Answer
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Hint: Redox reaction takes place here. Fe is reduced and NO is oxidised. Fe is in +1 state in the reactant and NO is neutral. The product is formed due to the charge transfer in $F{{e}^{+}}$ ion and NO.
Step by step solution: In the ring test, for nitrate ions when the freshly prepared ferrous sulphate is added to an aqueous solution of nitrate and sulphuric acid, the following reaction takes place-
$HN{{O}_{3}}+FeS{{O}_{4}}\to NO+F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+H_2O$
But the complex ${{[Fe{{({{H}_{2}}O)}_{3}}NO]}^{+}}$ is formed due to the following charge transfer,
$F{{e}^{2+}}+{{e}^{\_}}\to F{{e}^{+}}$
\[\]\[NO\to N{{O}^{+}}+{{e}^{-}}\]
It is a redox reaction. $F{{e}^{2+}}$ is reduced to $F{{e}^{+}}$and NO is oxidised to $N{{O}^{+}}$. Fe is considered to be in +1 oxidation state here as observed magnetic moment suggests of 3 unpaired electrons.
However, this is not a complex formed during the ring test.
${{[Fe{{({{H}_{2}}O)}_{5}}NO]}^{2+}}$ is the complex formed in the ring test for nitrate due to the charge transfer of
\[F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{\_}}\]
\[NO\to N{{O}^{+}}+{{e}^{-}}\]
High spin Fe (III) (S=5$\div $2) antiferromagnetically coupled with N (S=1) for an observed spin of S=3$\div $2.
Therefore, [C] is the correct answer.
Additional Information: The ring test is the test to detect the presence of nitrate in the given solution. Mixed acids i.e., Sulphuric acid and Nitric acid are used here. The nitrate present in the solution forms a brown ring hence the name ring test.
Note: Here it is important to note that some books say Fe is in +3 oxidation state whereas some say it is in +1 oxidation state. However, It is widely accepted that Fe in +1 oxidation state.
Step by step solution: In the ring test, for nitrate ions when the freshly prepared ferrous sulphate is added to an aqueous solution of nitrate and sulphuric acid, the following reaction takes place-
$HN{{O}_{3}}+FeS{{O}_{4}}\to NO+F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+H_2O$
But the complex ${{[Fe{{({{H}_{2}}O)}_{3}}NO]}^{+}}$ is formed due to the following charge transfer,
$F{{e}^{2+}}+{{e}^{\_}}\to F{{e}^{+}}$
\[\]\[NO\to N{{O}^{+}}+{{e}^{-}}\]
It is a redox reaction. $F{{e}^{2+}}$ is reduced to $F{{e}^{+}}$and NO is oxidised to $N{{O}^{+}}$. Fe is considered to be in +1 oxidation state here as observed magnetic moment suggests of 3 unpaired electrons.
However, this is not a complex formed during the ring test.
${{[Fe{{({{H}_{2}}O)}_{5}}NO]}^{2+}}$ is the complex formed in the ring test for nitrate due to the charge transfer of
\[F{{e}^{2+}}\to F{{e}^{3+}}+{{e}^{\_}}\]
\[NO\to N{{O}^{+}}+{{e}^{-}}\]
High spin Fe (III) (S=5$\div $2) antiferromagnetically coupled with N (S=1) for an observed spin of S=3$\div $2.
Therefore, [C] is the correct answer.
Additional Information: The ring test is the test to detect the presence of nitrate in the given solution. Mixed acids i.e., Sulphuric acid and Nitric acid are used here. The nitrate present in the solution forms a brown ring hence the name ring test.
Note: Here it is important to note that some books say Fe is in +3 oxidation state whereas some say it is in +1 oxidation state. However, It is widely accepted that Fe in +1 oxidation state.
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