Question

A compound is formed by atoms of A, B and C. Atoms of C form hcp lattice. Atoms of A occupy 50% of octahedral voids and atoms of B occupy $\frac{2}{3}$ of tetrahedral voids. What is the molecular formula of the solid?

• A. \[ \text{A}_3\text{B}_8\text{C}_6 \]
• B. \[ \text{A}_2\text{B}_8\text{C}_6 \]
• C. \[ \text{A}_4\text{B}_4\text{C}_3 \]
• D. \[ \text{A}_5\text{B}_8\text{C}_6 \]

Hint:

For close-packed lattices, remember that C atoms form the basic lattice structure. The number of atoms in voids (octahedral and tetrahedral) must be calculated based on the voids present in the unit cell.

Answer 1

The Correct Option is A

- Atoms of C form an hcp lattice, which means C atoms are arranged in hexagonal close packing.
- In an hcp lattice, each unit cell contains 2 atoms of C. Thus, there are 2 atoms of C per unit cell.
- Atoms of A occupy 50% of octahedral voids. The number of octahedral voids in an hcp unit cell is equal to the number of atoms of C, which is 2. Therefore, the number of A atoms in the unit cell is 2.
- Atoms of B occupy $\frac{2}{3}$ of the tetrahedral voids. The number of tetrahedral voids in an hcp unit cell is 8. So, the number of B atoms in the unit cell is $\frac{2}{3} \times 8 = 5.33$, approximately 5 atoms of B.
Therefore, the molecular formula of the solid is A$_3$B$_8$C$_6$.