ Practice | Questions with answers that are open-ended | Download FRQ PDF |
| AP Chemistry Unit 2 Worksheet | Worksheets for geometry and bonding practice | Download Worksheet |
| AP Chemistry Unit 2 Quick Revision Sheet | A summary of important formulas and ideas | Download Revision Sheet |
Sub-topics 2.1 to 2.7 of AP Chemistry Unit 2 cover the structure and properties of molecular and ionic compounds. It makes up 7–9% of the AP Chemistry exam and takes about 12–13 class periods. The main skills are drawing Lewis structures, using VSEPR to guess molecular geometry, figuring out hybridization, and explaining how bond type changes physical properties like melting point and conductivity.
Key Takeaways – AP Chemistry Unit 2 Review
| Category | Details |
| Official Unit Name | Structure and Properties of Molecular and Ionic Compounds |
| Sub-Topics Covered | 2.1 to 2.7 (seven sub-topics) |
| AP Exam Weight | 7–9% of the total score on the AP Chemistry exam |
| Recommended Class Periods | 12–13 periods (College Board CED, 2024–25) |
| Difficulty Level | Medium—can be accessed once you know how to bond |
| Hardest Sub-Topics | 2.5 (Formal Charge and Resonance) and 2.7 (Hybridization and VSEPR) |
| Most Tested on FRQs | Lewis structures, formal charge, molecular shape, and reasons for polarity |
| Calculator Required? | No, Unit 2 is mostly about ideas and diagrams. |
| Prerequisite | Unit 1: Atomic Structure, Electron Configuration, and Periodic Trends |
Every sub-topic from 2.1 to 2.7 is tested on the AP Chemistry exam. Use the table below to identify exactly what you need to know for each sub-topic, with the key skill and typical exam question type.
| Sub-Topic | Title | Key Skill Tested |
| 2.1 | Types of Chemical Bonds | Use electronegativity to guess the type of bond; describe ionic, covalent, and metallic bonds. |
| 2.2 | Intramolecular Force and Potential Energy | Read and understand graphs that show PE vs. bond distance; talk about bond length and bond energy. |
| 2.3 | Structure of Ionic Solids | Describe the structure of a crystal lattice and use Coulomb’s law to compare the energies of different lattices. |
| 2.4 | Structure of Metals and Alloys | Explain the sea-of-electrons model and the difference between substitutional and interstitial alloys. |
| 2.5 | Lewis Diagrams | Make Lewis structures, figure out the formal charge, and make resonance structures. |
| 2.6 | Resonance and Formal Charge | Put resonance structures in order of stability, and then use formal charge to find the best one. |
| 2.7 | VSEPR and Bond Hybridization | Guess the shape, bond angles, and polarity; figure out the hybridization; and find the sigma and pi bonds. |
The first and most important thing to do in the AP Chemistry Unit 2 review is to figure out what kind of bond forms between two atoms. The kind of bond affects the compound’s structure and physical properties.
| Bond Type | When It Forms | Key Properties |
| Ionic Bond | Metal + nonmetal | Electronegativity difference > 1.7 | It has a high melting point, is brittle, and conducts electricity when it is dissolved or melted, but not when it is solid. |
| Polar Covalent Bond | Nonmetal + nonmetal | Difference 0.4–1.7 | Molecule may be polar or nonpolar depending on its shape; electrons are not shared evenly. |
| Nonpolar Covalent Bond | Same nonmetal OR difference < 0.4 | Sharing electrons evenly; usually a lower melting point and bad conductor |
| Metallic Bond | Metal + metal | A lot of electrons that aren’t in one place; good at conducting electricity and heat; can be shaped and bent |
| Network Covalent Solid | Nonmetals in extended lattice (e.g., SiO₂, diamond) | Very hard, with a very high melting point and bad conductivity. |
Lattice energy is the amount of energy needed to completely break up one mole of an ionic solid into its gaseous ions. A stronger ionic bond and a higher melting point mean that the lattice energy is higher.
| Factor | Effect on Lattice Energy |
| Higher ion charge (e.g., +2/−2 vs. +1/−1) | Coulomb’s law says that F = kq₁q₂/r², which means that stronger electrostatic attraction increases lattice energy. |
| Smaller ion radius | Lattice energy goes up because the ions are closer together and the attraction is stronger. |
| Larger ion radius | Lattice energy goes down – there’s more space and less force. |
AP Exam Tip: You will NOT need to use a formula to figure out lattice energy. You will be asked to use Coulomb’s law to compare two ionic compounds and say which one has the higher lattice energy.
