Answer
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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.
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.
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