Atoms of element B form hcp lattice and those of the element A occupy \(\frac{2^{rd}}{3}\) of octahedral voids. What is the formula of the compound formed by the element A and B?

\(A_4B_3\)

The correct answer is option 2. \(A_4B_3\).

Given that, atom B forms hcp lattice. So, the atoms at the corners are shared by 6 unit cells and contribution to each unit cell is \(\frac{1}{6}\). The face centred atoms contribute \(\frac{1}{2}\) and the atom at the body contributes 1.

S, the effective number of atoms in the cell

\(= \frac{1}{6} × 12 + \frac{1}{2} × 2 + 1 × 3\)

\(= 2 + 1 + 3 =6\)

Thus, the number of tetrahedral voids \(= 2 ×\) Number of atoms forming the lattice \(= 2 × 6 = 12\)

Therefore, the number of atoms of A \(= \frac{2}{3} × 12 = 8\)

and the number of atoms of B \( = 6\)

Thus, \(A : B = 8 : 6 = 4 : 3\)

Hence the formula of the compound will be \(A_4B_3\)