Question

A compound having cubic structure is formed by the elements X, Y and Z.
Atoms of element X are present at corners and face centres of the cube . Atoms of element Y are present at edges and atoms of element Z are present at body centre.

What will be the simplest formula of the compound if all the atoms are removed from one of the plane which is parallel to either one of the faces of the cube and also passing through middle of the cube?

• A. $X_2{Y}_3$
• B. $XY$
• C. $X_{2}YZ$
• D. $X_{2}Y{Z}_3$

Answer 1

The correct option is B $XY$

In a cubical unit cell:

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

Atoms removed from one of the plane which is parallel to either one of the faces of the cube and also passing through middle of the cube are:
1 body centre (here Z)
4 edge centres (here Y)
4 face centres (here X)


For X :
Since no corners are removed, so 8 atoms are at corners and (6-4= 2) atoms are left at face centres
So,
$\text{Total X atoms in a unit cell}=(8\times \frac{1}{8})+(2\times \frac{1}{2})=2$

For Y:
Out of the 12 edges , only 8 are left.
$\text{Total Y atoms in a unit cell}=(8\times \frac{1}{4})=2$
For Z :
Since Z was only present at body centre and it is removed.
$\text{Total Z atoms in a unit cell}=0$

Formula :
$X_2{Y}_2{Z}_0\Rightarrow \text{Simplest Formula is}XY$