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Question: In \(K_{4}\left\lbrack Fe(CN)_{6} \right\rbrack,\) the E.A.N. of \(Fe\) is....

In K4[Fe(CN)6],K_{4}\left\lbrack Fe(CN)_{6} \right\rbrack, the E.A.N. of FeFe is.

A

33

B

35

C

36

D

26

Answer

36

Explanation

Solution

EAN =Atomic number – Oxidation state + 2 × number of Ligands = 26 – 2 + 2 (6) = 36.