If you have to create “one Schottky defect per unit cell of NaCl” by removing ions from least number of lattice points, then those suitable lattice points would be

Options

A. Edge centre and body centre
B. Face centre and body centre
C. Edge centre and face centre
D. Body centre and corners

Explanation

To solve the problem of creating one Schottky defect per unit cell of NaCl by removing ions from the least number of lattice points, we need to follow these steps:

### Step-by-Step Solution:

1. **Understanding the Structure of NaCl**:
- NaCl crystallizes in a face-centered cubic (FCC) lattice structure.
- In the FCC unit cell, there are 4 Na⁺ ions and 4 Cl⁻ ions. This is because there are 8 corners (each contributing 1/8) and 6 face centers (each contributing 1/2).

2. **Identifying the Number of Ions**:
- Total Na⁺ ions = 4 (from 8 corners and 6 face centers)
- Total Cl⁻ ions = 4 (from 8 corners and 6 face centers)

3. **Understanding Schottky Defect**:
- A Schottky defect occurs when an equal number of cations and anions are missing from the lattice, maintaining the stoichiometric ratio.
- To create one Schottky defect, we need to remove one Na⁺ ion and one Cl⁻ ion.

4. **Choosing Lattice Points to Remove Ions**:
- The least number of lattice points would be to remove one Na⁺ ion from the body center (which is an octahedral void) and two Cl⁻ ions from the face centers.
- This is because the face centers contain Cl⁻ ions, and removing two Cl⁻ ions from the face centers will maintain the required stoichiometric ratio.

5. **Calculating Remaining Ions**:
- After removing 1 Na⁺ from the body center, we have:
- Remaining Na⁺ = 4 - 1 = 3
- After removing 2 Cl⁻ from the face centers, we have:
- Remaining Cl⁻ = 4 - 2 = 2

6. **Checking the Stoichiometric Ratio**:
- Remaining Na⁺: 3
- Remaining Cl⁻: 2
- The ratio of Na⁺ to Cl⁻ is now 3:2, which does not maintain the 1:1 ratio required for a Schottky defect.
- Therefore, we need to ensure that we remove ions in a way that keeps the ratio balanced.

7. **Final Adjustment**:
- To maintain the 1:1 ratio, we can remove:
- 1 Na⁺ from the body center
- 1 Cl⁻ from one of the face centers
- This gives us:
- Remaining Na⁺ = 4 - 1 = 3
- Remaining Cl⁻ = 4 - 1 = 3
- Now the ratio is 3:3 or 1:1, which is acceptable.

### Conclusion:
The suitable lattice points to create one Schottky defect per unit cell of NaCl by removing ions from the least number of lattice points would be:
- Remove 1 Na⁺ from the body center.
- Remove 1 Cl⁻ from one of the face centers.