Shapes of Molecules and Ions

Molecular shape is the bridge between Lewis structures and almost every physical and chemical property a chemist cares about: polarity, intermolecular forces, dipole moments, reactivity, optical activity, and the geometry of transition-metal complexes. The framework you use to predict shape at A-Level is VSEPR theory — Valence Shell Electron Pair Repulsion theory — developed by Gillespie and Nyholm in 1957 and refined since. The central idea is mechanically simple: electron pairs in the valence shell of a central atom repel each other and arrange themselves in three-dimensional space to be as far apart as possible. From this single premise the shapes of molecules with two through six electron pairs follow, the bond-angle deviations caused by lone pairs become predictable, and the geometries of polyatomic ions slot neatly into the same pattern. This lesson develops the framework, drills the prediction algorithm, and exposes the subtleties — multiple-bond counting, AB₄E₂ ambiguity, and the empirical limits of VSEPR.

Spec mapping (AQA 7405): This lesson maps to §3.1.3 (bonding — shapes of simple molecules and ions). The repulsion-driven framework is extended in §3.1.3 (electronegativity and polarity — lesson 3 of this course, where shape determines whether bond dipoles cancel), §3.2.5 (transition-metal complex geometries — lesson 5 of the inorganic course, where octahedral, tetrahedral and square-planar arrangements arise from d-orbital splitting), and §3.3.7 (optical isomerism — depends on a tetrahedral asymmetric carbon, see lesson 9 of organic foundations). Refer to the official AQA specification document for the exact wording of each section.

Assessment objectives: AO1 — recall of shape names and the ideal bond angles for linear, bent, trigonal planar, trigonal pyramidal, tetrahedral, trigonal bipyramidal, octahedral, square pyramidal and square planar geometries. AO2 — predicting shape and bond angle from the number of bonded and lone pairs around a central atom, applied to molecules and to polyatomic ions where the charge alters the electron count. AO3 — rationalising deviations from ideal bond angles (NH₃ 107°, H₂O 104.5°, SF₄ see-saw, NF₃ ~102°) using the LP–LP > LP–BP > BP–BP repulsion hierarchy, and selecting between two possible AB₄E₂ geometries.

The VSEPR Premise

Key Definition: VSEPR theory states that electron pairs around a central atom adopt the spatial arrangement that minimises mutual repulsion, taking account of the fact that lone-pair electron clouds repel more strongly than bond-pair clouds.

Three operational rules follow:

Count regions of electron density on the central atom. A "region" is either a bonded pair (single, double, or triple bond) or a non-bonding lone pair. Multiple bonds count as a single region for the purposes of shape prediction, because the shared electrons occupy roughly the same spatial direction.
Place the regions at the maximum mutual separation: 2 → linear, 3 → trigonal planar, 4 → tetrahedral, 5 → trigonal bipyramidal, 6 → octahedral.
Adjust for lone pairs. Lone pairs are held by only one nucleus and so spread out further from the central atom than bonded pairs (which are attracted by two nuclei). The order of repulsion strength is:

Lone pair–lone pair  >  Lone pair–bond pair  >  Bond pair–bond pair

This hierarchy is the source of every bond-angle deviation you will be asked to explain at A-Level. Each lone pair compresses the bond angles around it by 2–3° relative to the ideal value for that arrangement.

Exam Tip: When asked to explain a bond angle (as opposed to merely stating one), always quote the repulsion order LP–LP > LP–BP > BP–BP and identify how many lone pairs are present. Marks are awarded for the reasoning, not just the numerical answer.

Shapes by Electron-Pair Count

Two Regions → Linear (180°)

Two regions of electron density on a central atom adopt a linear arrangement at 180°. Examples: BeCl₂ (Cl—Be—Cl), CO₂ (O═C═O — note that the two C═O double bonds count as two regions, not four), HCN (H—C≡N).

    Cl — Be — Cl
         180°

Three Regions → Trigonal Planar (120°) or Bent

With three regions and no lone pairs, the geometry is trigonal planar with 120° bond angles: BF₃, AlCl₃ (monomer), SO₃, NO₃⁻, CO₃²⁻.

With three regions of which one is a lone pair, the geometry is bent (sometimes called V-shaped) with a bond angle slightly less than 120°: SO₂ has S with two bonded oxygens and one lone pair, giving a bond angle of approximately 119°.

Four Regions → Tetrahedral, Pyramidal or Bent

Four electron-density regions tetrahedrally arranged give a base bond angle of 109.5°.

0 lone pairs → Tetrahedral, 109.5°. CH₄, CCl₄, SiH₄, NH₄⁺ (cation), BF₄⁻ (anion), SO₄²⁻, PO₄³⁻.
1 lone pair → Trigonal pyramidal, ~107°. NH₃ (107°), H₃O⁺ (~107°), PCl₃, NF₃ (~102°, smaller — see lesson 3 on electronegativity).
2 lone pairs → Bent, ~104.5°. H₂O (104.5°), H₂S (92.1°, drops further because S is larger and bond pairs are further from the nucleus).

