Question

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

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

Answer 1

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

Option (C) is correct.

The brown ring test for $N{O}_2^{-}$ and $N{O}_3^{-}$ is due to the formation of complex,$\left[Fe\right(H_{2}O{)}_{5}NO{]}^{2+}$.

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 ($FeS{O}_4.NO$). Nitric Acid (Nitrate) reduces Ferrous to Ferric and itself is reduced to nitrogen monoxide ($NO$). This reacts with $FeS{O}_4$ and forms the brown ring.