Count electron regions, minimise their repulsion and then name the positions of the atoms. Apply the same method to unfamiliar ions and organic centres.
Electron regions determine the arrangement; atoms determine the shape name
Electron pairs around a central atom repel and arrange themselves to minimise repulsion. First draw a valid outer-electron structure, then count bonding regions and lone pairs at the central atom. A double or triple bond is one directional region for shape, although it contains more than one electron pair and can repel more strongly than a single bond.
Lone pairs are held closer to only one nucleus and occupy more space around the central atom than bonding pairs. Their repulsions are generally greater: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. They can compress angles between bonds. The shape name describes atom positions and does not include lone pairs as extra atoms.
For many single-bonded species with terminal monovalent atoms, a quick central count is [central valence electrons + number of attached monovalent atoms − positive charge + magnitude of negative charge] ÷ 2. This is a useful checking method, not a substitute for the full electron structure when multiple bonds are present.
Two, three and four electron regions
Worked comparison: CH₄ has four bonding regions and no lone pair, so equal repulsions give tetrahedral 109.5° angles. NH₃ has one lone pair that repels the three bonding pairs more strongly, compressing the H–N–H angle to about 107°. H₂O has two lone pairs and a 104.5° H–O–H angle.
These numerical angles are characteristic of those named molecules. Do not automatically subtract 2.5° for every lone pair in every compound: atom identity, bond type and other effects also affect real angles. BeCl₂ here is the isolated gaseous molecule; some solid substances have different extended structures.
| Regions / lone pairs | Molecular shape | Angles | Examples and explanation |
|---|---|---|---|
| 2 / 0 | Linear | 180° | BeCl₂(g), CO₂; two regions farthest apart |
| 3 / 0 | Trigonal planar | 120° | BCl₃; three equal regions in one plane |
| 3 / 1 | Bent | Less than 120° in the simple model | A two-bond, one-lone-pair analogue |
| 4 / 0 | Tetrahedral | 109.5° | CH₄, NH₄⁺ |
| 4 / 1 | Trigonal pyramidal | About 107° in NH₃ | Three bonding pairs and one lone pair |
| 4 / 2 | Bent | 104.5° in H₂O | Two bonding pairs and two lone pairs |
Five and six regions, including lone-pair variants
In a trigonal bipyramid, an equatorial site has two close 90° interactions, while an axial site has three. A lone pair therefore usually occupies an equatorial position to reduce the number of strong close repulsions. In an octahedral arrangement, two lone pairs occupy opposite positions, producing square-planar atom positions.
The last two rows are model-transfer arrangements, not a requirement to memorise stable example compounds. For an unfamiliar question, draw the supplied or deduced electron-region arrangement and remove lone-pair positions from the molecular shape. Do not invent an exact measured bond angle where the simple model only predicts a distortion.
| Regions; bonds; lone pairs | Shape | Ideal or indicative angles | Example / placement |
|---|---|---|---|
| 5; 5; 0 | Trigonal bipyramidal | 90°, 120°, 180° | PCl₅(g): three equatorial, two axial |
| 5; 4; 1 | Seesaw | Distorted from 90° and 120° | SF₄: lone pair in an equatorial site |
| 5; 3; 2 | T-shaped | Near 90° and 180° | ClF₃: both lone pairs equatorial |
| 5; 2; 3 | Linear | 180° | I₃⁻: all three lone pairs equatorial |
| 6; 6; 0 | Octahedral | 90° and 180° | SF₆(g) |
| 6; 5; 1 | Square pyramidal | Near 90° and 180° | BrF₅ |
| 6; 4; 2 | Square planar | 90° and 180° | XeF₄: lone pairs opposite one another |
| 6; 3; 3 | T-shaped in the ideal octahedral model | Near 90° and 180° | Use a supplied electron-region arrangement |
| 6; 2; 4 | Linear in the ideal octahedral model | 180° | Two opposite bonding positions |
Worked deductions for ions and multiple bonds
For NH₄⁺: [5 + 4 − 1] ÷ 2 = 4 electron pairs, all bonding. It is tetrahedral, 109.5°. For I₃⁻: [7 + 2 + 1] ÷ 2 = 5 pairs at the central iodine; two are bonding and three are lone pairs. The lone pairs occupy equatorial positions, leaving the two bonds axial and the ion linear, 180°.
