Question

The oxidation state of Ni in the compound $K_4\left[Ni\right(CN{)}_4]$ is

• A. 1
• B. 2
• C. -1

Answer 1

How do we start here?
We know that potassium has a +1 charge always, so the negative ion i.e. the complex would have a total charge of -4 to balance out the 4 $K^{+}$ ions.
Let's look at the compex now. The complex has CN as well as Ni and as you know, the cyanide ion would have a charge of -1. So the entire complex needs to have a negative 4 charge, we can write this equation then: ($x$ is the oxidation state of nickel here)
$x+4\times \left(C{N}^{-}\right)=$ charge of the anionic complex
$x+4\times (–1)=–4$ ;
which gives the value of $x$ to be 0.