Question

The composition in brown ring test is:
(A) $[Fe{({H_2}O)_6}NO]S{O_4}$
(B) $[Fe{({H_2}O)_5}NO]S{O_4}$
(C) $[Fe{({H_2}O)_4}{(NO)_2}]S{O_4}$
(D) $[Fe{({H_2}O)_3}NO]S{O_4}$

Answer 1

Hint : A brown ring is formed by adding drops of concentrated tetraoxosulphate(VI) acid during brown ring test which is sandwiched between acidified trioxonitrate(V) $+$ Iron(II) tetraoxosulphate(VI) and concentrated tetraoxosulphate(VI) acid.

Complete step by step answer
First let us know why and how the brown ring test is conducted. Brown ring test is conducted to find out the presence of nitrate in any solution. Nitrate anion is an oxidizer. Determining nitrate anion through wet chemistry is not possible.
In the brown ring test, first the solution of nitrate is taken. Then iron sulphate is added in this solution where iron is in $+ 2$ oxidation state. Then to this mixture, sulphuric acid is added slowly so that the acid forms a layer below the aqueous solution. After some time a brown ring is formed sandwiched between two other layers. This brown ring indicates the presence of nitrate ions.
Now let us see the brown ring test chemically,
Reduction of nitrate anion, reduction of nitric oxide and oxidation of iron(II) takes place,
$2HN{O_3} + 3{H_2}S{O_4} + 6FeS{O_4} \to 3F{e_2}{(S{O_4})_3} + 2NO + 4{H_2}O$
$[Fe{({H_2}O)_6}]S{O_4} + NO \to [Fe{({H_2}O)_5}NO]S{O_4} + {H_2}O$
The oxidation state of iron in brown ring test is $+ 1$ . The complex formed in the last step is a brown ring complex and its name is pentaaqua nitrosyl iron(I) sulphate.
So option B is the correct answer.

Note
Brown ring test is used to detect presence of nitrate in many places such as soil, water and food samples. The reaction in the brown ring test is exothermic. Only freshly prepared ferrous sulphate is added in brown ring test because if it is old or stored then ferrous sulphate can react with oxygen in the atmosphere and oxidized to ferric sulphate forming a brown yellow corrosive coating.