Question

In the brown ring test for the nitrate ion, the brown colour is due to:

• A. ${\left[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}\right]}^{2+}$
• B. ${\left[{\text{Fe(H}}_2{\text{O)}}_5{\text{NO}}_2\right]}^{2+}$
• C. ${\left[{\text{Fe(H}}_2{\text{O)}}_2{\text{NO}}_2\right]}^{+}$
• D. ${\left[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}\right]}^{3+}$

Answer 1

The correct option is A ${\left[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}\right]}^{2+}$

When an aqueous solution of the nitrate salt is treated with freshly prepared solution of ferrous sulphate and conc. ${\text{H}}_2{\text{SO}}_4$, a brown ring is formed on account of the formation of complex $\left[{\text{Fe(H}}_2{\text{O)}}_5\text{NO}\right]{\text{SO}}_4$ at the interface of the two liquids.

${\text{NaNO}}_3+{\text{H}}_2{\text{SO}}_4\to {\text{NaHSO}}_4+{\text{HNO}}_3$

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

$\left[{\text{Fe(H}}_2{\text{O)}}_6\right]{\text{SO}}_4{\text{.H}}_2\text{O}+\text{NO}\to \left[\text{Fe (}\underset{\text{Brown ring}}{{\text{H}}_2{\text{O)}}_5\text{NO}}\right]{\text{SO}}_4+2{\text{H}}_2\text{O}$