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_2BO_4`
B. `AB_2O_4`
C. `A_2B_2O`
D. `A_4B_2O`

Explanation

To solve the question regarding the structure of the bimetallic oxide formed by elements A and B, 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 lattice, the number of atoms per unit cell is 4.

### Step 2: Determine the number of octahedral and tetrahedral voids
In a CCP structure:
- The number of octahedral voids (OV) is equal to the number of atoms, which is 4.
- The number of tetrahedral voids (TV) is double the number of atoms, which is 2 * 4 = 8.

### Step 3: Determine the number of atoms of A
According to the problem, element A occupies half of the octahedral voids. Since there are 4 octahedral voids:
- Number of A atoms = 1/2 * 4 = 2.

### Step 4: Determine the number of atoms of oxygen
The problem states that oxygen atoms occupy all the tetrahedral voids. Since there are 8 tetrahedral voids:
- Number of oxygen atoms = 8.

### Step 5: Write the empirical formula
Now we can summarize the number of atoms:
- A: 2 atoms
- B: 4 atoms
- O: 8 atoms

Thus, the empirical formula of the bimetallic oxide can be written as:
- A2B4O8

To simplify the formula, we can divide all the subscripts by 2:
- AB2O4

### Final Answer
The structure of the bimetallic oxide is represented by the formula **AB2O4**.

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