There is mixture of Cu(II) chloride and Fe(II) sulphate. The best way to separate the metal ions from the mixture in qualitative analysis is:

Options

A. hydrogen sulphide in acidic medium, where only Cu(II) sulphide will be precipitated
B. ammonium hydroxide buffer, where only Fe(II) hydroxide will be precipitated
C. hydrogen sulphide in acidic medium, where only Fe(II) sulphide will be precipitated
D. ammonium hydroxide buffer, where only Cu(II) hydroxide will be precipitated

Explanation

To separate the metal ions from a mixture of Cu(II) chloride (CuCl2) and Fe(II) sulfate (FeSO4) in qualitative analysis, follow these steps:

### Step-by-Step Solution:

1. **Identify the Metal Ions**:
- The mixture contains Cu²⁺ ions from CuCl2 and Fe²⁺ ions from FeSO4.

2. **Understand the Group Classifications**:
- Cu²⁺ is classified as a basic radical of Group 2.
- Fe²⁺ is classified as a basic radical of Group 3.

3. **Know the Reagents for Each Group**:
- Group 2 (Cu²⁺) can be precipitated using dilute hydrochloric acid (HCl) and hydrogen sulfide (H2S).
- Group 3 (Fe²⁺) can be precipitated using ammonium hydroxide (NH4OH) or ammonium chloride (NH4Cl).

4. **Determine the Precipitation Reactions**:
- When CuCl2 reacts with H2S in acidic medium, it forms copper(II) sulfide (CuS), which is a black precipitate.
- When FeSO4 reacts with NH4OH, it oxidizes to Fe³⁺ and forms iron(III) hydroxide (Fe(OH)3), which is a brown precipitate.

5. **Choose the Best Method for Separation**:
- Using H2S in acidic medium will selectively precipitate Cu²⁺ as CuS, while Fe²⁺ will remain in solution because it does not react under these conditions.

6. **Conclusion**:
- The best way to separate the metal ions from the mixture is to use hydrogen sulfide in acidic medium, where only Cu²⁺ will precipitate as CuS.

### Final Answer:
The best way to separate the metal ions from the mixture in qualitative analysis is to use hydrogen sulfide in acidic medium, where only Cu²⁺ sulfide will be precipitated.

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