The EAN of Fe in K_4[Fe(CN)_6] is | vedclass

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(D) The complex is $K_4[Fe(CN)_6]$.The atomic number $(Z)$ of $Fe$ is $26$.Calculate the oxidation state of $Fe$: $4(+1) + x + 6(-1) = 0 \implies 4 +

Vedclass · Accepted Answer

$36$

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(D) The complex is $K_4[Fe(CN)_6]$.The atomic number $(Z)$ of $Fe$ is $26$.Calculate the oxidation state of $Fe$: $4(+1) + x + 6(-1) = 0 \implies 4 + x - 6 = 0 \implies x = +2$.The coordination number $(CN)$ of $Fe$ is $6$ (since $CN^-$ is a monodentate ligand).The formula for $EAN$ is: $EAN = Z - (	ext{Oxidation state}) + 2 	imes (CN)$.$EAN = 26 - 2 + 2(6) = 24 + 12 = 36$.

The $EAN$ of $Fe$ in $K_4[Fe(CN)_6]$ is

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