# How to rationalise with MO theory that CO is a two-electron donor through carbon?

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?

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?

• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. If you receive useful answers, consider accepting one. I recommend you have a look at: How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory? My answer there contains a full MO diagram of CO which should make it easier to comprehend. – Martin - マーチン May 16 '16 at 11:08
• Hi Martin. Thank for your reply! I'm struggling to understand the explanation given in the linked thread. This question is actually an exam question, so I wouldn't have access to the data you've given there. Is there a way to see from the MO diagram alone, why CO is a two-electron donor through C? Also, how do I upvote you? – Adam May 16 '16 at 11:26
• Upvoting is a privilege you have to earn on this site. But it is not hard. I recommend you include your MO sketches into the post and you'll soon gather a few upvotes that get you there. When you consider the donor ability you usually look at the HOMO. From your graph (or the MO diagram linked) you should be able to tell which atom has the largest contribution to it. If you figure the solution out by yourself, you are more than welcome to answer your own question. This way you can also earn some reputation points. – Martin - マーチン May 16 '16 at 11:36
• I see. I have provided an image of the MO diagram and a suggestion! – Adam May 16 '16 at 11:49
• So overall, how could I tell exclusively from the MO diagram that HOMO electrons are centered on C and not on O? – Adam May 16 '16 at 23:24

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.

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.

• 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. – Jan Sep 21 '16 at 9:44
• 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. – orthocresol Sep 21 '16 at 10:06
• 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. – Martin - マーチン Sep 21 '16 at 10:19

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.

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 2σ valence orbital above the 1π orbitals.

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 nitrogen'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.

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.

• 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". – Tan Yong Boon Jun 7 '19 at 1:35