Coordination number of $N i$ in $[N i (C_{2} O_{4})_{3}]^{4 -}$ is :

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Solution

The correct option is B 6
Co-ordination number of $N i$ in $[N i (C_{2} O_{2})]^{4 -}$ is 6.
Because $C_{2} O_{4}^{- 3}$ is bidentate ligand and these ligands are 3 in number. So $2 \times 3 = 6$ be the co-ordination number.

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[N i (C_{2} O_{4})_{3}]^{4 -}

Co-ordination number of Ni in $[N i (C_{2} O_{4})_{3}]^{4 -}$ is:

In the coordination compound, $K_{4} [N i (C N)_{4}]$ , the oxidation state of nickel is :

K_{4} [N i (C N)_{4}]

N i

F e (C N)_{6}]^{4 -}

[C r (N H_{3})_{4} C l_{2}]^{+}

[N i (C N)_{4}] 2 -

F e = 26, C r = 74, N i = 28

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