Topic 3 of 12
Molecular shapes
Describe the local geometry around a carbon before the whole molecule.
A-Level 8873, revised syllabus (2026-2027)
Read shape from sigma and pi bonding
Describe the local geometry around a carbon before the whole molecule.
| Molecule | Carbon bonding and local shape | Angles and freedom of movement |
|---|---|---|
| Ethane, CH3CH3 | Each carbon forms four sigma bonds and has a tetrahedral arrangement. The C-C bond is one sigma bond from head-on overlap. | About 109.5° around each carbon. Rotation about C-C is possible without breaking the sigma bond; the whole molecule is not planar. |
| Ethene, CH2=CH2 | Each carbon has three regions of sigma bonding in a trigonal planar arrangement. Parallel p orbitals overlap sideways above and below the plane, forming one pi bond. | About 120° around each carbon. The molecule is planar; rotation about C=C is restricted. |
| Benzene, C6H6 | Each carbon has three sigma bonds in a planar hexagonal framework. The six parallel p orbitals overlap around the ring, giving delocalised pi electron density above and below it. | About 120° around every carbon. All six C-C bonds are equivalent; benzene is not a ring of alternating independent single and double bonds. |
Benzene: a planar ring with delocalised pi bonding
A hexagon with an internal circle is the preferred representation of benzene. Each corner represents a carbon attached to one hydrogen. The circle represents delocalised pi bonding, not an extra atom or separate small ring.
A multiple bond counts as one direction of electron density when predicting local shape. Ethene has a double bond but only three directions around each carbon, so it is trigonal planar. No hybridisation labels are needed here: use the number and arrangement of sigma bonds and explain where sideways pi overlap is possible.
Worked example
Apply the pattern to an unfamiliar chain
Predict the approximate angles around each carbon in propene, CH2=CHCH3.
- The two double-bonded carbons each have three sigma-bond directions. Their local arrangements are trigonal planar with angles near 120°.
- The CH3 carbon has four sigma-bond directions, so it is tetrahedral with angles near 109.5°.
- The C=C consists of one sigma and one pi bond; the adjacent C-C single bond is a sigma bond. Do not assign 120° to every carbon just because the molecule contains one double bond.
The alkene region is locally planar; the methyl carbon is tetrahedral.