Question

A compund has a cubical unit cell in which A atoms are present at face center(s) and body centre. B atoms are present at 4 corners and at one-sixth of the edge centers . The simplest formula for the compound is :

• A. $A_{8}B$
• B. $A{B}_8$
• C. $A{B}_4$
• D. $A_{4}B$

Answer 1

The correct option is D $A_{4}B$

In a cubical unit cell:

$\text{Contribution for a corner particle is}=\frac{1}{8}$
$\text{Contribution for a face centre particle is}=\frac{1}{2}$
$\text{Contribution for a edge centre particle is}=\frac{1}{4}$
$\text{Contribution for a body centre particle is}=1$
A is present at 6 face centres and 1 body centre.
Since there are 12 edge centres,
B is present at 4 corners and 2 edge centres

For A:
$\text{Total A atoms in a unit cell}=(6\times \frac{1}{2})+(1\times 1)=4$
For B:
$\text{Total B atoms in a unit cell}=(4\times \frac{1}{8})+(2\times \frac{1}{4})=1$
$\therefore \text{Formula is :}{A}_{4}B$