In K_4[Fe(CN)_6],the E.A.N. of Fe is | vedclass

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(C) The $E.A.N.$ (Effective Atomic Number) is calculated as: $E.A.N. = 	ext{Atomic number} - 	ext{Oxidation state} + 2 	imes 	ext{Coordination num

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(C) The $E.A.N.$ (Effective Atomic Number) is calculated as: $E.A.N. = 	ext{Atomic number} - 	ext{Oxidation state} + 2 	imes 	ext{Coordination number}$. For $K_4[Fe(CN)_6]$: $1.$ The oxidation state of $Fe$ is $x + 6(-1) = -4$,so $x = +2$. $2.$ The atomic number of $Fe$ is $26$. $3.$ The coordination number is $6$. $E.A.N. = 26 - 2 + 2(6) = 24 + 12 = 36$.

In $K_4[Fe(CN)_6]$,the $E.A.N.$ of $Fe$ is

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