# What is the oxidation state of zinc in "ZnO"_2?

Aug 17, 2017

This is zinc peroxide (not "zinc(IV)" oxide!), i.e. it contains $2 \times \text{O}$ in the $\textcolor{red}{- 1}$ oxidation state.

The way you would know that is by already knowing that zinc has a maximum oxidation state of $+ 2$. Its electron configuration is $\left[A r\right] 3 {d}^{10} 4 {s}^{2}$, so losing more than two electrons is nearly impossible (it would require freeing up a quantum state in the $3 d$ outer-core orbitals).

Hence, the zinc atom has an oxidation state of $\textcolor{b l u e}{+ 2}$ (not $+ 4$, and not $+ 1$).

You may accidentally say that this is simply $2 \times {\text{O}}^{2 -}$ for the anion(s). You may also miss that for the anion without prior knowledge or without looking it up, it is a non-innocent ligand that could either be ${\text{O}}_{2}^{-}$ (the superoxide anion) or ${\text{O}}_{2}^{2 -}$ (the peroxide anion).

Either misconception could lead you to accidentally say that zinc has an oxidation state of $+ 4$ or $+ 1$. Neither is correct!