Basically, I'm confused on when lone pairs can participate in aromaticity, by the Huckel definition. Why can lone pairs on a carboanion be aromatic? I thought carboanions are sp3 hybridized, and the lone pair is not in a p orbital. I understand that by resonance, you can bring the lone pair into the ring, but then you end up with another sp3 carboanion.
I have a similar question for oxygen and nitrogen. Why can their lone pairs be part of aromaticity if they are sp3 hybridized.
As a general rule, when counting $\pi$ electrons in a conjugated ring to assess aromaticity you need to fill the stable $\sigma$ orbitals first. That means (again, generally, maybe not always) if an atom has no ligand, it has to use a lone pair to fill that ligand orbital and only if there is a second lone pair, as with the sulfur atom in thiophene, can that second lone pair join the $\pi$ conjugation.
For instance, in pyrrole there is not this issue because every ring atom has a ligand, so the lone pair formally on the nitrogen atom is conjugated into the ring. In pyridine, by contrast, the nitrogen atom has no ligand and its lone pair has to occupy the ligand position, so all the conjugated electrons in the pyridine ring have to come from $\pi$ bonds. Unsurprisingly, in both cases the net result is just right to get six conjugated $\pi$ electrons in the ring.
Some carboanions are sp3 hybridized and some aren't.
The simplest carboanion is methide. Methane (CH4) is sp3 hybridized. Remove a proton (H+) to obtain the methide (CH3–) ion, and the central carbon is still sp3. It's just sp3 surrounded by three atoms and one lone pair instead of four atoms. The same thinking would apply to a cyclohexane ring. Whether an H+ is removed to obtain a cyclohexanide ion or or not, the carbons are all sp3 hybridized.
Molecules that obey Hückel's rule for π electrons will have carbon atoms that are sp2 or sp hybridized and aromatic. They are still aromatic after an H+ is removed to obtain a carboanion. Therefore, the benzenide ion (C6H5-) is aromatic.
