Which of the following reagents willnot be useful in separating a mixture of `Zn^(2+) and Cu^(2-)` ?

Options

A. `H_(2)S` in an acid medium
B. `H_(2)S` in an alkaline method
C. Excess of NaOH solution
D. `NH_(3)` solution

Explanation

To determine which of the given reagents will not be useful in separating a mixture of \( \text{Zn}^{2+} \) and \( \text{Cu}^{2+} \), we will analyze each option based on their chemical behavior in different conditions.

### Step-by-Step Solution:

1. **Identify the Groups:**
- \( \text{Zn}^{2+} \) belongs to the third group of cations.
- \( \text{Cu}^{2+} \) belongs to the second group of cations.

2. **Evaluate the Reagents:**
- **Option A: \( \text{H}_2\text{S} \) in Acid Medium**
- In acidic medium, \( \text{Cu}^{2+} \) reacts with \( \text{H}_2\text{S} \) to form \( \text{CuS} \) (black precipitate).
- \( \text{Zn}^{2+} \) does not precipitate as \( \text{ZnS} \) in acidic conditions. Thus, this reagent can help in separating \( \text{Cu}^{2+} \) from \( \text{Zn}^{2+} \).

3. **Option B: \( \text{H}_2\text{S} \) in Alkaline Medium**
- In alkaline medium, both \( \text{Zn}^{2+} \) and \( \text{Cu}^{2+} \) react with \( \text{H}_2\text{S} \) to form precipitates:
- \( \text{ZnS} \) (white precipitate)
- \( \text{CuS} \) (black precipitate)
- This reagent is also useful for separation as both ions precipitate.

4. **Option C: Excess \( \text{NaOH} \)**
- When \( \text{Zn}^{2+} \) is treated with excess \( \text{NaOH} \), it forms \( \text{Na}_2\text{ZnO}_2 \), which is soluble.
- \( \text{Cu}^{2+} \) reacts with \( \text{NaOH} \) to form \( \text{Cu(OH)}_2 \), which is insoluble.
- Therefore, this reagent can be used to separate \( \text{Zn}^{2+} \) and \( \text{Cu}^{2+} \).

5. **Option D: \( \text{NH}_3 \) Solution**
- In \( \text{NH}_3 \) solution, \( \text{Zn}^{2+} \) forms a soluble complex \( \text{[Zn(NH}_3\text{)}_4]^{2+} \).
- Similarly, \( \text{Cu}^{2+} \) also forms a soluble complex \( \text{[Cu(NH}_3\text{)}_4]^{2+} \).
- Since both ions remain soluble, this reagent will not help in separating \( \text{Zn}^{2+} \) from \( \text{Cu}^{2+} \).

### Conclusion:
The reagents that will not be useful in separating a mixture of \( \text{Zn}^{2+} \) and \( \text{Cu}^{2+} \) are:
- **Option B: \( \text{H}_2\text{S} \) in Alkaline Medium**
- **Option D: \( \text{NH}_3 \) Solution**

### Final Answer:
**Option B and Option D will not be useful in separating \( \text{Zn}^{2+} \) and \( \text{Cu}^{2+} \).**

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