Comparing strengths of σ-bonds formed by $\mathrm d_{z^2}$ and $\mathrm p_z$ orbitals

Question

According to my professor $\mathrm d_{z^2}–\mathrm d_{z^2}$ σ-bond is stronger than $\mathrm p_z–\mathrm p_z$ σ-bond as the extent of overlapping is greater in $\mathrm d_{z^2}–\mathrm d_{z^2}$ overlap, because of directional nature and of $\mathrm d_{z^2}$ and its greater distance from the nucleus. There is a assumption that principal quantum number of both $\mathrm d_{z^2}$ orbital and $\mathrm p_{z}$ orbitals are the same.

But according to my intuition $\mathrm p_z$ orbital is also directional in nature, and $\mathrm d_{z^2}$ orbital has a ring in $x–y$ plane which can cause repulsion and the biggest reason that $\mathrm d$ orbitals are very diffuse so overlapping is not very good.

1. Is there any way we can compare their bond energy?
2. Which σ-bond will be stronger: $\mathrm d_{z^2}–\mathrm d_{z^2}$ or $\mathrm p_z–\mathrm p_z?$

Answer 1

It is possible that your instructor meant $\sigma$ bonding -- one lobe pointed straight at another lobe -- is stronger because it has more overlap than $\pi$ bonding, which would be them vertically aligned, but contacting "shoulder-to-shoulder" instead.

The energy term contains orbital overlap, and so it can be reasoned qualitatively that an interaction with more overlap (direct, $\sigma$) would be more favorable than the sideways version, e.g. $\pi$ bonds in alkenes.

Also, one has to remember that both of these are happening in the molecule at the same time, and these two states don't really "exist" in this "one or the other" form, but rather in some hybrid (e.g. 80:20 in the ground state, but 20:80 in the excited state). Whether this basis needs $d$-like functions to make the energy lower, or perhaps more $p$ is situational. One can post-hoc dig through orbital contribution terms, but there's some black magic here.

But otherwise, I would agree if there is ambiguity because technically d and p come from the same shell, and may have different field effects so that one would not know just how contracted in the bonding direction an orbital is to make a determination without resorting to quantitative calculation.

In my opinion, the fundamental problem here is using spherical harmonic (isolated atoms) models in a situation that is not spherical. Its endemic to the chemistry curriculum, unfortunately.