Hybridisation of the negatively charged carbon atom in a carbanion

Question

Given the carbanion, $\ce{R3C-}$, the carbon is $\mathrm{sp^3}$ hybridized, unless it is participating in resonance. This is clear from its steric number.

However, I am having trouble in drawing its orbital diagram. I know that if this was a neutral carbon atom, it would look like this:

• $\mathrm{3s}$: _
• $\mathrm{2p}$: ↑ ↑ _
• $\mathrm{2s}$: ↑↓
• $\mathrm{1s}$: ↑↓

Which would hybridize to this:

• $\mathrm{3s}$: _
• $\mathrm{2sp^3}$: ↑ ↑ ↑ ↑
• $\mathrm{1s}$: ↑↓

Now if I was to examine the carbanionic state, would I simply be adding one electron to the top most diagram, then re-hybridize that? Or would I add two electrons because the carbon is not sharing the electrons with any other atom? That's another question I have with orbital diagrams. Do we break each bond homolytically, and base our diagram off of that? This would give:

• $\mathrm{3s}$: _
• $\mathrm{2p}$: ↑ ↑ ↑
• $\mathrm{2s}$: ↑↓
• $\mathrm{1s}$: ↑↓

I am not sure how I would rehybridize that. Have I drawn the radical form?

Answer 1

Hybridization is a concept, which is often misconceived. One has to understand, that it is a mathematical concept, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will be formed respecting steric and electronic interactions and possibly interactions with a medium or external field. The geometric arrangement will not be formed because a molecule is hybridized in a certain way.

Hybridisation is therefore only a result of a certain geometric arrangement of atoms in a molecule. A linear combination of all available (atomic) orbitals will form molecular orbitals (MO). These can then be rearranged (recombined) in a certain way to form localised molecular orbitals (LMO). These can then be used to interpret a bonding situation in a simpler theory.

Each LMO is expressed as a linear combination of the atomic orbitals, hence it is possible to calculate contributions of the atomic orbitals and describe these also as hybrid orbitals.

In a very simple case methane, the $\ce{C-H}$ bonds are formed from contributions of carbon's $\ce{2s, 2p}$ and hydrogen's $\ce{1s}$. For the carbon this may be interpreted as a $\ce{2sp^3}$ orbital.

In the simple case of $\ce{{}^{-}CH3}$, as Uncle Al already pointed out is isoelectronic to ammonia, the bonding situation may be also described in terms of $\ce{2sp^3}$ orbitals. As well as ammonia (or amines in general) it may also undergo (nitrogen) inversion. (source)

In this inversion scenario the hybridisation of the carbon (nitrogen) changes from $\ce{2sp^3}$ to $\ce{2sp^2}$ to $\ce{2sp^3}$.

With larger carbanions this becomes a little more complicated. As long as $\ce{R=Alkyl}$, hyper conjugation is not strong enough to prevent pyramidalisation. However, the inversion berrier is furthermore reduced.

If any of the $\ce{R}$ groups are different from alkyl substituents, then I would expect that the environment around the carbon becomes (more) planar, i.e. the contribution of carbon's $\ce{2s}$ orbital increases. That may have various reasons, but some kind of conjugation or hyperconjugation will be involved. When these $\ce{R}$ groups become pretty bulky, I would expect the same, due to steric interactions.

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Summary

Please understand hybridisation as a tool to understand bonding situations. Therefore it is a result of a geometric alignment of atoms in a molecule. It is not the driving force to form a certain geometric constellation.

Answer 2

$\ce{H3C^-}$ is isoelectronic with$\ce{NH3}$, as $\ce{CH4}$ and $\ce{NH4^+}$ correspond. Each of the four $\ce{sp^3}$ hybrid orbitals contain a pair of electrons, in all cases shown. Nuclear charge versus electron count and the source of the four electrons added to the naked $\ce{sp^3}$ hybrid orbitals are accessories.