Na_2S + Na_2[Fe(CN)_5NO] to Na_4[Fe(CN)_5NOS] | vedclass

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(B) In the reactant complex $Na_2[Fe(CN)_5NO]$,the oxidation state of $Fe$ is calculated as: $2(+1) + x + 5(-1) + 0 = 0$ (assuming $NO$ is neutral $NO

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$2, 2$

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(B) In the reactant complex $Na_2[Fe(CN)_5NO]$,the oxidation state of $Fe$ is calculated as: $2(+1) + x + 5(-1) + 0 = 0$ (assuming $NO$ is neutral $NO^+$ or $NO$ depending on convention,but standard nitroprusside has $Fe$ in $+2$ state). $x = +2$.In the product complex $Na_4[Fe(CN)_5NOS]$,the ligand $NOS$ is $NOS^{3-}$. The oxidation state of $Fe$ is calculated as: $4(+1) + x + 5(-1) + (-2) = 0$ $\Rightarrow x - 3 = 0$ $\Rightarrow x = +3$.However,in the specific reaction of sodium nitroprusside with sulfide,the $Fe$ center remains in the $+2$ oxidation state as the reaction involves the conversion of $NO^+$ to $NOS^-$. Thus,the oxidation state of $Fe$ in both complexes is $+2$.

$Na_2S + Na_2[Fe(CN)_5NO] \to Na_4[Fe(CN)_5NOS]$; what are the oxidation numbers of $Fe$ in the reactant complex and the product complex,respectively?

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