You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
Predicting the three-dimensional shapes of molecules is a fundamental skill in A-Level Chemistry. The theory that allows us to do this is called VSEPR — Valence Shell Electron Pair Repulsion theory. It is based on a single, elegant principle: electron pairs around a central atom arrange themselves as far apart as possible to minimise repulsion.
All electron pairs in the valence shell of the central atom — whether they are bonding pairs (shared with another atom) or lone pairs (not shared) — repel each other because they are all regions of negative charge. The shape adopted by the molecule is the one that places these electron pairs as far apart as possible.
However, there is a crucial hierarchy of repulsion strengths:
Lone pair–lone pair repulsion > Lone pair–bonding pair repulsion > Bonding pair–bonding pair repulsion
Lone pairs occupy more space than bonding pairs because they are held closer to the central atom (they are not being "pulled" towards another nucleus). This means lone pairs squash the bonding pairs closer together, reducing bond angles.
Use this decision tree to predict molecular shape systematically:
flowchart TD
A["Draw Lewis structure"] --> B["Count electron domains around central atom"]
B --> C{"How many total domains?"}
C -->|"2"| D["Linear base geometry (180°)"]
C -->|"3"| E["Trigonal planar base (120°)"]
C -->|"4"| F["Tetrahedral base (109.5°)"]
C -->|"5"| G["Trigonal bipyramidal base"]
C -->|"6"| H["Octahedral base (90°)"]
D --> D1{"Lone pairs?"}
D1 -->|"0"| D2["LINEAR (180°)"]
E --> E1{"Lone pairs?"}
E1 -->|"0"| E2["TRIGONAL PLANAR (120°)"]
E1 -->|"1"| E3["BENT (~117°)"]
F --> F1{"Lone pairs?"}
F1 -->|"0"| F2["TETRAHEDRAL (109.5°)"]
F1 -->|"1"| F3["PYRAMIDAL (~107°)"]
F1 -->|"2"| F4["BENT (~104.5°)"]
G --> G1{"Lone pairs?"}
G1 -->|"0"| G2["TRIGONAL BIPYRAMIDAL (90°, 120°)"]
G1 -->|"1"| G3["SEESAW (~117°, ~90°)"]
G1 -->|"2"| G4["T-SHAPED (~87.5°)"]
G1 -->|"3"| G5["LINEAR (180°)"]
H --> H1{"Lone pairs?"}
H1 -->|"0"| H2["OCTAHEDRAL (90°)"]
H1 -->|"1"| H3["SQUARE PYRAMIDAL (~85°)"]
H1 -->|"2"| H4["SQUARE PLANAR (90°)"]
When there are no lone pairs on the central atom, the shape depends purely on the number of bonding pairs (or more precisely, the number of electron domains — regions of electron density):
2 electron domains → Linear (180°) Example: BeCl₂, CO₂. The two bonding regions are on opposite sides of the central atom.
3 electron domains → Trigonal planar (120°) Example: BF₃, AlCl₃. Three bonding regions arranged at 120° in a flat plane.
4 electron domains → Tetrahedral (109.5°) Example: CH₄, CCl₄. Four bonding regions pointing to the corners of a tetrahedron.
5 electron domains → Trigonal bipyramidal (90° and 120°) Example: PCl₅. Five bonding regions — three in an equatorial plane at 120° to each other, and two in axial positions at 90° to the equatorial plane.
6 electron domains → Octahedral (90°) Example: SF₆. Six bonding regions pointing to the corners of an octahedron.
Lone pairs occupy electron domains but are not visible as bonds. The molecular shape is described by the positions of the atoms only, not the electron pairs. This means the name of the shape may be different from what the total electron domain count suggests.
3 bonding pairs + 1 lone pair → Pyramidal (approximately 107°) Example: NH₃. The four electron domains adopt a roughly tetrahedral arrangement, but one position is occupied by a lone pair. The visible shape (based on atom positions) is a triangular pyramid. The lone pair pushes the bonding pairs closer together, reducing the bond angle from 109.5° to about 107°.
