VSEPR and Bond Hybridization

AP Chemistry· AP Chemistry CED — Molecular and Ionic Compound Structure and Properties· 14 min read

1. VSEPR: Steric Number and Molecular Geometry★★☆☆☆⏱ 4 min

Valence Shell Electron Pair Repulsion (VSEPR) is the core predictive model for 3D molecular shape, built on the 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.

📘 Definition

Electron Domain

Any single region of electron density around a central atom; a single bond, double bond, triple bond, or lone pair all count as one domain, regardless of bond order.

Example:

A carbon-carbon double bond counts as exactly 1 electron domain for VSEPR calculations.

📘 Definition

Steric Number

$S N$

The total number of electron domains around the central atom, used to predict geometry and hybridization.

Example:

A central atom with 3 bonding domains and 1 lone pair has $S N = 4$ .

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.

📐 Worked Example

Predict the electron domain geometry and molecular geometry of the sulfite ion, $SO_{3}^{2 -}$ .

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 = $S N = 3 + 1 = 4$ .
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.

2. Bond Hybridization: Correlation to Steric Number★★★☆☆⏱ 4 min

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. 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.

$S N = 2 \to s p$ (1 s + 1 p mixed, 2 hybrids)
$S N = 3 \to s p^{2}$ (1 s + 2 p mixed, 3 hybrids)
$S N = 4 \to s p^{3}$ (1 s + 3 p mixed, 4 hybrids)
$S N = 5 \to s p^{3} d$ (1 s + 3 p + 1 d, 5 hybrids)
$S N = 6 \to s p^{3} d^{2}$ (1 s + 3 p + 2 d, 6 hybrids)

📐 Worked Example

Identify the hybridization of all unique carbon atoms in propyne, $CH_{3} CCH$ .

1
Draw the Lewis structure and label unique carbons: terminal $CH_{3}$ 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 → $S N = 4 + 0 = 4$ . C2 is bonded to C1 and C3, 0 lone pairs → $S N = 2 + 0 = 2$ . C3 is bonded to C2 and 1 H, 0 lone pairs → $S N = 2 + 0 = 2$ .
3
Match SN to hybridization: SN=4 corresponds to $s p^{3}$ , SN=2 corresponds to $s p$ .
4
Verify with pi bonding: C2 and C3 form a triple bond, which requires 2 unhybridized p orbitals per carbon, consistent with $s p$ hybridization (which leaves 2 unhybridized p orbitals per atom).

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.

3. VSEPR Exceptions: Expanded Octets and Lone Pair Placement★★★★☆⏱ 3 min

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 to minimize 90° repulsions.

📐 Worked Example

Predict the molecular geometry of $ClF_{3}$ 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 = $S N = 5$ , 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.

4. AP-Style Concept Check★★★☆☆⏱ 3 min

✓ Quick check

Evaluate the options below to test your understanding:

Which of the following correctly matches the species, its molecular geometry, and the hybridization of the underlined central atom?
- A) $\underline{O}$ in $O_{3}$ : Bent, $s p^{2}$
- B) $\underline{Xe}$ in $XeF_{4}$ : Square planar, $s p^{3} d$
- C) $\underline{P}$ in $PCl_{3}$ : Trigonal planar, $s p^{3}$
- D) $\underline{Be}$ in gaseous $BeCl_{2}$ : Linear, $s p^{2}$

5. Common Pitfalls

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 $S N > 4$ 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 highlight 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. Quick Reference Cheatsheet

Category	Formula/Rule	Notes
Steric Number	$S N = n (bonding domains) + m (lone pairs on A)$	1 domain per bonded group, regardless of bond order
Electron Domain Geometry (SN=2)	Linear	Ideal bond angle = 180°
Electron Domain Geometry (SN=3)	Trigonal Planar	Ideal bond angle = 120°
Electron Domain Geometry (SN=4)	Tetrahedral	Ideal bond angle = 109.5°
Electron Domain Geometry (SN=5)	Trigonal Bipyramidal	Ideal angles: 90° axial-equatorial, 120° equatorial-equatorial
Electron Domain Geometry (SN=6)	Octahedral	Ideal bond angle = 90° between adjacent domains
Hybridization Correlation	SN2=sp, SN3=sp², SN4=sp³, SN5=sp³d, SN6=sp³d²	Only period 3+ central atoms can have SN>4
Repulsion Strength Order	Lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding	Lone pairs compress adjacent bond angles
Lone Pair Placement (SN=5)	All lone pairs go to equatorial positions	Minimizes total 90° repulsions
Sigma/Pi Count	1 sigma per bond; 1 pi per double bond, 2 pi per triple bond	Sigma from hybrid overlap, pi from unhybridized p overlap

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Identify geometry/hybridization
2022 · FRQ
Justify bond angle deviation
2021 · MCQ
Match species to hybridization

What's Next

Mastery of VSEPR and bond hybridization is an essential prerequisite for all subsequent topics related to molecular properties on the AP Chemistry exam. 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 both multiple-choice and free-response sections. This topic also feeds into the study of resonance, delocalized pi bonding, and molecular orbital theory, which explain the unique properties of aromatic compounds and conductive materials.