Fixed parts of a colloidal solution of Agl are
respectively` [AgI]I^(-) and [AgI] Ag^(+)` in presence of

Options

A. Kl and `AgNO_(3)`
B. `AgNO_(3)` and Kl
C. Kl and `KIO_(3)`
D. `AgNO_(3) and Ba(NO_(3))_(2)`

Answer

To solve the question regarding the fixed parts of a colloidal solution of AgI, we need to analyze the options provided and understand the chemistry involved in the formation of the colloidal solution.

### Step-by-Step Solution:

1. **Understanding the Components**:
- We have a colloidal solution of AgI (Silver Iodide).
- The question mentions fixed parts as [AgI]I^(-) and [AgI]Ag^(+). This indicates that we are looking for conditions that lead to the formation of negatively charged and positively charged colloids.

2. **Analyzing the Options**:
- **Option A: KI and AgNO3**:
- When KI (Potassium Iodide) is in excess and reacts with AgNO3 (Silver Nitrate), AgI precipitate forms. The excess I^- ions preferentially adsorb onto the surface of AgI, creating a negatively charged colloid ([AgI]I^(-)).
- **Option B: AgNO3 and KI**:
- This is essentially the same as Option A, just the order of reactants is reversed. The result will still be the same.
- **Option C: KI and KIO3**:
- KIO3 does not provide Ag^+ ions, so it cannot form the positively charged colloid.
- **Option D: AgNO3 and Ba(NO3)2**:
- Ba(NO3)2 does not contribute to the formation of either Ag^+ or I^- ions in a way that would lead to the fixed layers described.

3. **Conclusion**:
- The correct answer is **Option A: KI and AgNO3**. In this scenario, we get the fixed layer of I^- ions on AgI, leading to the negatively charged colloid, and if AgNO3 is in excess, we can also form a fixed layer of Ag^+ ions around AgI, leading to the positively charged colloid.

### Final Answer:
**Option A: KI and AgNO3**