# Hybridisation of carbon atom in cyclopentadienyl anion

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?

• 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. – Ivan Neretin Jan 12 '16 at 16:40
• Related answer: chemistry.stackexchange.com/a/10936/4325 – jerepierre Jan 12 '16 at 16:54