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 $= 4 - \left(2 \times \frac{1}{8}\right) = \frac{1 5}{4}$
Number of Y atom left = 4 - (1 × 1) = 3
Number of Z atom left = 8 - (2 × 1) = 6
The simplest formula $= \text{X}_{\frac{1 5}{4}} \text{Y}_{3} \text{Z}_{6} \Rightarrow \text{X}_{1 5} \text{Y}_{1 2} \text{Z}_{2 4}$
$\Rightarrow \text{X}_{5} \text{Y}_{4} \text{Z}_{8}$