Which one among the following pairs of ions cannot be separated by `H_(2)S` in dilute hydrochloric acid ?

Options

A. `Bi^(3+) , Sn^(4+)`
B. `Al^(3+) , Hg^(2+)`
C. `Zn^(2+) , Cu^(2+)`
D. `Ni^(2+) , Cu^(2+)`

Explanation

To determine which pair of ions cannot be separated by H₂S in dilute hydrochloric acid, we need to analyze the behavior of each ion in the presence of H₂S and dilute HCl.

### Step-by-Step Solution:

1. **Identify the Ions in Each Option**:
- Option A: Bi³⁺ and Sn⁴⁺
- Option B: Hg²⁺ and Al³⁺
- Option C: Cu²⁺ and Zn²⁺
- Option D: Cu²⁺ and Ni²⁺

2. **Analyze Option A (Bi³⁺ and Sn⁴⁺)**:
- Bi³⁺ reacts with H₂S to form bismuth sulfide (Bi₂S₃), which is a black precipitate.
- Sn⁴⁺ reacts with H₂S to form stannous sulfide (SnS₂), which is a yellow precipitate.
- Since both ions precipitate with H₂S, they cannot be separated.

3. **Analyze Option B (Hg²⁺ and Al³⁺)**:
- Hg²⁺ reacts with H₂S to form mercuric sulfide (HgS), which is a black precipitate.
- Al³⁺ does not precipitate with H₂S in acidic conditions.
- Therefore, these can be separated.

4. **Analyze Option C (Cu²⁺ and Zn²⁺)**:
- Cu²⁺ reacts with H₂S to form copper(I) sulfide (CuS), which is a black precipitate.
- Zn²⁺ does not precipitate with H₂S in acidic conditions.
- Therefore, these can be separated.

5. **Analyze Option D (Cu²⁺ and Ni²⁺)**:
- Cu²⁺ reacts with H₂S to form CuS, which is a black precipitate.
- Ni²⁺ does not precipitate with H₂S in acidic conditions.
- Therefore, these can be separated.

6. **Conclusion**:
- The only pair of ions that cannot be separated by H₂S in dilute hydrochloric acid is **Option A: Bi³⁺ and Sn⁴⁺**.

### Final Answer:
**Option A: Bismuth (Bi³⁺) and Tin (Sn⁴⁺)** cannot be separated by H₂S in dilute hydrochloric acid.