Question

The oxidation number of nickel in ${{K}_{4}}[Ni{{(CN)}_{4}}]$ is:

• A. -2
• B. -1
• C. 2
• D. 0

Answer 1

Answer: (D)

0

Answer 2

Suppose, the oxidation number of Ni is x, then $4(+1)+x+4\,(-1)=0$ $x+4-4=0$ $x=0$ Hence, oxidation state of Ni in ${{K}_{4}}[Ni{{(CN)}_{4}}]$ is zero.