Question

A crystal made up of particles X, Y, and Z. X forms fcc packing. Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X. It all particles along one body diagonal are removed, then the formula of the crystal is

• 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

Answer: (D)

For fcc, number of $X$ atoms $=4/$ unit cell
Number of Tetrahedral Voids $=Z=8$
Number of Octahedral Voids $=Y=4$
Number of atoms removed along one body diagonal $=2X$ (corner) and $2Z$ (TVs) and $1Y(OV$ at body centre)
$\therefore$ Number of $X$ atoms left
$=4-2\times \frac{1}{8}=\frac{15}{4}$
Number of $Y$ atom left $=4-(1\times 1)=3$
Number of $Z$ atom left $=8-(2\times 1)=6$
The simplest formula
$=X_{\frac{15}{4}}{Y}_3{Z}_6\Rightarrow X_{15}{Y}_{12}{Z}_{24}$
$\Rightarrow X_5{Y}_4{Z}_8$