In a solid, oxide ions are arranged in ccp. Cations 'A' occupy one - sixth of the tetrahedral voids and cations 'B' occupy one - third of the octahedral voids. Which of the following is the correct formula of the oxide?

Options

A. `ABO_(3)`
B. `A_(2)BO_(3)`
C. `AB_(2)O_(3)`
D. None of these

Answer

To determine the correct formula of the oxide given the arrangement of cations in the ccp structure, we will follow these steps:

### Step 1: Determine the number of oxide ions (O²⁻)
In a cubic close-packed (ccp) structure, which is also known as face-centered cubic (fcc), the number of oxide ions can be calculated using the formula:
- Number of oxide ions (O²⁻) = 4

This is because in an fcc unit cell, there are 8 corner atoms contributing 1/8 each and 6 face-centered atoms contributing 1/2 each:
\[ \text{Total contribution} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Step 2: Calculate the number of tetrahedral voids
In a ccp structure, the number of tetrahedral voids is given by:
- Number of tetrahedral voids = 2 × Z, where Z is the number of atoms per unit cell.

For fcc, Z = 4, so:
\[ \text{Number of tetrahedral voids} = 2 \times 4 = 8 \]

### Step 3: Determine the number of cations A in tetrahedral voids
Cations A occupy one-sixth of the tetrahedral voids:
\[ \text{Number of cations A} = \frac{1}{6} \times 8 = \frac{8}{6} = \frac{4}{3} \]

### Step 4: Calculate the number of octahedral voids
In a ccp structure, the number of octahedral voids is given by:
- Number of octahedral voids = Z.

For fcc, Z = 4, so:
\[ \text{Number of octahedral voids} = 4 \]

### Step 5: Determine the number of cations B in octahedral voids
Cations B occupy one-third of the octahedral voids:
\[ \text{Number of cations B} = \frac{1}{3} \times 4 = \frac{4}{3} \]

### Step 6: Write the empirical formula
Now we have:
- Number of A = \( \frac{4}{3} \)
- Number of B = \( \frac{4}{3} \)
- Number of O = 4

To write the empirical formula, we need to find the simplest whole number ratio. We can multiply all coefficients by 3 to eliminate the fractions:
- A: \( 4 \)
- B: \( 4 \)
- O: \( 12 \)

Thus, the empirical formula becomes:
\[ \text{Formula} = A_4B_4O_{12} \]
This can be simplified to:
\[ \text{Formula} = AB_3O_3 \]

### Final Answer
The correct formula of the oxide is \( A_2B_3O_6 \).

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