Shapes of Molecules and Bond Angles — AQA A-level Chemistry

SPEC 3.1.3.5 BONDING

Shapes of Molecules & Ions

VSEPR theory, electron pair repulsion, molecular geometries, and bond angles for molecules and polyatomic ions

VSEPR Theory

Definition

VSEPR (Valence Shell Electron Pair Repulsion) — the shape of a molecule is determined by the repulsion between all electron pairs in the valence shell of the central atom.

Electron pairs (both bonding pairs and lone pairs) are negatively charged. They repel each other and arrange themselves as far apart as possible to minimise repulsion.

Two Types of Electron Pair

Bonding Pair (BP)

Shared between two atoms. Located between the nuclei. Less space needed. Lone Pair (LP)

Non-bonding electrons on central atom. Takes up more space — closer to nucleus, more diffuse.

Key Point

The shape is determined by the positions of the atoms, not the lone pairs. But lone pairs influence bond angles by repelling bonding pairs more strongly.

Repulsion Hierarchy

Not all repulsions are equal. Lone pairs occupy more space and exert greater repulsion than bonding pairs:

Repulsion Order

LP–LP > LP–BP > BP–BPStrongest repulsion Weakest repulsion

Why Lone Pairs Repel More

Lone pairs are held only by one nucleus (the central atom) and spread out more than bonding pairs, which are attracted by two nuclei. This greater spread = stronger repulsion on neighbouring pairs.

2 Electron Pairs — Linear

Two bonding pairs maximise separation at 180°, giving a linear shape.

MoleculeCentral AtomElectron PairsBond AngleShape

BeCl₂Be2 BP, 0 LP180°Linear

CO₂C2 BP (double bonds count as 1 pair), 0 LP180°Linear

Multiple Bonds

Double bonds and triple bonds each count as ONE electron pair for VSEPR purposes. CO₂ has two double bonds = two electron pairs = linear.

Worked Example01

Determine the shape and bond angle of the BeCl₂ molecule. Explain your reasoning using VSEPR theory.

1

Draw the Lewis structure: Be is in Group 2 (2 valence electrons), each Cl has 7. BeCl₂ uses all 4 shared electrons in 2 single bonds. Be has no lone pairs.

2

Count electron pairs around central atom (Be): 2 bonding pairs, 0 lone pairs = 2 electron pairs total.

3

To minimise repulsion, 2 electron pairs arrange at 180° from each other.

4

All electron pairs are bonding pairs, so the molecular shape matches the electron pair geometry.

Shape: Linear | Bond angle: 180°

Both bonding pairs repel equally (BP–BP repulsion), so the Cl–Be–Cl angle is exactly 180°.

3 Electron Pairs

Trigonal Planar

Bond angle: 120°

3 BP, 0 LP | e.g. BF₃, NO₃⁻

Bent (V-Shaped)

Bond angle: ~117°

2 BP, 1 LP | e.g. SO₂

Why Bent is Less Than 120°

In SO₂, the lone pair on S exerts greater repulsion on the two bonding pairs than they exert on each other. This pushes the X–A–X angle below 120° (approximately 117°).

Worked Example02

BF₃ and SO₂ both have three electron pairs around their central atoms. Explain why BF₃ has a bond angle of 120° while SO₂ has a bond angle of approximately 117°.

1

BF₃: B has 3 valence electrons, each F shares 1. Central B has 3 bonding pairs and 0 lone pairs. All repulsions are BP–BP (weakest type), so electron pairs are equally spaced at exactly 120°. All electron pairs are bonding = shape is trigonal planar, bond angle = 120°.

2

SO₂: S has 6 valence electrons. Forming 2 double bonds uses 4 electrons, leaving 2 as a lone pair. Central S has 2 bonding pairs and 1 lone pair.

3

In SO₂, the LP–BP repulsions are greater than BP–BP. The lone pair pushes the two bonding pairs closer together, compressing the O–S–O bond angle below 120° to approximately 117°.

4

The shape of SO₂ is described as bent (V-shaped) because we name shapes based on atom positions only, ignoring lone pairs.

BF₃: trigonal planar, 120° | SO₂: bent, ~117°

4 Electron Pairs

Four electron pairs arrange tetrahedrally. Lone pairs progressively compress bond angles:

Tetrahedral

109.5°

4 BP, 0 LP | CH₄, NH₄⁺, CCl₄

Trigonal Pyramidal

107°

3 BP, 1 LP | NH₃, PCl₃

Bent (V-shaped)

104.5°

2 BP, 2 LP | H₂O, H₂S

Effect of Lone Pairs

Each lone pair added reduces the bond angle by approximately 2–2.5°

CH₄: 4BP, 0LP → 109.5° NH₃: 3BP, 1LP → 107.0° H₂O: 2BP, 2LP → 104.5°

LP–BP repulsion > BP–BP repulsion

Worked Example03

Explain why the bond angles decrease in the order: CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°). Use VSEPR theory in your answer.

1

All three molecules have 4 electron pairs around the central atom. The base geometry (electron pair geometry) is tetrahedral for all three.

2

CH₄: 4 bonding pairs, 0 lone pairs. Only BP–BP repulsions (weakest type). All four bond pairs repel equally, giving bond angles of exactly 109.5°.

3

NH₃: 3 bonding pairs, 1 lone pair. There is now 1 LP–BP repulsion, which is stronger than BP–BP. The lone pair pushes the three bonding pairs closer together, reducing the H–N–H angle to 107°.

4

H₂O: 2 bonding pairs, 2 lone pairs. There are now 2 LP–BP repulsions and 1 LP–LP repulsion (strongest of all). The two lone pairs exert even greater repulsion on the bonding pairs, compressing the H–O–H angle further to 104.5°.

Summary: As the number of lone pairs increases, repulsion on bonding pairs increases, progressively compressing bond angles: 109.5° → 107° → 104.5°.

5 & 6 Electron Pairs

Trigonal Bipyramidal

90° and 120°

Equatorial bonds: 120° | Axial bonds: 90°

Octahedral

90°

6 BP, 0 LP | SF₆, PF₆⁻

All bond angles 90°

Square Planar — 6 EP, 2 LP

XeF₄ has 6 electron pairs (4 BP + 2 LP). The 2 lone pairs go to opposite positions (top and bottom) to maximise separation. Result: 4 X atoms in a square plane, all F–Xe–F angles = 90°. Shape is called square planar.

Polyatomic Ions

Apply VSEPR to ions exactly as for molecules. Adjust the total electron count for the charge first.

Method for Ions

Positive ion (+n charge): Remove n electrons from total count.

Negative ion (−n charge): Add n electrons to total count.

IonTotal EP around central atomBPLPShapeBond Angle

NH₄⁺4 (N has 5 e⁻, loses 1 for + charge; 4 bonds)40Tetrahedral109.5°

NO₃⁻3 (count around N with extra e⁻ from − charge)30Trigonal planar120°

CO₃²⁻3 (double bond counts as 1 EP)30Trigonal planar120°

SO₄²⁻4 (S forms 4 bonds)40Tetrahedral109.5°

Exam Tip

For AQA, focus on NH₄⁺, NO₃⁻, CO₃²⁻, and SO₄²⁻. Note that NO₃⁻ and CO₃²⁻ are both trigonal planar with 120° bond angles — the negative charge is delocalised across the O atoms.

04Worked Example

Determine the shape and bond angles of (a) NH₄⁺ and (b) NO₃⁻. Explain your reasoning in each case.

(a) NH₄⁺:

4 bonding pairs, 0 lone pairs