# Oxalate Anion MO Diagram

I figured I’d study $$\ce{C2O4^2-}$$ as an organometallic ligand, and found that it is weak field (pi donor). I hence decided to construct a MO diagram.

The oxalate anion is of D2h symmetry, so one can immediately use that, form irreducible representations and find the C-C orbitals:

Following which, I tried to use projection operator to look at all the group orbitals.

Unfortunately, this means that there are a lot of orbitals with Ag and B2u symmetry, and a couple with B1g and B3u symmetry. How would these orbitals interact?

• You've pretty much done the hard part already. I would focus on the sigma framework first - exclude the O px orbitals as nonbonding and all of the pz orbitals as pi and then match up what's left. For each pair that you match, make a bonding and an antibonding version (or bonding/nonbonding/antibonding if there are three going in). Bear in mind that the orbitals of same symmetry can mix, but for the first pass ignore that and just find symmetry matches. – Andrew Feb 15 at 21:38
• Thanks! Are the oxygen 2s orbitals mostly nonbonding too? – ANZGC FlyingFalcon Feb 16 at 0:57
• No, the s orbitals mix with the py orbitals, so both contribute to bonding and nonbonding. If you want to simplify things, you could represent the Os as having an sp-like orbital (oriented with large lobe toward C) that participates in sigma bonding and an sp-like orbital (oriented with large lobe away from C) that is nonbonding. – Andrew Feb 16 at 2:07
• Also, for the pi framework, you can simplify by treating each CO2- separately, since nonplanar conformations of oxalate are common. – Andrew Feb 16 at 2:08
• The ordering is quite difficult to get exactly right without quantitative approaches, but you can get pretty close by assuming that sigma interactions are more stabilizing than pi and that generally more nodes means higher energy. So group your orbitals as sigma bonding, pi bonding or nonbonding and then try to differentiate energy within each group. I might have time later to write up a longer answer with some of that info, but I'm not sure. – Andrew Feb 16 at 20:22

Here's how I would approach it. I would start by making some (over)simplifying assumptions:

1) O $$p_x$$ orbitals are mostly nonbonding, so I'm going to ignore interactions of those orbitals with carbon orbitals

2) Due to $$sp$$ mixing, the carbon $$p_y$$ and $$s$$ orbitals can be merged into $$sp$$ orbitals aligned on the $$y$$ axis.

3) Same for the O $$p_y$$ and $$s$$ orbitals. In this case, the mixed $$sp$$ orbital with its large lobe oriented away from C is mostly nonbonding, so I'll ignore interactions of it with carbon orbitals

With those assumptions in place, I focus on building and ordering the occupied molecular orbitals. The highest energy occupied orbitals will be some of those oxygen-based nonbonding orbitals, so let's start by putting all eight of them at the top of the diagram. The relative energy levels among them will be affected by the number of positive and negative interactions, so the highest energy one will likely be the $$b_{1g}$$ where all of the $$p_x$$ orbitals have anti-bonding interactions with their neighbors. I mention this orbital in particular because it is the HOMO, and I think that the $$\sigma$$ interaction between oxalate and a metal center would come from the interaction of this orbital with the metal $$d_{x^2-y^2}$$ or $$d_{xy}$$, for example.

Mixed in with the $$p_x$$ and $$sp_y$$ nonbonding orbitals will be the two $$\pi_{nb}$$ orbitals, also centered on the oxygens. These are the $$b_{2g}$$ and $$a_u$$ orbitals in your set above. Since there isn't a carbon $$\pi$$ orbital with the right symmetry to interact with either of these, the resulting MO's have a nodal plane aligned with the C-C bond. These are the orbitals that will have $$\pi$$ interactions with a metal center if the oxalate is acting as bidentate ligand.

Below the nonbonding $$\pi$$ orbitals, we have the two $$\pi$$ bonding orbitals formed from the $$b_{1u}$$ and $$b_{3g}$$ central and oxygen $$p_z$$ orbital combinations. The $$b_{3g}$$ is higher energy of the two because of the C=C $$\pi^*$$ interaction.

At the bottom of the occupied orbitals, we have the five primarily $$\sigma$$ orbitals. I think the highest energy of the group is the $$\sigma_{C-C}$$ formed from the mixed $$sp_y$$ carbon orbitals oriented towards each other, ie a mix of the two $$a_g$$ central orbitals. Since the density is primarily between the carbons, I'm going to ignore the interactions with the O orbitals.

The four $$\sigma_{C-O}$$ orbitals come from the pairings of the central $$b_{2u}$$, $$b_{1g}$$, $$b_{3u}$$ and $$a_g$$ orbitals with their ligand orbital counterparts. In energy, from low to high I would guess $$a_g$$ ($$\sigma_{C-O}$$ and $$\sigma_{C-C}$$), $$b_{3u}$$ ($$\sigma_{C-O}$$ and $$\pi_{C-C}$$), $$b_{1g}$$ ($$\sigma_{C-O}$$ and $$\pi^*_{C-C}$$), then $$b_{2u}$$ ($$\sigma_{C-O}$$ and $$\sigma^*_{C-C}$$).

• Thanks so much, Andrew! That definitely helped :D – ANZGC FlyingFalcon Feb 21 at 12:17