Mercury shows variable valency while zinc does not. Its electronic configuration is $\ce{[Xe]\:4f^14 5d^10 6s^2}$. So it can donate the $\ce{6s^2}$ electrons and should only be able to form $\ce{Hg^2+}$, right?

  1. How can it show $+1$ oxidation state?

  2. Why does not $\ce{Zn}$ show $+1$ oxidation state?


1 Answer 1


On the contrary, zinc(I) compounds do exist, though they are rare, and relatively unstable. Most zinc(I) compounds contain a $\ce{[Zn2]^{2+}}$ core, which is analogous to the $\ce{[Hg2]^{2+}}$ cation. The $\ce{[Zn2]^{2+}}$ ion does, however, rapidly disproportionate into zinc metal and zinc(II), and has only ever been obtained by cooling a solution of metallic zinc in molten $\ce{ZnCl2}$ and in a few compounds such as decamethyldizincocene.

This should show you that the chemistry of zinc and mercury really aren't all that different. The $\ce{[Hg2]^{2+}}$ ion's much greater stability relative to $\ce{[Zn2]^{2+}}$ is due to the atypically large ionization enthalpy of the $\ce{Hg}$ atom ($100.7\:\mathrm{kJ\:mol^{-1}}$ greater than $\ce{Zn}$), which in turn due to relativistic stabilization of the $\mathrm{6s}$ orbital. Relativistic effects account for a myriad of the properties that are characteristic of heavy metals, including why mercury is the only metal that is a liquid at room temperature, and even gold's color.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.