Which of the following pairs of cations can be separated by passing `H_(2)S` through the mixture in the presence of 0.2 M HCl?

Options

A. `Pb^(2+),Cu^(2+)`
B. `Ag^(+),Cu^(2+)`
C. `Cd^(2+),Bi^(3+)`
D. `Cu^(2+),Zn^(2+)`

Answer

To determine which pairs of cations can be separated by passing H₂S through the mixture in the presence of 0.2 M HCl, we need to analyze the solubility and precipitation behavior of the cations involved.

### Step-by-step Solution:

1. **Identify the Cations**: The pairs of cations given in the question are:
- (i) Lead (Pb²⁺) and Copper (Cu²⁺)
- (ii) Silver (Ag⁺) and Copper (Cu²⁺)
- (iii) Cadmium (Cd²⁺) and Bismuth (Bi³⁺)
- (iv) Copper (Cu²⁺) and Zinc (Zn²⁺)

2. **Understand Group Classification**: According to qualitative analysis, cations can be classified into different groups based on their behavior towards sulfide precipitation:
- Group 2 cations (which precipitate with H₂S) include: Ag⁺, Hg₂²⁺, Pb²⁺, Cd²⁺, Cu²⁺, and Bi³⁺.
- Group 4 cations (which do not precipitate with H₂S) include: Zn²⁺.

3. **Analyze Each Pair**:
- **Pair (i): Pb²⁺ and Cu²⁺**: Both belong to Group 2 and will precipitate as sulfides when H₂S is added. They cannot be separated.
- **Pair (ii): Ag⁺ and Cu²⁺**: Both belong to Group 2 and will also precipitate as sulfides. They cannot be separated.
- **Pair (iii): Cd²⁺ and Bi³⁺**: Both belong to Group 2 and will precipitate as sulfides. They cannot be separated.
- **Pair (iv): Cu²⁺ and Zn²⁺**: Cu²⁺ belongs to Group 2 and will precipitate, while Zn²⁺ belongs to Group 4 and will not precipitate. Therefore, they can be separated by filtration.

4. **Conclusion**: The only pair of cations that can be separated by passing H₂S through the mixture in the presence of 0.2 M HCl is **Copper (Cu²⁺) and Zinc (Zn²⁺)**.

### Final Answer:
The correct pair of cations that can be separated is **Copper (Cu²⁺) and Zinc (Zn²⁺)**.