Question

A compound consists of atoms A and B. The atoms of B form hcp lattice. The atoms of A occupy \(\frac{1}{3}\) rd of octahedral voids and \(\frac{1}{3}\) rd of tetrahedral voids. What is the molecular formula of the compound?

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

Hint:

Calculate molecular formula by considering atoms in lattice and occupancy of voids.

Answer 1

The Correct Option is D

- Number of atoms of B in hcp lattice per unit cell = 6.
- Number of octahedral voids per atom = 1.
- Number of tetrahedral voids per atom = 2.

Atoms of A occupy:
\[ \frac{1}{3} \times 6 = 2 \quad \text{octahedral voids} \]
\[ \frac{1}{3} \times (2 \times 6) = 4 \quad \text{tetrahedral voids} \]

Total atoms of A = \(2 + 4 = 6\). So ratio of A to B is \(6:6 = 1:1\).

Hence, molecular formula is \(\mathrm{AB}\).