For example, which has more lattice energy: MgO or NaCl? Give a reason for using Coulomb’s law.
Answer: MgO has higher lattice energy. Mg²⁺ and O²⁻ have charges of +2/−2, while Na⁺ and Cl⁻ have charges of +1/−1. Per Coulomb’s law (F = kq₁q₂/r²), greater charge produces stronger electrostatic attraction. Additionally, Mg²⁺ and O²⁻ have smaller ionic radii, placing ions closer together and further increasing the force.
Lewis structures and formal charge are among the most tested FRQ skills in the entire AP Chemistry Unit 2 review. Every year, the AP Chemistry exam includes at least one FRQ that asks students to draw a Lewis structure, calculate formal charge, or rank resonance structures
| Step | Action |
| Step 1 – Count valence electrons | Put together all of the valence electrons from all of the atoms. Take away 1 for each positive charge and add 1 for each negative charge. |
| Step 2 – Draw the skeleton | Put the atom with the least electronegativity in the middle. Use single bonds to link atoms. |
| Step 3 -Complete octets on outer atoms | First, give the remaining electrons to the outer atoms (F, O, Cl, N, etc.). |
| Step 4 -Place remaining electrons on central atom | After the outer atoms have eight electrons, add lone pairs to the central atom. |
| Step 5 – Check and adjust | If the central atom doesn’t have enough octets, change lone pairs into double or triple bonds. |
| Step 6 – Calculate formal charge | FC = Valence e⁻ − Lone pair e⁻ − ½(Bonding e⁻). The best structure is one where all the FC values are as close to zero as possible. |
Formula: Formal Charge = Valence electrons − Lone pair electrons − ½ (Bonding electrons)
Goal: The most stable Lewis structure has all formal charges equal to zero OR the smallest formal charges, with negative FC on the more electronegative atom.
Resonance structures are two or more valid Lewis structures for the same molecule that differ only in the placement of electrons (not atoms). The actual molecule is a hybrid of all resonance structures.
| Resonance Concept | What It Means for the AP Exam |
| Resonance hybrid | The actual molecule is the average of all the valid resonance structures, not just one of them. |
| Equal bond lengths | In a molecule that has resonance, all the bonds that are the same length are either single or double. |
| Ranking resonance structures | Sort by: (1) FC values that are closest to zero, (2) a negative FC on the atom that is more electronegative, and (3) no like charges next to each other. |
| Example: SO₃ | Three resonance structures that are the same: all of the S–O bonds are the same length (between S=O and S–O). |
| Example: CO₃²⁻ | Three resonance structures that are the same: all C–O bonds are the same length and strength. |
Practice Question: Draw the two resonance structures of NO₂⁻ and identify which formal charges are present on each atom.
Answer: NO₂⁻ has 18 valence electrons (5+6+6+1 = 18). Draw N in the center with a single bond to one O and a double bond to the other O, with lone pairs completing each octet. FC on N: in first structure, 5−2−½(6) = 0. The two resonance structures are equivalent – the actual N–O bond length is between a single and double bond
VSEPR (Valence Shell Electron Pair Repulsion) theory is the most heavily tested concept in AP Chemistry Unit 2. It predicts the three-dimensional shape of a molecule by minimizing repulsion between electron domains around the central atom.
| Electron Domains | Lone Pairs | Molecular Shape | Bond Angle |
| 2 | 0 | Linear | 180° |
| 3 | 0 | Trigonal Planar | 120° |
| 3 | 1 | Bent (V-shaped) | < 120° |
| 4 | 0 | Tetrahedral | 109.5° |
| 4 | 1 | Trigonal Pyramidal | < 109.5° |
| 4 | 2 | Bent (V-shaped) | < 109.5° |
| 5 | 0 | Trigonal Bipyramidal | 90° and 120° |
| 5 | 1 | See-Saw | < 90° and < 120° |
| 5 | 2 | T-Shaped | < 90° |
| 5 | 3 | Linear | 180° |
| 6 | 0 | Octahedral | 90° |
| 6 | 1 | Square Pyramidal | < 90° |
| 6 | 2 | Square Planar | 90° |
Critical Rule: Lone pairs repel more strongly than bonding pairs. Every lone pair on the central atom compresses the bond angles below the ideal value.
