Question

Brown ring in the test of $N{O}_3^{-}$ is formed due to the formation of :

• A. $FeS{O}_4\cdot NO$
• B. $[Fe{\left(S{O}_4\right)}_2\cdot HO]\cdot H_{2}O$
• C. $F{e}_2{\left(S{O}_4\right)}_3\cdot NO$
• D. none of these

Answer 1

Answer: (A)

$FeS{O}_4\cdot NO$

Option (A) is correct.
Brown ring in the test of $N{O}_3^{-}$ is formed due to the formation of $FeS{O}_4\cdot NO$.
Brown ring test is the test to find out the presence of nitrate radical. For this equal volume of the freshly prepared $FeS{O}_4$ is mixed with solution to be analysed. To this concentrated $H_{2}S{O}_4$ is added along the sides of the test tube. A brown ring will forms at the junction of the two liquids. This is Nitroso ferrous sulphate $(FeS{O}_4.NO)$