Q.Benzene, $\text{C}_6\text{H}_6$, is represented by two Kekulé resonance structures. Explain what a 'resonance hybrid' means here and why benzene's actual carbon–carbon bond length ($139\ \text{pm}$) lies between that of a pure single bond ($154\ \text{pm}$) and a pure double bond ($134\ \text{pm}$).

[!TLDR] Benzene's two Kekulé structures are equally valid resonance contributors, differing only in which three bonds are drawn as double bonds. [!ANSWER] The real molecule is neither Kekulé structure but their resonance hybrid, with the six pi electrons delocalised evenly around the ring, giving one unifo …

Benzene's ring can be drawn with alternating single and double C–C bonds in two distinct ways (the two Kekulé structures), which differ from each other only in which three of the six bonds are drawn double and which three are drawn single — the ring skeleton and all atom positions are identical in both. Because these two structures are equally valid and equally low in energy, the real molecule is best described as their resonance hybrid: a single, real structure in which the six pi electrons are delocalised evenly over all six carbons, rather than being confined to three fixed double bonds. This delocalisation is why benzene's actual carbon–carbon bond length, measured experimentally, is exactly the same all the way around the ring — every bond is $139\ \text{pm}$ — rather than alternating between short (double-bond-like) and long (single-bond-like) values as either individual Kekulé structure alone would suggest. The measured $139\ \t …

Draw both Kekulé structures, explain they are equally valid resonance contributors describing one real delocalised hybrid, and connect …

Treating benzene as literally alternating between the two Kekulé structures rather than being one …