(a) Develop a bonding scheme for a ring made up of six \(\mathrm{CH}\) groups, \((\mathrm{CH})_{6}\). Each carbon is hybridized \(s p^{2}\). (b) This part is harder. The molecule you built for part (a) exists, but all the carbon-carbon bond lengths are equal. Is this consistent with the structure you wrote for part (a), which almost certainly requires two different carbon-carbon bond lengths? How would you resolve the problem?

In cyclic structures like \((CH)_6\), delocalized π bonds play a crucial role. Each carbon in this ring has an \(sp^2\) hybridization, forming three sigma bonds: one with a hydrogen atom and two with neighboring carbons. This leaves each carbon with an unhybridized p orbital that is perpendicular to the plane. These unhybridized p orbitals overlap side-by-side to create a network of delocalized π bonds. Delocalization means that the π electrons do not stay between any specific pair of carbon atoms but are spread out over the entire ring. This continuous overlap of p orbitals results in a stable electron cloud that circulates above and below the plane of the ring.
Such delocalization is significant because it increases the stability of the molecule. Instead of having distinct double and single bonds, the electrons are shared equally, leading to uniform bond lengths. This property is akin to what we see in benzene, even though in our case the structure is (CH)_6 instead of the classic C_6H_6.

Resonance structures are a way to represent delocalized electrons within molecules that can’t be adequately described by a single Lewis structure. In the case of (CH)_6, one might initially expect alternating double and single bonds if we only consider sp2 hybridization. However, the true picture is more complex due to resonance.

In (CH)_6, multiple resonance structures can represent the possible locations of the π electrons. No single resonance structure accurately depicts the molecule’s electron distribution. Instead, these structures are hybrids, meaning the real structure is an average of all possible resonance forms. This hybrid reflects a more stable configuration where the π electrons are evenly distributed across the ring. Hence, each C-C bond has the same strength and length, showing none of the characteristics of a localised double bond. Resonance is a powerful concept because it helps explain the equal bond lengths observed in cyclic structures like benzene and (CH)_6, providing a deeper insight into molecular stability and bonding.

(CH)_6 exhibits equal carbon-carbon bond lengths due to the effects of delocalization and resonance. Normally, one might expect different bond lengths if alternating single and double bonds were considered. However, due to electron delocalization, a uniform π electron cloud forms over the ring structure.

This electron distribution results in identical C-C bonds throughout the molecule. In benzene, all C-C bond lengths are approximately 1.39 Å, a value intermediate between typical single and double bond lengths. Similarly, in (CH)_6, the C-C bond lengths will all be equal and reflect the averaged value contributed by delocalized electrons. This uniformity explains why we do not see varied bond lengths typically expected in a structure with discrete single and double bonds. The concept of equal carbon-carbon bond lengths due to delocalization and resonance underscores the principles of stability and uniformity brought about by these molecular phenomena.