# A forms ccp lattice, B occupy half of the octahedral voids and ‘O’ occupy all the tetrahedral voids. Calculate formula –

a. A2BO4

b. ABO4

c. A2B2O

d. A2B2O

CCP lattice of A → no of atom of A per unit cell → 4.

N oct. voids = 2 N tetra.Voids. But B occupies only half of oct. voids

&therefore; No of atoms of B =

$$\begin{array}{l}4\times \frac{1}{2}=2\end{array}$$

No of atoms of C → 4 × 2 = 8

&therefore; The formula → A4 B2 C8