Out of `Cu^(2+),Ni^(2+),Co^(2+)` and `Mn^(2+)` of those that dissolve in dil HCI only one give precipitate when `H_(2)S` is passed. Identify the corresponding order which do not give precipitation :

Options

A. `Ni^(2+)`
B. `Cu^(2+)`
C. `Co^(2+)`
D. `Mn^(2+)`

Answer

To solve the problem, we need to analyze the behavior of the cations \(Cu^{2+}\), \(Ni^{2+}\), \(Co^{2+}\), and \(Mn^{2+}\) when they are treated with dilute hydrochloric acid (HCl) and hydrogen sulfide (H₂S).

### Step-by-step Solution:

1. **Identify the Cations**: The cations we are dealing with are:
- \(Cu^{2+}\) (Copper)
- \(Ni^{2+}\) (Nickel)
- \(Co^{2+}\) (Cobalt)
- \(Mn^{2+}\) (Manganese)

2. **Solubility in Dilute HCl**:
- All four cations (\(Cu^{2+}\), \(Ni^{2+}\), \(Co^{2+}\), and \(Mn^{2+}\)) are soluble in dilute HCl. This means they will dissolve when HCl is added.

3. **Reaction with H₂S**:
- When H₂S is passed through a solution containing these cations, certain cations will form precipitates as sulfides.
- The group reagents for sulfide precipitation (Group II cations) include \(Cu^{2+}\), which forms a black precipitate of copper sulfide (\(CuS\)).
- The other cations, \(Ni^{2+}\), \(Co^{2+}\), and \(Mn^{2+}\), do not form precipitates with H₂S in the presence of dilute HCl.

4. **Identify the Cation that Forms Precipitate**:
- From the cations listed, only \(Cu^{2+}\) gives a precipitate when H₂S is passed. The others do not precipitate.

5. **Conclusion**:
- The cations that do not give precipitation when H₂S is passed are \(Ni^{2+}\), \(Co^{2+}\), and \(Mn^{2+}\).
- Therefore, the answer is \(Ni^{2+}\), \(Co^{2+}\), and \(Mn^{2+}\).

### Final Answer:
The corresponding order of cations that do not give precipitation when H₂S is passed is \(Ni^{2+}, Co^{2+}, Mn^{2+}\).

---