A solution containing a group-IV cation gives a precipitate on passing, `H_(2)S`. A solution of this precipitate in dil. `HCl` produces a white precipitate with NaOH solution and bluish-white precipitate with basic potassium ferrocyanide. The cation is :

Options

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

Answer

To solve the question, we need to identify the group-IV cation based on the reactions described. Let's break down the steps:

### Step-by-Step Solution:

1. **Identify the Group-IV Cation**:
The problem states that a solution containing a group-IV cation gives a precipitate when H₂S is passed. The group-IV cations include Zn²⁺, Mn²⁺, Ni²⁺, and Co²⁺.

2. **Reaction with H₂S**:
When H₂S is passed through the solution, it forms precipitates with certain cations. The relevant precipitates are:
- Zn²⁺ forms ZnS (zinc sulfide) which is a white precipitate.
- Mn²⁺ forms MnS (manganese sulfide) which is also a precipitate but typically not white.
- Ni²⁺ and Co²⁺ form NiS and CoS respectively, which are insoluble.

Since we are looking for a precipitate, we focus on Zn²⁺ and Mn²⁺.

3. **Solubility in Dilute HCl**:
The problem states that the precipitate formed is soluble in dilute HCl.
- ZnS is soluble in dilute HCl, while MnS is not. This indicates that the cation must be Zn²⁺.

4. **Reaction with NaOH**:
The solution of the precipitate in dilute HCl produces a white precipitate with NaOH.
- Zn²⁺ reacts with NaOH to form Zn(OH)₂, which is a gelatinous white precipitate.

5. **Reaction with Potassium Ferrocyanide**:
The precipitate with potassium ferrocyanide produces a bluish-white precipitate.
- The reaction of Zn²⁺ with potassium ferrocyanide (K₄[Fe(CN)₆]) yields Zn₂[Fe(CN)₆], which is a bluish-white precipitate.

6. **Conclusion**:
Based on the reactions and the solubility characteristics, the cation is confirmed to be Zn²⁺.

### Final Answer:
The cation is **Zn²⁺**.