In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one eight of tetrahedral voids are occupied by divalent `(X^(2+))` ions, while one - half of the octahedral voids are occupied by trivalent ions `(Y^(3+))`, then the formula of the oxide is

Options

A. `XY_(2)O_(4)`
B. `X_(2)YO_(4)`
C. `X_(4)Y_(5)O_(10)`
D. `X_(5)Y_(4)O_(10)`

Answer

To determine the formula of the mixed oxide in a cubic close-packed (CCP) structure, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure**:
- The cubic close-packed structure is equivalent to a face-centered cubic (FCC) structure. In this structure, the oxide ions (O²⁻) form the lattice.

2. **Calculate the Number of Oxide Ions**:
- In an FCC unit cell, there are 4 oxide ions per unit cell. This is calculated as follows:
- There are 8 corner atoms, each contributing 1/8 to the unit cell (8 * 1/8 = 1).
- There are 6 face-centered atoms, each contributing 1/2 to the unit cell (6 * 1/2 = 3).
- Total = 1 + 3 = 4 oxide ions (O²⁻) per unit cell.

3. **Determine the Number of Tetrahedral and Octahedral Voids**:
- The number of tetrahedral voids in an FCC structure is twice the number of atoms per unit cell. Thus, there are:
- Tetrahedral voids = 2 * 4 = 8.
- The number of octahedral voids is equal to the number of atoms per unit cell, which is:
- Octahedral voids = 4.

4. **Occupancy of Tetrahedral Voids**:
- It is given that one-eighth of the tetrahedral voids are occupied by divalent ions (X²⁺).
- Therefore, the number of tetrahedral voids occupied by X²⁺ ions is:
- Occupied tetrahedral voids = 1/8 * 8 = 1.

5. **Occupancy of Octahedral Voids**:
- It is given that one-half of the octahedral voids are occupied by trivalent ions (Y³⁺).
- Therefore, the number of octahedral voids occupied by Y³⁺ ions is:
- Occupied octahedral voids = 1/2 * 4 = 2.

6. **Determine the Ratio of Ions**:
- From the above calculations, we have:
- Number of X²⁺ ions = 1
- Number of Y³⁺ ions = 2
- Number of O²⁻ ions = 4
- Thus, the ratio of X:Y:O is 1:2:4.

7. **Write the Formula**:
- The formula of the oxide can be written based on the ratio of the ions:
- The formula is \( X_1Y_2O_4 \) or simply \( XY_2O_4 \).

### Final Answer:
The formula of the oxide is \( XY_2O_4 \).