Question

If$$FeS{O}_{\mathbf{4}}$$ is used in brown ring tests for nitrates and nitrites. In this test, a freshly prepared $$FeS{O}_{\mathbf{4}}$$​ solution is mixed with a solution containing $N{O}_2^{-}$ and $N{O}_3^{-}$ and the conc. $$H_{\mathbf{2}}S{O}_{\mathbf{4}}$$ ​ is run down the side of the test tube. If the mixture gets hot or is shaken,
(I) the brown colour disappear (II) $$NO$$ is evolved (III) a yellow solution of $$F{e}_{\mathbf{2}}{\left(S{O}_{\mathbf{4}}\right)}_{\mathbf{3}}$$ appears​
A. I, II, III correct
B. I, III correct
C. II, III correct
D. only I correct

Answer 1

Hint: We need to know that the Brown Ring Test is the confirmatory test for nitrate anions, although nitrite ions does interfere with the test. We have to remember that only the elements present in d-block give colour to compounds due to presence of energy difference of d-orbitals while in case of organic compound due to conjugation which appears as coloured substance.

Complete step by step answer:
We have to remember that a brown ring test also known as the Nitrate test, iron (II) sulphate is added to a solution containing nitrate ions. To this solution, concentrated sulphuric acid is added slowly through the sides of the test tube such that the sulphuric acid forms a layer above the aqueous solution. A brown ring forms at the junction of the two layers indicating the presence of nitrate ions. The overall reaction is the reduction of nitrate ions to nitric oxide by iron (II) which is oxidised to iron (III) followed by the formation of nitrosyl complex which forms the brown ring. Let us understand the reaction by taking the example of nitric acid.
$$2HN{O}_3+3H_{2}S{O}_4+6FeS{O}_4\to 3F{e}_2(S{O}_4{)}_3+NO+4H_{2}O$$(Reduction of $$\mathbf{NO}{3}^{-}$$ )
$$[Fe(H_{2}O{)}_6]S{O}_4+NO\to [Fe(H_{2}O{)}_5\left(NO\right)]S{O}_4+H_{2}O$$ (Brown ring formation)
From the understanding of the above reactions, it is clear that option (II) is correct, i.e $$NO$$ is evolved. In addition, when the mixture containing the brown ring is heated or shaken vigorously, the brown nitrosyl complex breaks down and forms a yellow colored precipitate of $$F{e}_{\mathbf{2}}{\left(S{O}_{\mathbf{4}}\right)}_{\mathbf{3}}$$. Hence option (I) and (III) are also correct.

So, the correct answer is Option A.

Note: It must be noted that the brown ring test is the confirmatory test for nitrate ions only although nitrate ions may interfere with the reaction but does not really confirms its presence. Adding sulphuric acid slowly to the solution containing nitrate and nitrite ions is the crucial part of the experiment to obtain the brown ring. As concentrated sulphuric acid is used in this experiment, it is likely for the reaction mixture to get heated up if not added slowly which will lead to the immediate degradation of the brown ring complex.