Question

A brown ring is formed in the ring test for $\text{NO}_{3}^{-}$ ion. It is due to the formation of: A ) $\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}\text{ }$B ) $\text{FeS}{{\text{O}}_{4}}\text{N}{{\text{O}}_{2}}$C ) ${{\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{4}}{{\left( \text{NO} \right)}_{2}} \right]}^{2+}}$D ) $\text{FeS}{{\text{O}}_{4}}\text{HN}{{\text{O}}_{3}}$

Hint: The brown ring test for nitrate ion confirms the presence of nitrate ion. A freshly prepared ferrous sulphate solution is added to the solution containing nitrate ions. A ring containing a brown coloured complex is formed.

The brown ring test for nitrate ion is a qualitative test that confirms the presence of nitrate ion. In the ring test for $\text{NO}_{3}^{-}$ ion, a freshly prepared ferrous sulphate solution is added to the solution containing nitrate ion. A ring containing a brown coloured complex is formed. The formation of this brown coloured complex confirms the ring test for the presence of nitrate ion. The brown coloured complex is $\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}$.
${\text{NO}_{3}^{-}\text{ + 3F}{{\text{e}}^{2+}}\text{ + 4 }{{\text{H}}^{+}}\text{ }\to \text{ NO + 3F}{{\text{e}}^{3+}}\text{ +2 }{{\text{H}}_{2}}\text{O}}$
${{{\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{6}} \right]}^{2+}}\text{ + NO }\to \text{ }\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}\text{ + }{{\text{H}}_{2}}\text{O}}$
Hence, the option A ) $\left[ \text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO} \right]\text{S}{{\text{O}}_{4}}$ is the correct option.