Concept: Resonance and Bond Order

Benzene ($C_6{H}_6$) is a classic example of a molecule exhibiting resonance. Its structure cannot be accurately represented by a single Lewis structure but rather by a resonance hybrid of two Kekulé structures. In these structures, the carbon-carbon bonds are depicted as alternating single and double bonds.

Why (C) is correct:

Due to resonance, the electrons in benzene are delocalized over all six carbon atoms. This delocalization means that none of the carbon-carbon bonds are purely single or purely double. Instead, all carbon-carbon bonds are identical in length and strength, and their character is intermediate between a single bond and a double bond. The bond order for each carbon-carbon bond in benzene is calculated as the average of the bond orders from the contributing resonance structures. In one Kekulé structure, there are three C=C double bonds (bond order 2) and three C-C single bonds (bond order 1). Averaging these over the six bonds gives $(3\times 2+3\times 1)/6=9/6=1.5$. Therefore, the bond order is between 1 and 2.

Option Analysis:

• A) one: This would imply all bonds are single bonds, which is incorrect as benzene has delocalized pi electrons.
• B) two: This would imply all bonds are double bonds, which is incorrect as benzene is not cyclohexatriene.
• D) one and two alternately: This describes the individual Kekulé structures, but not the actual resonance hybrid, where all C-C bonds are equivalent.

Correct Answer: (C)