In the coordination compound${{K}_{4}}[Ni{{(CN)}_{4}}]$ the oxidation state of nickel is
(A) -1
(B) 0
(C) +1
(D) 2

Hint: To solve this problem, let’s suppose the oxidation state of Nickel to be x. We must first add the oxidation states of the remaining elements until the total is equal to the net charge in the molecule.

Complete Step by Step Solution:
In the given question we have to find the oxidation state of Nickel. We first must know that the given compound is a coordination compound in which the $CN$ molecule is acting as a ligand.

The hypothetical charge that an atom would hold if all of its links to other atoms were entirely ionic in nature is what is known as an atom's oxidation state. To proceed further let’s suppose the oxidation number of Ni to be x.

Now if we assume the complete dissociation of the compound into its ionic form then the compounds that will form would be:
${{K}_{4}}[Ni{{(CN)}_{4}}]4{{K}^{+}}+{{[Ni{{(CN)}_{4}}]}^{4-}}$

So, we can see that the complex has a net charge of -4 and we know that $C{{N}^{-}}$ is a negatively charged ligand and has a charge of -1. Since, there are 4 CN molecules then the total charge would be -4.

So, oxidation state of Ni can be given as
$x+4(-1)=-4$ Where, x is the oxidation state of Ni
After solving it we get $x=0$ and hence, the oxidation state of Nickel in the given compound is $0$.
Hence, the correct option is B. 0 i.e. the oxidation state of Ni is 0.

Note: In the given coordination compound the ligand is $C{{N}^{-}}$which is a negative charged ligand. Hence we need to remember that the charge on the same is -1. Another thing to take care of is to properly distribute the charge when dissociating the compound into the ions. If there is no attacking group (like K in this case) then the compound is neutral and neutral compounds have a net charge of 0.