Question

$FeS{O}_4$ solution gives a brown coloured ring during the test for nitrates or nitrites. This is due to the formation of :

• A. $\left[Fe\right(H_{2}O{)}_{5}NO{]}^{2+}$
• B. $\left[Fe\right(H_{2}O{)}_{5}N{O}_2{]}^{2+}$
• C. $\left[Fe\right(H_{2}O{)}_4(NO{)}_2{]}^{2+}$
• D. $\left[Fe\right(H_{2}O{)}_{4}NO{]}^{2+}$

Answer 1

The correct option is A $\left[Fe\right(H_{2}O{)}_{5}NO{]}^{2+}$

$FeS{O}_4$ solution gives a brown coloured ring during the test for nitrates or nitrites. This is due to the formation of $\left[Fe\right(H_{2}O{)}_{5}NO{]}^{2+}$ ion.

This is the brown ring test for $N{O}_3^{-}$ in which nitrates are treated with freshly prepared ferrous sulphate solution in the presence of concentrated sulphuric acid. The brown colour is due to the charge transfer. $NO$ transfers an electron to $F{e}^{2+}$ reducing it to $F{e}^{+}$ and itself gets oxidised to $N{O}^{+}$.

$NO\to N{O}^{+}+e^{-}$

$F{e}^{2+}+e^{-}\to F{e}^{+}$