Oxidation number of Fe in the violet coloured | vedclass

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(A) In the complex $Na_4[Fe(CN)_5(NOS)]$,the ligand $NOS$ is the nitroprusside-related species,specifically $NO^+$. Let the oxidation state of $Fe$ be

Vedclass · Accepted Answer

$0$

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(A) In the complex $Na_4[Fe(CN)_5(NOS)]$,the ligand $NOS$ is the nitroprusside-related species,specifically $NO^+$. Let the oxidation state of $Fe$ be $x$. The oxidation state of $Na$ is $+1$,$CN$ is $-1$,and $NOS$ is $0$ (as it is a neutral ligand $NO$ attached to $S$ or acting as $NO^+$ depending on the complex structure,but in this specific complex,$NOS$ is treated as a neutral ligand $NO$ or $NOS$ group). However,for the complex $Na_4[Fe(CN)_5(NOS)]$,the charge balance is: $4(+1) + x + 5(-1) + 0 = 0$. $4 + x - 5 = 0$. $x - 1 = 0$. $x = +1$. Wait,let us re-evaluate: The complex is $Na_4[Fe(CN)_5(NOS)]$. The $NOS$ ligand is $NO^+$. $4(+1) + x + 5(-1) + (+1) = 0$. $4 + x - 5 + 1 = 0$. $x = 0$. Therefore,the oxidation state of $Fe$ is $0$.

Oxidation number of $Fe$ in the violet coloured complex $Na_4[Fe(CN)_5(NOS)]$ is:

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