Hybridization of carbanion in rings

AI Thread Summary
The discussion revolves around determining the aromaticity and hybridization of a specific compound, particularly focusing on a carbanion. Initial thoughts suggested the carbanion was sp3 hybridized due to having four groups of electrons. However, it was clarified that the lone pair actually resides in a p-orbital, indicating that the carbanion is sp2 hybridized, contributing to a total of 8 pi electrons. The presence of resonance structures that allow electron delocalization around the ring supports this hybridization. The conversation also touches on the distinction between valence bond theory and VSEPR, noting that while VSEPR is derived from valence bond concepts, terms like hybridization and resonance are not part of VSEPR theory. The stability of resonance structures is discussed, suggesting that similar but rotated structures indicate enhanced stability.
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So this question asked me to determine whether this compound was aromatic, antiaromatic, or nonaromatic and also asked how many pi electrons there were. I figure since there are four groups of electrons the carbanion is sp3 hybridized and I think I'm correct because that is what this thread says.

https://www.physicsforums.com/threads/question-regarding-hybridization-of-carbon.256312/

But apparently the lone pair on this carbanion in this specific compound lies in a p-orbital (as opposed to sp3)? thus making this carbanion sp2 hybridized? for a total of 8 pi electrons. This makes no sense to me any help would be greatly appreciated.
 

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mnmman said:
So this question asked me to determine whether this compound was aromatic, antiaromatic, or nonaromatic and also asked how many pi electrons there were. I figure since there are four groups of electrons the carbanion is sp3 hybridized and I think I'm correct because that is what this thread says.

https://www.physicsforums.com/threads/question-regarding-hybridization-of-carbon.256312/

But apparently the lone pair on this carbanion in this specific compound lies in a p-orbital (as opposed to sp3)? thus making this carbanion sp2 hybridized? for a total of 8 pi electrons. This makes no sense to me any help would be greatly appreciated.

In this particular compound, you can draw resonance structures that move the pair of electrons around the ring. Each of the carbons has one hydrogen bonded to it, and the charge is delocalized around the ring. The only way that this could happen is if each carbon is sp^2.

Remember that Lewis structures are approximations. How about Benzene? The single Lewis structure would predict alternating short and long C-C bonds. The measurement reveals six identical bonds with bond length that is intermediate between C-C and C=C. In the present case, the Lewis structure is incorrect because it predicts a lone pair (with negative charge) on a single carbon -- the lower energy form is with the charge spread out over the entire ring, which is what the resonance structures would predict.
 
Usually this kind of questions is not about hybridization, which is a concept from valence bond theory, as in school aromaticity is introduced using molecular orbital theory. The pi orbitals are formed from p orbitals on the atoms.
 
Quantum Defect said:
draw resonance structures that move the pair of electrons around the ring

ok I can see it now that was very helpful, so if you draw the resonance structures and they end up looking the same but rotated, is that a way you can tell they are especially stable?

DrDu said:
concept from valence bond theory

Does this theory have anything in common with VSEPR? My book only mentions VSEPR and nothing about valence bond theory
 
mnmman said:
Does this theory have anything in common with VSEPR? My book only mentions VSEPR and nothing about valence bond theory
Yes, VSEPR is a scheme that was derived from valence bond ideas, although it is highly simplified.
However terms like hybridization or resonance are from valence bond theory and not VSEPR.
 
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