In which case, van't Hoff factor I remains unchanged ? ( Assume common complexes of these ions)

Options

A. `PtCl_(4)` reacts with `KCl`
B. `aq. ZnCl_(2)` reacts with `aq. NH_(3)`
C. `aq. FeCl_(3)` reacts with `aq. K_(4)[Fe(CN)]_(6)`
D. `KMnO_(4)` reduced to `MnO_(2)` in alkaline medium `(MnO_(2)` a black ppt)

Explanation

To determine in which case the van't Hoff factor (i) remains unchanged, we need to analyze the number of ions present before and after the reactions in each of the provided options. The van't Hoff factor is defined as the number of particles into which a solute dissociates in solution. If the number of ions remains the same before and after the reaction, then the van't Hoff factor remains unchanged.

### Step-by-Step Solution:

1. **Option A: PtCl4 reacts with KCl**
- Before the reaction:
- PtCl4 dissociates into 1 Pt ion and 4 Cl ions, totaling 5 ions.
- After the reaction:
- It forms K2PtCl6, which dissociates into 2 K+ ions and 1 PtCl6^2- ion, totaling 3 ions.
- **Conclusion**: The number of ions changes from 5 to 3. Therefore, i changes.

2. **Option B: Aqueous ZnCl2 reacts with aqueous ammonia**
- Before the reaction:
- ZnCl2 dissociates into 1 Zn^2+ ion and 2 Cl^- ions, totaling 3 ions.
- After the reaction:
- It forms [Zn(NH3)4]Cl2, which dissociates into 1 [Zn(NH3)4]^2+ ion and 2 Cl^- ions, totaling 3 ions.
- **Conclusion**: The number of ions remains the same at 3. Therefore, i remains unchanged.

3. **Option C: Aqueous FeCl3 reacts with aqueous K4[Fe(CN)6]**
- Before the reaction:
- FeCl3 dissociates into 1 Fe^3+ ion and 3 Cl^- ions, totaling 4 ions.
- After the reaction:
- It forms [Fe(CN)6]^-4, which dissociates into 4 Fe^3+ ions and 3 [Fe(CN)6]^-4 ions, totaling 7 ions.
- **Conclusion**: The number of ions changes from 4 to 7. Therefore, i changes.

4. **Option D: KMnO4 reduces to MnO2 in alkaline medium**
- Before the reaction:
- KMnO4 dissociates into 1 K^+ ion and 1 MnO4^- ion, totaling 2 ions.
- After the reaction:
- It forms MnO2 and dissociates into 1 MnO4^- ion and 2 O^2- ions, totaling 3 ions.
- **Conclusion**: The number of ions changes from 2 to 3. Therefore, i changes.

### Final Answer:
The only case where the van't Hoff factor (i) remains unchanged is **Option B: Aqueous ZnCl2 reacts with aqueous ammonia**.