Question

In the separation of $Cu^{2+}$ and $Cd^{2+}$ of Ilnd group in qualitative analysis of cations, tetrammine copper (II) sulphate and tetrammine cadmium (II) sulphate react with KCN to form the corresponding cyano complexes, which one of the following pairs of the complexes and their relative stability enables the separation of $Cu^{2+} $ and $Cd^{2+}$?

• A. $K_3[Cu(CN)_4] $: less stable and $K_2[Cd(CN)_4] $: more stable
• B. $K_3[Cu(CN)_4] $: more stable and $K_2[Cd(CN)_4] $: less stable
• C. $K_2[Cu(CN)_4] $: less stable and $K_2[Cd(CN)_4] $: more stable
• D. $K_2[Cu(CN)_4] $: more stable and $K_2[Cd(CN)_4] $: less stable

Answer 1

The Correct Option is B

To determine which complexes and their relative stability allow for the separation of $Cu^{2+}$ and $Cd^{2+}$ ions in the second group of qualitative analysis, we need to understand the chemistry behind cyanide complexes.

Step 1: Formation of Cyanide Complexes

Both $Cu^{2+}$ and $Cd^{2+}$ can form complexes with cyanide ions. The formation of these complexes can be represented by the following reactions:

• $Cu^{2+} + 4CN^- \rightarrow [Cu(CN)_4]^{3-}$, forming $K_3[Cu(CN)_4]$
• $Cd^{2+} + 4CN^- \rightarrow [Cd(CN)_4]^{2-}$, forming $K_2[Cd(CN)_4]$

Step 2: Evaluating Stability of Complexes

The stability of these complexes is influenced by the metal's ability to form strong bonds with the cyanide ion. Copper generally forms more stable complexes than cadmium due to its higher charge density and preference for forming complexes with ligands such as cyanide.

Therefore, the complex $K_3[Cu(CN)_4]$ is more stable than $K_2[Cd(CN)_4]$.

Conclusion

Given the relative stabilities of the cyanide complexes, $K_3[Cu(CN)_4]$ is more stable, and $K_2[Cd(CN)_4]$ is less stable. This difference in stability enables the selective separation of $Cu^{2+}$ and $Cd^{2+}$ ions during qualitative analysis.

Therefore, the correct option is: $K_3[Cu(CN)_4]$: more stable and $K_2[Cd(CN)_4]$: less stable.