Question

In a crystalline soild anion $B$ are arranged in a $ccp$ lattice. Cation $A$ are equally distributed between octahedral and tetrahedral voids. If all octaahedral vods are occupied. What is the formula of the soild.

Answer 1

No. of atoms in a ccp unit cell$=4$

$B$ is in ccp so $B=4$Cation $A$ are equally distributed between octahedral and tetrahedral voids and all octahedral voids are occupied-i.e in ccp octahedral voids =4 since octahedral and tetrahedral are equal that means-

Tetrahedral voids $=4$ so,

$A=4tetrahedral+4octahedral$
$A=8$
Therefore Formula is-

$A_8{B}_4$ or $A_{2}B$