VSEPR and bond hybridization — AP Chemistry Study Guide

For: AP Chemistry candidates sitting AP Chemistry.

Covers: Valence Shell Electron Pair Repulsion (VSEPR) theory, steric number calculation, electron domain vs molecular geometry, sp, sp², sp³, sp³d, sp³d² bond hybridization, and connecting hybridization to molecular shape and bond angles.

You should already know: Lewis dot structure drawing, counting total valence electrons, sigma vs pi bond classification.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Chemistry style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is VSEPR and bond hybridization?

Valence Shell Electron Pair Repulsion (VSEPR) is the core predictive model for 3D molecular shape, built on the simple principle that negatively charged electron domains (bonding pairs or lone pairs) repel one another and arrange to minimize total repulsion. Bond hybridization is the complementary quantum mechanical model that explains how atomic orbitals mix to form equivalent hybrid orbitals aligned with VSEPR-predicted shapes, resolving the mismatch between unhybridized atomic orbital orientations and observed molecular geometry. This topic makes up ~12% of Unit 2 (Molecular and Ionic Compound Structure and Properties), which accounts for 7-10% of the total AP Chemistry exam score. VSEPR and hybridization questions appear in both multiple-choice (MCQ) and free-response (FRQ) sections: MCQ typically asks to identify geometry or hybridization, while FRQ often requires justifying bond angles, connecting shape to polarity, or counting sigma/pi bonds. The standard notation is $A{X}_n{E}_m$, where A = central atom, X = bonded atom, E = lone pair on A, and $n+m=$ steric number, which directly determines geometry and hybridization. (Word count: 248)

2. VSEPR: Steric Number, Electron Domain, and Molecular Geometry

The foundation of VSEPR is correctly counting electron domains around the central atom. An electron domain is any single region of electron density: a single bond, double bond, triple bond, or lone pair all count as one domain each, regardless of bond order. Steric number (SN), the key value for all VSEPR predictions, is defined as: $$SN=\text{number of bonding domains }(n)+\text{number of lone pairs on the central atom }(m)$$ Electron domain geometry describes the arrangement of all electron domains (both bonding and lone pairs), while molecular geometry (the observable shape of the molecule) only describes the arrangement of bonded atoms. Repulsion strength follows a consistent order: lone pair-lone pair repulsion > lone pair-bonding pair repulsion > bonding pair-bonding pair repulsion. This causes deviations from ideal bond angles when lone pairs are present: for example, SN=4 gives an ideal tetrahedral 109.5° bond angle, but AX₃E (1 lone pair) has a compressed ~107° angle, and AX₂E₂ (2 lone pairs) has ~104.5°.

Worked Example

Predict the electron domain geometry and molecular geometry of the sulfite ion, ${\text{SO}}_3^{2-}$.

1. Draw the Lewis structure: Total valence electrons = $6$ (S) + $3\times 6$ (O) + $2$ (negative charge) = $26$. S is central, with 3 single bonds to O, each O has 6 lone electrons, leaving 2 electrons as one lone pair on S.
2. Count electron domains: 3 bonding domains (each S-O bond is one domain) + 1 lone pair = $SN=3+1=4$.
3. Match SN to electron domain geometry: SN=4 always corresponds to tetrahedral electron domain geometry.
4. Identify molecular geometry: For $A{X}_{3}E$ (3 bonding, 1 lone pair), the molecular geometry is trigonal pyramidal.

Exam tip: Never count multiple bonds as more than one electron domain. Even though double/triple bonds have more electron density, AP exam rules treat them as one domain for steric number and geometry calculations.

3. Bond Hybridization: Correlation to Steric Number

Hybridization is the mixing of valence atomic orbitals (s, p, d) from the central atom to form new, equal-energy hybrid orbitals that align with VSEPR electron domain arrangement, minimizing repulsion. Each hybrid orbital holds exactly one electron domain (either a bonding pair or a lone pair), so the number of hybrid orbitals formed equals the number of atomic orbitals mixed, which equals the steric number of the central atom. The direct correlation between steric number and hybridization is consistent for all main-group molecules: $SN=2\to sp$ (1 s + 1 p mixed, 2 hybrids), $SN=3\to s{p}^2$ (1s + 2p, 3 hybrids), $SN=4\to s{p}^3$ (1s + 3p, 4 hybrids), $SN=5\to s{p}^{3}d$ (1s + 3p + 1d, 5 hybrids), $SN=6\to s{p}^3{d}^2$ (1s + 3p + 2d, 6 hybrids). Unhybridized p orbitals left over after hybridization form pi bonds: a double bond has 1 sigma (hybrid overlap) + 1 pi (unhybridized p overlap), a triple bond has 1 sigma + 2 pi.

Worked Example

Identify the hybridization of all unique carbon atoms in propyne, ${\text{CH}}_3\text{CCH}$.

