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

Options

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

Answer

To determine which pair of ions cannot be separated by H₂S in dilute HCl, we need to understand the behavior of different metal ions when treated with H₂S.

### Step-by-Step Solution:

1. **Identify the Groups**:
- H₂S is a group reagent for the second and fourth groups of metal ions in qualitative analysis.
- The second group includes ions like Cu²⁺ (copper), while the fourth group includes ions like Ni²⁺ (nickel) and Zn²⁺ (zinc).

2. **Reactions with H₂S**:
- When H₂S is added to a solution containing metal ions, it precipitates certain sulfides.
- For example:
- Cu²⁺ + H₂S → CuS (black precipitate)
- Ni²⁺ + H₂S → NiS (black precipitate)
- Zn²⁺ + H₂S → ZnS (white precipitate)

3. **Analyzing the Precipitates**:
- CuS and NiS both form black precipitates, while ZnS forms a white precipitate.
- This means that if we have a mixture of Cu²⁺ and Ni²⁺, both will produce black precipitates, making it impossible to distinguish between them based solely on the color of the precipitate.

4. **Conclusion**:
- Therefore, the pair of ions that cannot be separated by H₂S in dilute HCl is **Nickel (Ni²⁺) and Copper (Cu²⁺)**, as both produce black precipitates.

### Final Answer:
The pair of ions that cannot be separated by H₂S in dilute HCl is **Nickel (Ni²⁺) and Copper (Cu²⁺)**.

---