QJEEM19Ch152e: JEE Main 2019 Chemistry - Class 12 JEEM19Ch152e Solution: Element 'B' forms ccp structure and 'A' occupies half of the…

We know, for cubic unit cell, only FCC has octahedral and tetrahedral voids.

B forms ccp structure means B forms FCC structure.

For FCC, z = 4

We know for octahedral voids z = 4. In this lattice, A present in half of octahedral voids.

$$\therefore$$ For A, z = 2

For tetrahedral voids, 2z = 8. In this lattice, oxygen present in all the tetrahedral voids.

$$\therefore$$ Simplest formula :

$$A_{4\times \frac{1}{2}}{B}_4{O}_8$$

= $$A_2{B}_4{O}_8$$

= $$A{B}_2{O}_4$$