AP Exam Tip: You must memorize all geometries through 4 electron domains perfectly. 5- and 6-domain geometries appear less frequently but are still tested.
If the bond dipoles cancel out because of symmetry, a molecule can have polar bonds and still be nonpolar as a whole. This is one of the most common questions on the AP Chemistry Unit 2 test.
| Molecule | Shape | Polar or Nonpolar? |
| H₂O | Bent (2 lone pairs on O) | Polar — dipoles do NOT cancel due to bent geometry |
| CO₂ | Linear (no lone pairs on C) | Nonpolar — dipoles cancel due to linear symmetry |
| NH₃ | Trigonal pyramidal | Polar — asymmetric arrangement, dipoles do not cancel |
| CH₄ | Tetrahedral (no lone pairs) | Nonpolar — perfectly symmetric, all dipoles cancel |
| BF₃ | Trigonal planar (no lone pairs on B) | Nonpolar — symmetric geometry, dipoles cancel |
| SO₂ | Bent (1 lone pair on S) | Polar — asymmetric, dipoles do not cancel |
If you have the right study materials, learning AP Chemistry is easier. TestprepKart offers a number of free e-books that students can download to help them understand the material better and do better on tests. These e-books cover all the important ideas and formulas needed to do well in AP Chemistry and other AP science classes.
Hybridization explains how atomic orbitals mix to form new orbitals for bonding. It is directly connected to VSEPR — the number of electron domains around the central atom determines the hybridization.
| Electron Domains | Hybridization | Example Molecule |
| 2 | sp | CO₂, BeCl₂, C₂H₂ (acetylene) |
| 3 | sp² | BF₃, SO₂, C₂H₄ (ethylene) |
| 4 | sp³ | CH₄, NH₃, H₂O, CCl₄ |
| 5 | sp³d | PCl₅, SF₄ |
| 6 | sp³d² | SF₆, XeF₄ |
| Bond Type | How It Forms and What It Means |
| Sigma bond (σ) | Forms from direct (head-on) overlap of orbitals. EVERY bond starts with one sigma bond. Allows free rotation. |
| Pi bond (π) | Forms from side-by-side (lateral) overlap of p-orbitals. Only in double and triple bonds. Does NOT allow free rotation. |
| Single bond | 1 sigma bond (1σ) |
| Double bond | 1 sigma + 1 pi bond (1σ + 1π) |
| Triple bond | 1 sigma + 2 pi bonds (1σ + 2π) |
Practice Question: How many sigma bonds and pi bonds are in a molecule of acetylene (C₂H₂)?
Answer: Acetylene: H–C≡C–H. Two C–H single bonds = 2 sigma bonds. One C≡C triple bond = 1 sigma + 2 pi bonds. Total: 3 sigma bonds and 2 pi bonds.
One important skill for reviewing AP Chemistry Unit 2 is being able to connect the type of bond to its physical properties, like melting point, hardness, and electrical conductivity. Every year, this shows up on both MCQ and FRQ.
| Bond Type | Melting Point | Electrical Conductivity | Examples |
| Ionic | Very high | Only when dissolved (aq) or molten — NOT as solid | NaCl, MgO, CaCO₃ |
| Metallic | Varies (low to very high) | Excellent in all states — delocalized electrons | Fe, Cu, Al, brass |
| Polar Covalent (molecular) | Low to moderate | Very poor (no free ions or electrons) | H₂O, NH₃, HCl |
| Nonpolar Covalent (molecular) | Very low | Poor | CH₄, Cl₂, CO₂ |
| Network Covalent | Extremely high | Poor (exception: graphite conducts) | Diamond, SiO₂, graphite |
AP Exam Tip: The most commonly missed MCQ fact: ionic solids are poor conductors as solids because ions are locked in the crystal lattice and cannot move. They only conduct when dissolved or melted. Metallic solids conduct in all states because electrons are always free to move.
