Question

A and B atoms crystallize in cubic lattice. A atoms are present at the corners while B atoms are at face centers. What would be the simplest formula of the compound if two of the A atoms are missing from each unit cell?

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

Answer 1

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

Each of the 8 corner atoms of a cube is shared by 8 such cubes so that each corner atom contribution to one cube will be $\frac{1}{8}$.
After the removal of 2 corner A atoms,
$\text{No. of A atom per unit cell}=6\times \frac{1}{8}=\frac{3}{4}$

Each atom at the face centre is shared by two unit cell so that each face centred atom contributes half of it to each cube.
$\text{No. of B atom per unit cell}=6\times \frac{1}{2}=3$

So, the number of atoms present or the formula of the compound is
$A_{\frac{3}{4}}{B}_3$ or $A{B}_4$