Final answer:
Benzene has two common resonance structures, representing electron delocalization across the hexagonal ring. While it's conceivable to draw a third resonance structure, such a depiction would not accurately represent the equivalence of all six C-C bonds observed experimentally. Benzene's actual structure is a hybrid of resonance structures, displaying delocalized π electrons rather than fixed double bonds.
Step-by-step explanation:
You're correct that the molecule benzene is typically represented with two resonance structures. These structures show alternating single and double bonds around the hexagonal ring.
However, if you were to draw a third resonance structure with fixed double bonds, you would end up with a structure that is not representative of the actual electron delocalization in benzene.
All six carbon-carbon bonds in benzene are equivalent in length and strength, which is only possible due to the delocalized nature of the π electrons.
In benzene, each carbon atom is sp² hybridized, with three sigma bonds (one to hydrogen and two to adjacent carbon atoms) and one unhybridized p orbital containing an electron, which overlaps with the p orbitals of adjacent atoms.
This overlapping forms a delocalized π system above and below the plane of the carbon atoms. The true electronic structure of benzene is often represented by a hexagon with a circle inside, indicating the delocalized nature of the π electrons.
Chemists use the concept of resonance to suggest that the actual structure of a molecule like benzene is a hybrid of all valid resonance structures.
In the case of benzene, since all the C-C bonds are experimentally found to be of the same length and energy, this indicates that the electronic structure of benzene is an average of the two classical resonance structures, not the result of individual localized double bonds.