`O_(2-)` ions are arranged in `ccp` in spinel strructure. `A^(2+)` ions occupy `1//8` of `TV_(s)` and `B^(o+)` ions occupy half of `OV `. The void volume of unit cell `= 0.11 Å`. Find the value of `A`.

Explanation

To solve the problem, we need to analyze the arrangement of ions in the spinel structure and apply the given information to find the value of \( A \).

### Step-by-Step Solution:

1. **Understanding the Spinel Structure**:
- In a spinel structure, \( O^{2-} \) ions are arranged in a cubic close-packed (ccp) manner.
- The unit cell contains \( 8 \) tetrahedral voids (TV) and \( 4 \) octahedral voids (OV).

2. **Identifying the Occupation of Ions**:
- \( A^{2+} \) ions occupy \( \frac{1}{8} \) of the tetrahedral voids.
- \( B^{3+} \) ions occupy half of the octahedral voids.

3. **Calculating the Number of Ions**:
- Since there are \( 8 \) tetrahedral voids, the number of \( A^{2+} \) ions is:
\[ \text{Number of } A^{2+} = 8 \times \frac{1}{8} = 1 \]
- Since there are \( 4 \) octahedral voids, the number of \( B^{3+} \) ions is:
\[ \text{Number of } B^{3+} = 4 \times \frac{1}{2} = 2 \]

4. **Calculating the Void Volume**:
- The void volume of the unit cell is given as \( 0.11 \, \text{Å}^3 \).
- The relationship between the void volume and the unit volume of the unit cell is given by:
\[ \text{Void Volume} \times \text{Unit Volume} = 0.22 \]

5. **Finding the Unit Volume**:
- Let \( V \) be the unit volume of the unit cell. We can set up the equation:
\[ 0.11 \times V = 0.22 \]
- Solving for \( V \):
\[ V = \frac{0.22}{0.11} = 2 \, \text{Å}^3 \]

6. **Finding the Value of \( A \)**:
- The unit volume \( V \) is related to the edge length \( a \) of the cubic unit cell by the formula:
\[ V = a^3 \]
- Therefore, we can find \( a \):
\[ a^3 = 2 \implies a = 2^{1/3} \]
- Since we are looking for the value of \( A \) in terms of the unit cell edge length, we can conclude that:
\[ A = 2 \]

### Final Answer:
The value of \( A \) is \( 2 \).