Silver nitrate solution is gradually added to an aqueous solution containing `0.01 M` each of chloride, bromide and iodide ions. The correct sequence in which the halides will be precipitated is:

Options

A. `Br^-,CI^-,I^-`
B. `I^-,CI^-,Br^-`
C. `I^-,Br^-,CI^-`
D. `CI^-,Br^-,I^-`

Answer

To determine the correct sequence in which the halides will be precipitated when silver nitrate (AgNO₃) is added to a solution containing chloride (Cl⁻), bromide (Br⁻), and iodide (I⁻) ions, we need to consider the solubility products (Ksp) of the silver halides formed.

### Step-by-Step Solution:

1. **Identify the Halides Formed**:
When silver nitrate is added to the solution, it reacts with the halide ions to form silver halides:
- Ag⁺ + Cl⁻ → AgCl (s)
- Ag⁺ + Br⁻ → AgBr (s)
- Ag⁺ + I⁻ → AgI (s)

2. **Understand the Concept of Ksp**:
The solubility product constant (Ksp) is a measure of the solubility of a sparingly soluble salt. The lower the Ksp value, the less soluble the salt is, and thus it precipitates first.

3. **Order of Solubility of Silver Halides**:
The solubility of the silver halides decreases in the following order:
- AgCl > AgBr > AgI
This means AgCl is the most soluble, and AgI is the least soluble.

4. **Order of Precipitation**:
Since the precipitation occurs when the ions exceed their solubility limits, the order of precipitation will be the reverse of their solubility:
- AgI will precipitate first (least soluble),
- followed by AgBr,
- and finally AgCl (most soluble).

5. **Conclusion**:
Therefore, the correct sequence in which the halides will be precipitated is:
- Iodide (I⁻) > Bromide (Br⁻) > Chloride (Cl⁻)

### Final Answer:
The correct sequence in which the halides will be precipitated is: **Iodide (I⁻), Bromide (Br⁻), Chloride (Cl⁻)**.