Question

A brown ring is formed in the ring test for $N{O}_3^{-}ion$. It is due to the formation of:

• A. $\left[Fe\right(H_{2}O{)}_5\left(NO\right){]}^{2+}$
• B. $FeS{O}_4.N{O}_2$
• C. $FeS{O}_4.HN{O}_3$
• D. $\left[Fe\right(H_{2}O{)}_4(NO{)}_2{]}^{2+}$

Answer 1

The correct option is A $\left[Fe\right(H_{2}O{)}_5\left(NO\right){]}^{2+}$

$\text{Brown ring test}$

• In the ring test for $N{O}_3^{-}$ ion, a freshly prepared ferrous sulphate solution is added to the solution containing the nitrate ion.
• A ring containing a brown−coloured complex is formed.
• The brown-coloured complex is $\left[Fe\right(H_{2}O{)}_{5}NO]S{O}_4$.
• In the first step, ferrous ions $\left(F{e}^{2+}\right)$ react with $N{O}_3^{-}$ ions in the presence of $H^{+}$ ions to form ferric ions and nitric oxide $\left(NO\right)$. $N{O}_3^{-}+3f{e}^{2+}+4H^{+}\to NO+3F{e}^{3+}+2H_{2}O$
• In the second step, the nitric oxide reacts with hydrated ferrous ions to form a brown-coloured complex, $\left[Fe\right(H_{2}O{)}_5\left(NO\right){]}^{2+}$

$\left[Fe\right(H_{2}O{)}_6{]}^{2+}+NO\to \left[Fe\right(H_{2}O{)}_5\left(NO\right){]}^{2+}+H_{2}O$

So, option $\left(A\right)$ is the correct answer.

$\text{Additional information}$

• The oxidation state of iron in the brown ring complex is $+1$.