Question

Three cations $A,B,C$ are allowed to react with ammonium hydroxide solution. $A$ and $B$ produce a dirty green precipitate and a reddish brown precipitate respectively, both of which do not dissolve on further addition of ${NH}_{4}OH$. $C$ gives pale blue colour precipitate with less amount of ammonium hydroxide and on further addition of ${NH}_{4}OH$, precipitate becomes soluble and deep blue colour solution is formed:

• A. ${Fe}^{+2},{Fe}^{+3},{Cu}^{+2}$
• B. ${Fe}^{+3},{Fe}^{+2},{Cu}^{+2}$
• C. ${Cu}^{+2},{Fe}^{+2},{Fe}^{+3}$
• D. ${Cu}^{+2},{Fe}^{+3},{Fe}^{+2}$

Answer 1

The correct option is A ${Fe}^{+2},{Fe}^{+3},{Cu}^{+2}$

The answer is option $\left(A\right)$. Cation $A$ is $F{e}^{2+}$ as it gives green coloured precipitate of ferrous hydroxide when reacted with $N{H}_{4}OH$ . When $N{H}_{4}OH$ is added to a solution of $F{e}^{3+}$ions it results in the formation of ferric hydroxide which is reddish brown in colour.Hence cation $B$ is $F{e}^{3+}$. When $N{H}_{4}OH$ is added to a solution of copper ions the copper ions will react with the few $O{H}^{-}$ to make insoluble $Cu(OH{)}_2$. The solid exists in a suspension. As more $N{H}_{4}OH$ is added the insoluble $Cu(OH{)}_2$ will dissolve as ${\left[{Cu\left(N{H}_3\right)}_4\right]}^{2+}$ is formed. The copper-ammonia complex ion is a deep purple, and it will be difficult to see through the solution, but there won't be any solid precipitate remaining. Thus the cation $C$ is $C{u}^{2+}$