On passing `H_(2)S` black ppt. of II group is obtained. The mixture may not contain

Options

A. `Pb^(++)`
B. `Cd^(++)`
C. `Hg^(++)`
D. `Cu^(++)`

Explanation

To solve the problem, we need to analyze the behavior of different metal ions when hydrogen sulfide (H₂S) is passed through a solution containing them. The question specifically asks for a metal ion that does not produce a black precipitate when H₂S is introduced.

### Step-by-Step Solution:

1. **Understanding Group II Cations**:
Group II cations (also known as the second group of cations in qualitative analysis) typically include ions like Pb²⁺, Hg²⁺, and Cu²⁺, which form black precipitates when H₂S is passed through their solutions.

2. **Identifying the Precipitate**:
When H₂S is passed through a solution containing these cations, they react to form metal sulfides. For example:
- Pb²⁺ + H₂S → PbS (black precipitate)
- Hg²⁺ + H₂S → HgS (black precipitate)
- Cu²⁺ + H₂S → CuS (black precipitate)

3. **Considering Cadmium (Cd²⁺)**:
Cadmium is also a Group II cation, but it behaves differently. When cadmium ions (Cd²⁺) react with H₂S, they form cadmium sulfide (CdS), which is not black but yellow:
- Cd²⁺ + H₂S → CdS (yellow precipitate)

4. **Conclusion**:
Since the question asks for a cation that does not produce a black precipitate upon the addition of H₂S, the answer is cadmium (Cd²⁺), which produces a yellow precipitate instead.

### Final Answer:
The mixture may not contain **Cadmium (Cd²⁺)**.

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