In a crystalline solid, anions `B` are arranged in ccp structure. Cations `A` are equally distributed between octahedral and tetrahedral voids if all the octahedral voids are occupied the formula of the ionic solids will be

Explanation

To determine the formula of the ionic solid based on the arrangement of anions and cations, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Arrangement of Anions**:
- The anions `B` are arranged in a cubic close-packed (CCP) structure. In a CCP structure, the number of anions per unit cell is 4.

2. **Determine the Number of Voids**:
- In a CCP structure, there are two types of voids:
- **Octahedral voids**: There are 4 octahedral voids per unit cell.
- **Tetrahedral voids**: There are 8 tetrahedral voids per unit cell.

3. **Distribution of Cations**:
- The cations `A` are distributed equally between the octahedral and tetrahedral voids. This means that half of the cations will occupy octahedral voids and the other half will occupy tetrahedral voids.

4. **Occupancy of Octahedral Voids**:
- According to the problem, all octahedral voids are occupied. Since there are 4 octahedral voids, this means there are 4 cations `A` in the octahedral voids.

5. **Occupancy of Tetrahedral Voids**:
- Since the cations are equally distributed, and there are 4 octahedral voids occupied by `A`, there will also be 4 cations `A` occupying the tetrahedral voids.

6. **Calculate Total Cations**:
- The total number of cations `A` is the sum of those in octahedral and tetrahedral voids:
\[ \text{Total } A = \text{Cations in octahedral voids} + \text{Cations in tetrahedral voids} = 4 + 4 = 8 \]

7. **Write the Formula**:
- The formula for the ionic solid can be written as:
\[ A_8B_4 \]

8. **Simplify the Formula**:
- To simplify the formula, divide the subscripts by the greatest common divisor:
\[ A_8B_4 \rightarrow A_2B \]

### Final Answer:
The formula of the ionic solid is \( A_2B \).