$\text{FeS}{{\text{O}}_{4}}$ solution gives a brown coloured ring during the test for nitrates or nitrites. This due to the formation of:
A. ${{[\text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO }\!\!]\!\!\text{ }}^{+2}}$
B. ${{[\text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{N}{{\text{O}}_{2}}\text{ }\!\!]\!\!\text{ }}^{+2}}$
C. ${{[\text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{4}}\text{N}{{\text{O}}_{2}}\text{ }\!\!]\!\!\text{ }}^{+2}}$
D. ${{[\text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{4}}\text{NO }\!\!]\!\!\text{ }}^{+2}}$

Hint:The ring test is a wet chemistry method that is used in the laboratory to check the presence of a nitrate ion. It undergoes a wet test because of its solubility in the water. It is also known as the brown ring test. The reactants used in this test are nitric acid, sulphuric acid and ferrous sulphate.

Complete Step-by-Step Answer:
-To test the presence of nitrate ion we have to take any unknown solution in the test tube and we have to add iron (II) sulphate.
-Now, add sulfuric acid slowly you will notice the formation of a brown ring at the bottom of the test tube.
-In this process, the nitrate ion will be reduced to the nitric oxide by iron (II).
-The reaction of the above process is:
\[\begin{align}
  & \text{2HN}{{\text{O}}_{3}}\text{ + 3}{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\text{ + 6FeS}{{\text{O}}_{4}}\text{ }\to \text{ 3F}{{\text{e}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}+\text{ 2NO + 4}{{\text{H}}_{2}}\text{O} \\
 & \\
 & \text{ }\!\![\!\!\text{ Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{6}}]\text{S}{{\text{O}}_{4}}\text{ + NO }\to \text{ }\!\![\!\!\text{ Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO }\!\!]\!\!\text{ S}{{\text{O}}_{4}}+\text{ }{{\text{H}}_{2}}\text{O} \\
\end{align}\]
-Through the above reaction, we can see that end product formed is \[\text{ }\!\![\!\!\text{ Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO }\!\!]\!\!\text{ S}{{\text{O}}_{4}}\].
-So, ${{[\text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{NO }\!\!]\!\!\text{ }}^{+2}}$ is responsible for the formation of the brown coloured ring at the bottom of the test tube.
-The formation of ${{[\text{Fe}{{\left( {{\text{H}}_{2}}\text{O} \right)}_{5}}\text{N}{{\text{O}}_{2}}\text{ }\!\!]\!\!\text{ }}^{+2}}$ is not possible because after the reduction of the nitrate ion there is a loss of an electron which gained by the iron (II) and converts into iron (III).
-So, option B. and C. are an incorrect answer.
 -From the above reaction, it is proved that a total of 5 molecules of water combines with the iron along with a single molecule of nitric oxide.
-So, option D is an incorrect answer.

Therefore, option A is the correct answer.

Note: The sensitivity of the ring test is 2.5 micrograms. It means the minimum quantity of nitric acid that can be detected through this test is about 2.5 micrograms. The complex which is responsible for the formation of a brown ring is commonly known as nitrosyl ion or nitroso ferrous sulphate.