In certain solid, the oxide ions are arranged in ccp. Cations A occupy `(1)/(6)` of the tetrahedral voids and cations B occupy one third of the octahedral voids. The probable formula of the compound is

Options

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

Answer

To determine the probable formula of the compound based on the arrangement of oxide ions and the occupancy of cations A and B in the tetrahedral and octahedral voids, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the number of oxide ions in the unit cell:**
- Since the oxide ions are arranged in a cubic close-packed (ccp) structure, which is equivalent to face-centered cubic (FCC), the number of oxide ions (O²⁻) per unit cell (Z) is 4.

2. **Identify the number of tetrahedral and octahedral voids:**
- In a ccp structure, the number of tetrahedral voids is 8 and the number of octahedral voids is 4.

3. **Calculate the number of cations A in tetrahedral voids:**
- Cation A occupies \( \frac{1}{6} \) of the tetrahedral voids.
- Total tetrahedral voids = 8.
- Therefore, the number of A cations = \( \frac{1}{6} \times 8 = \frac{8}{6} = \frac{4}{3} \).

4. **Calculate the number of cations B in octahedral voids:**
- Cation B occupies \( \frac{1}{3} \) of the octahedral voids.
- Total octahedral voids = 4.
- Therefore, the number of B cations = \( \frac{1}{3} \times 4 = \frac{4}{3} \).

5. **Summarize the composition:**
- From the calculations:
- Number of A cations = \( \frac{4}{3} \)
- Number of B cations = \( \frac{4}{3} \)
- Number of O²⁻ ions = 4

6. **Formulate the empirical formula:**
- The formula can be expressed as \( A_{\frac{4}{3}} B_{\frac{4}{3}} O_{4} \).
- To simplify, multiply all coefficients by 3 to eliminate the fractions:
- \( A_{4} B_{4} O_{12} \).

7. **Further simplification:**
- Divide all coefficients by 4:
- \( A_{1} B_{1} O_{3} \).
- Thus, the empirical formula is \( ABO_{3} \).

### Final Answer:
The probable formula of the compound is \( ABO_{3} \).