Why are the C-C bonds in benzene closer in length to double bonds than single bonds?

Question

The $\ce{C-C}$ bonds in benzene are $\mathrm{140~pm}$ long which puts them in between single bonds and double bonds (given at $\mathrm{147-154~pm}$ and $\ce{134-135~pm}$ respectively, sourced from here and here) as expected. However, since the two resonance structures for benzene contribute equally, leading to six identical bonds, one might expect that the bond lengths would be exactly intermediate between a single and a double bond, but in fact the bond lengths seem to be closer to a double bond than a single bond. Why is this the case? My teacher asked me this a while ago when I was preparing for university interviews and I have never thought of a satisfactory answer.

Answer 1

one might expect that the bond lengths would be exactly intermediate between a single and a double bond

They actually are. The relevant reference structure, that can provide us lengths of single and double bond for $sp^2$ carbons with not conjugation involved is cyclooctatetraene. The article on the X-ray structure of cycloocta-1,3,5,7-tetrene gives lengths 1.47 and 1.33 angstrom, with their average (1.40) being the exactly value of C-C bond in benzene you provided.

Answer 2

Why should they be bang in the middle? Or to better phrase the question: What reasoning are you applying to assume an averaged bond length?

Bonds don’t work as we laymen like to write them, with either a single line or a double line; and a double line being equivalent to two single lines. Rather, bonds — and most importantly, their lengths — are the result of equilibrated forces: The repulsion of two atoms pushing them apart and the bond-forming energy drawing them together.

Most notably, a double bond is not shorter than a single bond because more or less or whatever energy is involved, but because the molecular orbitals are best (i.e. lowest energy) at that particular length.

When comparing the aromatic benzene bonds to single or double bonds, we have to notice:

• benzene’s carbons are sp² hybridised while single bonds’ carbons are sp³
• there is a significant $\pi$-bonding across the benzene ring still happening, even though that amounts to only $0.5$ additional bond order.

Therefore, benzene’s bonds are much closer to double bonds in nature by the orbitals forming them. It is only reasonable that the resulting bond length is closer to that of double bonds than that of single bonds.

Remember finally, that ‘bond order’ is only a concept counting bonding and anti-bonding electrons. So even formal single bonds could be much shorter than expected (even though I do not have an example on me at the moment).

Answer 3

Relevant to this discussion is the prediction coming from the Hückel model.

As many readers know, for a single homoatomic cycle the Hückel model resolves into the Frost circle, which has a radius equivalent to two isolated pi bonds. Putting in the orbital positions for a six-membered ring and filling the lower half, modeling benzene, we find that the total pi- bonding energy on this model is equivalent to

$2+2×(2\cos60°)=4$

isolated pi bonds, not three. The delocalization of the optimal number of conjugated pi electrons around the ring has effectively created an extra bond beyond those formally drawn into the structure.

This effect not only accounts for the stability of the benzene ring (and aromatic rings in general), it also means that in effect there is more than half a pi bond to go with the sigma bond at each carbon-carbon linkage. So the observed properties of the linkages will be closer to double bonds than to single bonds.