   CH₄ (109.5°)  →  NH₃ (107°)  →  H₂O (104.5°)
   0 LP             1 LP            2 LP

Each additional lone pair compresses the H—X—H angle by ~2–2.5°. This pattern is one of the most heavily examined facts in §3.1.3 and you must be fluent with it.

Five Regions → Trigonal Bipyramidal Family

Five regions give the trigonal bipyramidal parent shape: three equatorial positions at 120° to each other in a plane, plus two axial positions at 90° above and below the plane.

0 lone pairs → Trigonal bipyramidal. PCl₅, PF₅. Equatorial bond angles 120°; axial–equatorial 90°.
1 lone pair → See-saw (AB₄E). SF₄. The lone pair occupies an equatorial position (where it has only two 90° neighbours instead of three), giving bond angles of approximately 89° and 117°.
2 lone pairs → T-shaped (AB₃E₂). ClF₃. Both lone pairs equatorial; the three F atoms form a distorted T with ~87.5° angles.
3 lone pairs → Linear (AB₂E₃). XeF₂. Three equatorial lone pairs leave two axial F atoms at 180°.

Lone pairs always occupy the equatorial sites in the trigonal bipyramid because each equatorial site has only two 90° neighbours, whereas axial sites have three 90° neighbours — placing the bulky lone pair equatorially minimises 90° repulsions.

Six Regions → Octahedral Family

Six regions give an octahedral parent shape with all bond angles 90°.

0 lone pairs → Octahedral, 90°. SF₆, [Cu(H₂O)₆]²⁺, [Fe(CN)₆]³⁻.
1 lone pair → Square pyramidal (AB₅E). BrF₅, IF₅. Bond angles approximately 84.8°.
2 lone pairs → Square planar (AB₄E₂). XeF₄, [Ni(CN)₄]²⁻ (a d⁸ complex; geometry here arises from crystal-field stabilisation rather than VSEPR alone). The two lone pairs occupy opposite (trans) positions to minimise LP–LP repulsion, leaving four F atoms in a square at 90°.

Master Reference Table

Total regions	Bond pairs	Lone pairs	Parent geometry	Actual shape	Ideal bond angle	Example
2	2	0	Linear	Linear	180°	CO₂, BeCl₂
3	3	0	Trigonal planar	Trigonal planar	120°	BF₃, CO₃²⁻
3	2	1	Trigonal planar	Bent	~119°	SO₂
4	4	0	Tetrahedral	Tetrahedral	109.5°	CH₄, NH₄⁺, SO₄²⁻
4	3	1	Tetrahedral	Trigonal pyramidal	~107°	NH₃, H₃O⁺
4	2	2	Tetrahedral	Bent	~104.5°	H₂O
5	5	0	Trigonal bipyramidal	Trigonal bipyramidal	120°, 90°	PCl₅
5	4	1	Trigonal bipyramidal	See-saw	~117°, ~89°	SF₄
5	3	2	Trigonal bipyramidal	T-shaped	~87.5°	ClF₃
5	2	3	Trigonal bipyramidal	Linear	180°	XeF₂
6	6	0	Octahedral	Octahedral	90°	SF₆, [Cu(H₂O)₆]²⁺
6	5	1	Octahedral	Square pyramidal	~84.8°	BrF₅
6	4	2	Octahedral	Square planar	90°	XeF₄, [Ni(CN)₄]²⁻

The Prediction Algorithm

Use these four steps for every shape question, regardless of whether the species is a neutral molecule or a polyatomic ion.

Count valence electrons on the central atom (group number, with the post-2011 IUPAC numbering: group 1 = 1 valence electron, group 14 = 4, group 17 = 7, etc.).
Adjust for charge: add one electron for each negative charge, subtract one for each positive charge. (Add 2 for SO₄²⁻; subtract 1 for NH₄⁺.)
Count regions of electron density: each single, double or triple bond is one region; each remaining non-bonding pair is one region. Total = bonded regions + lone pairs.
Look up the geometry in the master table and quote the bond angle.

Worked Example 1 — Phosphate ion, PO₄³⁻

P is in group 15 → 5 valence electrons. Add 3 for the 3− charge → 8 electrons. P forms 4 bonds to O. That accounts for 8 electrons (one from P + one from O per bond, with the formal charges on O), so all 8 electrons of P are involved in bonding and there are no lone pairs on P. 4 bonded regions, 0 lone pairs → tetrahedral, 109.5°.

Worked Example 2 — Sulfate ion, SO₄²⁻

S has 6 valence electrons; add 2 for the 2− charge → 8. S forms 4 bonds to O, so 4 bonded regions, 0 lone pairs → tetrahedral, 109.5°. (The Lewis structure with two S═O and two S—O⁻ is a common representation; for VSEPR the formal-charge distribution does not matter, only the number of bonded regions.)

