AgCl is crystallized from molten AgCl containing a little `CdCl_(2)`. The solid obtained will have

Options

A. cationic vacancies equal to number of `Cd^(2+)` ions incorporated
B. cationic vacancies equal to double the number of `Cd^(2+)` ions
C. anionic vacancies
D. neither cationic nor anionic vacancles

Answer

To solve the question regarding the crystallization of AgCl from molten AgCl containing a little CdCl₂, we need to analyze the situation step by step.

### Step-by-Step Solution:

1. **Understanding the Doping Process**:
- When AgCl is doped with CdCl₂, the Cd²⁺ ions from CdCl₂ will replace some of the Ag⁺ ions in the crystal lattice of AgCl.

2. **Substitution of Ions**:
- Each Cd²⁺ ion has a +2 charge, while each Ag⁺ ion has a +1 charge. To maintain charge neutrality in the crystal lattice, one Cd²⁺ ion will replace two Ag⁺ ions.

3. **Formation of Vacancies**:
- When one Cd²⁺ ion replaces two Ag⁺ ions, it creates a vacancy for the Ag⁺ ions that were replaced. Therefore, for every Cd²⁺ ion incorporated into the lattice, one vacancy is created for the Ag⁺ ions.

4. **Counting Vacancies**:
- If 'n' is the number of Cd²⁺ ions incorporated into the crystal, then the number of vacancies created will also be 'n' because each Cd²⁺ replaces two Ag⁺ ions but creates only one vacancy.

5. **Conclusion**:
- Thus, the solid obtained will have cationic vacancies equal to the number of Cd²⁺ ions incorporated. This leads us to conclude that the correct answer is option A: "Cationic vacancies equal to the number of Cd²⁺ ions incorporated."

### Final Answer:
The solid obtained will have cationic vacancies equal to the number of Cd²⁺ ions incorporated.

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