In a solid, oxide ions are arrnged in ccp, cations A occupy A occupy `(1/8)^th` of the tetrahedral voids and cation B occupy `(1/4)^(th)` of the octahedral voids. The formula of the compound is:

Options

A. `ABO_(4)`
B. `AB_(2)O_(3)`
C. `A_(@)BO_(4)`
D. `AB_(4)O_(4)`

Answer

To determine the formula of the compound formed by oxide ions, cation A, and cation B in a solid where oxide ions are arranged in cubic close packing (CCP), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Arrangement**:
- The oxide ions (O²⁻) are arranged in a cubic close packing (CCP) structure, which is also known as face-centered cubic (FCC).

2. **Calculating the Number of Oxide Ions**:
- In a CCP structure, the number of ions per unit cell is 4 (1 from each of the 8 corners and 3 from the 6 face centers).
- Therefore, the number of oxide ions (O²⁻) in the unit cell is 4.

3. **Identifying Tetrahedral and Octahedral Voids**:
- The number of tetrahedral voids in a CCP structure is given by the formula:
\[ \text{Number of Tetrahedral Voids} = 2 \times \text{Number of Atoms in CCP} = 2 \times 4 = 8 \]
- The number of octahedral voids is equal to the number of atoms in the CCP structure, which is 4.

4. **Calculating Cations Occupying Voids**:
- Cation A occupies \( \frac{1}{8} \) of the tetrahedral voids:
\[ \text{Cation A} = \frac{1}{8} \times 8 = 1 \]
- Cation B occupies \( \frac{1}{4} \) of the octahedral voids:
\[ \text{Cation B} = \frac{1}{4} \times 4 = 1 \]

5. **Writing the Formula**:
- Now we have:
- Number of A cations = 1
- Number of B cations = 1
- Number of O²⁻ ions = 4
- Therefore, the formula of the compound can be written as:
\[ \text{Formula} = A^1B^1O^{2-}_4 = ABO_4 \]

### Final Answer:
The formula of the compound is **ABO₄**.