Sub-topic 2.2 talks about how the distance between atoms affects the potential energy when bonds are made. This idea comes up in multiple-choice questions that ask you to read a graph of PE vs. bond distance.
| Concept | What It Means on the AP Exam |
| Atoms too far apart | Additionally, there isn’t much of an attraction—high PE and almost no interaction. |
| Atoms approaching | PE goes down as attractive forces take over and a bond starts to form. |
| Optimal bond length | The minimum PE on the curve is the length of the bond at which the two sides are equal. |
| Atoms too close | PE goes up quickly because nuclear and electron repulsion are stronger. |
| Bond energy | The depth of the PE well is the amount of energy needed to separate bonded atoms to an infinite distance. |
| Stronger bond | A deeper PE well (a more negative minimum) and a shorter equilibrium bond length |
| Triple bond vs. single bond | Triple bond: shorter bond length, deeper minimum PE, and higher bond energy |
Practice Question: On a PE vs. bond distance graph, what does the minimum point represent?
Answer: The minimum point represents the equilibrium bond length — the distance at which the attractive and repulsive forces between the two atoms are balanced, and the potential energy is at its lowest value. This is where the stable bond exists.
All questions below are written at the AP Chemistry exam level and cover the most frequently tested concepts in the Unit 2 review. Work through each question before reading the answer.
Q1 (Bond Type): Which of the following pairs of elements is most likely to form an ionic compound?
Answer: B) Sodium and Chlorine. Na is a metal (low electronegativity) and Cl is a nonmetal (high electronegativity). The electronegativity difference > 1.7 — satisfying the condition for ionic bond formation. The other pairs involve two nonmetals forming covalent bonds.
Q2 (Formal Charge): In a Lewis structure of SO₄²⁻ with two double bonds and two single bonds, what is the formal charge on sulfur?
Answer: Sulfur has 6 valence electrons. In this structure: bonding electrons = 4×2(for 2 singles) + 4×2(for 2 doubles) — let us count directly. S has 4 bonds: 2 single (2 electrons each) + 2 double (4 electrons each) = 12 bonding electrons total, 0 lone pairs. FC(S) = 6 − 0 − ½(12) = 6 − 6 = 0. Formal charge on S = 0.
Q3 (VSEPR Geometry): What is the molecular geometry and bond angle of SF₄?
Answer: SF₄ has 5 electron domains around S: 4 bonding pairs + 1 lone pair. The electron geometry is trigonal bipyramidal. The molecular geometry (shape) is See-Saw because the lone pair occupies one of the equatorial positions. Bond angles are slightly less than 120° (equatorial) and slightly less than 90° (axial) due to lone pair compression.
Prompt: Consider the molecule SO₂.
(a) Lewis Structure: SO₂ has 18 valence electrons (6+6+6 = 18). S is the central atom. Most stable structure has one S=O double bond and one S–O single bond, with a lone pair on S. Alternatively, a resonance structure with a double bond on the other O is also valid. Formal charge on S: 6 − 2 − ½(6) = 6 − 2 − 3 = +1 in the single/double structure. The two resonance structures are equivalent – actual SO₂ is a resonance hybrid.
(b) Molecular Geometry: SO₂ has 3 electron domains around S: 2 bonding domains (one to each O) + 1 lone pair. By VSEPR, 3 electron domains = trigonal planar electron geometry. With 1 lone pair, the molecular geometry is bent. Ideal bond angle for 3 domains = 120°, but the lone pair compresses it – actual O–S–O bond angle ≈ 119°.
(c) Polarity: SO₂ is POLAR. The bent geometry means the two S–O bond dipoles do NOT cancel – they add together to give a net dipole moment pointing away from the lone pair. Unlike CO₂ (linear, nonpolar), SO₂’s asymmetric shape creates a net molecular dipole.
(d) Hybridization: 3 electron domains around S → sp² hybridization. The unhybridized p-orbital on S participates in the pi bond with one of the oxygen atoms (and is delocalized across both S–O bonds due to resonance).
