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An octahedral complex is formed when

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Excess $KCN$ is added to aqueous ${CdCl}_{2}$ solution.

Excess $NaOH$ is added to aqueous ${ZnCl}_{2}$ solution.

Excess $KCN$ is added to aqueous suspension of $Fe {(CN)}_{2}$

All the above cases yield octahedral complexes.

answer is C.

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Detailed Solution

$(A) 4 KCN + {CdCl}_{2} \overset{water}{⇌} 4 K^{+} + [Cd {(CN)}_{4}]^{2 -} + 2 {Cl}^{-} (B) {ZnCl}_{2} + 4 NaOH ⇌ 4 {Na}^{+} + [Zn {(OH)}_{4}]^{2 -} + 2 {Cl}^{-} (C) 4 KCN + Fe {(CN)}_{2} ⇌ 4 K^{+} + [Fe {(CN)}_{6}]^{4 -}$

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