Which one of the following compounds shows both, Frenkel as well as Schottky defects?

Options

A. AgBr
B. AgI
C. NaCl
D. ZnS

Answer

To determine which compound shows both Frenkel and Schottky defects, we need to understand the characteristics of each type of defect and the conditions under which they occur.

### Step-by-Step Solution:

1. **Understanding Schottky Defect**:
- In a Schottky defect, equal numbers of cations and anions are missing from the crystal lattice. This results in vacancies that maintain the stoichiometry of the compound.
- Conditions for Schottky defects:
- The ionic sizes of cations and anions should be similar.
- The coordination number should be relatively low.

2. **Understanding Frenkel Defect**:
- In a Frenkel defect, a cation is displaced from its normal position to an interstitial site, creating a vacancy where the cation was originally located.
- Conditions for Frenkel defects:
- There should be a significant size difference between the cation and anion.
- The compound should have interstitial sites available for the displaced cation.

3. **Identifying Suitable Compounds**:
- We need to find a compound that can exhibit both types of defects. A good candidate would be a compound where the cation and anion sizes are similar enough to allow Schottky defects, but also have a significant size difference to allow for Frenkel defects.

4. **Analyzing AgBr (Silver Bromide)**:
- AgBr is a compound that meets the criteria:
- The ionic radii of Ag⁺ and Br⁻ are similar enough to allow for Schottky defects.
- The size difference between Ag⁺ and Br⁻ is significant enough to allow for Frenkel defects.
- The coordination number of AgBr is 6, which is suitable for both defects.

5. **Conclusion**:
- Therefore, the compound that shows both Frenkel and Schottky defects is **AgBr (Silver Bromide)**.