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  4. The oxidation number of nickel in K4[Ni(CN)4] is:

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

JIPMERJIPMER 1999

Solution:

Let the oxidation state of $Ni$ be $x$ in $K _4\left[ Ni ( CN )_4\right]$. 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, $4+x-4=0$
$ \text { or, } x=0 $