In the ‘ring test’ of $N{{O}_{3}}^{-}$ion, there is formation of a brown colour ring. What is its formula?
${{[Fe{{({{H}_{2}}O)}_{5}}N{{O}_{3}}]}^{2+}}$
${{[Fe{{({{H}_{2}}O)}_{4}}NOS{{O}_{4}}]}^{2+}}$
${{[Fe{{({{H}_{2}}O)}_{5}}NO]}^{2+}}$
None of these

Hint: $N{{O}_{3}}^{-}$is the chemical formula of the nitrate ion.
The brown colour ring is formed in the identification test of nitrate ion in the ‘ring test’. Reagents used are a nitrating mixture (concentrated ${{H}_{2}}S{{O}_{4}}$(sulphuric acid) + concentrated $HN{{O}_{3}}$) and $FeS{{O}_{4}}$(ferrous sulphate or iron (II) sulphate). A solution containing nitrate ion is added to that and hence the brown ring is formed.
Iron exists in (+3) oxidation state and nitrosyl exists in (-1) oxidation state in the complex.

Complete step by step answer:
It is the confirmation test of the nitrate ion.
In a test tube, a solution containing $N{{O}_{3}}^{-}$ion is taken as a sample such as $KN{{O}_{3}}$.
Then, concentrated ${{H}_{2}}S{{O}_{4}}$is added dropwise by keeping the test tube tilted. It is heated in a Bunsen burner and after that freshly prepared greenish $FeS{{O}_{4}}$is added,
Due to the interfering presence of nitrate ion, a brown ring is formed at the separation layer of two compounds which testifies the presence of the ion.
The reactions involved are:
\[KN{{O}_{3}}\text{ }+\text{ }{{H}_{2}}S{{O}_{4}}\text{ }\to \text{ }KHS{{O}_{4}}\text{ }+\text{ }HN{{O}_{3}}\text{ }\]
\[6FeS{{O}_{4}}\text{ + 3}{{H}_{2}}S{{O}_{4}}\text{ + }HN{{O}_{3}}\to \text{ 3F}{{\text{e}}_{2}}{{(S{{O}_{4}})}_{3}}\text{ + 4}{{H}_{2}}O\text{ + }2NO\]
\[FeS{{O}_{4}}\text{ + }NO\text{ + 5}{{H}_{2}}O\to \text{ }\!\![\!\!\text{ Fe(}{{H}_{2}}O{{\text{)}}_{5}}\text{(NO) }\!\!]\!\!\text{ }S{{O}_{4}}\text{ + }{{H}_{2}}O\]
                                          (brown ring complex)
The brown ring formula is ${{\text{ }\!\![\!\!\text{ Fe(}{{H}_{2}}O{{\text{)}}_{5}}\text{(NO) }\!\!]\!\!\text{ }}^{2+}}$.
So, the correct option is C.

Additional information:
The complex shows anti-ferromagnetism as found from Mossbauer spectroscopy.
Name of the complex – iron(III) pentaaquanitrosyl sulphate

Note: In the second step, it is assumed that oxidation is carried by nascent oxygen [O] which is produced during the reaction. And in the final step, $FeS{{O}_{4}}$exists in aqueous solution as $\text{ }\!\![\!\!\text{ Fe(}{{H}_{2}}O{{\text{)}}_{6}}\text{ }\!\!]\!\!\text{ }S{{O}_{4}}$which further reacts with $NO$ to form the complex. Concentrated ${{H}_{2}}S{{O}_{4}}$must be added dropwise by keeping the test tube tilted or else the reaction will be vigorous and the test tube will blast.