# What is the metal oxidation state in Ni(CO)_4?

We have zerovalent nickel, i.e. $N {i}^{0}$
$\text{Oxidation number}$ is the charge left on the central atom when all the bonding pairs of electrons are removed, with the charge devolved to the most electronegative atom.
For $N i {\left(C O\right)}_{4}$, we break the $N i - C O$ bond and we get $N {i}^{0}$ +4xx""^(-):C-=O^+. And thus nickel has a formal oxidation state of $\text{ZERO} .$
Because $N i {\left(C O\right)}_{4}$ is quite volatile, its formation forms the basis for the Mond process for nickel purification.