The question asks for the total bond order between adjacent carbon atoms in a benzene molecule.

Benzene Structure

Benzene ($\text{C}_6\text{H}_6$) is a cyclic organic molecule with a ring of six carbon atoms. Each carbon atom is bonded to one hydrogen atom and two adjacent carbon atoms. The structure of benzene is unique due to the presence of alternating single and double bonds, which are not fixed in position but are delocalized over the entire ring.

Resonance in Benzene

Benzene exhibits resonance. This means its actual structure is a hybrid of two equally contributing structures, often called Kekulé structures. In one Kekulé structure, there are three single bonds and three double bonds arranged alternately around the ring. In the other Kekulé structure, the positions of the single and double bonds are reversed.

Due to resonance, the pi electrons are delocalized over all six carbon atoms in the ring. This delocalization results in all the carbon-carbon bonds in the benzene ring being equivalent, with bond lengths intermediate between a typical single bond ($\text{C-C}$) and a typical double bond ($\text{C=C}$).

Calculating Bond Order

The bond order between two specific atoms is calculated as the total number of bonds (including both sigma and pi bonds) shared between those atoms divided by the number of resonance structures contributing to that bond.

Let's consider the bond order between any two adjacent carbon atoms in the benzene ring:

• In one Kekulé structure, this bond is a single bond (bond order = 1).
• In the other Kekulé structure, this bond is a double bond (bond order = 2).

There are two equally contributing resonance structures.

The average bond order between adjacent carbon atoms is:

\text{Bond Order} = \frac{\text{Sum of bond orders in all resonance structures}}{\text{Number of resonance structures}}

\text{Bond Order} = \frac{1 (\text{from single bond}) + 2 (\text{from double bond})}{2 (\text{number of resonance structures})}

\text{Bond Order} = \frac{3}{2} = 1.5

Alternatively, we can think about the total number of bonds shared between the 6 pairs of adjacent carbon atoms. There are 6 sigma bonds forming the hexagonal ring and 3 pi bonds delocalized over the ring. These 3 pi bonds are shared among the 6 C-C links.

Each C-C link consists of one sigma bond (bond order 1) plus a share of the pi bonds.

\text{Share of pi bond per link} = \frac{\text{Total pi bonds}}{\text{Number of C-C links}} = \frac{3}{6} = 0.5

\text{Total bond order per C-C link} = \text{Sigma bond order} + \text{Share of pi bond order}

\text{Total bond order per C-C link} = 1 + 0.5 = 1.5

Both methods show that the total bond order between adjacent carbon atoms in benzene is 1.5.