Practicing Success
Atoms of electrons B form hcp lattice and those of the element A occupy 2/3 rd of tetrahedral voids. What is the formula of the compound formed by the elements A and B? |
\(A_3B_4\) \(A_4B_3\) \(A_2B_3\) \(A_3B_2\) |
\(A_4B_3\) |
Given that the atoms of \(B\) forms hcp lattice. So, the atoms at the corners are shared by \(6\) unit cells. So, its contribution is \(\frac{1}{6}\)th, fcc contributes \(1\) atom each. So effective number of atoms in the unit cell \(= \frac{1}{6} × 12 + \frac{1}{2} × 2 + 1 × 3\) \(= 2 + 1 + 3 = 6\) Number of tetrahedral voids \(= 2 × \text{Number of atoms forming the lattice}\) \(= 2 × 6 = 12\) So, the number of atoms of A \(\frac{2}{3} × 12 = 8\) Number of atoms of B \(= 6\) So, \(A : B ⇒ 8 : 6\) or, \(A : B = 4 : 3\) So, the formula of the compound is \(A_4B_3\) |