Question

A crystal is made up of particles X, Y and Z; X forms FCC packing, Y occupies all the octahedral voids of X and Z occupies all the tetrahedral voids of X. If all the 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 $X_5{Y}_4{Z}_8$

Since X forms FCC, the number of X particles = 4
The number of octahedral voids = number of particles in the unit cell.
The number of tetrahedral voids = twice the number of particles in the unit cell.
Number of Y particles = 4
Number Z particles = 8
After removing the particles along a body diagonal, two corner particles from the lattice are removed (X), one octahedral void is removed (Y - the body centre) and two tetrahedral voids are removed (Z - each at a distance of 1/4th the length of the body diagonal from the corner particles)
So, we get:

of X atoms in unit cell = 4 - $\frac{1}{8}=\frac{15}{4}$
Y atoms in unit cell = 4 - 1 = 3
No. of Z atoms in unit cell = 8 - 2 = 6
Hence, the formula will be $X_{15/4}{Y}_3{Z}_6$
On simplifying this, we get - $X_{15}{Y}_{12}{Z}_{24}$= $X_5{Y}_4{Z}_8$