A complex of platinum, ammonia and chloride produces four ions per molecule in the solution. The structure consistent with the observation is:

Options

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

Answer

To solve the problem of determining which complex of platinum, ammonia, and chloride produces four ions per molecule in solution, we will analyze the possible structures of the complexes and their ionization behavior.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find a complex that, when dissolved in solution, produces a total of four ions. This means we need to consider how the complex dissociates into ions.

2. **Identifying Possible Complexes**:
We can consider different coordination complexes of platinum with ammonia (NH3) and chloride (Cl-). The general formula for such complexes can be represented as [Pt(NH3)x(Cl)y]z, where x, y, and z are the number of ammonia and chloride ligands and the charge of the complex, respectively.

3. **Analyzing Complexes**:
- **Complex 1**: [Pt(NH3)5Cl]Cl2
- Upon ionization, it would produce:
- 1 complex ion: [Pt(NH3)5Cl]^+
- 2 chloride ions (Cl^-)
- Total ions = 1 + 2 = 3 ions (not suitable).

- **Complex 2**: [Pt(NH3)6]Cl4
- Upon ionization, it would produce:
- 1 complex ion: [Pt(NH3)6]^4+
- 4 chloride ions (Cl^-)
- Total ions = 1 + 4 = 5 ions (not suitable).

- **Complex 3**: [Pt(NH3)4Cl2]
- Upon ionization, it would produce:
- 1 complex ion: [Pt(NH3)4Cl]^2+
- 2 chloride ions (Cl^-)
- Total ions = 1 + 2 = 3 ions (not suitable).

- **Complex 4**: [Pt(NH3)5Cl]Cl
- Upon ionization, it would produce:
- 1 complex ion: [Pt(NH3)5Cl]^+
- 1 chloride ion (Cl^-)
- Total ions = 1 + 1 = 2 ions (not suitable).

- **Complex 5**: [Pt(NH3)4Cl]Cl2
- Upon ionization, it would produce:
- 1 complex ion: [Pt(NH3)4Cl]^+
- 2 chloride ions (Cl^-)
- Total ions = 1 + 2 = 3 ions (not suitable).

- **Complex 6**: [Pt(NH3)4Cl2]
- Upon ionization, it would produce:
- 1 complex ion: [Pt(NH3)4Cl]^2+
- 2 chloride ions (Cl^-)
- Total ions = 1 + 2 = 3 ions (not suitable).

4. **Final Analysis**:
After analyzing the possible complexes, we find that the complex [Pt(NH3)5Cl]Cl2 produces a total of 4 ions when it dissociates in solution.

5. **Conclusion**:
The structure consistent with the observation of producing four ions per molecule in solution is **[Pt(NH3)5Cl]Cl2**.