In certain solid, the oxide ions are arranged in ccp. Cations A occupy `(1)/(6)` of the tetrahedral voids and cations B occupy one third of the octahedral voids. The probable formula of the compound is

Options

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

Answer

To determine the probable formula of the compound in which oxide ions are arranged in a close-packed structure, we can follow these steps:

### Step 1: Determine the number of oxide ions in the unit cell
In a close-packed structure (CCP), which is equivalent to face-centered cubic (FCC), the number of oxide ions can be calculated as follows:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) to the unit cell.

Calculating the contributions:
\[ \text{Total contribution from corners} = 8 \times \frac{1}{8} = 1 \]
\[ \text{Total contribution from faces} = 6 \times \frac{1}{2} = 3 \]
\[ \text{Total number of oxide ions} = 1 + 3 = 4 \]

### Step 2: Calculate the number of tetrahedral voids
In a face-centered cubic structure, the number of tetrahedral voids is given by:
\[ \text{Number of tetrahedral voids} = 2n \]
where \( n \) is the number of atoms in the unit cell. Since we have 4 oxide ions:
\[ \text{Number of tetrahedral voids} = 2 \times 4 = 8 \]

### Step 3: Determine the number of cation A
Cation A occupies \( \frac{1}{6} \) of the tetrahedral voids. Thus, the number of cation A is:
\[ \text{Number of cation A} = \frac{1}{6} \times 8 = \frac{8}{6} = \frac{4}{3} \]

### Step 4: Calculate the number of octahedral voids
In a face-centered cubic structure, the number of octahedral voids is equal to the number of atoms in the unit cell:
\[ \text{Number of octahedral voids} = n = 4 \]

### Step 5: Determine the number of cation B
Cation B occupies \( \frac{1}{3} \) of the octahedral voids. Thus, the number of cation B is:
\[ \text{Number of cation B} = \frac{1}{3} \times 4 = \frac{4}{3} \]

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

The formula can be expressed as:
\[ \text{Formula} = A_{\frac{4}{3}} B_{\frac{4}{3}} O_4 \]

### Step 7: Convert to integer ratio
To convert the formula to an integer ratio, we multiply all subscripts by 3:
\[ \text{Formula} = A_4 B_4 O_{12} \]

### Step 8: Simplify the formula
This simplifies to:
\[ \text{Formula} = ABO_3 \]

Thus, the probable formula of the compound is **ABO3**.

---