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

Answer

To determine the formula of the mixed oxide based on the given information, we will follow these steps:

### Step 1: Determine the number of oxide ions in the cubic close-packed (CCP) structure.
In a CCP structure, oxide ions occupy the lattice points. The structure consists of:
- 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom.
- 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom.

Calculating the total number of oxide ions:
\[ \text{Number of oxide ions} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]
Thus, there are 4 oxide ions in the unit cell.

### Step 2: Determine the number of divalent metal A ions in the tetrahedral voids.
In a CCP structure, there are a total of 8 tetrahedral voids. Given that one-fourth of these voids are occupied by divalent metal A:
\[ \text{Number of A ions} = \frac{1}{4} \times 8 = 2 \]

### Step 3: Determine the number of monovalent metal B ions in the octahedral voids.
In a CCP structure, there are a total of 4 octahedral voids. The octahedral voids consist of:
- 1 body-centered atom (contributes 1).
- 12 edge-centered atoms, each contributing \( \frac{1}{4} \).

Calculating the total number of octahedral voids:
\[ \text{Number of octahedral voids} = 1 + 12 \times \frac{1}{4} = 1 + 3 = 4 \]
Since all octahedral voids are occupied by monovalent metal B, we have:
\[ \text{Number of B ions} = 4 \]

### Step 4: Write the empirical formula of the mixed oxide.
Now we can compile the number of each type of ion:
- Number of A ions = 2
- Number of B ions = 4
- Number of oxide ions = 4

The empirical formula can be written as:
\[ \text{Formula} = A_2B_4O_4 \]

### Step 5: Simplify the formula.
To simplify the formula \( A_2B_4O_4 \), we can divide all subscripts by 2:
\[ \text{Simplified formula} = AB_2O_2 \]

### Final Answer:
The formula of the oxide is \( AB_2O_2 \).

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