Question

The anions (A) of \(C_pA_q\) molecule forms an fcc lattice. Cations (C) are positioned at the body center and half of the edge centers. The formula of the molecule is:

• A. \(CA\)
• B. \(CA_2\)
• C. \(C_3A_4\)
• D. \(C_5A_8\)

Hint:

In crystal structure problems, always remember: \[ \text{Corner atom contribution}=\frac{1}{8} \] \[ \text{Face center contribution}=\frac{1}{2} \] \[ \text{Edge center contribution}=\frac{1}{4} \] \[ \text{Body center contribution}=1 \]

Answer 1

The Correct Option is D

Step 1: Count the number of anions \(A\).
The anions form an FCC lattice.
In an FCC unit cell: Corner atoms contribute: \[ 8\times \frac{1}{8}=1 \] Face-centered atoms contribute: \[ 6\times \frac{1}{2}=3 \] Hence, total number of anions: \[ 1+3=4 \] Therefore, \[ A=4 \]

Step 2: Count the number of cations at body center.
There is one body center atom completely inside the unit cell.
Contribution: \[ 1 \]

Step 3: Count the cations at edge centers.
A cube has \(12\) edge centers.
Only half of them are occupied: \[ \frac{12}{2}=6 \] Each edge-centered atom contributes: \[ \frac{1}{4} \] Hence, total contribution from edge centers: \[ 6\times \frac{1}{4}=\frac{6}{4}=\frac{3}{2} \]

Step 4: Calculate total number of cations.
Total cations: \[ 1+\frac{3}{2}=\frac{5}{2} \] Thus, \[ C=\frac{5}{2} \]

Step 5: Find the simplest ratio.
Ratio of cations to anions: \[ \frac{5}{2}:4 \] Multiply by \(2\): \[ 5:8 \] Hence, the formula is \[ \boxed{C_5A_8} \]