# Why do heavier post-transition metals have smaller common oxidation states than lighter post-transition metals?

Jun 23, 2017

This has to do with the inert pair effect (i.e. with respect to the $n s$ orbital) for POST-transition metals, i.e. $S e , T e , P o$ are the most relevant.

For context...

• $S e$ has been known to have common oxidation states of $- 2$, $+ 2$, $+ 4$, and $+ 6$.
• $T e$ has been known to have common oxidation states of $- 2$, $+ 2$, $+ 4$, and $+ 6$. We expect $+ 6$ to be less common than $+ 4$.
• $P o$ has been known to have common oxidation states of $- 2$, $+ 2$, $+ 4$, but $\textcolor{red}{+ 6}$ is not as common (not in bold).

These oxidation states arise due to hypothetical full transfers of SIX (FOUR) valence electrons for a $+ 6$ ($+ 4$) oxidation state. The trend implies that two of these valence electrons are not used as easily, namely the $n s$ electrons.

This can be explained by considering the radial density distributions of the $6 s , 6 p , 5 d$, and $4 f$ orbitals:

Notice how as the angular momentum of the orbital increases, the bumps near the nucleus move further away from the nucleus. This indicates decreasing ability to penetrate the core orbitals to be near the nucleus.

We can thus see the poor penetrating ability of the $\left(n - 2\right) f$ and $\left(n - 1\right) d$ orbitals, whereas the $n s$ orbitals are quite penetrating, i.e. core-like, thus rendering themselves less accessible for bonding.

That means the $n s$ orbitals in general are more inert as we go down group $16$ for the post-transition metals; the less core-like $\left(n - 2\right) f$ and $\left(n - 1\right) d$ orbitals "shield" the $n s$ electrons from incoming atoms' valence electrons, and $\boldsymbol{+ 4}$ oxidation states are more stabilized.