Oxygen atom forms FCC unit cell with 'A ' atoms occupying all tetrahedral voids and 'B ' atoms occupying all octahedral voids. If atoms are removed from two of the body diagonals then determine the formula of resultant compound formed .

Options

A. `A_4B_4O_7`
B. `A_8B_6O_7`
C. `A_8B_8O_7`
D. `A_6B_8O_6`

Answer

To solve the problem, we need to analyze the structure of the FCC unit cell and the distribution of atoms in the tetrahedral and octahedral voids, as well as the effect of removing atoms from the body diagonals.

### Step-by-Step Solution:

1. **Understanding the FCC Structure**:
- In a Face-Centered Cubic (FCC) unit cell, oxygen (O) atoms are located at the corners and face centers of the cube.
- There are 8 corner atoms (each contributes 1/8) and 6 face-centered atoms (each contributes 1/2).
- Total contribution of O atoms = \(8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4\) O atoms per unit cell.

2. **Identifying Tetrahedral and Octahedral Voids**:
- In an FCC structure, there are 8 tetrahedral voids and 4 octahedral voids.
- Tetrahedral voids are located at positions that are \( \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \) and its equivalents.
- Octahedral voids are located at the center of the cube and at the midpoints of the edges.

3. **Filling the Voids**:
- 'A' atoms occupy all tetrahedral voids. Since there are 8 tetrahedral voids, there will be 8 A atoms.
- 'B' atoms occupy all octahedral voids. Since there are 4 octahedral voids, there will be 4 B atoms.

4. **Removing Atoms from Body Diagonals**:
- The body diagonal of the cube connects opposite corners. Each body diagonal contains 2 tetrahedral voids.
- If we remove atoms from 2 body diagonals, we are effectively removing 4 A atoms (2 from each diagonal).
- After removing these, the number of A atoms left = \(8 - 4 = 4\).

5. **Counting Remaining Atoms**:
- The number of B atoms remains unchanged at 4 since we are not removing any B atoms.
- The number of O atoms remains unchanged at 4.

6. **Determining the Formula**:
- After the removal of atoms, we have:
- A = 4
- B = 4
- O = 4
- Therefore, the formula of the resultant compound can be expressed as \( A_4B_4O_4 \).
- This can be simplified to \( A_4B_4O_4 \) or \( A_1B_1O_1 \) if we divide by 4.

### Final Result:
The formula of the resultant compound formed is \( A_4B_4O_4 \).

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