Question

The oxidation number of $Ni$ in $K_4\left[Ni\right(CN{)}_4]$ is :

Answer 1

Let $x$ be the oxidation state of $Ni$ in $K_4\left[Ni\right(CN{)}_4]$.
Since the overall charge on the complex is $0$, the sum of oxidation states of all elements in it should be equal to $0$.
Therefore,
$+4+x-4=0$
or, $x=0$.
Hence,
the oxidation state of $Ni$ in $K_4\left[Ni\right(CN{)}_4]$ is $0$.