Question

The brown ring for $N{O}_2^{-}$ and $N{O}_3^{-}$ is due to the formation of complex ion with formula :

• A. ${\left[Fe{\left(H_{2}O\right)}_6\right]}^{2+}$
• B. ${\left[Fe\left(NO\right){\left(CN\right)}_5\right]}^{2-}$
• C. ${\left[Fe{\left(H_{2}O\right)}_{5}NO\right]}^{2+}$
• D. ${\left[Fe{\left(H_{2}O\right)}_5\right]}^{2+}$

Answer 1

The correct option is C ${\left[Fe{\left(H_{2}O\right)}_{5}NO\right]}^{2+}$

The brown ring test for $N{O}_2$and $N{O}_3$ is due to the formation of complex, ${\left[Fe{\left(H_{2}O\right)}_{5}NO\right]}^{2+}$
Brown ring 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 form at the juction of the two liquids. This is nitroso ferrous sulphate $(FeS{O}_4.NO)$. Nitric acid (Nitrate) reduces ferrous to ferric and itself gets reduced to nitrogen monoxide (NO). This reacts with $FeS{O}_4$ and forms the brown ring.