Question

In a crystalline solid atoms $B$ are arranged in cubic close-packing. Cations $A$ are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, the formula of the solid is $A_{2}B$.

Answer 1

In a cubic close packing (CCP), the number of atoms $=4$.

Since $B$ are arrange in CCP, therefore no. of atoms of $B$ in one unit cell $=4$.

In CCP, no. of octahedral voids $=4=$ atoms in octahedral voids

Since atom $A$ are equally distributed between octahedral voids and tetrahedral voids.

Atoms of a $A$ tetrahedral voids $=$ atoms of $A$ in octahedral voids $=4.$

Total no. of $A$ atoms $=4+4=8.$

So, formula $=A_8{B}_4=A_{2}B$

Hence, it is true.