In a compound,oxide ions are arranged in cubic close packing arrangement. Cations A occupy one-sixth of the tetrahdral voids and cations B occupy one-third of the octahedral voids. The formula of the compound is

Options

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

Answer

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

### Step-by-Step Solution:

1. **Identify the Arrangement of Oxide Ions**:
- The oxide ions are arranged in a cubic close packing (CCP) arrangement, which is also known as face-centered cubic (FCC). In this arrangement, there are 4 oxide ions per unit cell.

2. **Determine the Number of Tetrahedral and Octahedral Voids**:
- In a face-centered cubic (FCC) structure, the number of tetrahedral voids is 8, and the number of octahedral voids is equal to the number of atoms, which is 4.

3. **Calculate the Number of Cations A**:
- Cations A occupy one-sixth of the tetrahedral voids.
- Total tetrahedral voids = 8
- Cations A = (1/6) × 8 = 8/6 = 4/3

4. **Calculate the Number of Cations B**:
- Cations B occupy one-third of the octahedral voids.
- Total octahedral voids = 4
- Cations B = (1/3) × 4 = 4/3

5. **Write the Empirical Formula**:
- From the calculations, we have:
- Cations A = 4/3
- Cations B = 4/3
- Oxide ions = 4
- Thus, the formula can be represented as:
- A(4/3) B(4/3) O(4)

6. **Simplify the Formula**:
- To simplify the formula, multiply all the coefficients by 3 to eliminate the fractions:
- A(4) B(4) O(12)
- This can be factored to:
- 4(A + B) O(3)

7. **Final Formula**:
- The final formula of the compound is:
- A4B4O12, which can be simplified to:
- AB3O4

### Final Answer:
The formula of the compound is **AB3O4**.

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