For CO₂, do not treat the four shared electron pairs as four separate directions. The two C=O double bonds form two regions, producing a linear 180° molecule. Around each carbon in ethene there are three regions, so the arrangement is trigonal planar, approximately 120°. Around each carbon in ethyne there are two, so it is linear, 180°.
For a shape explanation, state the count, explain how repulsion is minimised, identify any extra lone-pair repulsion and give the resulting shape/angle. Naming the shape without the electron-pair reasoning answers ‘state’ but not a full ‘explain’ request.
Show three dimensions honestly
Use ordinary lines for bonds in the paper’s plane, a solid wedge for a bond towards the viewer and a dashed wedge for a bond behind. Place lone-pair symbols around the central atom and label the relevant bond angle. A tetrahedron is not a flat cross; an octahedron has four bonds in one plane and two perpendicular to it.
Diagram placeholder
Three-dimensional tetrahedral, trigonal-bipyramidal and octahedral sketches
Labels to include:
- Central atom in each structure
- Tetrahedral: two planar bonds, one solid wedge, one dashed wedge; 109.5°
- Trigonal bipyramid: three equatorial bonds at 120°, two axial bonds at 90° to that plane
- Octahedron: four square-planar bonds plus two opposite axial bonds; 90°/180°
- Lone pairs shown when deriving NH₃, H₂O, SF₄, ClF₃ and XeF₄
This placeholder identifies the remaining drawing work. The tables and worked deductions specify all counts, orientations and angles needed to construct each sketch.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.
Q1. Explain why NH₃ and NH₄⁺ have different shapes.Show answer
NH₃ has three bonding pairs and one lone pair: trigonal pyramidal, about 107°. NH₄⁺ has four bonding pairs and no lone pair: tetrahedral, 109.5°. The ammonia lone pair compresses the bond angles; the positive charge must be included in the electron count.
Q2. Predict the shape of PF₆⁻ using an electron count.Show answer
[5 + 6 + 1] ÷ 2 = 6 pairs around P, all bonding. Equal repulsions give an octahedral shape with 90° between adjacent bonds and 180° between opposite bonds.
Q3. Why is CO₂ linear although it contains four shared electron pairs?Show answer
Each double bond occupies one direction, so there are two bonding regions around carbon and no lone pair. They repel to 180°. Four pairs do not mean four separate regions here.
Q4. Deduce the shape of BrF₅ and distinguish it from SF₆.Show answer
BrF₅ has [7 + 5] ÷ 2 = 6 pairs: five bonding and one lone pair. The electron-region arrangement is octahedral, but atom positions are square pyramidal. SF₆ has six bonding pairs and no lone pair, so its molecular shape is octahedral.
Q5. Where should the lone pair be placed in SF₄, and why should its exact angles not simply be labelled 90° and 120°?Show answer
SF₄ has five regions and takes a seesaw shape with the lone pair equatorial, minimising close 90° repulsions. The lone pair repels bonding regions more strongly and distorts the ideal angles. The simple model predicts distortion, not exact experimental angles.
Sources
Sources and examiner guidance (reviewed 9 October 2026)
- Pearson Edexcel 9CH0 specification, Issue 3 — Topic 2, printed pp. 9–11, checked against 8CH0 printed pp. 7–9. Reviewed 9 October 2026.
- Chemrevise — Edexcel Bonding — All 14 pages reviewed as a secondary coverage check. Explanations, worked applications and exercises are original.
- Pearson 9CH0/01 June 2023 mark scheme — Q3(c)(ii), PDF p. 11: four bonding regions and two lone pairs give square-planar shape. The broader method here is original Finesse teaching.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.