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 at. `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 dissociation of the compounds in each option provided. The van't Hoff factor (i) is defined as the ratio of the number of particles in solution after dissociation to the number of formula units initially dissolved. For i to remain unchanged, the number of ions before the reaction must equal the number of ions after the reaction.

Let's analyze each option step by step:

### Step 1: Analyze the first option - PtCl4 + KCl
- **Reactants**: PtCl4 (1 unit) + KCl (1 unit)
- **Dissociation**: PtCl4 dissociates into Pt^4+ and 4 Cl^-. KCl dissociates into K^+ and Cl^-.
- **Total ions before reaction**: 1 (PtCl4) + 1 (KCl) = 2 units
- **Total ions after reaction**: 1 (Pt^4+) + 4 (Cl^-) + 1 (K^+) + 1 (Cl^-) = 7 ions
- **Conclusion**: The number of ions is not the same before and after the reaction. Therefore, i is not unchanged.

### Step 2: Analyze the second option - ZnCl2 + NH3
- **Reactants**: ZnCl2 (1 unit) + NH3 (1 unit)
- **Dissociation**: ZnCl2 dissociates into Zn^2+ and 2 Cl^-. NH3 does not dissociate further.
- **Total ions before reaction**: 1 (ZnCl2) + 1 (NH3) = 2 units
- **Total ions after reaction**: Zn(NH3)4Cl2 does not dissociate further, so it remains as 1 unit.
- **Conclusion**: The number of ions is the same before and after the reaction. Therefore, i remains unchanged. This option is correct.

### Step 3: Analyze the third option - FeCl3 + K4[Fe(CN)6]
- **Reactants**: FeCl3 (1 unit) + K4[Fe(CN)6] (1 unit)
- **Dissociation**: FeCl3 dissociates into Fe^3+ and 3 Cl^-. K4[Fe(CN)6] does not dissociate further.
- **Total ions before reaction**: 1 (FeCl3) + 1 (K4[Fe(CN)6]) = 2 units
- **Total ions after reaction**: The complex remains intact, so we have 1 unit.
- **Conclusion**: The number of ions is not the same before and after the reaction. Therefore, i is not unchanged.

### Step 4: Analyze the fourth option - KMnO4 reduced to MnO2
- **Reactants**: KMnO4 (1 unit)
- **Dissociation**: KMnO4 dissociates into K^+ and MnO4^-. MnO2 is a solid precipitate and does not dissociate in solution.
- **Total ions before reaction**: 1 (KMnO4) = 2 ions (K^+ and MnO4^-)
- **Total ions after reaction**: MnO2 does not contribute to ions in solution, so we have only 1 ion (K^+).
- **Conclusion**: The number of ions is not the same before and after the reaction. Therefore, i is not unchanged.

### Final Conclusion
The only case where the van't Hoff factor (i) remains unchanged is the second option: **Aqueous ZnCl2 reacts with NH3 to form Zn(NH3)4Cl2**.