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: Set up the FCC anion framework.
Anions form an FCC lattice. In an FCC unit cell: corner contributions = $8 \times \frac{1}{8} = 1$; face contributions = $6 \times \frac{1}{2} = 3$. Total anions per unit cell = $1 + 3 = 4$.
Step 2: Count the body center cation.
One cation at the body center of the unit cell. The body center is entirely within the unit cell, so its contribution = $1 \times 1 = 1$.
Step 3: Count edge center cations.
A cube has 12 edges. Half of the edges have cations, so 6 edges have cations. Each edge is shared by 4 unit cells. Contribution per unit cell = $6 \times \frac{1}{4} = 1.5$.
Step 4: Calculate total cations per unit cell.
Total cations = body center + edge centers = $1 + 1.5 = 2.5$.
Step 5: Determine the empirical formula.
Ratio of cations (C) to anions (A) = 2.5 : 4. Multiply both by 2 to get whole numbers: $2.5 \times 2 = 5$ and $4 \times 2 = 8$. The formula is C₅A₈.
Step 6: State the answer.
With 4 anions per unit cell (FCC), 1 cation at body center, and 6 edge-center cations each shared by 4 cells giving 1.5 cations, total cations = 2.5. The simplest whole-number ratio is C:A = 5:8.
\[ \boxed{C_5A_8 \text{ (option 4)}} \]