The correct option(s) to distinguish nitrate salts of `Mn^(2+) and Cu^(2+)` taken separately is (Are)
(1) `Mn^(2+)` shows the characteristic green colour in the flame test
(2) only `Cu^(2+)` shows the the formation of precipitate by passing `H_(2)S` in acidic medium
(3) only `Mn^(2+)` shows the formation of precipitate by passing `H_(2)S` in faintly basic medium
(4) `Cu^(2+)|"Cu has higher reduction potential then "Mn^(2+)|Mn` (measured under similar conditions)

Options

A. 1, 2
B. 1, 3
C. 2, 4
D. 1, 2, 4

Answer

To determine the correct options to distinguish between the nitrate salts of \( \text{Mn}^{2+} \) and \( \text{Cu}^{2+} \), we will analyze each statement provided in the question.

### Step-by-Step Solution:

1. **Statement 1: \( \text{Mn}^{2+} \) shows the characteristic green color in the flame test.**
- The flame test for \( \text{Mn}^{2+} \) typically shows a yellowish-green color, not a pure green. Therefore, this statement is **incorrect**.

2. **Statement 2: Only \( \text{Cu}^{2+} \) shows the formation of precipitate by passing \( \text{H}_2\text{S} \) in acidic medium.**
- In acidic medium, \( \text{Cu}^{2+} \) can react with \( \text{H}_2\text{S} \) to form a precipitate of \( \text{CuS} \). The solubility product (Ksp) of \( \text{CuS} \) is very low, making it precipitate readily. Thus, this statement is **correct**.

3. **Statement 3: Only \( \text{Mn}^{2+} \) shows the formation of precipitate by passing \( \text{H}_2\text{S} \) in faintly basic medium.**
- In faintly basic medium, \( \text{H}_2\text{S} \) can also lead to the precipitation of \( \text{MnS} \). However, the precipitation of \( \text{MnS} \) is less favorable compared to \( \text{CuS} \) due to the higher Ksp of \( \text{MnS} \). Therefore, this statement is **incorrect**.

4. **Statement 4: \( \text{Cu}^{2+} \) has a higher reduction potential than \( \text{Mn}^{2+} \).**
- The standard reduction potential for \( \text{Cu}^{2+}/\text{Cu} \) is positive, while that for \( \text{Mn}^{2+}/\text{Mn} \) is negative. This means \( \text{Cu}^{2+} \) indeed has a higher reduction potential than \( \text{Mn}^{2+} \). Thus, this statement is **correct**.

### Summary of Correct Statements:
- Statement 2 and Statement 4 are correct.

### Final Answer:
The correct options are **(2) and (4)**.