Arrange the following in increasing order of oxidation state of $N i$ .
$K_{2} [N i (C N)_{4}], K_{2} [N i F_{6}], N i (C O)_{4}$

N i (C O)_{4}, K_{2} [N i (C N)_{4}], K_{2} [N i F_{6}]

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K_{2} [N i (C N)_{4}], N i (C O)_{4}, K_{2} [N i F_{6}]

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N i (C O)_{4}, K_{2} [N i F_{6}], K_{2} [N i (C N)_{4}]

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K_{2} [N i F_{6}], K_{2} [N i (C N)_{4}], K_{2} [N i (C N)_{4}] "

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Solution

The correct option is C $N i (C O)_{4}, K_{2} [N i (C N)_{4}], K_{2} [N i F_{6}]$
Let the oxidation number of $N i$ be $x$ .
Oxidation number of $K = + 1$ , $C N = F = - 1$ and of $C O = 0$ .
(A) $K_{2} [N i (C N)_{4}]$ : $2 \times 1 + x + 4 \times (- 1) = 0$
$2 + x - 4 = 0$ or $x = + 2$
(B) $K_{2} [N i F_{6}]$ : $2 \times 1 + x + 6 \times (- 1) = 0$
$2 + x - 6 = 0$ or $x = + 4$
(C) $N i (C O)_{4}]$ : $x + 4 \times 0 = 0$
$x + 0 = 0$ or $x = 0$
thus the increasing order of oxidation state of $N i$ is $0 < + 2 < + 4$ in:
$N i (C O)_{4} < K_{2} [N i (C N)_{4}] < K_{2} [N i F_{6}]$

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