In a solid between A and B atoms of A are arranged in ccp array and atoms of B occupy all the octahedral voids and half of the tetrahedral voids. The formula of the compound is

Options

A. `PQ_(2)`
B. `P_(2)Q`
C. PQ
D. `P_(2)Q_(2)`

Answer

To determine the formula of the compound formed by atoms A and B, we can follow these steps:

### Step 1: Determine the number of A atoms in the CCP array
- A atoms are arranged in a close-packed structure, specifically a cubic close-packed (CCP) or face-centered cubic (FCC) arrangement.
- In an FCC unit cell, the number of atoms can be calculated as follows:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.

Calculation:
\[ \text{Total A atoms} = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \]

### Step 2: Determine the number of B atoms occupying voids
- In a CCP structure, the number of octahedral voids is equal to the number of atoms, which is 4.
- The number of tetrahedral voids is double the number of octahedral voids, so there are \( 2 \times 4 = 8 \) tetrahedral voids.
- B occupies all the octahedral voids and half of the tetrahedral voids:
- All octahedral voids: 4
- Half of the tetrahedral voids: \( \frac{8}{2} = 4 \)

Calculation:
\[ \text{Total B atoms} = 4 + 4 = 8 \]

### Step 3: Write the empirical formula
- Now, we have 4 A atoms and 8 B atoms in the formula.
- The empirical formula can be simplified by dividing both subscripts by their greatest common divisor, which is 4.

Calculation:
\[ \text{Empirical formula} = A_4B_8 \rightarrow \frac{A_4B_8}{4} = AB_2 \]

### Final Answer
The formula of the compound is \( \text{AB}_2 \).

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