Question

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

• A. $\left[Fe\right(H_{2}O\left)5\right(NO\left)\right]S{O}_4$
• B. $FeS{O}_{4}N{O}_2$
• C. $\left[Fe\right(H_{2}O{)}_4(NO{)}_2{]}^{2+}$
• D. $FeS{O}_4.HN{O}_3$

Answer 1

The correct option is A $\left[Fe\right(H_{2}O\left)5\right(NO\left)\right]S{O}_4$

The identification and conformation test of nitrate ion ( $N{O}_3^{–}$) using conc. $H_2SO_4 and freshly prepared FeSO4 , is called as 'Brown Ring Test. In a test tube take the $N{O}_3^{–}$ containing sample .Then add conc.$H_{2}S{O}_4$ dropwise with gentle shaking and then heat it shortly using Bunsen burner & add freshly prepared greenish $FeS{O}_4$. Then, we can observe a brown ring formed in that test tube due to the formation of brown colored $\left[Fe\right(H_{2}O\left)5\right(NO\left)\right]S{O}_4$ complex.

Hence option A is correct.