Question

A compound is formed by two elements A and B.Atoms of element B ( as anions) make ccp and those of element A (as Cations) occupies all the octahedral voids and half of the tetrahedral voids. What is the formula of the compound? *

Answer 1

Atoms in $B$ form ccp( cubic close parking) and is analogous to fcc( face centred cubic.

The effective number of atoms of $B$ in lattice is thus;

$($ $1\over 8$ $*8)$ $+($ $1\over 2$ $*6)$ $=4$

Number of tetrahedral voids in $A$ which forms (ccp or fcc) is $n$ atoms or anions.

The number of tetrahedral voids will thus be $1\over 2$ $(2n)$ $=n$

$n=4$

The compound is thus $A_4B_4$

It's simplified ration $A:B=4:4$ $=1:1$

The compound has formula $AB$