If `H_(2)S` gas is passed into a solution of `Cu^(2+), Cd^(2+)` having excess of KCN

Options

A. CuS and Cds both are precipitated.
B. Soluble complex `[Cu(CN_(4)]^(3-)` and `[Cd(CN)_(4)]^(2-)` are formed and no effect of passing `H_(2)S` gas
C. Soluble complex `[Cu(CN)_(4)]^(3-)` and `[Cd(CN)_(4)]^(2-)` are formed, of which CdS is precipitated as yellow ppt.
D. Soluble complex `[Cu(CN)_(4)]^(3-)` and `[Cd(CN)_(4)]^(2-)` are formed of which CuS is precipitated as black ppt.

Explanation

To solve the problem of what happens when H₂S gas is passed into a solution of Cu²⁺ and Cd²⁺ ions in the presence of excess KCN, we can break down the steps as follows:

### Step-by-Step Solution:

1. **Understanding the Reactants:**
- We have a solution containing Cu²⁺ and Cd²⁺ ions.
- There is an excess of KCN present in the solution.
- H₂S gas is being passed into this solution.

**Hint:** Identify the ions and compounds involved in the reaction.

2. **Effect of KCN on Cu²⁺ and Cd²⁺:**
- KCN is a strong ligand and will form stable complexes with Cu²⁺ and Cd²⁺ ions.
- Cu²⁺ will be reduced to Cu⁺ in the presence of KCN, forming a complex:
\[ \text{Cu}^{2+} + 2 \text{CN}^- \rightarrow \text{K}_2[\text{Cu(CN)}_4] \]
- For Cd²⁺, it will also form a complex:
\[ \text{Cd}^{2+} + 4 \text{CN}^- \rightarrow \text{K}_2[\text{Cd(CN)}_4] \]

**Hint:** Consider how ligands interact with metal ions to form complexes.

3. **Passing H₂S into the Solution:**
- When H₂S is introduced, it can react with metal ions to form sulfides.
- The sulfide precipitates are formed as follows:
- Cu⁺ will react with S²⁻ from H₂S to form CuS (black precipitate).
- Cd²⁺ will react with S²⁻ to form CdS (yellow precipitate).

**Hint:** Identify the possible precipitates formed from the reaction of sulfide ions with metal ions.

4. **Final Outcomes:**
- The formation of CuS and CdS indicates that both sulfides are precipitated from the solution.
- The soluble complexes formed earlier (K₂[Cu(CN)₄] and K₂[Cd(CN)₄]) remain in solution.

**Hint:** Summarize the products formed and their states (solid or soluble).

5. **Conclusion:**
- The final answer is that both CuS and CdS are precipitated when H₂S is passed into the solution containing Cu²⁺ and Cd²⁺ with excess KCN.

### Final Answer:
- The correct option is **C**: Soluble complex K₃[Cu(CN)₄] and K₂[Cd(CN)₄] are formed, of which CdS is precipitated as a yellow PPT.