A question I am looking at is as follows:

$\ce{CO}$ is isoelectronic with $\ce{N2}$. Sketch MO diagrams for $\ce{CO}$ and $\ce{N2}$. Point out key differences between the diagrams and use the diagram to explain why $\ce{CO}$ acts as a two-electron donor through carbon rather than through oxygen.

Understandably, the key difference between these molecules is that $\ce{CO}$ is heteronuclear, and thus will have differences in energy between the molecular orbital and the atoms.

But I can't explain why $\ce{CO}$ is a two-electron donor using MO theory, even though I can with Lewis?

Does anyone have any thoughts?

MO of CO

In the above MO diagram, the 5σ is the HOMO. But it is closest in energy to oxygen's 2p orbitals, so why is it centered on carbon?


2 Answers 2


I had to give this a good, long thought, after I initially thought this is a question which is hardly answerable with a molecular orbitals diagram by itself. Now that I have spent quite some time thinking about it, I must admit I enjoy this exercise a lot.

The key to explaining that carbon monoxide is a two electron donor through carbon is to start with with a very fundamental MO scheme. This is basically to get you dimensions right and not to mix things up later. It is also one of the things I would consider a prerequisite for these kinds of exercises.

generic s-p valence MO diagram for a symmetric diatomic molecule

The above diagram is for a symmetric diatomic molecule with no s-p mixing. This is true for the heavier main group elements; have a look at this explanation by Wildcat.

For dinitrogen we will have to adapt this scheme to include s-p mixing. Here it is important to know, that the interaction is still strong enough to raise the 3σ valence orbital above the 1π orbitals.

generic s-p valence MO diagram for dinitrogen N2

With pen and paper it is a bit too difficult to draw the MO, so it is probably easier to exaggerate the respective contributions to the orbitals. For example the 1σ orbital in the non mixing case is comprised of pure s-orbitals. When it is mixing, there has to be some s-p character which is difficult to draw. As a general guideline electron density concentrates along the bonding axis for bonding orbitals, while for anti-bonding orbitals in the lone-pair region. The π orbitals largely remain untouched by this.

From there we can use the fact, that carbon monoxide is isoelectronic with dinitrogen. We have to skew the energies of the atomic orbitals. The energy of the atomic orbitals is decreasing from left to right in the period; therefore carbon will have slightly elevated levels and oxygen's are lowered.
In this case it is easier to not tinker with the ordering of the MO as this is just a pen and paper exercise. However, one should be aware, that the levels will overall change in this molecule.

generic s-p valence MO diagram for carbon monoxide CO

The atomic orbitals that are closer to the energy of the molecular orbital will have a larger coefficient. Therefore 1σ and 1π will be polarised towards oxygen. On the other hand 2σ and 3σ will be polarised towards carbon.
One can also argue this with Bent's rule (gold book), emphasis mine:

In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

The rule itself (is an observation, which) applies to more complex molecules, but can be applied to this situation, too. Oxygen is more electronegative than carbon therefore the bonding hybrid orbitals will have more p-character directed towards oxygen. At carbon itself will be more s-character for the lone pair, which is what we observe for the HOMO, when we apply the polarisation, too.

From this deductive chain one can reasonably explain, that the HOMO of carbon monoxide must be of σ symmetry (two electron donor) and have the largest coefficient at the carbon atom and will therefore rather coordinate there than through the oxygen.

  • 1
    $\begingroup$ It seems reasonable but slightly awkward to apply Bent's rule here, considering that CO is a diatomic and thus, there are no "ligands" and "central atom". $\endgroup$ Jun 7, 2019 at 1:35
  • $\begingroup$ Very detailed answer but are there references of having $(\sigma ^* 2\text s)^2$ as HOMO? $\endgroup$
    – Apurvium
    Dec 16, 2021 at 16:46
  • $\begingroup$ @Apurvium I don't understand your question. I don't use this numbering and I don't see in what way it would be applied here, i.e. I don't know what your numbering is referring to. $\endgroup$ Dec 16, 2021 at 17:28
  • $\begingroup$ @Martin-マーチン Sigma 2s antibonding molecular orbital $\endgroup$
    – Apurvium
    Dec 16, 2021 at 17:36
  • 1
    $\begingroup$ @Apurvium I can only hope so, because it is qualitatively correct in the framework of MO theory. || Don't think about mixing of orbitals in a time manner, there is no before and no after. This is only a mathematical solution. The molecular orbitals do or don't exist just as the atomic orbitals do or don't exist. We can only observe electron density. So it's getting more cloudy. But from all the definitions of 'bonding' I know, the HOMO of CO fits these descriptions. $\endgroup$ Dec 17, 2021 at 19:21

The electrons in the frontier orbital(s) play a special role for the chemical reactivity. In $\ce{CO}$, the HOMO is the $ 5 \sigma $ orbital (ref: your diagram), and has mainly $\ce{C}$ character. The two electrons in it act like a lone pair on the carbon.

The predominant $\ce{C}$ character of the HOMO accounts for the reactivity via carbon ($\sigma$ donor

Side note: $\ce{CO}$ is also a $\pi ^*$ acceptor due to the LUMO (which is also predominantly $\ce{C}$ in character.

To answer your comment:

So overall, how could I tell exclusively from the MO diagram that HOMO electrons are centered on C and not on O?

Don't think you can. The contribution to MOs is determind by the coefficients in the linear combination. You should see Martin's calculated values in the answer link he provided.

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    $\begingroup$ Well, there is a hint that the lower orbitals will generally be centred on the electronegative partner and the higher ones generally on the electropositive partner, but that’s a weak indication at best. $\endgroup$
    – Jan
    Sep 21, 2016 at 9:44
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    $\begingroup$ Agreed. It has its uses, say for something like HF where there are very few orbitals overlapping, but in CO it's really difficult (more precisely, I found it difficult) to use "intuition" because the C2s, C2p, O2p all interact significantly. Even the usual adage of "AOs close in energy to the MO contribute more" fails here because the HOMO is very close to O2p energy level. $\endgroup$ Sep 21, 2016 at 10:06
  • 1
    $\begingroup$ Not only that, but there are still sources out there stating that the HOMO of CO is antibonding in character. If we can't even get that right by intuition, it is not hard to imagine that we will fail at predicting more complicated features. $\endgroup$ Sep 21, 2016 at 10:19

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