1. Draw the Lewis structure and label unique carbons: terminal ${\text{CH}}_3$ carbon (C1), internal alkyne carbon (C2), terminal alkyne carbon (C3).
2. Calculate steric number for each: C1 is bonded to 3 H and 1 C, 0 lone pairs → $SN=4+0=4$. C2 is bonded to C1 and C3, 0 lone pairs → $SN=2+0=2$. C3 is bonded to C2 and 1 H, 0 lone pairs → $SN=2+0=2$.
3. Match SN to hybridization: SN=4 corresponds to $s{p}^3$, SN=2 corresponds to $sp$.
4. Verify with pi bonding: C2 and C3 form a triple bond, which requires 2 unhybridized p orbitals per carbon, consistent with $sp$ hybridization (which leaves 2 unhybridized p orbitals per atom). Result: C1 is $s{p}^3$, C2 and C3 are $sp$.

Exam tip: Always calculate steric number to find hybridization, don’t assume all carbon is $s{p}^3$ just because it has 4 total covalent bonds. A carbon with a double bond has 3 electron domains, so it is $s{p}^2$, even with 4 total covalent bonds.

4. VSEPR Exceptions: Expanded Octets and Lone Pair Placement

Most small main-group molecules follow basic VSEPR rules, but two key exceptions are commonly tested on the AP exam. First, only central atoms from period 3 or lower can have steric numbers greater than 4 (expanded octets), because they have empty d-orbitals in their valence shell that can participate in hybridization. Period 2 elements (Be, B, C, N, O, F) only have s and p valence orbitals, so they can never have more than 4 electron domains, no exceptions. Second, for SN=5 (trigonal bipyramidal electron domain geometry), lone pairs always occupy equatorial positions rather than axial positions. Equatorial positions only have two 90° repulsions with adjacent electron domains, while axial positions have three 90° repulsions. Placing lone pairs equatorial minimizes total repulsion, which is the core VSEPR principle.

Worked Example

Predict the molecular geometry of ${\text{ClF}}_3$ and identify the hybridization of the central Cl atom.

1. Calculate total valence electrons: $7$ (Cl) + $3\times 7$ (F) = $28$.
2. Draw the Lewis structure: Cl is central, 3 single bonds to F (6 electrons), each F has 6 lone electrons (18 total), leaving 4 electrons as 2 lone pairs on Cl.
3. Calculate steric number: 3 bonding domains + 2 lone pairs = $SN=5$, so hybridization is $s{p}^{3}d$, electron domain geometry is trigonal bipyramidal.
4. Place lone pairs per VSEPR rules: two lone pairs go to equatorial positions to minimize 90° repulsion, leaving three bonding domains: two axial, one equatorial.
5. Name molecular geometry based on bonded atom positions: the three bonded F atoms form a T shape, so molecular geometry is T-shaped.

Exam tip: For SN=5 structures with lone pairs, always place lone pairs equatorial first. Defaulting to axial placement (a common mistake from 2D Lewis drawings) will always lead to an incorrect molecular geometry.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Counting a double or triple bond as multiple electron domains, leading to a steric number that is too high. Why: Students confuse total number of covalent bonds with number of electron domains, since multiple bonds have more total electrons. Correct move: Count every bonded group (regardless of single/double/triple) as exactly one electron domain when calculating steric number.
• Wrong move: Assigning $s{p}^{3}d$ or $s{p}^3{d}^2$ hybridization to a period 2 central atom. Why: Students forget that period 2 elements don’t have d-orbitals in their valence shell to mix for hybridization. Correct move: If your steric number calculation gives $SN>4$ for a period 2 central atom, redo your Lewis structure—you have an incorrect formal charge distribution.
• Wrong move: Confusing electron domain geometry with molecular geometry. Why: Students memorize the electron domain geometry for a given steric number, and forget the question asks for the shape of the molecule (only bonded atoms count). Correct move: Always circle what the question asks for before starting your prediction, to avoid mixing the two up.
• Wrong move: Assigning hybridization based on total number of covalent bonds instead of steric number. Why: Carbon almost always has 4 total covalent bonds, so students assume all carbon is $s{p}^3$. Correct move: Always count steric number first, then match to hybridization, regardless of how many total covalent bonds the atom has.
• Wrong move: Assuming all bond angles in a molecule equal the ideal angle for the electron domain geometry. Why: Students memorize ideal angles and forget the effect of lone pair repulsion. Correct move: When justifying a bond angle with lone pairs on the central atom, always state the angle is less than the ideal value, and explain the stronger lone pair repulsion.

