(B) The compound given is $Na_{4}[Fe(CN)_{5}(NOS)]$. Let the oxidation state of $Fe$ be $x$. The oxidation state of $Na$ is $+1$,$CN$ is $-1$,and $NOS

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(B) The compound given is $Na_{4}[Fe(CN)_{5}(NOS)]$. Let the oxidation state of $Fe$ be $x$. The oxidation state of $Na$ is $+1$,$CN$ is $-1$,and $NOS$ is $-1$. The sum of oxidation states in a neutral complex is $0$. $4(+1) + x + 5(-1) + 1(-1) = 0$. $4 + x - 5 - 1 = 0$. $x - 2 = 0$. $x = +2$.