2 bonding pairs + 2 lone pairs → Bent / V-shaped (approximately 104.5°) Example: H₂O. Two of the four electron domains are lone pairs. The visible shape is bent. The two lone pairs compress the bond angle further, from 109.5° to about 104.5°.
4 bonding pairs + 1 lone pair → Seesaw (approximately 117° and 90°) Example: SF₄. The lone pair occupies an equatorial position (where it has more space), giving a seesaw shape.
3 bonding pairs + 2 lone pairs → T-shaped (approximately 87.5°) Example: ClF₃. Two lone pairs occupy equatorial positions, leaving bonds in a T-shape.
2 bonding pairs + 3 lone pairs → Linear (180°) Example: XeF₂. Three lone pairs in equatorial positions, two bonds in axial positions.
5 bonding pairs + 1 lone pair → Square pyramidal (approximately 85°) Example: BrF₅. One lone pair occupies one position of the octahedron.
4 bonding pairs + 2 lone pairs → Square planar (90°) Example: XeF₄. Two lone pairs are on opposite sides of the central atom (to minimise lone pair–lone pair repulsion), giving a flat square arrangement of the four bonds.
The key pattern to remember:
This progressive reduction occurs because each additional lone pair exerts more repulsion on the bonding pairs, squashing them closer together.
In VSEPR theory, a double bond or triple bond counts as a single electron domain. This is because all the electron density in a multiple bond is concentrated in the same region of space.
For example, in CO₂:
In SO₂:
| Bonding pairs | Lone pairs | Total domains | Shape | Bond angle |
|---|---|---|---|---|
| 2 | 0 | 2 | Linear | 180° |
| 3 | 0 | 3 | Trigonal planar | 120° |
| 2 | 1 | 3 | Bent | ~117° |
| 4 | 0 | 4 | Tetrahedral | 109.5° |
| 3 | 1 | 4 | Pyramidal | ~107° |
| 2 | 2 | 4 | Bent | ~104.5° |
| 5 | 0 | 5 | Trigonal bipyramidal | 90°, 120° |
| 4 | 1 | 5 | Seesaw | ~117°, ~90° |
| 3 | 2 | 5 | T-shaped | ~87.5° |
| 6 | 0 | 6 | Octahedral | 90° |
| 5 | 1 | 6 | Square pyramidal | ~85° |
| 4 | 2 | 6 | Square planar | 90° |
Common exam mistake: Students often forget that a double bond counts as a single electron domain. In CO₂, there are only 2 electron domains (not 4). In SO₂, there are 3 domains (not 5). Always count regions of electron density, not individual bonds or electrons.
Edexcel 9CH0 specification Topic 2 — Bonding and Structure, sub-topic 2.4 addresses predicting the shapes of simple molecules and ions using Valence Shell Electron Pair Repulsion (VSEPR) theory, including the relative repulsion of bonding pairs and lone pairs and the consequent effect on bond angles. Shape questions surface across all three papers. Paper 1 Topic 4 (Inorganic chemistry) demands the prediction of period 3 chloride shapes — bent SCl₂, pyramidal PCl₃, tetrahedral SiCl₄. Paper 2 (Organic chemistry) relies on shape implicitly: sp³ tetrahedral carbons in alkanes, sp² trigonal planar around the C=C of alkenes (preventing free rotation and underpinning E/Z isomerism), and sp linear carbons in alkynes are all VSEPR predictions in disguise. Paper 3 (synoptic) routinely asks for the shape of an unfamiliar species — xenon halides, interhalogens, oxoanions. Candidates must predict shapes for species containing up to six electron pairs on the central atom (refer to the official Pearson Edexcel specification document for exact wording).
Question (8 marks): Sulfur, chlorine and xenon all form fluorides in which the central atom carries one or more lone pairs in addition to bonding pairs.
(a) Predict the shape and bond angle of SF₄, justifying your reasoning. (3)
(b) Predict the shape and bond angle of ClF₃, justifying your reasoning. (3)
(c) Predict the shape and bond angle of XeF₂, justifying your reasoning. (2)
Solution with mark scheme:
(a) Step 1 — count electron pairs around sulfur. Sulfur has 6 valence electrons; each F supplies one electron to a shared pair. Total around S = 6 + 4 = 10 electrons = 5 pairs. Of these, 4 are bonding pairs (one per S–F bond) and 1 is a lone pair.
