Question

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

What will be the simplest formula of the compound if all the atoms are removed from one body diagonal of the cube?

• A. $X_3{Y}_2$
• B. $X_2{Y}_3$
• C. $X_3{Y}_4$
• D. $X_4{Y}_3$

Answer 1

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

In a cubical unit cell:

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

When a one body diagonal is removed, two atoms at corners and one atom at body center is removed.
Now for X :
2 corner particles are removed. so,
$\text{Total X atoms in a unit cell}=(6\times \frac{1}{8})+(3\times \frac{1}{2})=\frac{9}{4}$
For Y :
Here, body centre is removed.
$\text{Total Y atoms in a unit cell}=(3\times \frac{1}{2})+0=\frac{3}{2}$

Formula for the compound:
$X_{\frac{9}{4}}{Y}_{\frac{3}{2}}\text{Simplest formula :}{X}_3{Y}_2$