Question

In a cubic lattice, atom X occupies the corners of the cube and atom Y occupies the end-centred positions and Z occupies the edge centres of the cube.

What will be the simplest formula of the unit cell if one of the face diagonals containing end centre atom is removed?

• A. $X_5{Y}_6{Z}_3$
• B. $X_{12}{Y}_4{Z}_3$
• C. $X_5{Y}_5{Z}_6$
• D. $X_3{Y}_2{Z}_{12}$

Answer 1

The correct option is D $X_3{Y}_2{Z}_{12}$

In a cubical unit cell:

$\text{Contribution for a corner particle is}=\frac{1}{8}$
$\text{Contribution for a end particle is}=\frac{1}{2}$
$\text{Contribution for a edge centre particle is}=\frac{1}{4}$

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.
$\text{Total Z atoms in a unit cell}=(12\times \frac{1}{4})=3$
Formula for the compound:
$X_{\frac{3}{4}}{Y}_{\frac{1}{2}}{Z}_3\text{Simplest formula :}{X}_3{Y}_2{Z}_{12}$