Shapes & bond angles
Shapes & bond angles
VSEPR: count charge clouds, then minimise repulsion
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The shape comes from electron regions repelling one another around the central atom. They arrange as far apart as possible. Count bonding regions and lone pairs before naming the shape.
- Find the central atom's outer-shell electron number: for example, 5 for N or 6 for O.
- For a suitable species with single bonds to H or halogens, add one electron for each attached atom.
- Add electrons for a negative charge and subtract them for a positive charge.
- Divide by two to get the total pairs. Subtract the bonding pairs to find the lone pairs.
- Choose the parent electron arrangement. Then name the molecular shape using the positions of the atoms.
The shortcut has limits. With multiple bonds or a supplied electron diagram, count regions directly: a single, double or triple bond is one bonding region. Lone pairs count as regions too.
- VSEPR
- Valence Shell Electron Pair Repulsion: outer-shell electron pairs repel and arrange as far apart as possible.
- Charge cloud
- One region of electron density: each bond region and each lone pair counts as one.
Ammonia versus ammonium
For NH₃ and NH₄⁺, calculate the total number of electron pairs around nitrogen, state how many are bonding pairs and lone pairs, and predict the shape.
Worked answer
- NH3: (5 + 3)/2 = 4 pairs; 3 bonding and 1 lone.
- NH4+: (5 + 4 − 1)/2 = 4 pairs; all 4 bonding.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
- Explain why electron pairs around a central atom adopt particular positions in space. [2]
- Place bond pair–bond pair, lone pair–bond pair and lone pair–lone pair repulsions in decreasing strength. [2]
- Explain why a lone pair repels more strongly than a bonding pair. [2]
Total: 6 marks
Show answer & marking points
- Charge clouds repel [1] and arrange as far apart as possible to minimise repulsion [1].
- Lone pair–lone pair > lone pair–bond pair [1] > bond pair–bond pair [1].
- A lone pair is attracted to only one nucleus [1], so its electron density occupies more space near the central atom and repels more strongly [1].
Two and three charge clouds
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With two electron regions and no lone pairs, the greatest separation is opposite sides of the central atom. The shape is linear, 180°.
With three electron regions and no lone pairs, the regions spread into one plane. The shape is trigonal planar, 120°.
Count regions, not individual shared pairs. Each double bond in CO₂ counts as one region, so CO₂ has two regions around carbon and is linear.
| Species | Bonding regions | Lone pairs | Shape | Angle |
|---|---|---|---|---|
| CO2 | 2 | 0 | Linear | 180° |
| BeCl2 | 2 | 0 | Linear | 180° |
| BF3 | 3 | 0 | Trigonal planar | 120° |
| SO3 | 3 | 0 | Trigonal planar | 120° |
Sulfur trioxide
Deduce SO3’s shape from three bonding regions and no lone pairs.
State the bond angle and explain the arrangement of the bonding regions.
Worked answer
The three charge clouds repel equally into one plane.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
- Deduce CS2’s shape and bond angle around carbon. [2]
- Deduce BCl3’s shape and angle around boron. [2]
Total: 4 marks
Show answer & marking points
- Carbon forms two C=S double bonds. Each double bond is one bonding region; carbon has no lone pairs. The two regions repel as far apart as possible, giving a linear shape [1] and 180° bond angle [1].
- Three bonding regions and no lone pairs: trigonal planar [1], 120° [1].
Four bonding regions: tetrahedral shapes
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Four bonding regions spread out in three dimensions, giving a tetrahedral shape with angles of 109.5°. Four flat bonds at right angles would put the regions too close together.
CH₄, SiCl₄, NH₄⁺ and BF₄⁻ all have four bonding regions and no lone pairs around the central atom.
- An ordinary line shows a bond in the plane of the paper.
- A solid wedge shows a bond coming towards you.
- A hashed wedge shows a bond going away from you.
- For an ion, put brackets and the overall charge around the whole species.
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Use solid and hashed wedges where needed to show three-dimensional arrangement.
- Draw CH4. [2]
- Draw NH3, including its lone pair. [3]
- Draw H2O, including both lone pairs, and label the H–O–H angle. [3]
Total: 8 marks
Show answer & marking points
- Four bonds in a tetrahedral arrangement [1], using valid wedge-and-dash representation [1].
- Three bonds in a trigonal-pyramidal arrangement [1], one N lone pair [1], valid 3D representation [1].
