The acidic aqueous solution of ferrous ion forms a brown complex in the presence of $${\text{NO}}_3^{-}$$, by the following two steps:

$3{\left(\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right)}^{2+}+{NO}_3^{-}\mathrm{ }+\mathrm{ }4{\mathrm{H}}^{+}\mathrm{ }\to A+ 3{\left(\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right)}^{3+}\mathrm{ }+ 2{\mathrm{H}}_2\mathrm{O}$

and

${\left(\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right)}^{2+}+ A\to \mathrm{ }B+ {\mathrm{H}}_2\mathrm{O}$

A and B are-

The given reaction is performed to test whether nitrate ions are present in the solution or not and this test is known as the ring test.
In the ring test, a complex of ferrous (II) sulphate was added to a solution containing nitrate ions.
Along with ferrous (II) sulphate, concentrated sulfuric acid is also added slowly to the solution because it produces Hexaaquairon (II) ions and nitric oxide.
A balanced response will be:

$3(\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{{)}_2}^{+}+ {{\mathrm{NO}}_3}^{-} + 4{\mathrm{H}}^{+}\to \mathrm{NO} + 3(\mathrm{Fe}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6{{)}_3}^{+} + 2{\mathrm{H}}_2\mathrm{O}$

Now, in the second reaction, nitric oxide will react with the iron complex. Also, we can see that one water molecule is released so in the ligand one molecule will drop.
- Therefore, the final product will have one ligand of nitric oxide so the reaction will be:

$(\mathrm{Fe}{\left(\mathrm{H}2\mathrm{O}\right)}_6{)}^{2+}+ \mathrm{NO} \to (\mathrm{Fe}{\left(\mathrm{H}2\mathrm{O}\right)}_5\mathrm{NO}{)}^{2+} + {\mathrm{H}}_2\mathrm{O}$

Hence, option A is correct.