Question

Atoms A and B crystallize in a cubic lattice where atoms of A are present at the corners and atoms of B are present at edge centres. If two of the atoms of A are missing from a unit cell, then the simplest formula for the compound will be:

• A. $A{B}_4$
• B. $A{B}_2$
• C. $A_3{B}_2$
• D. $A_2{B}_3$

Answer 1

The correct option is A $A{B}_4$

In a cube, there are 8 corners and 6 faces.
A occupies the corner and B occupies the edge centres.
Corner atoms contribute $\frac{1}{8}$ to the unit cells.
Edge centre atoms contribute $\frac{1}{4}$ to the unit cells.
As two of the A atoms are missing,
Number of A per unit cell: $6\times \frac{1}{8}=\frac{3}{4}$
Number of B per unit cell: $12\times \frac{1}{4}=3$
so, $A_{3/4}{B}_3$
or $A_3{B}_{12}$ i.e $A{B}_4$.