JEE Main 2022 — Practical Chemistry Question with Solution

Brown ring test is usually carried out by adding dilute ferrous sulphate solution to an aqueous solution containing nitrate ion, and then carefully adding concentrated sulphuric acid along the sides of the test tube. A brown ring at the interface between the solution and sulphuric acid layers indicates the presence of nitrate ion in solution.

${\mathrm{NO}}_3^{-}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{HNO}}_3+{\mathrm{HSO}}_4^{-}$

$6{\mathrm{FeSO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4+2{\mathrm{HNO}}_3\to 3{\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3+2\mathrm{NO}+4{\mathrm{H}}_2\mathrm{O}$

$\mathrm{NO}+{\mathrm{FeSO}}_4+5{\mathrm{H}}_2\mathrm{O}\to \left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{NO}\right]{\mathrm{SO}}_4$                                                    (Brown ring)

Brown ring complex formed is $\left[\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{NO}\right]{\mathrm{SO}}_4$

IUPAC name of the above complex is  PentaaquanitrosylIron(I ) Sulphate.

Nitrate ions reacts with sulphuric acid to form nitrogen dioxide which is responsible for brown fumes.