Element `'B'` forms `ccp` structure and `'A'` occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is:

Options

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

Answer

To solve the problem, we need to analyze the information given about the elements and their positions in the crystal structure. Let's break it down step by step:

### Step 1: Identify the structure of element B
Element B forms a cubic close-packed (CCP) structure. In a CCP structure, also known as face-centered cubic (FCC), the number of atoms per unit cell (z) is 4.

### Step 2: Determine the number of voids
In a face-centered cubic structure, the number of octahedral voids is equal to the number of atoms in the unit cell, which is 4. The number of tetrahedral voids is double that of octahedral voids, so there are 8 tetrahedral voids in a CCP structure.

### Step 3: Analyze the occupancy of voids
According to the problem:
- Element A occupies half of the octahedral voids.
- Oxygen atoms occupy all the tetrahedral voids.

Since there are 4 octahedral voids in total, half of that would mean:
- A occupies 2 octahedral voids.

### Step 4: Determine the number of oxygen atoms
Since oxygen occupies all the tetrahedral voids, and there are 8 tetrahedral voids, this means:
- There are 8 oxygen atoms.

### Step 5: Write the formula for the bimetallic oxide
Now we can summarize the contributions from each element:
- A contributes 2 (from octahedral voids).
- B contributes 4 (from the CCP structure).
- O contributes 8 (from tetrahedral voids).

Thus, the empirical formula for the bimetallic oxide can be written as:
\[ \text{A}_2\text{B}_4\text{O}_8 \]

### Step 6: Simplify the formula
To simplify the formula, we can divide all the subscripts by 2:
\[ \text{A}_1\text{B}_2\text{O}_4 \]
This gives us the final formula:
\[ \text{AB}_2\text{O}_4 \]

### Conclusion
The structure of the bimetallic oxide is \( \text{AB}_2\text{O}_4 \).