Question

A crystal made up of particles X, Y, and Z. X forms fcc structure for packing. Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X. If all particles along one body diagonal are removed, then the formula of the crystal would be:

• A. $XY{Z}_2$
• B. $X_{2}Y{Z}_2$
• C. $X_8{Y}_4{Z}_5$
• D. $X_5{Y}_4{Z}_8$

Answer 1

The correct option is D $X_5{Y}_4{Z}_8$

In FCC, unit cell number of atoms present $=4$

Now, $2\times$ Number of octahedral voids $=$ number of tetrahedral voids.

And number of octahedral voids $=$ Number of atoms present in cubic unit cell.

So, $X=4,Y=4,Z=8$

On the body diagonal there is one octahedral void (at body center) and two tetrahedral voids (at ${\frac{1}{4}}^{th}$ of the distance from each corner). So, after removal of atoms on body diagonal, we have:-

$X=6\times \frac{1}{2}+\frac{\theta -2}{8}=3+\frac{3}{4}=\frac{15}{4}$

(face) (corners)

$Y=4-1=3;Z=8-2=6$

So, $X$ $Y$ $Z$ $\Rightarrow$ $X$ $Y$ $Z$

$\frac{15}{4}\times 4$ $3\times 4$ $6\times 4$ $\frac{15}{5}$ $\frac{12}{4}$ $\frac{24}{8}$

So, formula of crystal is $X_5{Y}_4{Z}_8$