In a cubic packed structure of mixed oxides the lattice is made up of oxide ions. One fifth 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_(3)YZ`

Explanation

To find the formula of the oxide in the given cubic packed structure, we will follow these steps:

### Step 1: Determine the number of oxide ions
In a cubic close-packed (CCP) structure, which is equivalent to face-centered cubic (FCC), the number of oxide ions (O²⁻) can be calculated as follows:
- There are 8 atoms at the corners and 6 atoms at the face centers.
- Each corner atom contributes 1/8 to the unit cell, and each face-centered atom contributes 1/2.

Total number of oxide ions:
\[ \text{Number of O}^{2-} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Step 2: Calculate the number of tetrahedral voids
In a cubic close-packed structure, the number of tetrahedral voids is given by:
\[ \text{Number of tetrahedral voids} = 2 \times \text{Number of atoms in the unit cell} = 2 \times 4 = 8 \]

### Step 3: Calculate the number of octahedral voids
The number of octahedral voids in a cubic close-packed structure is given by:
\[ \text{Number of octahedral voids} = \text{Number of atoms in the unit cell} = 4 \]

### Step 4: Determine the number of divalent X²⁺ ions in tetrahedral voids
According to the problem, one-fifth of the tetrahedral voids are occupied by divalent X²⁺ ions:
\[ \text{Number of X}^{2+} = \frac{1}{5} \times \text{Number of tetrahedral voids} = \frac{1}{5} \times 8 = \frac{8}{5} \]

### Step 5: Determine the number of trivalent Y³⁺ ions in octahedral voids
According to the problem, one-half of the octahedral voids are occupied by trivalent Y³⁺ ions:
\[ \text{Number of Y}^{3+} = \frac{1}{2} \times \text{Number of octahedral voids} = \frac{1}{2} \times 4 = 2 \]

### Step 6: Write the empirical formula
Now, we can write the empirical formula based on the number of each ion and the oxide ions:
- For X²⁺: \(\frac{8}{5}\)
- For Y³⁺: \(2\)
- For O²⁻: \(4\)

The formula can be expressed as:
\[ \text{Formula} = \left( \frac{8}{5} \text{X}^{2+} \right) \left( 2 \text{Y}^{3+} \right) \left( 4 \text{O}^{2-} \right) \]

### Step 7: Simplify the formula
To simplify, we can multiply through by 5 to eliminate the fraction:
\[ \text{Formula} = \text{X}^{8} \text{Y}^{10} \text{O}^{20} \]

Now, dividing each term by 2 to simplify further:
\[ \text{Formula} = \text{X}^{4} \text{Y}^{5} \text{O}^{10} \]

### Final Answer
Thus, the formula of the oxide is:
\[ \text{X}_4\text{Y}_5\text{O}_{10} \]