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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