A Pt complex of NH3 and chlorine produces four ions per molecule in the solution is :-

Options

A. `[Pt(NH_(3))_(4)Cl_(2)]Cl_(2)`
B. `[Pt(NH_(3))_(6)]Cl`
C. `[Pt(NH_(3))_(2)]Cl_(4)`
D. `[Pt(NH_(3))_(5)Cl]Cl_(3)`

Answer

To solve the question regarding a platinum complex of ammonia (NH3) and chlorine (Cl) that produces four ions per molecule in solution, we will analyze each option systematically.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We need to find a platinum complex that, when dissolved in solution, dissociates to produce exactly four ions.

2. **Analyzing Each Option:**
We will evaluate each given option to see how many ions they produce in solution.

3. **Option A: [Pt(NH3)4Cl2]**
- This complex will dissociate as follows:
\[ [Pt(NH3)4Cl2] \rightarrow Pt^{2+} + 4NH3 + 2Cl^- \]
- Total ions produced: 1 (Pt) + 4 (NH3) + 2 (Cl) = 7 ions.
- This option does not satisfy the condition.

4. **Option B: [Pt(NH3)6Cl2]**
- This complex will dissociate as follows:
\[ [Pt(NH3)6Cl2] \rightarrow Pt^{2+} + 6NH3 + 2Cl^- \]
- Total ions produced: 1 (Pt) + 6 (NH3) + 2 (Cl) = 9 ions.
- This option does not satisfy the condition.

5. **Option C: [Pt(NH3)2Cl4]**
- This complex will dissociate as follows:
\[ [Pt(NH3)2Cl4] \rightarrow Pt^{2+} + 2NH3 + 4Cl^- \]
- Total ions produced: 1 (Pt) + 2 (NH3) + 4 (Cl) = 7 ions.
- This option does not satisfy the condition.

6. **Option D: [Pt(NH3)5Cl3]**
- This complex will dissociate as follows:
\[ [Pt(NH3)5Cl3] \rightarrow Pt^{2+} + 5NH3 + 3Cl^- \]
- Total ions produced: 1 (Pt) + 5 (NH3) + 3 (Cl) = 9 ions.
- This option does not satisfy the condition.

7. **Revisiting the Options:**
It seems we need to consider the possibility of different oxidation states or coordination numbers. Let's analyze a different complex that could yield four ions.

8. **Correct Complex: [Pt(NH3)4Cl]**
- This complex will dissociate as follows:
\[ [Pt(NH3)4Cl] \rightarrow Pt^{2+} + 4NH3 + Cl^- \]
- Total ions produced: 1 (Pt) + 4 (NH3) + 1 (Cl) = 6 ions.
- This option does not satisfy the condition.

9. **Final Complex: [Pt(NH3)3Cl3]**
- This complex will dissociate as follows:
\[ [Pt(NH3)3Cl3] \rightarrow Pt^{2+} + 3NH3 + 3Cl^- \]
- Total ions produced: 1 (Pt) + 3 (NH3) + 3 (Cl) = 7 ions.
- This option does not satisfy the condition.

10. **Conclusion:**
After careful consideration, we find that the only complex that can yield exactly four ions is [Pt(NH3)3Cl].

### Final Answer:
The correct platinum complex that produces four ions per molecule in solution is **[Pt(NH3)3Cl]**.