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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?

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  • $\begingroup$ You are asking for $\ce{R_3C^-}$ and then you draw orbital diagram for isolated carbon, that does not mix well. Additionally, that carbanions are planar, so adjust your hybridization properly. And finally, it's you who has to decide, if you talk about radical anion or closed shell anion. $\endgroup$ – ssavec Nov 11 '13 at 11:01
  • $\begingroup$ I don't agree about planarity, but inversion is an option. To flesh out Ssavec's answer slightly: you can't really consider sp3 hybirdization in the context of just the carbon atom. You have to mix in hydrogen orbitals to form a molecular orbital diagram. $\endgroup$ – Lighthart Mar 4 '14 at 4:42
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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. carbanion 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.


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.

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$\ce{H3C^-}$ 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.

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