Hybridized Orbital Question

[powersellingmom]

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Why does the lone pair on an sp3 hybridized nitrogen atom (such as NH3) exist in a hybridized orbital while the lone pair on a carbanion (such as CR2H^-) exist in a p-orbital? I really don't get it and I can't find any similar questions asked about it, which makes me believe it's probably a really dumb question.

What prompted this question was a problem in TBR about the acidity of 1,3 cyclopentadienyl. It loses it's proton more readily than cyclopentane because the electrons gained from the heterolytic cleave of the hydrogen results in aromaticity following huckel's rule. But this means that they must be within p-orbitals to be pi electrons.

 

[LoLCareerGoals]

I have no idea what the second part of the question is, but for the first, I just looked up methyl anion and it is the same shape/hybridization as NH3. The non-bonding pair completes the tetrahedron.
I can't imagine 2 alkyl substitutions would change the shape, but who knows.

Now I do see a p orbital on a neutral free radical. But here it is only a single el-on, not a pair so it wouldn't surprise me if the most stable configuration is a planar triangle with a perpendicular el-on.

 

[V5RED]

Why does the lone pair on an sp3 hybridized nitrogen atom (such as NH3) exist in a hybridized orbital while the lone pair on a carbanion (such as CR2H^-) exist in a p-orbital? I really don't get it and I can't find any similar questions asked about it, which makes me believe it's probably a really dumb question.

What prompted this question was a problem in TBR about the acidity of 1,3 cyclopentadienyl. It loses it's proton more readily than cyclopentane because the electrons gained from the heterolytic cleave of the hydrogen results in aromaticity following huckel's rule. But this means that they must be within p-orbitals to be pi electrons.

The reason they exist in p-orbitals is because of exactly what the book said. Them being in a p-orbital means the molecule gains aromaticity which means it gains stability.

The carbon in question has 3 bonds to other atoms. This could lead to having sp2 hybridization with an unhybridized p-orbital or sp3 with a lone pair sp3 orbital depending on which is more stable. Having sp2 hybridization is more stable for the aforementioned reason.

This is not a dumb question, it tests an important concept that ties together conjugation, stability, and acidity.