Question

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

What will be the simplest formula of the compound if all the atoms are removed from two different body diagonals of the same cube?

• A. $X_3{Y}_{6}Z$
• B. $X_2{Y}_3{Z}_6$
• C. $X_{6}Y{Z}_6$
• D. $X{Y}_6{Z}_6$

Answer 1

The correct option is D $X{Y}_6{Z}_6$

In a cubical unit cell:

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

When a one body diagonal is removed, two atoms at corners (here X) and one atom at body center (here Z) is removed.
When another body diagonal is removed, only two more corner atoms are removed, since body centre particle is already removed.

For X : Since 4 corner particles are removed.
$\text{Total X atoms in a unit cell}=(4\times \frac{1}{8})=\frac{1}{2}$

For Y : No particle removed.
$\text{Total Y atoms in a unit cell}=(6\times \frac{1}{2})=3$

For Z : Body centre removed.
$\text{Total Z atoms in a unit cell}=(12\times \frac{1}{4})+0=3$
Formula for the compound:
$X_{\frac{1}{2}}{Y}_3{Z}_3\text{Simplest formula :}X{Y}_6{Z}_6$