The total number of ions per unit cell in sphalerite (zinc blend) structure if there is a Schottky defect of one anion per unit cell, is:

Explanation

To determine the total number of ions per unit cell in sphalerite (zinc blend) structure with a Schottky defect of one anion per unit cell, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Structure**:
Sphalerite has a face-centered cubic (FCC) structure. In this structure, both cations (Zn²⁺) and anions (S²⁻) are present.

2. **Calculate the Number of Anions (S²⁻)**:
In a face-centered cubic structure:
- There are 8 corner atoms, each contributing 1/8 to the unit cell.
- There are 6 face-centered atoms, each contributing 1/2 to the unit cell.

Therefore, the total number of S²⁻ ions in the unit cell without defects is:
\[ \text{Total S}^{2-} = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \]

3. **Account for the Schottky Defect**:
A Schottky defect involves the removal of an anion (S²⁻). Since one anion is missing, we subtract one from the total number of S²⁻ ions:
\[ \text{Remaining S}^{2-} = 4 - 1 = 3 \]

4. **Calculate the Number of Cations (Zn²⁺)**:
The number of Zn²⁺ ions in the unit cell is the same as the number of S²⁻ ions in a perfect structure, which is also 4. However, since the Schottky defect maintains the ratio of cations to anions, we also remove one Zn²⁺ ion to keep the ratio 1:1:
\[ \text{Remaining Zn}^{2+} = 4 - 1 = 3 \]

5. **Total Number of Ions in the Unit Cell**:
Now, we sum the remaining ions:
\[ \text{Total ions} = \text{Remaining S}^{2-} + \text{Remaining Zn}^{2+} = 3 + 3 = 6 \]

### Final Answer:
The total number of ions per unit cell in sphalerite with one Schottky defect is **6**.