Question

In a cubic lattice of X and Y, X atoms are present at the corners and body centre while Y atoms are at face centres and edge centres. What would be the simplest formula of the compound be if two corner atoms of X and 4 edge centre atoms of Y are replaced by Z atoms?

• A. $X_7{Y}_{19}{Z}_3$
• B. $X_3{Y}_{19}{Z}_7$
• C. $X_{20}{Y}_5{Z}_7$
• D. $X_7{Y}_{20}{Z}_5$

Answer 1

The correct option is D $X_7{Y}_{20}{Z}_5$

In a cubical unit cell:

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

Here, X is present at corners and body centre and Y is present at face centre and edge centre.

After the replaceent of 2 corner X atoms,
$\text{No. of X atom per unit cell}=6\times \frac{1}{8}+1=\frac{7}{4}$

After the replacement of 4 edge centred atoms,
$\text{No. of Y atom per unit cell}=6\times \frac{1}{2}+8\times \frac{1}{4}=5$

Atoms in two corners and four edge centres were replace by Z atoms
$\therefore$
$\text{No. of Z atom per unit cell}=2\times \frac{1}{8}+4\times \frac{1}{4}=\frac{5}{4}$

Ratio of X, Y and Z will be
$X:Y:Z=\frac{7}{4}:5:\frac{5}{4}$

$X:Y:Z=7:20:5$

Therefore, the number of atoms present or the formula of the compound is
$X_7:Y_{20}:Z_5$
Simplest formula is ;
$X_7{Y}_{20}{Z}_5$