'A' has fcc arrangement, 'B' is present in `2//3^(rd)` of tetrahedral voids. The formula of the compound will be

Options

A. `A_(2)B_(3)`
B. `A_(4)B_(3)`
C. `A_(3)B_(4)`
D. `A_(3)B_(2)`

Answer

To find the formula of the compound formed by 'A' and 'B', we will follow these steps:

### Step 1: Determine the number of 'A' atoms in the FCC arrangement.
In a face-centered cubic (FCC) arrangement:
- 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 total number of 'A' atoms:
\[ \text{Total A atoms} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]
So, there are 4 'A' atoms in the unit cell.

### Step 2: Determine the number of tetrahedral voids.
In an FCC structure, the number of tetrahedral voids is given by:
\[ \text{Number of tetrahedral voids} = 2n \]
where \( n \) is the number of atoms per unit cell. Since we have determined that there are 4 'A' atoms:
\[ \text{Number of tetrahedral voids} = 2 \times 4 = 8 \]

### Step 3: Calculate the number of 'B' atoms.
According to the problem, 'B' occupies \( \frac{2}{3} \) of the tetrahedral voids. Therefore, the number of 'B' atoms is:
\[ \text{Number of B atoms} = \frac{2}{3} \times 8 = \frac{16}{3} \]

### Step 4: Write the empirical formula.
We have:
- Number of 'A' atoms = 4
- Number of 'B' atoms = \( \frac{16}{3} \)

To express the formula in whole numbers, we can multiply both coefficients by 3 to eliminate the fraction:
\[ \text{A: } 4 \times 3 = 12 \quad \text{and} \quad \text{B: } \frac{16}{3} \times 3 = 16 \]

Thus, the formula becomes:
\[ \text{A}_{12} \text{B}_{16} \]

### Step 5: Simplify the formula.
To simplify \( \text{A}_{12} \text{B}_{16} \), we divide both subscripts by 4:
\[ \text{A}_{3} \text{B}_{4} \]

### Final Answer:
The formula of the compound is \( \text{A}_{3} \text{B}_{4} \).

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