On adding `AgNO_(3)` solution to KI solution, a negatively charged colloidal sol will be formed in which of the following conditions ?

Options

A. 100 mL of 0.1 M `AgNO_(3)` + 100 mL of 0.1 M KI
B. 100 mL of 0.1 M `AgNO_(3)` + 50 mL of 0.2 M KI
C. 100 mL of 0.2 M `AgNO_(3)` + 100 mL of 0.1 M KI
D. 100 mL of 0.1 M `AgNO_(3)` + 100 mL of 0.15 M KI

Answer

To determine the conditions under which a negatively charged colloidal sol is formed when adding AgNO₃ solution to KI solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
When silver nitrate (AgNO₃) is added to potassium iodide (KI), a reaction occurs that produces silver iodide (AgI) and potassium nitrate (KNO₃). The reaction can be represented as:
\[ \text{AgNO}_3 + \text{KI} \rightarrow \text{AgI (s)} + \text{KNO}_3 \]

2. **Understand Colloidal Formation**:
Silver iodide (AgI) can form a colloidal solution. The charge of the colloidal particles depends on the presence of excess ions. In this case, if there are excess iodide ions (I⁻) from KI, the AgI particles will acquire a negative charge.

3. **Determine Conditions for Negatively Charged Colloidal Sol**:
For the colloidal sol to be negatively charged, there must be an excess of iodide ions (I⁻) in the solution. This means that the concentration of KI must be greater than that of AgNO₃.

4. **Calculate Moles in Each Option**:
We need to analyze the given options to find out where KI is in excess:
- **Option A**: 100 mL of 0.1 M AgNO₃ + 100 mL of 0.1 M KI
- Moles of AgNO₃ = 0.1 M × 0.1 L = 0.01 moles (10 millimoles)
- Moles of KI = 0.1 M × 0.1 L = 0.01 moles (10 millimoles)
- **Result**: Equal moles, not in excess.

- **Option B**: 100 mL of 0.1 M AgNO₃ + 50 mL of 0.2 M KI
- Moles of AgNO₃ = 0.1 M × 0.1 L = 0.01 moles (10 millimoles)
- Moles of KI = 0.2 M × 0.05 L = 0.01 moles (10 millimoles)
- **Result**: Equal moles, not in excess.

- **Option C**: 100 mL of 0.2 M AgNO₃ + 100 mL of 0.1 M KI
- Moles of AgNO₃ = 0.2 M × 0.1 L = 0.02 moles (20 millimoles)
- Moles of KI = 0.1 M × 0.1 L = 0.01 moles (10 millimoles)
- **Result**: AgNO₃ is in excess.

- **Option D**: 100 mL of 0.1 M AgNO₃ + 100 mL of 0.15 M KI
- Moles of AgNO₃ = 0.1 M × 0.1 L = 0.01 moles (10 millimoles)
- Moles of KI = 0.15 M × 0.1 L = 0.015 moles (15 millimoles)
- **Result**: KI is in excess.

5. **Conclusion**:
The correct condition for forming a negatively charged colloidal sol is found in **Option D**, where the moles of KI are greater than the moles of AgNO₃.

### Final Answer:
**Option D** is the correct answer, as it provides the condition where KI is in excess, leading to the formation of a negatively charged colloidal sol.