The oxidation number of $\mathrm{Fe}$ in $\mathrm{Na}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{5}…
The oxidation number of $\mathrm{Fe}$ in $\mathrm{Na}_{2}\left[\mathrm{Fe}(\mathrm{CN})_{5} \mathrm{NO}^+ight]$ is
$+2$
$+1$
$+3$
$-2$
Solution
Let \(\mathrm{X}\) be the oxidation number of \(\mathrm{Fe}\) in \(\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}^+ight]\).
The Na atoms have oxidation number of \(+1\).
The $\mathrm{NO}+$ group has oxidation number of 1.
The \(\mathrm{CN}\) group has oxidation number of \(-1\).
\(\begin{aligned} & 2(+1)+X+5(-1)+1=0 \\ & 2+X-5+1=0 \\ & X-2=0 \\ & X=+2 \end{aligned}\)
The oxidation number of \(\mathrm{Fe}\) in \(\mathrm{Na}_2\left[\mathrm{Fe}(\mathrm{CN})_5 \mathrm{NO}ight]\) is \(+2\).