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

Options

A. AB
B. `AB_2`
C. `A_2B`
D. `A_2B_3`

Answer

To find the formula of the ionic solid given the arrangement of anions and cations, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**:
- The anions (B) are arranged in a cubic close-packed (CCP) structure.

2. **Determine the Number of Anions**:
- In a CCP structure, the number of anions present is 4. This is because there are 4 atoms per unit cell in a CCP arrangement.

3. **Calculate the Number of Octahedral Voids**:
- The number of octahedral voids in a CCP structure is equal to the number of atoms present, which is also 4. Therefore, there are 4 octahedral voids.

4. **Calculate the Number of Tetrahedral Voids**:
- The number of tetrahedral voids is always twice the number of octahedral voids. Hence, there are \(2 \times 4 = 8\) tetrahedral voids.

5. **Distribution of Cations**:
- The cations (A) are distributed equally between the octahedral and tetrahedral voids. Since all octahedral voids are occupied, we have 4 cations in the octahedral voids.

6. **Determine the Number of Cations in Tetrahedral Voids**:
- Since the cations are equally distributed, there will also be 4 cations in the tetrahedral voids.

7. **Total Number of Cations**:
- The total number of cations (A) is the sum of those in octahedral and tetrahedral voids:
\[ \text{Total Cations} = 4 \text{ (from octahedral)} + 4 \text{ (from tetrahedral)} = 8 \]

8. **Write the Ratio of Cations to Anions**:
- Now we have 8 cations (A) and 4 anions (B). The ratio of cations to anions is:
\[ \text{Ratio of A to B} = \frac{8}{4} = 2:1 \]

9. **Formulate the Ionic Compound**:
- Based on the ratio, the formula of the ionic solid can be written as \(A_2B\).

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

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