To the mixture of aqueous solution of a Nitrate salt and Ferrous sulphate, con. $$\text{H}_2\text{SO}_4$$ is slowly added through the slanted walls of the test tube. A brown ring is formed little above (nearly middle) the bottom of the test tube.<br>a) What is the significance of this test?<br>b) Which acts as reductant?<br>c) What is the change of oxidation state of oxidant?<br>d) What is the chemical composition of the Brown's ring?<br>e) Why con. $$\text{H}_2\text{SO}_4$$ is added only through the slanted walls of the test tube?<br>f) Write the related chemical equations for above chemical changes.

PLACEHOLDER ANSWER - the question has six written parts. a) It is the brown ring test, used to detect the nitrate ion. b) Ferrous sulphate ($$\text{Fe}^{2+}$$) is the reductant. c) The oxidant, nitrate, is reduced from N in $$+5$$ state to NO with N in $$+2$$ state, a change of 3. d) The ring is the complex $$[\text{Fe(H}_2\text{O)}_5\text{NO}]\text{SO}_4$$, i.e. $$[\text{Fe(H}_2\text{O)}_5\text{NO}]^{2+}$$. e) The dense acid runs down the wall and forms a separate layer at the bottom without mixing, so the ring forms at the junction of the layers; slow addition also avoids heat that would decompose the complex. f) $$\text{NO}_3^- + 3\text{Fe}^{2+} + 4\text{H}^+ \rightarrow \text{NO} + 3\text{Fe}^{3+} + 2\text{H}_2\text{O}$$, then $$[\text{Fe(H}_2\text{O)}_6]^{2+} + \text{NO} \rightarrow [\text{Fe(H}_2\text{O)}_5\text{NO}]^{2+} + \text{H}_2\text{O}$$.