The nature of the bonding in a benzene ring, limited to planar structure…

The idea

Benzene, C₆H₆, is a flat regular hexagon with all bond angles 120°. Each carbon uses three of its four outer electrons in σ bonds, two to neighbouring carbons and one to a hydrogen. The fourth electron stays in a p orbital that points perpendicular to the plane of the ring. Those six p orbitals overlap sideways all the way round, so the electrons are no longer tied to any one pair of atoms but are delocalised in a ring of electron density above and below the plane. Because every carbon-carbon bond has an identical share of that density, all six are the same length, 0.139 nm, between a single bond at 0.154 nm and a double bond at 0.134 nm, which is exactly what X-ray studies measure.

What earns the marks

• The planar regular hexagon with bond angles of 120°.
• Each carbon contributing one electron from a p orbital perpendicular to the ring.
• Sideways overlap of those orbitals giving delocalised electrons above and below the plane.
• All six carbon-carbon bonds being identical and intermediate in length between a single and a double bond.

Common mistakes and how to put them right

• Drawing alternating single and double bonds and then claiming all the bonds are the same length.

• Saying the delocalised electrons lie in the plane of the ring rather than above and below it.

• Stating that the delocalised electrons come from the C-H bonds.

• Calling benzene an alkene because it is shown with double bonds in the Kekulé structure.

Exam tips

• Use the circle-in-a-hexagon symbol whenever you are asked about bonding; save Kekulé structures for mechanisms, where you need a specific C-H bond.
• Quote the bond length evidence if the question says 'explain, using evidence'; a number scores where a description may not.