A complex containing `K^(+),Pt(IV)` and `Cl^(--)` is `100%` ionised giving `i=3`. Thus, complex is:

Options

A. `K_(2)[PtCl_(4)]`
B. `K_(2)[PtCl_(6)]`
C. `K_(43)[PtCl_(5)]`
D. `K[PtCl_(3)]`

Explanation

To solve the problem, we need to identify the complex that contains \( K^+ \), \( Pt(IV) \), and \( Cl^{--} \) and results in 100% ionization with \( i = 3 \).

### Step-by-Step Solution:

1. **Understanding Ionization**:
- The ionization factor \( i \) indicates the number of particles (ions) produced when the complex dissociates in solution. Given \( i = 3 \), the complex must yield 3 ions upon dissociation.

2. **Identifying Possible Complexes**:
- We have the following complexes to consider:
1. \( K_2[PtCl_4] \)
2. \( K_2[PtCl_6] \)
3. \( K_2[PtCl_5] \)
4. \( K[PtCl_3] \)

3. **Calculating Ionization for Each Complex**:
- For \( K_2[PtCl_4] \):
- Dissociates into \( 2K^+ + [PtCl_4]^{2-} \) → Total ions = 3
- For \( K_2[PtCl_6] \):
- Dissociates into \( 2K^+ + [PtCl_6]^{2-} \) → Total ions = 3
- For \( K_2[PtCl_5] \):
- Dissociates into \( 2K^+ + [PtCl_5]^{-} \) → Total ions = 3
- For \( K[PtCl_3] \):
- Dissociates into \( K^+ + [PtCl_3]^{-} \) → Total ions = 2

4. **Determining the Oxidation State of Platinum**:
- For \( K_2[PtCl_4] \):
- Let the oxidation state of Pt be \( x \).
- Equation: \( x + 4(-1) = -2 \) → \( x = +2 \)
- For \( K_2[PtCl_6] \):
- Equation: \( x + 6(-1) = -2 \) → \( x = +4 \)
- For \( K_2[PtCl_5] \):
- Equation: \( x + 5(-1) = -2 \) → \( x = +3 \)
- For \( K[PtCl_3] \):
- Equation: \( x + 3(-1) = 0 \) → \( x = +3 \)

5. **Identifying the Correct Complex**:
- The only complex that has \( Pt(IV) \) (i.e., \( +4 \) oxidation state) and gives \( i = 3 \) is \( K_2[PtCl_6] \).

### Conclusion:
The complex is \( K_2[PtCl_6] \).