Question

A solid contains elements A and B. Anions of B form ccp lattice. Cations of A occupy 50% of octahedral voids and 50% of tetrahedral voids. What is the molecular formula of the solid?

• A. \(\text{AB}_3\)
• B. \(\text{A}_3\text{B}_2\)
• C. \(\text{A}_2\text{B}_3\)
• D. \(\text{AB}\)

Hint:

Always remember: Number of packing atoms = \(N\) Octahedral voids = \(N\) Tetrahedral voids = \(2N\) The locations of these voids in an FCC cell are: OV at body center + edge centers; TV at body diagonals.

Answer 1

The Correct Option is B

Step 1: Determine the number of atoms of B:
The anions B form a Cubic Close Packed (ccp) lattice. Let the number of atoms of B in the unit cell be \(N\). (For a standard unit cell calculation, ccp corresponds to FCC, so \(N=4\), but utilizing \(N\) is general).
Step 2: Determine number of Voids:
In a close packing of \(N\) atoms: - Number of Octahedral Voids (OV) = \(N\) - Number of Tetrahedral Voids (TV) = \(2N\)
Step 3: Determine the number of atoms of A:
Cations A occupy: - 50% of Octahedral Voids = \(50% \text{ of } N = 0.5N\) - 50% of Tetrahedral Voids = \(50% \text{ of } 2N = 0.5 \times 2N = N\) Total atoms of A = \(0.5N + N = 1.5N = \frac{3}{2}N\).
Step 4: Find the simplest ratio (Formula):
Ratio \( \text{A} : \text{B} \) \( = 1.5N : N \) \( = \frac{3}{2} : 1 \) Multiply by 2 to get whole numbers: \( = 3 : 2 \) So, the empirical formula is \(\text{A}_3\text{B}_2\). Final Answer:
\(\text{A}_3\text{B}_2\).