- Bent arrangement [1], two O lone pairs [1], 104.5° [1].
Lone pairs and compressed bond angles
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A lone pair takes up more space near the central atom than a bonding pair. It is attracted to one nucleus rather than shared between two nuclei. It repels nearby bonding regions more strongly and pushes them closer together.
CH₄, NH₃ and H₂O each have four electron regions around the central atom. Replacing bonds with lone pairs changes the positions of the atoms and therefore the molecular shape.
- CH₄: four bonds, no lone pairs; tetrahedral, 109.5°.
- NH₃: three bonds, one lone pair; pyramidal, about 107°.
- H₂O: two bonds, two lone pairs; bent/non-linear, about 104.5°.
Subtracting about 2.5° for each lone pair is a memory aid for this familiar comparison. It is not a universal formula for every molecule or parent arrangement.
| Species | Bonding regions | Lone pairs | Shape | Angle |
|---|---|---|---|---|
| SO2 | 2 | 1 | Bent | Less than 120° |
| NH3 | 3 | 1 | Trigonal pyramidal | 107° |
| H2O | 2 | 2 | Bent | 104.5° |
| H3O+ | 3 | 1 | Trigonal pyramidal | Use supplied data; do not automatically assume 107° |
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
CH4, NH3 and H2O each have four electron pairs around the central atom.
- State the shape and bond angle of each molecule. [6]
- Explain the angle decrease from CH4 to NH3 to H2O. [2]
Total: 8 marks
Show answer & marking points
- CH4: tetrahedral, 109.5° [2]; NH3: trigonal pyramidal, 107° [2]; H2O: bent, 104.5° [2].
- Lone pair–bond pair repulsion is stronger than bond pair–bond pair repulsion [1]. Water has two lone pairs, compared with ammonia’s one, compressing the bonding pairs further [1].
Five and six charge clouds
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Five regions arrange in a trigonal bipyramid. Three equatorial positions lie in one plane; two axial positions point above and below it. Six regions arrange octahedrally.
Lone pairs still occupy positions in those arrangements, even though they are left out when naming the molecular shape. That gives the seesaw, T-shaped, linear, square-pyramidal and square-planar shapes in the table.
With five regions, lone pairs take equatorial positions to reduce the number of 90° repulsions. With six regions and two lone pairs, as in XeF₄, the lone pairs sit opposite one another.
Use the table's angles and any data supplied. Do not apply the methane/ammonia/water subtraction shortcut to these shapes.
| Total regions | Bonding + lone | Example | Molecular shape | Angles |
|---|---|---|---|---|
| 5 | 5 + 0 | PCl5 | Trigonal bipyramidal | 90°, 120°, 180° |
| 5 | 4 + 1 | SF4 | Seesaw | Distorted below ideal 90°, 120°, 180° |
| 5 | 3 + 2 | ClF3 | T-shaped | Approximately 90°, 180°, slightly compressed |
| 5 | 2 + 3 | XeF2 | Linear | 180° |
| 6 | 6 + 0 | SF6 | Octahedral | 90°, 180° |
| 6 | 5 + 1 | BrF5 | Square pyramidal | Approximately 90°, 180° |
| 6 | 4 + 2 | XeF4 | Square planar | 90°, 180° |
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Complete all seven rows with the molecular shape and characteristic angle(s). Treat each multiple bond as one bonding region. [14]
| Species | Bonding regions | Lone pairs | Shape | Angle(s) |
|---|---|---|---|---|
| CO2 | 2 | 0 | … | … |
| BF3 | 3 | 0 | … | … |
| CH4 | 4 | 0 | … | … |
| NH3 | 3 | 1 | … | … |
| H2O | 2 | 2 | … | … |
| PCl5 | 5 | 0 | … | … |
| SF6 | 6 | 0 | … | … |
Total: 14 marks
Show answer & marking points
| Species | Bonding regions | Lone pairs | Shape | Angle(s) |
|---|---|---|---|---|
| CO2 | 2 | 0 | Linear | 180° |
| BF3 | 3 | 0 | Trigonal planar | 120° |
| CH4 | 4 | 0 | Tetrahedral | 109.5° |
| NH3 | 3 | 1 | Trigonal pyramidal | 107° |
| H2O | 2 | 2 | Bent / non-linear | 104.5° |
| PCl5 | 5 | 0 | Trigonal bipyramidal | 90°, 120° (also 180° between axial bonds) |
| SF6 | 6 | 0 | Octahedral | 90° (also 180° between opposite bonds) |
One mark per correct shape and one per correct angle entry: 14 marks.
