Question

Bond order of Benzene is $1.5$. Explain how.

Answer 1

Firstly draw the molecular orbitals. Then add electrons and count the number of bonding and anti-bonding electrons.

The bond order of a bond is half the difference between the number of bonding and antibonding electrons.

BO =$½(B–A)$

The $C-C$ sigma bonds

Each $C-C\sigma$ bond is a localized bond. It has 2 bonding electrons and 0 nonbonding electrons.

σ BO = $½(B–A)=½(2–0)=1$

The $C-C$ pi bonds

Benzene has 6 molecular $\pi$ orbitals.

Of these, three are bonding and three are anti-bonding. The six $\pi$ electrons go into the three bonding orbitals.

π BO = $½(B–A)=½(6–0)=3$

This is the $\pi$ bond order for 6 $C-C$ bonds.

For one $C-C\pi$ bond, BO = $3/6=\mathrm{0.5.}$

For a single $C-C$ bond in benzene, the total BO = $\sigma +\pi =1+0.5=\mathrm{1.5.}$