Question

Atoms X,Y,Z crystallize in a cubic lattice where atom X occupies the corners of the cube and atom Y occupies the end-centred positions and Z occupies the body centre positions
What will be the simplest formula of the unit cell if one of the face diagonals containing end centre atom is removed?

• A. $X_3{Y}_2{Z}_4$
• B. $X_3{Z}_4$
• C. $X{Y}_2{Z}_3$
• D. None of the above.

Answer 1

The correct option is A $X_3{Y}_2{Z}_4$

In a cubical unit cell:

$\text{Contribution for a corner particle is}=\frac{1}{8}$
$\text{Contribution for a end particle is}=\frac{1}{2}$
$\text{Contribution for a body centre particle is}=1$

When one of the face diagonals containing end centre atom is removed:
1 end centred atom is removed.
2 corner atoms are removed.

For X:
Since 2 corner atoms are removed ,
$\text{Total X atoms in a unit cell}=(6\times \frac{1}{8})=\frac{3}{4}$
For Y:
Since 1 end centred atom is removed ,
$\text{Total Y atoms in a unit cell}=(1\times \frac{1}{2})=\frac{1}{2}$
For Z :
No atoms removed.
$\text{Total Z atoms in a unit cell}=1$
Formula for the compound:
$X_{\frac{3}{4}}{Y}_{\frac{1}{2}}{Z}_1\text{Simplest formula :}{X}_3{Y}_2{Z}_4$