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Question: In the ring test for \(N{{O}_{3}}^{-}\)removal of \(B{{r}^{-}}\) and \({{I}^{-}}\) is done by adding...

In the ring test for NO3−N{{O}_{3}}^{-}removal of Br−B{{r}^{-}} and I−{{I}^{-}} is done by adding AgNO3AgN{{O}_{3}} solution
A) True
B) False

Explanation

Solution

The molecular formulae of the brown ring formed is [Fe(H2O)5NO]SO4\left[ Fe{{\left( {{H}_{2}}O \right)}_{5}}NO \right]S{{O}_{4}}
The complex formation is through the combining of ferrous ions and nitric oxide.

Complete step by step answer:
We have state that whether the given statement i.e., the removal of Br−B{{r}^{-}} and I−{{I}^{-}} is done with the usage of AgNO3AgN{{O}_{3}} solution.
The brown ring test is done for the detection of nitrate ion (NO3−N{{O}_{3}}^{-}) in the test solution.
The nitrate test is done by adding freshly prepared saturated solution of iron (II) Sulphate with the nitrate solution. Then conc.H2SO4{{H}_{2}}S{{O}_{4}} Is added drop wise through the side of the test tube so that the acid will form a layer above the aqueous layer. And a brown ring will start appearing within a few seconds of the addition of acid in between the acid and aqueous layer, indicating that the nitrate ion is present in the test solution.
In total, brown ring reaction is a reduction reaction of nitrate ion to nitric oxide by the ferrous ion present in the solution which gets oxidized to the ferric ions. And this is followed by the formation of nitrosyl complex, which is formed between the nitric oxide and the ferrous ions present in the reaction mixture. And the nitric oxide in the complex is again getting reduced to NO−N{{O}^{-}} ion.
The reactions involved in the brown ring formation,
2NO3−+4H2SO4+6Fe2+→6Fe3++2NO↑+4SO42−+4H2O2N{{O}_{3}}^{-}+4{{H}_{2}}S{{O}_{4}}+6F{{e}^{2+}}\to 6F{{e}^{3+}}+2NO\uparrow +4S{{O}_{4}}^{2-}+4{{H}_{2}}O
FeSO4+6H2O→[Fe(H2O)6]SO4FeS{{O}_{4}}+6{{H}_{2}}O\to \left[ Fe{{\left( {{H}_{2}}O \right)}_{6}} \right]S{{O}_{4}}
[Fe(H2O)6]SO4+NO↑→[Fe(H2O)NO]SO4+H2O\left[ Fe{{\left( {{H}_{2}}O \right)}_{6}} \right]S{{O}_{4}}+NO\uparrow \to \left[ Fe\left( {{H}_{2}}O \right)NO \right]S{{O}_{4}}+{{H}_{2}}O
In this reaction Br−B{{r}^{-}} and I−{{I}^{-}} interfere due to the liberation of halogens hence these anions are separated using AgSO4AgS{{O}_{4}} which is nitrate free and the insoluble salts of Ag formed is removed by filtration.

So the above given statement is false and the correct option for the question is option (B).

Note: The bromide and iodide ions are removed from the test solution by the adding of dil. HNO3HN{{O}_{3}} along with a few drops ofAgNO3AgN{{O}_{3}}.
The bromide ion precipitates as AgBr, which will be a cream precipitate
The iodide ion precipitates as AgI, which will be a yellow precipitate