AQA A-Level Chemistry 7405 · 3.1.3 Bonding

Part 3: Shapes of molecules & ions

All four parts available · diagram placeholders included. Reviewed 1 October 2026.

1. Electron pair repulsion

The shape of a molecule or ion depends on the electron pairs around the central atom. These pairs repel each other and spread out as far apart as possible to minimise repulsion. It is the electron pairs that repel, not the atoms.

A lone pair occupies more space around the central atom than a bonding pair, so its repulsion is stronger:

lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair

This squeezes bonding pairs closer together, so bond angles become smaller than the ideal angle.

2. Counting regions

Count the regions of electron density around the central atom: each single, double or triple bond counts as one region, and each lone pair counts as one.

Then separate two ideas:

  • The electron arrangement uses all regions (bonds and lone pairs).
  • The molecular shape describes only where the atoms are.

For example, CO2 has two C=O double bonds, so two regions around carbon, giving a linear shape. Each double bond is two shared pairs, so CO2 has four shared pairs in total, but only two regions.

3. The core shapes

Core shapes and bond angles
ExampleBonding regionsLone pairsShapeBond angle
CO220Linear180°
BF330Trigonal planar120°
CH4, NH4+40Tetrahedral109.5°
NH331Trigonal pyramidalabout 107°
H2O22Bent (V-shaped)about 104.5°
SO22 (each S=O is one region)1Bent (from trigonal planar)just under 120°
PCl550Trigonal bipyramidal120° (equatorial), 90° (axial–equatorial); axial–axial 180°
SF660Octahedral90° (opposite bonds 180°)

4. A counting shortcut (and its limits)

When the central atom is joined only by single bonds to monovalent atoms (H, F, Cl, Br, I), you can count electron pairs like this:

pairs = (outer electrons of central atom + number of bonded atoms + negative charge − positive charge) ÷ 2

Bonding pairs = number of bonded atoms; lone pairs = total pairs − bonding pairs.

Using the shortcut
SpeciesCalculationPairsBonding / loneShape
NH4+(5 + 4 − 1) ÷ 244 / 0Tetrahedral, 109.5°
XeF4(8 + 4) ÷ 264 / 2Square planar, 90°

This shortcut does not work for molecules with double bonds (such as CO2) or with oxygen bonded to the central atom (such as SO42−). For those, draw a dot-and-cross or Lewis structure and count regions directly.

5. Five and six regions with lone pairs

You may meet unfamiliar molecules. Work them out from the number of bonding pairs and lone pairs. The names are useful but you can always describe the shape in words.

Shapes with five or six regions and lone pairs
ExampleBonding pairsLone pairsShapeNotes
SF441Seesaw (distorted tetrahedral)Lone pair sits equatorial; angles reduced below 90° and 120°
ClF332T-shapedBoth lone pairs equatorial; adjacent F–Cl–F angles slightly below 90°, the two axial F near 180°
KrF2, I3−23LinearThree equatorial lone pairs; F–Kr–F 180°
XeF442Square planarLone pairs opposite each other; 90°
BrF551Square pyramidalAdjacent F–Br–F angles slightly below 90°; the apical F roughly opposite the lone pair

With five regions, lone pairs go in the equatorial positions, where they have more room. With six regions and two lone pairs, the lone pairs go opposite each other to keep as far apart as possible.

Diagram placeholder

3D shape chart: 2 to 6 regions, with lone-pair variants

Labels to include:

  • Linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral
  • Pyramidal and bent derived from tetrahedral
  • Seesaw, T-shaped, linear derived from trigonal bipyramidal (equatorial lone pairs)
  • Square pyramidal, square planar derived from octahedral (opposite lone pairs)
  • Bond angles on each

6. Worked predictions

XeF4

Xenon has 8 outer electrons. Four are used in bonds with four F atoms; the remaining four form two lone pairs. Total: 6 regions (4 bonding, 2 lone). The lone pairs sit opposite each other, so the four F atoms lie in a square around Xe: square planar, 90°.

KrF2

Krypton has 8 outer electrons. Two are used in bonds with F; six form three lone pairs. Total: 5 regions (2 bonding, 3 lone). The three lone pairs take the equatorial positions, leaving the F atoms at the two axial positions: linear, 180°. Show the three lone pairs on Kr in your diagram.

IF4+

Iodine has 7 outer electrons; the + charge removes one, leaving 6. Four are used in bonds; two form one lone pair. Total: 5 regions (4 bonding, 1 lone): seesaw, like SF4.

7. Drawing shapes

  • Solid lines: bonds in the plane of the page.
  • Wedges: bonds coming towards you. Dashed wedges: bonds going away.
  • Draw lone pairs on the central atom (two dots or a lobe). Leaving them out of a shape drawing is a common error.
  • For an ion, put the structure in square brackets with the charge outside.
  • Label the bond angle.

Diagram placeholder

Lone-pair examples in 3D: NH₃, H₂O, KrF₂, XeF₄

Labels to include:

  • Wedges and dashed wedges
  • Lone pairs drawn on the central atom
  • Bond angles labelled
  • Equatorial lone pairs in KrF₂; opposite lone pairs in XeF₄

Every shape can be described fully in words: number of bonding pairs, number of lone pairs, shape name and angle.

8. Quick checks

Finesse practice: indicative answers to check your reasoning, not official mark allocations.

Q1. State the shape and bond angle of NH₃, and explain the angle.Show answer

3 bonding pairs and 1 lone pair: trigonal pyramidal, about 107°. Lone pair–bonding pair repulsion is stronger than bonding pair–bonding pair repulsion, so the N–H bonds are pushed closer than 109.5°.

Q2. Why is the bond angle in H₂O smaller than in CH₄?Show answer

Both have four electron pairs around the central atom, but water has two lone pairs. Lone pairs repel more strongly than bonding pairs, so the bonding pairs are pushed closer together.

Q3. Predict the shape of ClF₃.Show answer

Cl has 7 outer electrons: 3 in bonds, 4 as two lone pairs. 5 regions, both lone pairs equatorial: T-shaped.

Q4. How many electron regions are around carbon in CO₂, and why?Show answer

Two. Each C=O double bond counts as one region, so CO2 is linear, 180°.

Q5. Use the shortcut to find the shape of BrF₄⁻.Show answer

(7 + 4 + 1) ÷ 2 = 6 pairs; 4 bonding, 2 lone. Lone pairs opposite: square planar, 90°.

9. Sources

Sources and examiner guidance (reviewed 1 October 2026)

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