Question

The brown ring test for $NO_2^ - $ and $NO_3^ - $ is due to the formation of complex ion with formula:
A. ${[Fe{({H_2}O)_6}]^{2 + }}$
B. ${[Fe(NO){(CN)_5}]^{2 - }}$
C. ${[Fe{({H_2}O)_5}NO]^{2 + }}$
D. ${[Fe({H_2}O){(NO)_5}]^{2 + }}$

Answer 1

Hint:The brown ring test is the most common nitrate test. In this test solution is treated with iron sulphate and concentrated sulphuric acid. The presence of nitrate ion is detected by the brown ring forms below the aqueous solution and between two layers.

Complete answer:
To analyse the presence of nitrate ions in a solution various methods are used, brown ring test is one of the nitrate tests. Brown ring test is the most common nitrate test which is used for the analysis of nitrate ions.
In the brown ring test, the solution is reacted with iron(ll) sulphate and then concentrated sulphuric acid is added to that solution very slowly. This reaction forms a complex ion, which has the formula ${[Fe{({H_2}O)_5}NO]^{2 + }}$ . This complex ion forms a brown ring around the test tube. This brown ring formed below the aqueous layer and between the two layers. The formation of this brown ring indicates the presence of nitrate ions in solution.
The reaction takes place as:
\[\
  2HN{O_3}\; + {\text{ }}3{H_2}S{O_4}\; + {\text{ }}6FeS{O_4}\; \to {\text{ }}3F{e_2}{\left( {S{O_4}} \right)_3}\; + {\text{ }}2NO{\text{ }} + {\text{ }}4{H_2}O \\
  \left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]S{O_4}\; + {\text{ }}NO{\text{ }} \to \mathop {[Fe{{({H_2}O)}_5}NO]S{O_4}}\limits_{BrownRing} \; + {\text{ }}{H_2}O \\
\ \]
In the reaction, nitrate ions get reduced to nitric oxide by iron(ll). And iron(ll) get oxidized to iron(lll) and nitrosyl complex that is ${[Fe{({H_2}O)_5}NO]^{2 + }}$ get formed between nitric oxide and the remaining iron(ll). And nitric oxide is reduced to $N{O^ - }$ . This nitrosyl compound is the reason for brown ring formation.

Hence, option C is correct.

Note:
Diphenylamine test is also used to test the presence of the nitrate ion. In this test, a solution of diphenylamine and ammonium chloride in sulfuric acid is used. Diphenylamine gets oxidized in the presence of nitrate ions and gives a blue color to the solution.