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  4. In K4[Fe(CN)6] the EAN of Fe is

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Q. In $K_{4}[Fe(CN)_{6}]$ the EAN of $Fe$ is

Coordination Compounds

Solution:

EAN = Atomic number - Oxidation state + 2 $\times$ number of
Ligands $= 2 6 - 2 + 2 (6) = 36.$