In an ionic compound `XY_(2)O_(4)` oxide ions are arranged in ccp. Cations `X` are present in octahedral voids.Cations `Y` are equally distributed between octahedral voids occupied is

Options

A. `1/2`
B. `1/4`
C. `1/8`
D. `1/6`

Answer

To solve the question regarding the ionic compound \( XY_2O_4 \) where oxide ions are arranged in cubic close packing (CCP) and cations \( X \) and \( Y \) occupy octahedral voids, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Structure**:
- In cubic close packing (CCP), the oxide ions \( O^{2-} \) are arranged in a close-packed structure. Each unit cell of CCP contains 4 oxide ions.

2. **Identify the Types of Voids**:
- In a CCP structure, there are two types of voids: tetrahedral and octahedral.
- The number of octahedral voids per unit cell is equal to the number of atoms (or ions) in the unit cell, which is 4 in this case.

3. **Cation Occupancy**:
- According to the question, cation \( X \) occupies octahedral voids.
- It is stated that cation \( Y \) is also distributed among the octahedral voids.

4. **Calculate the Occupancy**:
- Since there are 4 octahedral voids in the unit cell and cation \( X \) occupies some of them, we need to determine how many are occupied by \( Y \).
- The question implies that both \( X \) and \( Y \) are present in the octahedral voids.

5. **Distribution of Cations**:
- The total number of octahedral voids is 4. If \( X \) occupies some of these voids and \( Y \) is equally distributed, we can assume that \( X \) occupies 2 voids (as there are 2 \( X \) ions in the formula \( XY_2O_4 \)) and \( Y \) occupies the remaining voids.
- Therefore, if \( X \) occupies 2 out of 4 octahedral voids, then \( Y \) must occupy the remaining 2 octahedral voids.

6. **Conclusion**:
- Since 2 out of 4 octahedral voids are occupied by \( X \) and \( Y \), we conclude that 50% of the octahedral voids are occupied.

### Final Answer:
- The answer is that half of the octahedral voids are occupied.