Structure of a mixed oxide is cubic closed - packed (ccp) .The cubic unit cell of mixed oxide is composed of oxide ions .One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovelent metal B .The formula of the oxide is

Options

A. `A_(2)BO_(2)`
B. `A_(2)B_(3)O_(4)`
C. `AB_(2)O_(2)`
D. `ABO_(2)`

Explanation

To solve the problem, we need to determine the formula of the mixed oxide based on the information provided about the arrangement of ions in the cubic closed-packed (ccp) structure. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the number of oxide ions in the unit cell
In a cubic closed-packed (ccp) structure, the oxide ions are arranged in a face-centered cubic (FCC) lattice.

- There are 8 oxide ions located at the corners of the cube, each contributing \( \frac{1}{8} \) to the unit cell.
- There are 6 oxide ions located at the center of each face, each contributing \( \frac{1}{2} \) to the unit cell.

**Calculation:**
\[ \text{Total oxide ions} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Step 2: Determine the number of tetrahedral and octahedral voids
In a ccp structure:
- The number of tetrahedral voids is 8.
- The number of octahedral voids is 4.

### Step 3: Determine the occupancy of the voids
According to the problem:
- One-fourth of the tetrahedral voids are occupied by divalent metal A.
- All octahedral voids are occupied by monovalent metal B.

**Calculation:**
- Tetrahedral voids occupied by A:
\[ \text{Occupied tetrahedral voids} = \frac{1}{4} \times 8 = 2 \]
- Octahedral voids occupied by B:
\[ \text{Occupied octahedral voids} = 4 \]

### Step 4: Establish the ratio of ions
Now we can summarize the number of ions:
- Number of oxide ions (O) = 4
- Number of divalent metal ions (A) = 2
- Number of monovalent metal ions (B) = 4

The ratio of A : B : O can be expressed as:
\[ \text{A : B : O} = 2 : 4 : 4 \]
This can be simplified to:
\[ \text{A : B : O} = 1 : 2 : 2 \]

### Step 5: Write the empirical formula
From the ratio established, we can write the empirical formula of the oxide as:
\[ \text{AB}_2\text{O}_2 \]

### Conclusion
The formula of the mixed oxide is \( \text{AB}_2\text{O}_2 \).