Variation in the energy of the $\ce{\sigma_{2pz}}$ orbital

Is there a specific reason why the energy of the $\ce{\sigma\_{2pz}}$ orbital varies relative to the $\ce{\pi\_{2px}}$ and $\ce{\pi\_{2py}}$ orbitals, depending on the atomic number of the atom?


[![enter image description here](https://i.stack.imgur.com/hyzRq.png)](https://i.stack.imgur.com/hyzRq.png)




Orbital mixing is the general answer. Molecular orbitals with the same symmetry, like $\ce{\sigma\_g(2s)}$ and $\ce{\sigma\_g(2p)}$ can overlap. The result is that $\ce{\sigma\_g(2s)}$ will decrease in energy while $\ce{\sigma\_g(2p)}$ will increase in energy. The same thing happens for the antibonding orbitals with $\sigma\_u$ symmetry.


[![enter image description here](https://i.stack.imgur.com/LhjRl.jpg)](https://i.stack.imgur.com/LhjRl.jpg)
**sp-mixing**: Left diagram shows the MO orbital diagram for s and p orbitals without sp-mixing, while the right diagram shows one possible configuration after considering sp-mixing.


As the effective nuclear charge increases, the sp mixing is smaller due to the larger energy difference between the $\ce{2p}$ and $\ce{2s}$ orbitals. Therefore, for $\ce{O2}$, $\ce{F2}$ and theoretical $\ce{Ne2}$, you see the characteristic molecular orbital diagram shown on the left. However, with diatomic elements like $\ce{N2}$ and theoretical diatomic molecules with lower atomic numbers such as $\ce{C2}$ and $\ce{B2}$, you get more orbital mixing. That means the $\ce{\sigma\_u(2p)}$ would increase more in energy, which is why you may see the right molecular orbital diagram representing certain diatomic molecules.


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