Question

A compound formed by elements A and B has a cubic structure in which A atoms are corners of the cube B atoms are at face centers.Derive the formula of the compound.

Answer 1

Atoms at the corner share themselves with 8 other unit cells. Since there are 8 corners in a a unit cell,
No of atoms of A in one unit cell =
$\frac{1}{8}\times 8=1$

​Atoms at the faces share themselves with 2 other unit cells. Since there are 6 faces in a unit cell,
No of atoms of B in one unit cell = $\frac{1}{2}\times 6=3$

Therefore, the formula of compound is AB₃.