In AXmEn, A is the central atom, X a bonded atom and E a lone pair on A.
| Species | AXₘEₙ |
|---|---|
| A SF4 | AX4E |
| B ClF3 | AX3E2 |
| C XeF2 | AX2E3 |
| D BrF5 | AX5E |
- Determine the molecular shape of A–D. [4]
- Explain why the lone pairs in A–C occupy equatorial positions. [2]
- Give characteristic angles, using “less than” where lone-pair repulsion compresses ideal angles. [4]
- Predict whether C and D have permanent dipoles and explain. [3]
Total: 13 marks
Show answer & marking points
- A seesaw; B T-shaped; C linear; D square pyramidal. One mark each.
- Equatorial positions [1] have fewer 90° interactions and minimise strong lone-pair repulsion [1].
- SF4: <90°, <120° and <180° [1]; ClF3: <90° and <180° [1]; XeF2: 180° [1]; BrF5: approximately 90° and 180°, with distortions from lone-pair repulsion [1].
- XeF2 is linear [1], so its equal opposite dipoles cancel [1]. BrF5 is asymmetric square pyramidal, so its dipoles do not cancel and it is polar [1].
Deducing unfamiliar molecular shapes
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An unfamiliar formula does not need a new method. Work out its electron regions, place any lone pairs, then look at where the atoms sit.
- Count the bonding regions and lone pairs around the central atom.
- Choose the parent electron arrangement from the total number of regions.
- Place lone pairs where repulsion is minimised.
- Ignore the lone-pair positions when naming the molecular shape, but keep their effect on the arrangement.
For XeF₄, six regions include four bonds and two opposite lone pairs. The atoms left in the middle plane form a square, so the shape is square planar.
| Species | Bonding regions | Lone pairs | Reasoning | Shape |
|---|---|---|---|---|
| NO2+ | 2 | 0 | Opposite regions | Linear |
| SO2 | 2 | 1 | Three-region parent; omit lone-pair position | Bent |
| XeF4 | 4 | 2 | Six-region parent; opposite lone pairs | Square planar |
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Use the supplied counts; each multiple bond counts as one bonding region.
| Species | Bonding regions | Lone pairs |
|---|---|---|
| NH4+ | 4 | 0 |
| SO2 | 2 | 1 |
| I3- | 2 | 3 |
| XeF4 | 4 | 2 |
- Determine each molecular shape. [4]
- State one characteristic angle for each. [4]
- Draw XeF4, showing the two Xe lone pairs relative to the square plane. [2]
Total: 10 marks
Show answer & marking points
- NH4+: tetrahedral; SO2: bent; I3-: linear; XeF4: square planar. One mark each.
- 109.5°; slightly less than 120°; 180°; 90°, respectively. One mark each.
- Four F atoms in a square plane [1]; opposite lone pairs perpendicular to it [1].
U is ClF3 (AX3E2, 44 electrons per molecule). V is XeF4 (AX4E2, 90 electrons per molecule). V has the higher melting point. A is the central atom, X a bonded atom and E a lone pair.
- Determine U’s shape and the positions of its two lone pairs. [2]
- Determine V’s shape and the relative positions of its lone pairs. [2]
- Predict each molecule’s polarity and explain. [2]
- Suggest why V can have the higher melting point despite being non-polar. [4]
- Draw both molecules with all central lone pairs and appropriate 3D conventions. [3]
Total: 13 marks
Show answer & marking points
- U is T-shaped [1], with two equatorial lone pairs in a trigonal-bipyramidal arrangement [1].
- V is square planar [1], with two opposite lone pairs perpendicular to the plane [1].
- U is polar: its T-shaped dipoles do not cancel [1]. V is non-polar: its square-planar dipoles cancel [1].
- XeF4 has more electrons [1], a more polarisable cloud [1] and stronger van der Waals forces [1], so more energy is needed [1]. This explains why a non-polar substance can melt higher; crystal packing can also affect melting points.
- ClF3: AX3E2 with two equatorial lone pairs [1]. XeF4: AX4E2 with opposite lone pairs [1]. Correct 3D conventions and atom labels [1].