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. If all particles along one body diagonal are removed, then the formula of the crystal is

• A. XYZ₂
• B. X₂YZ₂
• C. X₈Y₄Z₅
• D. X₅Y₄Z₈

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 1 Y (OV at body centre)
$\therefore$ Number of X atoms left $=4-\left(2\times \frac{1}{8}\right)=\frac{15}{4}$
Number of Y atom left = 4 - (1 × 1) = 3
Number of Z atom left = 8 - (2 × 1) = 6
The simplest formula $={\text{X}}_{\frac{15}{4}}{\text{Y}}_3{\text{Z}}_6\Rightarrow {\text{X}}_{15}{\text{Y}}_{12}{\text{Z}}_{24}$
$\Rightarrow {\text{X}}_5{\text{Y}}_4{\text{Z}}_8$