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 find the formula of the mixed oxide in a cubic close packed (CCP) structure, we need to analyze the contributions of the different ions in the structure.

### Step-by-Step Solution:

1. **Identify the Structure**:
- The cubic close packed (CCP) structure is also known as face-centered cubic (FCC). In an FCC unit cell, the number of atoms (Z) is 4.

2. **Determine the Number of Tetrahedral and Octahedral Voids**:
- In an FCC structure, the number of tetrahedral voids is given by \(2n\) and the number of octahedral voids is given by \(n\), where \(n\) is the number of atoms per unit cell.
- Since \(Z = 4\) for FCC, the number of tetrahedral voids is \(2 \times 4 = 8\) and the number of octahedral voids is \(4\).

3. **Calculate the Contribution of Ions**:
- **Divalent ions (X²⁺)**: It is given that one-eighth of the tetrahedral voids are occupied by X²⁺ ions.
\[ \text{Number of X²⁺ ions} = \frac{1}{8} \times 8 = 1 \]
- **Trivalent ions (Y³⁺)**: It is given that one-half of the octahedral voids are occupied by Y³⁺ ions.
\[ \text{Number of Y³⁺ ions} = \frac{1}{2} \times 4 = 2 \]

4. **Count the Oxide Ions**:
- The lattice is made up of oxide ions (O²⁻). Since there are 4 atoms in the unit cell, there will be 4 oxide ions.
\[ \text{Number of O²⁻ ions} = 4 \]

5. **Write the Empirical Formula**:
- Now we can combine the contributions of the ions to write the empirical formula. The formula will be based on the number of each type of ion:
\[ \text{Formula} = X^{2+}Y^{3+}O^{2-} \Rightarrow X_1Y_2O_4 \]
- This simplifies to:
\[ \text{Formula} = XY_2O_4 \]

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