Answer: The oxidation state of Ni in $Ni(CO)_4$ is 0.
Metal carbonyl s have symmetrical structures and are charge-neutral, resulting in their high volatility. In $Ni(CO)_4$, the nickel atom has a formal oxidation number of zero.
The sum o fthe oxidaton states = the charge on the complex $Ni(CO)_4$ being neutral has a zero charge. CO is a neutral ligand having zero charge as well. Therefore, Ni mush have an oxidation state of 0.