# Which atom bears the negative charge in cyanide? I am confused by the MO diagram

In heteronuclear species, whichever atomic orbital a molecular orbital is closer to in energy contributes more to that molecular orbital, right? In other words, if the orbital is occupied, more electron density is around the atom whose atomic orbital is closer in energy to the molecular orbital. This feature of MO diagrams can predict dipole behavior and which atom bears charge in ions. If this is the case, and we know that carbon bears the negative charge in cyanide (and hence is a carbon nucleophile), why does the MO diagram for cyanide appear to have the HOMO (the "attacking" orbital) concentrated on the nitrogen rather than the carbon atom? Based soley on the diagram and not the accompanying image of the HOMO, I thought, since the HOMO is closer in energy to the nitrogen 2pz atomic orbital rather than the carbon 2pz, the HOMO should have more contribution from nitrogen and the LUMO should have more contribution from the carbon.

• There's some negative charge on both atoms. HOMO on C is more diffuse means something contrary to what you seem to say. – Mithoron Feb 11 '19 at 2:21
• @Mithoron My original post may have been a bit ambiguous. I have come to associate the position of a MO relative to those its parent AOs with the contribution that orbital has from its parent AOs. As such, I would expect the HOMO in cyanide to primarily reside on the nitrogen and the LUMO primarily on the carbon. The pictures show what I would expect based on my knowledge of the reactivity of cyanide, but I would have predicted the opposite just by looking at the energy diagram. – Adonis Pugh Feb 11 '19 at 8:28
• Carbon bears a negative formal charge, but formal charges do not necessarily represent the actual charge distribution. – orthocresol Feb 11 '19 at 12:56

As commenters have noted, charge isn't the best choice of a word here, but you're right that the HOMO is more centered on the C, and that's really your question.

The answer is a bit complicated. It relates to those dashed lines on the MO diagram. If we naively construct an MO diagram of cyanide ion, we have two favorable sigma interactions. First, the $$2s$$ orbitals can interact positively, which I'll call $$\sigma_s$$. That would be the lowest energy MO on our diagram. Second, we can have an end-to-end interaction of $$p$$ orbitals, which I'll call $$\sigma_p$$. Since that $$\sigma$$ interaction is stronger than a side-by-side $$\pi$$ interaction, we would expect the $$\sigma_p$$ MO to be lower in energy than the $$\pi$$ bonding MOs.

If we look at your MO diagram, though, we see that $$\sigma_p$$ is actually higher than the $$\pi$$ MOs, not lower as we predicted. It's the HOMO in the diagram. The reason it is higher than we naively expected is because there is mixing between it and the $$\sigma_s$$ (and also between their anti-bonding counterparts, but ignore that for now). These two MOs can interact positively or negatively. In the positive interaction, we get the (normalized) sum of the two MOs, which is an $$sp$$ orbital on C with its large lobe toward N and an $$sp$$ orbital on N with its large lobe towards C (see the lowest MO in the drawings below). Both large lobes have the same phase, so this is a strong bonding interaction that is the lowest MO on your diagram. Both of the starting bonding MOs are biased in favor of contribution from N, since both the $$s$$ and $$p$$ AOs of N are lower in energy than their counterparts on C. Thus, this sum orbital has a stronger N contribution as well.

In the negative interaction, we subtract $$\sigma_s$$ from $$\sigma_p$$ and again get $$sp$$-type orbitals on both atoms, but now the large lobes are facing away from each other. That makes the bonding interaction much weaker, so much so that it moves higher in energy than the $$\pi$$ bonding orbitals. Furthermore, subtraction takes away more N-character than it does C, and you end up with a greater contribution to the new orbital from C than N. As you get more comfortable with MOs, you can verify this mathematically. Another way to think about it is that the final MO is higher than the either the average of a $$Cs$$ and a $$Cp$$ orbital or the average of an $$Ns$$ and an $$Np$$ orbital, but it's closer to the C $$sp$$ average (which is higher than the N $$sp$$ average), so it has more C character by the same logic you used to predict that it would have more N character (which it would if it were just made up of $$p$$ orbitals).

Below is a drawing of the five occupied orbitals where you can see that the $$\sigma$$ ones have become $$sp$$-like and that the carbon contribution is greater in the HOMO. C is on the left and N on the right.

I'm trying to figure out which way the dipole moment on nitric oxide (NO) points, and wondering if a similar argument to the one above holds. Oxygen is more electronegative, but has a full valence shell when there is a double bond between the two, while the Nitrogen's is incomplete, the same effect which gives rise to the negative charge in CN⁻ being localised on the carbon.

I am aware that a more correct picture involves 2-centre-3-electron bonding, but I'm trying to figure out what that implies. According to the MO diagram, there's an unpaired electron in the $$\pi^*$$, which is closer in energy to the Nitrogen's contributing p-orbitals.

This is in line with the electron density distribution I found here.

Is there an explanation similar to the one above for how to think about what is going on here?

Does the dipole have the negative end on the nitrogen?

Is it possible for the electron density to be localised on one atom, but negative charge on the other, if that atom also has a higher $$Z_{eff}$$ to balance out the electron density? (I would think so) Meaning, in general, is electron density a bad way to check which way a dipole points?

Thanks for any replies.