A spinel is an important class of oxide consisting two types of metal ions withs the oxide ions arranged in ccp layers . The normal spinel has one -eight of the tetrahendral holes occupied by one type of metal ions and one- half of the octaherdral holes occupied by another type of metal ion. Such a spine is formed by `Mg^(2+), Al^(3+) and O^(2-)` . The netutrality of the crystal is benig maintained.
If oxide ion is replaced by `X^(-8//3)`, the number of anionic vacancy per unit cells is :

Options

A. 1
B. 2
C. 3
D. `3//4`

Answer

To solve the problem regarding the spinel structure and the replacement of oxide ions with `X^(-8/3)`, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Spinel Structure**:
- A normal spinel structure consists of oxide ions arranged in a cubic close-packed (CCP) or face-centered cubic (FCC) lattice.
- In this structure, there are tetrahedral and octahedral voids where metal ions can occupy.

2. **Calculating the Number of Oxide Ions**:
- In a CCP structure, the total number of oxide ions (O^2-) can be calculated as follows:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) to the unit cell.
- Therefore, the total number of oxide ions in a unit cell is:
\[ \text{Total O}^{2-} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

3. **Charge Balance**:
- The total charge contributed by the 4 oxide ions is:
\[ \text{Total charge from O}^{2-} = 4 \times (-2) = -8 \]
- The spinel structure maintains electrical neutrality, which means the positive charges from the metal ions must balance this negative charge.

4. **Replacement of Oxide Ions**:
- If one oxide ion (O^2-) is replaced by `X^(-8/3)`, we need to determine how many such replacements occur.
- The charge contributed by one `X^(-8/3)` ion is:
\[ \text{Charge from X}^{-8/3} = -\frac{8}{3} \]
- If we replace 3 oxide ions with 3 `X^(-8/3)` ions, the total charge from these ions would be:
\[ 3 \times \left(-\frac{8}{3}\right) = -8 \]

5. **Creating Anionic Vacancies**:
- Originally, we had 4 oxide ions contributing a total charge of -8.
- After replacing 3 oxide ions with 3 `X^(-8/3)` ions, we still have a total charge of -8, but we have lost one oxide ion.
- This means there is now one anionic vacancy created in the unit cell because we have one less oxide ion than before.

6. **Conclusion**:
- Therefore, the number of anionic vacancies per unit cell is **1**.

### Final Answer:
The number of anionic vacancies per unit cell is **1**.