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Question

The oxidation number of $$Ni$$ in $$K_4[Ni(CN)_4]$$ is :


Solution

Let $$x$$ be the oxidation state of $$Ni$$ in $$K_4[Ni(CN)_4]$$.
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 $$K_4[Ni(CN)_4]$$ is $$0$$.

Chemistry

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