Energy level of molecular orbitals in heteronuclear diatomic compounds

Question

I have been studying the molecular orbital theorem. I read that the energy levels for the $\ce{p}$ orbitals in the first series of $\ce{p}$ block molecules are something to the tune of $$\pi_x=\pi_y<\sigma_z<\pi_x^*=\pi_y^*<\sigma_z^*$$

except for $\ce{O2 and F2}$, which have $$\sigma_z<\pi_x=\pi_y<\pi_x^*=\pi_y^*<\sigma_z^*$$

But it got me thinking: what order would the orbitals take in a compound such as $\ce{CO, NO}$?

(Also, out of curiosity, what about $\ce{p-p}$ bonds in higher series of the $\ce{p}$ block? Does $\ce{Cl2}$ have the same order as $\ce{O2}$?)

My guess was that since in heteronuclear orbitals more electronegative elements have lower energy orbitals, compounded with the fact that the $\sigma$ orbitals it forms tend to be lower energy, it should follow the latter’s order.

Answer 1

This question requires idea of the Linear Combination of Atomic Orbitals (LCAO). It involves the following equation:

$\psi_n = \sum\limits_{i} c_{ni}\phi_{i} = c_{n1}\phi_{1} + c_{n2}\phi_{2} + ... c_{ni}\phi_{i}$

Here $\psi_n$ represents the wavefunction for the resulting molecular orbital, $\phi_{i}$ represents the atomic orbitals that contribute to the bond, and $c_{ni}$ represents the weight coefficient that can basically tell us how much of each atomic orbital is involved in the molecular orbital. If we look at the molecular orbital diagram for carbon monoxide, you are right, the bonding pi orbitals are lower in energy than the bonding sigma orbital:

The reason for this is due to the contributions from carbon. This comes from the relative energies of the atomic orbitals. From photoelectron spectroscopy, the energies of the carbon 2p orbitals and 2s can interact with the oxygen 2p orbitals because they are close enough in energy while the 2s oxygen orbital is a little too far to interact with the 2p orbitals of carbon. See image below

The important thing to note here is that the 2s orbital can interact as well. This means when we solve for the wave function using LCAO and the equation above, carbon can give a large contribution because two of its atomic orbitals interact, and it turned out it did. The same thing can be done for NO (which in fact is also similar to CO where the pi bonding orders are lower energy than the sigma bonding ones) and other molecules. However, it gets extremely complicated with larger and larger molecules.

I got this information from this public PowerPoint https://www.chem.uci.edu/~lawm/10-9.pdf which come from lectures from UCIrvine which were released: https://www.youtube.com/watch?v=1CH6CeORL_g&list=PLaLOVNqqD-2GDVq2xFUuAltjIXPgQNNTL&index=7&ab_channel=UCIOpen

Answer 2

I found this exact point discussed indirectly in JD LEE's inorganic chemistry 5th ed in page 110. In it, the author discusses the CO molecule system where for C it's on order and O it's another, the following point is noted in the last sentence to assume the order was of the lighter element (in this case, C).

Then it is noted that if CO is is ionized to $CO^+$ by removal of an electron from $\sigma2p_x$ then bond order reduces and hence bond length increases (by the above approximation) however this is not what happens that is the bond length of 1.128 angstrom turns to a length of 1.115 angstrom.

It is then noted that same problem remains if we assume the order follows for heavier elements. Finally it is concluded that the order of MO energy levels applicable for simple homonuclear diatomic molecules is not automatically applicable when two different types of atoms are bonded together.

Answer 3

I finally found a practical answer on Quora, the reason of difference in energy levels is due to something known as s-p mixing and it occurs for diatomic molecules having less than or equal to 14 electrons. Meaning the following order:

$$\pi_x=\pi_y<\sigma_z<\pi_x^*=\pi_y^*<\sigma_z^*$$

Is for molecules with $ \leq 14$ electrons.

For CO vs NO, CO has 14 electrons hence takes the order with $\pi_x = \pi_y < \sigma_z$ and NO takes the order $\sigma_z < \pi_x = \pi_y$ since it has 15 electrons

Answer 4

For homonuclear diatomics at and below period 3 the diagram is always the latter one, regardless of the group. See Chem. Rev. 2019, 119, 8781−8845.