10. Top Mistakes on AP Chemistry Unit 2 TestThese are the most common student errors on AP Chemistry Unit 2 — identified from College Board Chief Reader Reports and AP Chemistry teacher analysis.
| Mistake | Why It Loses Points | Fix It |
| Ionic solids conduct as solids | It is incorrect to say that ionic compounds conduct in all states; they do NOT conduct as solids. | Ionic solids: ions are stuck in a lattice, so they can’t conduct electricity. Only work when they are melted or dissolved. |
| Wrong formal charge calculation | Getting the number of lone pairs and bonding electrons wrong in the FC formula. | Before each calculation, write FC = valence e⁻ − lone pair e⁻ − ½(bonding e⁻) out in full. |
| Forgetting lone pairs compress bond angles | Giving the perfect angle without mentioning lone pair compression. | Every lone pair on the central atom makes the bond angles smaller than they should be. |
| Power Rule error on electron domains | For VSEPR, only counting bonding pairs and not lone pairs. | Add up all the electron domains, which include bonding pairs and lone pairs. Both decide on the shape. |
| Polarity confusion: polar bonds ≠ polar molecule | Because C=O is polar, CO₂ is also polar. | Check the geometry: if the bond dipoles cancel each other out because of symmetry, the molecule is nonpolar, even if it has polar bonds. |
| Pi bonds = separate bonds misconception | Believing that a double bond has two separate bonds. | One sigma (direct overlap) and one pi (side overlap) make up a double bond. A single bond has only one sigma. |
| Wrong hybridization for expanded octets | Using sp³ on all 4-domain molecules, even if they have an expanded octet. | For five areas: sp³d. For 6 domains, sp³d². These only use d-orbitals that are available to elements in period 3 and higher. |
Use this priority-based study plan to master all 7 Unit 2 sub-topics efficiently. Total recommended study time: 8–12 hours.
| Phase | Focus | Time Estimate |
| Phase 1 – Foundation (Sub-topics 2.1–2.4) | Bond types, electronegativity, lattice energy, ionic solid properties, metallic bonds, alloys. Master the conductivity table and lattice energy comparisons. | 2–3 hours |
| Phase 2 – Core Skills (Sub-topics 2.5–2.6) | Lewis structure method (6 steps), formal charge formula and calculation, resonance structure ranking. Draw at least 10 Lewis structures from scratch. | 3–4 hours |
| Phase 3 – VSEPR and Hybridization (Sub-topic 2.7) | Memorize complete VSEPR table. Practice predicting geometry, bond angle, polarity, and hybridization for 15+ molecules. Master sigma/pi bond counting. | 3–4 hours |
| Phase 4 – FRQ Practice | Complete 2–3 full FRQ prompts with Lewis structure + VSEPR + polarity parts. Score using College Board rubric. Review Chief Reader Report for Unit 2 FRQ errors. | 1–2 hours |
| Final Day Review | Re-attempt all wrong practice questions. Review conductivity table, formal charge formula, and VSEPR table one final time. | 30–60 min |
What topics are covered in AP Chemistry Unit 2?
There are 7 sub-topics (2.1–2.7) in AP Chemistry Unit 2 that talk about the structure and properties of molecular and ionic compounds. Some of the most important topics are types of chemical bonds, intramolecular forces, ionic solid structure, metallic bonding, Lewis structures, formal charge, resonance, VSEPR theory, hybridization, molecular geometry, and bond polarity.
How much of the AP Chemistry exam is Unit 2?
Unit 2 of AP Chemistry makes up 7–9% of the total score on the AP Chemistry exam. This means that there are about 4 to 5 multiple-choice questions on the full AP exam. FRQ questions that cover more than one unit often ask about concepts from Unit 2, like Lewis structures and VSEPR.
What is the formal charge in AP Chemistry?
Formal charge (FC) is a way to keep track of how electrons are spread out in a Lewis structure. FC = Valence electrons − Lone pair electrons − ½(Bonding electrons). The Lewis structure that is most stable has formal charges that are closest to zero, and any negative formal charges are on the atoms that are more electronegative.
How do you predict molecular geometry using VSEPR?
To use VSEPR to guess the shape of a molecule: (1) Make a drawing of the Lewis structure of the molecule. (2) Count all the electron domains around the central atom, including both bonding pairs and lone pairs. (3) Use the VSEPR table to figure out the molecular shape based on the total number of domains and lone pairs. (4) Remember that lone pairs make bond angles smaller than they should be.
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