Which of the following mixtures cannot be separated by passing `H_(2)S` through their solutions containing dilute HCl ?

Options

A. `Cu^(2+) and Sb^(3+)`
B. `Pb^(2+) and Cd^(2+)`
C. `Pb^(2+) and Al^(3+)`
D. `Zn^(2+) and Mn^(2+)`

Answer

To determine which of the following mixtures cannot be separated by passing H₂S through their solutions containing dilute HCl, we need to analyze the cations present in each option and their behavior in the presence of H₂S in an acidic medium.

### Step-by-Step Solution:

1. **Identify the Cations in Each Option:**
- **Option 1:** Cu²⁺ and Sb³⁺
- **Option 2:** Pb²⁺ and Cd²⁺
- **Option 3:** Pb²⁺ and Al³⁺
- **Option 4:** Zn²⁺ and Mn²⁺

2. **Understand the Group Behavior of Cations:**
- Cations can be classified into groups based on their precipitation behavior with H₂S in acidic solutions.
- H₂S acts as a reagent for group separation, particularly for group II cations.

3. **Analyze Each Option:**
- **Option 1 (Cu²⁺ and Sb³⁺):**
- Both Cu²⁺ and Sb³⁺ belong to the same group (Group II) and cannot be separated by H₂S.
- **Option 2 (Pb²⁺ and Cd²⁺):**
- Both Pb²⁺ and Cd²⁺ also belong to the same group (Group II) and cannot be separated by H₂S.
- **Option 3 (Pb²⁺ and Al³⁺):**
- Pb²⁺ belongs to Group II, while Al³⁺ belongs to Group III. They can be separated because they precipitate differently with H₂S.
- **Option 4 (Zn²⁺ and Mn²⁺):**
- Zn²⁺ and Mn²⁺ are both in Group II and generally remain unaffected by H₂S in an acidic medium, but they can be separated under specific conditions.

4. **Conclusion:**
- The mixtures that cannot be separated by passing H₂S through their solutions containing dilute HCl are:
- **Option 1:** Cu²⁺ and Sb³⁺
- **Option 2:** Pb²⁺ and Cd²⁺
- **Option 4 (Zn²⁺ and Mn²⁺)** can also be considered as they do not precipitate with H₂S in acidic medium.

Thus, the answer to the question is that **options 1, 2, and 4 cannot be separated**.