Triplet carbene has a carbon with 6 electrons, of which 2 are unpaired. I would expect the carbon to be $\ce{sp}$ or $\ce{sp^3}$ hybridised and the two singly occupied orbitals to be degenerate. However, Clayden et al. (p. 1011) say that, since the triplet carbenes are almost always bent, they are best considered to be $\ce{sp^2}$ hybridised, with the unpaired electrons occupying a $\ce{p}$ and an $\ce{sp^2}$ hybrids .

The very basic explanation I see why the two SOMOs are not degenerate is simply that $5E_{\ce{sp^2}} + E_{\ce{p}} + 2E_{\ce{sp^2},\ \text{pairing}}$ is less than $4E_{\ce{sp}} + 2E_{\ce{p}} + 2E_{\ce{sp},\ \text{pairing}}$ or $6E_{\ce{sp^3}} + 2E_{\ce{sp^3},\ \text{pairing}}$. However, I wonder if there's a better explanation.

  • $\begingroup$ It’s a little hard to tell $\mathrm{sp^2}$ and $\mathrm{sp^3}$ hybrid orbitals apart without exact angle measurements … $\endgroup$ – Jan Dec 18 '17 at 16:46
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    $\begingroup$ The key basically lies in symmetry. Also it is wrong to assume that changing the hybridisation changes the geometry and/or energy. The molecule is bent, because that is the most stable arrangement, everything else follows. Also, atoms cannot be hybridised, only orbitals can. $\endgroup$ – Martin - マーチン Dec 18 '17 at 17:40
  • $\begingroup$ @Martin, do you mean that you start with the bent geometry and then use symmetry to deduce hybridisation? But what I'm looking for is to rationalise the geometry. Surely in this simple case there should be a way to rationalise the preferred geometry, without resorting to intensive computations? $\endgroup$ – GingerBadger Dec 18 '17 at 18:30
  • $\begingroup$ chem.libretexts.org/Core/Organic_Chemistry/Fundamentals/… $\endgroup$ – Mithoron Dec 18 '17 at 22:06

That's actually a good question. I was lucky enough that Dietrich Gudat, who wrote a long chapter on this type of chemistry in the Riedel - Moderne Anorganische Chemie was my professor in inorganic molecular chemistry and we discussed these things in detail back then but it's been a while.

I think it's due to how you want your final structure to look like or VSEPR. You have bascially three ligands, two rests and the lone-pair on the carbon. And by having it sp² you get quite some distance with the 120° angles. This splits your possible energy states in one which is called a1, that is basically the sp² and an b1 (the p-orbital). As the a1 has more s-character it is lower in energy and thus you can, depending on how big the gap is, either get a triplett or a singulett carbene.

This will change however if you move to the isolobal and isoelectronic nitrenes. Here we have no sp² but an sp which is automatically filled with two electrons. The other two electrons occupy the degenerate sets of px and py and thus you always end up with a triplett.


You can put the electrons in whatever orbitals you would like to. The only difference is that a specific electronic structure produce least electronic energy. You can in a relatively simple manner picture this.

Let us consider the sp(linear) and sp2(bent) structures of carbene(Methylene).

Imagine the carbene's triplet state. In the early days it was actually believed that the structure is actually linear (Methylene story).

In such a case the two orbitals will be p and degenerate. Upon bending, the in-plane p orbital starts mixing with the higher lying 2s orbital and gets stabilized (look at the walsh diagram D∞h to C2v). This stabilization reduces the energy of the system stabilizing the carbene. On the other hand, bending increases the electron electron repulsion. A competition between these two factors decides the faith of the carbene's angle.

For example in the singlet state, which has both electrons in the same orbital and electron-electron repulsion between the two orbitals doesn't exist, has a more bent structure.


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