Question

The complex ${\left[Fe{\left(H_{2}O\right)}_{5}NO\right]}^{2+}$ is formed in the ring-test for nitrate when freshly prepared $FeS{O}_4$ solution is added to aqueous solution of $N{O}_3^{-}$ followed by addition of conc. $H_{2}S{O}_4$. This complex is formed by charge transfer in which :

• A. ${Fe}^{2+}$ changes to ${Fe}^{3+}$ and ${NO}^{+}$ changes to $NO$
• B. ${Fe}^{2+}$ changes to ${Fe}^{3+}$ and $NO$ changes to ${NO}^{+}$
• C. ${Fe}^{2+}$ changes to ${Fe}^{+}$ and $NO$ changes to ${NO}^{+}$
• D. no charge transfer takes place

Answer 1

Answer: (C)

${Fe}^{2+}$ changes to ${Fe}^{+}$ and $NO$ changes to ${NO}^{+}$

${Fe}^{2+}+e^{-}⟶{Fe}^{+}$
$NO⟶{NO}^{+}+e^{-}$
Based on magnetic properties of the complex, it is found that iron has three unpaired electrons. Thus, it is as ${Fe}^{+}$ formed by reduction of ${Fe}^{2+}$ by $NO$ which is oxidised to ${NO}^{+}$.