The complex compound formed with KCN solution is added to solution containing both `Cu^(2+)` and `Cd^(2+)` ions are

Options

A. `K_2[Cu(CN)_4],K_2[Cd(CN)_4]`
B. `K_3[Cu(CN)_4],K_2[Cd(CN)_4]`
C. `K_3[Cu(CN)_4],K_3[Cd(CN)_4]`
D. `K_2[Cu(CN)_4],K_3[Cd(CN)_4]`

Answer

To solve the problem, we need to analyze the reactions that occur when KCN solution is added to a solution containing both Cu²⁺ and Cd²⁺ ions.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We have KCN (potassium cyanide) and a solution containing Cu²⁺ (copper ions) and Cd²⁺ (cadmium ions).

2. **Reaction with Cu²⁺**:
- When KCN is added to Cu²⁺ ions, a complex is formed. The reaction can be represented as:
\[ \text{Cu}^{2+} + 4 \text{CN}^- \rightarrow \text{[Cu(CN)}_4]^{2-} \]
- This results in the formation of the complex ion \([Cu(CN)_4]^{2-}\). Since KCN provides K⁺ ions, the complete formula for the complex will be \(K_3[Cu(CN)_4]\).

3. **Reaction with Cd²⁺**:
- Similarly, when KCN is added to Cd²⁺ ions, another complex is formed:
\[ \text{Cd}^{2+} + 2 \text{CN}^- \rightarrow \text{[Cd(CN)}_2]^{2-} \]
- This results in the formation of the complex ion \([Cd(CN)_2]^{2-}\). The complete formula for this complex will be \(K_2[Cd(CN)_2]\).

4. **Final Products**:
- Therefore, the products formed when KCN is added to the solution containing both Cu²⁺ and Cd²⁺ ions are:
- \(K_3[Cu(CN)_4]\) (from Cu²⁺)
- \(K_2[Cd(CN)_2]\) (from Cd²⁺)

5. **Conclusion**:
- The correct answer to the question is that the complex compounds formed are \(K_3[Cu(CN)_4]\) and \(K_2[Cd(CN)_2]\).