Heteronuclear diatomics

Consider a metal oxide bond XO, where X is a 4th period metal. We need to construct the MOs from the metal $3d$ orbitals and the $2s$ and $2p$ orbitals of oxygen. The oxygen $2s$ orbital of oxygen is too low in energy compared to the metal $3d$ orbitals, so there is no overlap. The $2s$ oxygen orbital is, therefore, a $\sigma^{\rm nb}$ orbital. Denote this as $1\sigma^{\rm nb}$.

The oxygen $2p$ orbitals do overlap with the $3d$ orbitals of the metal. The $3d_{z^2}$ orbital overlaps with the $2p_z$ orbital of oxygen to give $\sigma$ and $\sigma^*$ bonds (see figure below):

Figure: MOs from combining $3d$ metal orbitals and $2p$ oxygen orbitals

Label these as $2\sigma$ and $4\sigma^*$.

The overlap of the $3d_{xz}$ and the $2p_x$ orbitals (see figure) or the $3d_{yz}$ and $2p_y$ orbitals (see figure) give $\pi$ and $\pi^*$ orbitals, which we will denote as $1\pi$ and $2\pi^*$. No orbitals of the oxygen are left to overlap with $3d_{x^2-y^2}$ or the $3d_{xy}$ orbitals, so these become nonbonding, and we denote them as $1\delta^{nb}$ (there are two of them). Finally, the $4s$ orbitals do not overlap with any oxygen orbitals, so this is a $\sigma^{\rm nb}$ orbital, which we denote as $3\sigma^{\rm nb}$.

The ordering of these orbitals is shown in the figure below:

Figure: Correlation diagram for a metal oxide

As an example, consider ScO. Sc has the electronic configuration [Ar]$3d^14s^2$ and O has $1s^2 2s^2 2p_x^2 2p_y 2p_z$. Thus, there are 9 valence electrons. Filling the above diagram gives us the electronic configuration for ScO of $(1\sigma^{\rm nb})^2 (2\sigma)^2 (1\pi)^4 (3\sigma^{\rm n})^1$.