Worked Example 3 — Ammonium cation, NH₄⁺

N has 5 valence electrons; subtract 1 for the + charge → 4. N forms 4 bonds to H. 4 bonded regions, 0 lone pairs → tetrahedral, 109.5°. Notice that NH₃ (pyramidal, 107°) becomes tetrahedral (109.5°) when protonated to NH₄⁺ — the lone pair on N has become a fourth bonding pair, removing the LP–BP repulsion.

Worked Example 4 — Hydronium cation, H₃O⁺

O has 6 valence electrons; subtract 1 for the + charge → 5. O forms 3 bonds to H, leaving 2 non-bonding electrons (1 lone pair). 3 bonded regions + 1 lone pair = 4 regions → trigonal pyramidal, ~107°, isoelectronic with NH₃.

Worked Example 5 — Bond angle comparison, NF₃ vs NH₃

Both have 3 bonded pairs + 1 lone pair around N → trigonal pyramidal. But NH₃ has a bond angle of 107° while NF₃ has ~102°. F is more electronegative than H, so it pulls bonding electrons further away from N. The N—F bond pairs are therefore further from the central atom and occupy less angular space around it. The lone pair on N is comparatively more dominant and can compress the F—N—F angle further. (This synoptic argument links VSEPR to electronegativity — see lesson 3.)

Polyatomic Ions: Shape Catalogue

Ion	Central atom	Bond pairs	Lone pairs	Shape	Bond angle
NH₄⁺	N	4	0	Tetrahedral	109.5°
H₃O⁺	O	3	1	Trigonal pyramidal	~107°
NO₃⁻	N	3	0	Trigonal planar	120°
CO₃²⁻	C	3	0	Trigonal planar	120°
SO₃²⁻	S	3	1	Trigonal pyramidal	~107°
SO₄²⁻	S	4	0	Tetrahedral	109.5°
PO₄³⁻	P	4	0	Tetrahedral	109.5°
ClO₄⁻	Cl	4	0	Tetrahedral	109.5°
ClO₃⁻	Cl	3	1	Trigonal pyramidal	~107°

Note that some species with the same VSEPR formula but different central atoms have slightly different actual angles because of differences in the size and electronegativity of the central atom (compare H₃O⁺ ~107° with NH₃ 107° — coincidentally similar).

Transition-Metal Complexes — Forward Signpost

The same shape vocabulary appears in inorganic chemistry. The d-block course (lesson 5, §3.2.5) introduces:

Octahedral complexes such as [Cu(H₂O)₆]²⁺ (the familiar pale-blue copper(II) aqua complex) and [Fe(H₂O)₆]³⁺ — 6 ligands, 90° bond angles.
Tetrahedral complexes such as [CuCl₄]²⁻ (yellow-green) — 4 large ligands, 109.5° bond angles, geometry favoured for bulky donors.
Square planar complexes such as [Ni(CN)₄]²⁻ and the platinum-based anticancer drug cisplatin, cis-[PtCl₂(NH₃)₂] — characteristic of d⁸ metal centres where crystal-field stabilisation favours square planar over tetrahedral.

For the d-block, VSEPR is a useful first pass but the full story requires crystal-field theory: the splitting of d-orbitals in an octahedral or tetrahedral ligand field provides extra stabilisation that VSEPR alone does not capture.

Synoptic Links

Lesson 0 (ionic bonding): VSEPR does not apply to ionic lattices. There is no discrete "shape" of NaCl in the solid state — only a repeating 6:6 coordination geometry of the lattice. Shape is a property of discrete covalent molecules and complex ions.
Lesson 1 (covalent bonding): VSEPR builds on the shared-pair model. Counting bond pairs and lone pairs depends on a correct Lewis structure first; if your dot-and-cross diagram is wrong, your VSEPR prediction will be wrong.
Lesson 3 (electronegativity and polarity, next lesson): Whether a molecule has a net dipole moment depends on the vector sum of bond dipoles, which is determined by molecular shape. CO₂ (linear) has zero net dipole despite polar C═O bonds; H₂O (bent) has a substantial dipole. Without VSEPR, polarity cannot be predicted.
§3.2.5 (transition-metal complexes, lesson 5 inorganic): Octahedral, tetrahedral and square-planar geometries reappear; crystal-field theory supplements VSEPR for the d-block.
§3.3.7 (optical isomerism, organic): Chirality requires a tetrahedral carbon bearing four different substituents. VSEPR justifies why sp³ carbon is tetrahedral, which is the prerequisite for chirality.

Specimen Exam Question — modelled on the AQA A-Level Chemistry paper format

Question 1. [12 marks total]

(a) State the name of the shape of SF₆ and give the bond angle. [2 marks]

(b) Predict and explain the shape and bond angle of NF₃. [3 marks]

(c) Explain why the bond angle in H₂O (104.5°) is smaller than the bond angle in CH₄ (109.5°), referring to electron-pair repulsion. [4 marks]

Shapes of Molecules and VSEPR Theory