Question

The complex $\left[Fe\right(H_{2}O{)}_s\left(NO\right){]}^{2+}$ 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. $H_2{S}_4$. The complex is formed by charge transfer, which of the following change occurs?

• A. $F{e}^{3+}\to F{e}^{2+};NO\to N{O}^{+}$
• B. $F{e}^{2+}\to F{e}^{3+};NO\to N{O}^{+}$
• C. $F{e}^{2+}\to F{e}^{+};NO\to N{O}^{+}$
• D. $F{e}^{2+}\to F{e}^{3+};N{O}^{+}\to NO$

Answer 1

Answer: (C)

$F{e}^{2+}\to F{e}^{+};NO\to N{O}^{+}$

When to the aqueous solution of the nitrate salt freshly prepared $FeS{O}_4$ is added and then conc. $H_{2}S{O}_4$ is run along with wall of test tube; there is formation of the brown ring at the junction.

$6FeS{O}_4+2HN{O}_3+3H_{2}S{O}_4\to 3F{e}_2(S{O}_4{)}_3+4H_{2}O+2NO$

$FeS{O}_4+NO\to FeS{O}_4.NO$ (Brown ring)

Here the complex is formed by charge transfer, so following changes has taken place :

$C:F{e}^{2+}\to F{e}^{+}$ and $NO\to N{O}^{+}$