The pair (s) where both the ions are precipitated upon passing `H_(2)S` gas in the presence of dilute `HCl` is (are) :

Options

A. `Ba^(2+) , Zn^(2+)`
B. `Ba^(2+), Fe^(3+)`
C. `Cu^(2+) , Pb^(2+)`
D. `Hg^(2+), Bi^(3+)`

Explanation

To solve the question of identifying the pairs where both ions are precipitated upon passing H₂S gas in the presence of dilute HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reagents**: We are using H₂S gas and dilute HCl. These reagents are known to precipitate certain metal ions as sulfides.

2. **Identify the Groups**: The ions that can be precipitated by H₂S in the presence of dilute HCl belong to specific groups in qualitative analysis. H₂S is typically used for Group 2 cations.

3. **Check the Given Options**: We need to analyze the pairs of ions provided in the question to see if both belong to Group 2.

4. **Evaluate Each Pair**:
- **Pair 1**: If the ions are from Group 2, they will precipitate as sulfides. For example, Cu²⁺ and Pb²⁺ both belong to Group 2.
- **Pair 2**: If one ion is from Group 3 (like Fe³⁺), it will not precipitate with H₂S in the presence of dilute HCl, so this pair can be eliminated.
- **Pair 3**: Check if both ions belong to Group 2. If both do, they will precipitate.

5. **Confirm Precipitation**:
- For example, Cu²⁺ reacts with H₂S to form CuS (black precipitate).
- Pb²⁺ reacts with H₂S to form PbS (also a black precipitate).
- Hg²⁺ and Bi³⁺ also form precipitates with H₂S.

6. **Final Selection**: After evaluating all pairs, select the pairs where both ions precipitate with H₂S in the presence of dilute HCl.

### Conclusion:
The pairs where both ions are precipitated upon passing H₂S gas in the presence of dilute HCl are:
- Copper (Cu²⁺) and Lead (Pb²⁺)
- Mercury (Hg²⁺) and Bismuth (Bi³⁺)