Question

Brown ring test is used for detection of:
A) Ferrous
B) Nitrite
C) Nitrate
D) Ferric

Answer 1

In the Brown ring test, a freshly prepared ferrous sulphate solution is added to the solution containing ion to be tested. A ring containing a brown coloured complex is formed.

Complete answer:
The brown ring test is a qualitative test that confirms the presence of nitrate $\left( {{\text{NO}}_3^ - } \right)$ ion. In the Brown ring test, the formation of a brown coloured complex confirms the ring test for the presence of nitrate ions. The brown coloured complex is $\left[ {{\text{Fe}}{{\left( {{{\text{H}}_2}{\text{O}}} \right)}_5}{\text{NO}}} \right]{\text{S}}{{\text{O}}_4}$.

Now, let us write the balanced chemical equation for the brown ring test for nitrate ions.

$${\text{NO}}_3^ - {\text{ + 3F}}{{\text{e}}^{2 + }}{\text{ + 4 }}{{\text{H}}^ + }{\text{ }} \to {\text{ NO + 3F}}{{\text{e}}^{3 + }}{\text{ + 2 }}{{\text{H}}_2}{\text{O}} \\\ {\left[ {{\text{Fe}}{{\left( {{{\text{H}}_2}{\text{O}}} \right)}_6}} \right]^{2 + }}{\text{ + NO }} \to {\text{ }}\left[ {{\text{Fe}}{{\left( {{{\text{H}}_2}{\text{O}}} \right)}_5}{\text{NO}}} \right]{\text{S}}{{\text{O}}_4}{\text{ + }}{{\text{H}}_2}{\text{O}} \\\ {\text{ brown ring}} \\\$$

In the first step, ferrous ions react with nitrate ions in presence of protons to form ferric ions and nitric oxide. In the second step, nitric oxide reacts with hydrated ferrous ions to form a brown ring complex. In the first step, the oxidation state of nitrogen decreases by $3$ . In the second step for the reaction, there is no change in the oxidation state of ferrous ions. Thus, the brown ring test for nitrate ion confirms the presence of nitrate ion.

Hence, the option (C ) nitrate ion is the correct answer.

Note: Oxidation number is the total number of electrons gained or lost by an atom in a molecule, to form a chemical bond. In a neutral molecule, the sum of the oxidation states of all the elements/ions is zero. The oxidation states of water, nitrosonium ion and sulphate ion are $0,{\text{ }} + {\text{1 and }} - {\text{2}}$ respectively. The oxidation number of ${\text{Fe}}$ in the brown ring complex $\left[ {{\text{Fe}}{{\left( {{{\text{H}}_{\text{2}}}{\text{O}}} \right)}_{\text{5}}}{\text{N}}{{\text{O}}^{\text{ + }}}} \right]{\text{S}}{{\text{O}}_{\text{4}}}$ is ${\text{ + 1}}$.