In the benzene molecule, carbon atoms form a ring with alternating single and double bonds in between each of them. Each carbon atom forms one σ bond with one carbon atom and one σ and one π bond with another. The π bonding electrons are delocalized over the entire molecule. Thus, benzene is a resonance hybrid of two equivalent structures, and the single and double bonds oscillate from one position to the other. Required:
Calculate the bond order for one carbon-carbon bond in the benzene molecule, taking σ and π bonding into consideration.

1.5

Explanation:

From;

BO = ½(B – A)

BO= Bond order

B= Number of bonding electrons

A = Number of antibonding electrons

For the sigma bonds, there are 2 bonding electrons and 0 nonbonding electrons.

σ BO = ½(B – A) = ½(2 – 0) = 1

π bond order for 6 C-C bonds;

π BO = ½(B – A) = ½(6 – 0) = 3

For each  π bond, BO = 3/6 = 0.5.

Hence, for a single C-C bond in benzene, the total Bond order = σ + π = 1 + 0.5 = 1.5.

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