A complex containing `K,Pt(IV)` and `Cl^(-)` is `100%` ionized giving `i=3` the complext is

Options

A. `K_(2)[PtCl_(4)]`
B. `K_(2)[PtCl_(5)]`
C. `K[PtCl_(3)]`
D. `K_(2)[PtCl_(6)]`

Explanation

To solve the problem, we need to determine the complex containing potassium (K), platinum (Pt) in the +4 oxidation state, and chloride ions (Cl^(-)) that results in 100% ionization with a van 't Hoff factor (i) of 3.

### Step-by-Step Solution:

1. **Understanding the Van 't Hoff Factor (i)**:
- The van 't Hoff factor (i) indicates the number of particles the solute dissociates into when it dissolves in a solution. Here, i = 3 means that the complex ionizes to produce a total of 3 ions.

2. **Analyzing Possible Complexes**:
- We need to consider the possible complexes that can form with K, Pt(IV), and Cl^(-) and check how many ions they produce upon dissociation.
- The complexes we can consider are:
- K2[PtCl4]
- K[PtCl5]
- K[PtCl3]
- K2[PtCl6]

3. **Calculating Ionization for Each Complex**:
- **K2[PtCl4]**:
- Dissociates into 2 K^+ and 1 [PtCl4]^2- → Total = 3 ions.
- **K[PtCl5]**:
- Dissociates into 1 K^+ and 1 [PtCl5]^- → Total = 2 ions.
- **K[PtCl3]**:
- Dissociates into 1 K^+ and 1 [PtCl3]^2- → Total = 2 ions.
- **K2[PtCl6]**:
- Dissociates into 2 K^+ and 1 [PtCl6]^2- → Total = 3 ions.

4. **Determining the Oxidation State of Platinum**:
- For the complexes that yield 3 ions (K2[PtCl4] and K2[PtCl6]), we need to check if Pt is in the +4 oxidation state.
- **For K2[PtCl4]**:
- Let the oxidation state of Pt be X.
- The equation is: 2(+1) + X + 4(-1) = 0 → 2 + X - 4 = 0 → X = +2 (not +4).
- **For K2[PtCl6]**:
- The equation is: 2(+1) + X + 6(-1) = 0 → 2 + X - 6 = 0 → X = +4 (correct).

5. **Conclusion**:
- The complex that meets the criteria of 100% ionization (i = 3) and has platinum in the +4 oxidation state is **K2[PtCl6]**.

### Final Answer:
The complex is **K2[PtCl6]**.