A-Level Chemistry May/June 2024 Q6(c): Describe the shape of delocalised benzene. Include the geometry of each carbon, the C-C…

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  Benzene is a planar, hexagonal molecule. The geometry around each of the six carbon atoms is trigonal planar, and the C-C-H bond angle is 120°.

  Each carbon atom forms a sigma (σ) bond to two other carbon atoms and to one hydrogen atom. The remaining p-orbitals on each carbon atom overlap sideways, above and below the plane of the ring, to form a delocalised pi (π) system. Therefore, the bonding between adjacent carbon atoms consists of both σ and delocalised π bonds, while the bonding between carbon and hydrogen atoms consists of σ bonds only.

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  How the marks are awarded

  • M1 — Correctly stating that the C-C-H bond angle is 120° and that the geometry around each carbon is trigonal planar, resulting in a planar hexagonal shape for the molecule.
  • M2 — Correctly identifying the types of bonds: sigma (σ) and delocalised pi (π) bonds between the carbon atoms, and only sigma (σ) bonds between the carbon and hydrogen atoms.

  Common mistakes

  • Stating the bond angle is 109.5°, which is characteristic of tetrahedral (sp³) carbon atoms, not the trigonal planar (sp²) carbons in benzene.
  • Describing the Kekulé structure with alternating single and double C-C bonds, instead of the delocalised pi system requested.
  • Forgetting to mention the sigma (σ) bonds that form the framework of the molecule, and only describing the pi (π) system.
  • Incorrectly stating that there are pi (π) bonds between carbon and hydrogen atoms.

  Examiner tip: For questions on molecular shape, always connect the type of hybridisation of the central atom(s) to the resulting geometry and bond angles for a complete answer.

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