Question

Correct formula of the complex formed in the brown ring test for nitrates is-
[A] $FeS{{O}_{4}}NO$
[B] $Fe{{[{{({{H}_{2}}O)}_{5}}NO]}^{2+}}$
[C] $Fe{{[{{({{H}_{2}}O)}_{5}}NO]}^{+}}$
[D] $Fe{{[{{({{H}_{2}}O)}_{5}}NO]}^{3+}}$

Answer 1

Hint: The ring test gives us a nitrosyl complex which is positively charged. A freshly prepared ferrous sulphate solution in presence of mixed acid gives this test as a detection of nitrate ions. It undergoes a redox reaction and forms a nitrosyl complex in the end.

Complete step by step solution:
As we know the brown ring is a common nitrate test that is used for a detection of nitrate ions as a brown ring is formed at the junction of the two layers I.e. of the acid and the sulphate solution, which indicates the presence of the nitrate ions.
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-
$2HN{{O}_{3}}+3{{H}_{2}}S{{O}_{4}}+6FeS{{O}_{4}}\to 3F{{e}_{2}}{{(S{{O}_{4}})}_{3}}+2NO+4{{H}_{2}}O$
$[Fe{{({{H}_{2}}O)}_{6}}]S{{O}_{4}}+NO\to [Fe{{({{H}_{2}}O)}_{5}}NO]S{{O}_{4}}+{{H}_{2}}O$
This is a redox reaction where the nitrate ion is reduced to nitric oxide by oxidation of iron (II) to iron(III), which is shown in the first part of the above reaction, followed by the formation of a nitrosyl complex by the remaining iron(II) and nitric oxide, which is shown in the second part of the reaction.
As we can see from the above reactions that the complex formed is $Fe[{{({{H}_{2}}O)}_{5}}NO]S{{O}_{4}}$
As we know, the charge on a sulphate ion is (2- ) therefore, if we replace the sulphate ion, the complex will gain a positive charge of two. Therefore, the correct answer will be the complex with a charge of (+2).
Therefore, option [B] $Fe{{[{{({{H}_{2}}O)}_{5}}NO]}^{2+}}$ is the correct answer.

Additional information:
The ring test is a test to detect the presence of nitrate in the given solution. Mixed acids i.e. Sulphuric acid and Nitric acid are used here. When the acid mixture is poured down in the test tube which contains ferrous sulphate solution such that the acid forms a layer beneath it, the nitrate present in the solution forms a brown ring hence the name brown ring test.

Note: For the reaction, it is important here to remember that it gives a positively charged complex. In the final complex, the charge on Fe is +1 and not +3.