Question

Atoms of an element 'A' occupy $\frac{2}{3}$ tetrahedral voids in the hexagonal close-packed (hcp) unit cell lattice formed by the element 'B'. The formula of the compound formed by 'A ' and 'B' is :

• A. $A{B}_2$
• B. $A_4{B}_3$
• C. $A_2{B}_3$
• D. $A_{2}B$

Answer 1

The correct option is B $A_4{B}_3$

The number of atoms per unit cell for Hexagonal Closest Packed Structure is 6.

The number of atoms of element B in the unit cell is 6.

Number of tetrahedral voids is $2\times 6=12$

Two third of tetrahedral voids are occupied. Total number of atoms of element A is :

$=12\times \frac{2}{3}=8$

The ratio of the number of atoms of element A to the number of atoms of element B is :

$8:6$ or 4:3

Hence, the formula of the compound is $A_4{B}_3$

Hence the correct option is B.