M1 — correct count of 5 electron pairs (4 bonding + 1 lone). Common slip: students forget to halve the shared electrons or miscount sulfur's group.
Step 2 — choose the parent geometry and place the lone pair. Five electron pairs adopt a trigonal bipyramidal arrangement. The lone pair is placed in an equatorial position because equatorial sites suffer only two 90° interactions with axial pairs (versus three 90° interactions for an axial site), minimising lone-pair repulsion.
A1 — identifies trigonal bipyramidal parent geometry and places the lone pair in the equatorial position with justification.
Step 3 — name the shape and quote bond angles. With the lone pair equatorial, the four bonding pairs form a see-saw (or "disphenoidal") shape. Axial F–S–F bond angle is reduced from 180° to ~173°; equatorial F–S–F angle is reduced from 120° to ~102°.
B1 — names "see-saw" and quotes both compressed angles (or quotes "less than 90°" and "less than 120°" with explicit reference to lone-pair compression).
(b) Step 1 — count electron pairs around chlorine. Cl has 7 valence electrons; each F contributes 1. Total = 7 + 3 = 10 electrons = 5 pairs. Bonding pairs = 3, lone pairs = 2.
M1 — correct count of 5 electron pairs (3 bonding + 2 lone).
Step 2 — place both lone pairs in equatorial positions. Both lone pairs occupy equatorial sites of the trigonal bipyramidal arrangement, again because equatorial placement minimises 90° lone-pair–bond-pair repulsions. The three F atoms sit at the two axial sites and one equatorial site.
A1 — identifies trigonal bipyramidal parent and equatorial placement of both lone pairs.
Step 3 — name the shape and quote bond angles. With three F atoms in this arrangement, the molecule is T-shaped. The F–Cl–F bond angle between an axial and the equatorial F is compressed from 90° to approximately 87.5° because of the two adjacent lone pairs.
B1 — names "T-shaped" and quotes ~87.5° (or "slightly less than 90°" with reference to lone-pair compression).
(c) Step 1 — count. Xe has 8 valence electrons; each F contributes 1. Total = 8 + 2 = 10 electrons = 5 pairs. Bonding pairs = 2, lone pairs = 3.
M1 — correct count of 5 electron pairs (2 bonding + 3 lone).
Step 2 — place all three lone pairs equatorially. The three lone pairs occupy the three equatorial positions; the two F atoms sit at the axial positions. Result: linear, with F–Xe–F bond angle of exactly 180° (the symmetric placement of lone pairs cancels their angular distortion of the F–Xe–F axis).
A1 — names "linear" and gives 180°.
Total: 8 marks (M3, A3, B2 distributed as shown).
Highlighted slip: the most frequent error across all three parts is placing lone pairs in axial positions of the trigonal bipyramidal arrangement. An axial lone pair has three 90° interactions with the equatorial pairs; an equatorial lone pair has only two 90° interactions (plus less-repulsive 120° interactions with the other equatorial pairs). Equatorial siting is always energetically preferred.
Question (6 marks): Methane (CH₄), ammonia (NH₃) and water (H₂O) all have four electron pairs around the central atom but exhibit different bond angles: 109.5°, 107° and 104.5° respectively.
(a) Identify the shape of each molecule. (3)
(b) Explain, in terms of electron pair repulsion, why the H–X–H bond angle decreases in the order CH₄ > NH₃ > H₂O. (3)
Mark scheme decomposition by AO:
(a)
(b)
Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. Bond-angle questions of this style consistently weight AO1 most heavily — recall of the parent shape and the repulsion ordering — but reserve the final mark for candidates who can quantify or justify the magnitude of the compression rather than simply naming "more lone pairs → smaller angle".
VSEPR-derived shape underpins much of the rest of the 9CH0 specification. Key forward connections:
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.