How do I figure out the hybridisation of the carbon with the negative charge?

One logic is that it is an $\mathrm{sp^3}$ carbon because there are 3 sigma bonds and 1 lone pair around it.

Another reasoning, though, is that it is $\mathrm{sp^2}$; we don't count the lone pair, because it is in conjugation with the double bonds.

What is the correct explanation?

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  • 3
    $\begingroup$ There is no carbon with negative charge, nor is there a lone pair. All carbons are in fact identical; the negative charge is distributed over the whole ring. So your second option is closer to the reality. $\endgroup$ – Ivan Neretin Jan 12 '16 at 16:40
  • $\begingroup$ Related answer: chemistry.stackexchange.com/a/10936/4325 $\endgroup$ – jerepierre Jan 12 '16 at 16:54

It entirely depends on what that electron pair is doing.

If the lone pair is hanging out by itself in its own orbital, then the carbon is probably sp3 hybridized, as greater hybridization is typically more stable. This is what you'll see on an alcohol or an amine lone pair. Those lone pairs aren't really interacting with anything else in the molecule, hence the "standard" logic of the rule you invoke.

However, in certain compounds you're able to stabilize the system further by delocalizing the electron pair. In this sort of case, the sp3 hybridized lone pair orbital is not able to participate in the delocalization, and thus will be higher in energy than an electron pair that is in the remaining p orbital of an sp2 hybridized carbon. This is the sort of thing you see with amide groups, where the "lone pair" on the nitrogen is sp2-like, which allows the electron pair to delocalize into the carbonyl, allowing for resonance stabilization.

Note that delocalization is not always going to happen. You can potentially draw up examples where delocalization would decrease the stability of the compound. For example, 1,4-Dihydropyrazines would be anti-aromatic if both lone pairs from each nitrogen were to participate in the ring system. This is highly disfavorable, and the molecule contorts itself to avoid it, putting the nitrogens into sp3 (pyramidal) hybridization, except where it needs to be planar to conjugate with an exocyclic double bond.

In this particular case - the one of the cyclopentadienyl anion - you would be able to form an aromatic system if the lone pair were to delocalize into the ring - much like pyrrole does. The lowered energy obtained from this aromatic delocalization more than makes up for the small amount of energy gained by going from sp2 to sp3, so the carbon should be sp2 hybridized.

The important thing here is not simply to invoke rote-memorized rules to determine hybridization states, but think about what the electrons are doing and what the comparative stability of the various options are. The hybridization will change to be the most stable (lowest energy) state. -- To be perfectly honest, neither sp2 nor sp3 is the "true" answer. The actual answer to these sorts of questions lies more toward Molecular Orbital Theory, where the molecule is taken as a whole. Per-atom hybridization is a convenient shorthand, but still one which ultimately rests on the basis of the true molecular stability.


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