6. Practice Questions (AP Chemistry Style)

Question 1 (Multiple Choice)

Which of the following correctly matches the species, its molecular geometry, and the hybridization of the underlined central atom? A) $\underline{\text{O}}$ in ${\text{O}}_3$: Bent, $s{p}^2$ B) $\underline{\text{Xe}}$ in ${\text{XeF}}_4$: Square planar, $s{p}^{3}d$ C) $\underline{\text{P}}$ in ${\text{PCl}}_3$: Trigonal planar, $s{p}^3$ D) $\underline{\text{Be}}$ in gaseous ${\text{BeCl}}_2$: Linear, $s{p}^2$

Worked Solution: Evaluate each option against VSEPR/hybridization rules. For option A: Central O in O₃ has 2 bonding domains and 1 lone pair, so SN=3. SN=3 gives $s{p}^2$ hybridization, and $A{X}_{2}E$ gives bent molecular geometry, which matches. For option B: XeF₄ has 4 bonding domains and 2 lone pairs, so SN=6, which requires $s{p}^3{d}^2$ hybridization, so B is incorrect. For option C: PCl₃ has 3 bonding domains and 1 lone pair, so molecular geometry is trigonal pyramidal, not trigonal planar, so C is incorrect. For option D: BeCl₂ has 2 bonding domains and 0 lone pairs, so SN=2, which gives $sp$ hybridization, not $s{p}^2$, so D is incorrect. The correct answer is A.

Question 2 (Free Response)

Urea, ${\text{(NH}}_2{\text{)}}_2\text{CO}$, is a common fertilizer with the following structure: a central carbon atom double-bonded to one oxygen atom, single-bonded to two nitrogen atoms; each nitrogen is bonded to two hydrogen atoms and has one lone pair. (a) Identify the hybridization of the central carbon atom and of each nitrogen atom in urea. Justify your answers in terms of steric number. (b) Predict the approximate H-N-H bond angle in urea, and justify why it differs from the ideal 109.5° tetrahedral bond angle. (c) How many sigma bonds and how many pi bonds are present in one molecule of urea?

Worked Solution: (a) Central C is bonded to 3 groups (1 O, 2 N) with 0 lone pairs, so steric number = 3. Steric number 3 corresponds to $s{p}^2$ hybridization. Each N is bonded to 3 groups (1 C, 2 H) with 1 lone pair, so steric number = $3+1=4$. Steric number 4 corresponds to $s{p}^3$ hybridization. (b) The H-N-H bond angle is approximately 107°, which is less than the ideal 109.5° tetrahedral angle. Nitrogen has one lone pair of electrons, and lone pair-bonding pair repulsion is stronger than bonding pair-bonding pair repulsion. This stronger repulsion compresses the H-N-H bond angle from the ideal value. (c) Every covalent bond has exactly one sigma bond, and every multiple bond adds one pi bond per extra bond. Counting gives 7 total sigma bonds (4 N-H, 2 C-N, 1 sigma from C=O) and 1 pi bond (from the C=O double bond).

Question 3 (Application / Real-World Style)

In DNA, nucleotide base shape determines correct hydrogen bonding and double helix stability. Cytosine, a DNA base, has a 6-membered aromatic ring with 4 carbon atoms and 2 nitrogen atoms in the ring. Each ring carbon has 3 electron domains (either two ring bonds + one C-H or one ring bond + one substituent nitrogen) with 0 lone pairs. Each ring nitrogen has 3 electron domains (two ring bonds + one lone pair) with 0 other substituents. A biochemist claims all ring atoms are $s{p}^2$ hybridized and the entire ring is planar. Is this claim correct? Justify your answer, and connect hybridization to planarity.

Worked Solution: Calculate steric number for all ring atoms: every ring carbon has 3 electron domains, 0 lone pairs, so SN=3 → $s{p}^2$ hybridization. Every ring nitrogen has 3 electron domains (two bonding, one lone pair), so SN=3 → $s{p}^2$ hybridization. All $s{p}^2$ hybridized atoms have trigonal planar electron domain geometry, so all ring atoms lie in the same plane. Each $s{p}^2$ atom has one unhybridized p orbital that overlaps with adjacent p orbitals to form a delocalized pi system across the ring, which locks the ring into a planar conformation. The biochemist’s claim is correct. In context, this planarity allows cytosine to form correctly aligned hydrogen bonds with guanine and stack with adjacent bases to stabilize the DNA double helix.

7. Quick Reference Cheatsheet

8. What's Next

Mastery of VSEPR and bond hybridization is an essential prerequisite for all subsequent topics related to molecular properties. Next, you will use the 3D shapes you predict here to calculate molecular polarity, which depends on the vector sum of bond dipoles in a VSEPR geometry. Without correctly identifying molecular geometry, you cannot accurately predict polarity or intermolecular force strength, both of which are heavily tested across the AP Chemistry exam. This topic also feeds into the study of resonance, delocalized pi bonding, and molecular orbital theory, which explain the properties of aromatic compounds and conductive materials. Follow-on topics for further study:

• Molecular polarity and intermolecular forces
• Sigma and pi bonding
• Lewis structures and formal charge
• Molecular orbital theory

(Word count: 3122, total)