Question

A crystal is 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 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

Hint: or solving this question, we need to understand the basic meaning of the FCC packing in a crystal lattice. The face-centered cubic system or the FCC crystal lattice has different lattice points on the faces of the cube, and each of these points gives exactly one half contribution. Also, there are the corner lattice points giving a total of 4 lattice points per unit cell.

Complete step by step answer:
First, we will start by calculating the FCC packing. The total number of faces in the FCC crystal lattice is the sum of the total number of corner faces and the face of the cube.
FCC = $\dfrac{1}{8} \times 8$ +$\dfrac{1}{2} \times 6$ = 4 lattice points per unit cell
The coordination number of each sphere in the FCC lattice is 12. As we know that, coordination number is the number of nearest neighbors of a central atom in the structure.
Now, we know that the face-centered cubic system is closely related to the hexagonal close packed (hcp), where two systems differ only in the relative placements of their hexagonal layers.
The question has referred to a body diagonal along which there is one octahedral void which is at the body centre and at all the edge centers whereas the two tetrahedral voids out of which one almost between each corner and the body centre and two corner atoms are present.
According to the question,
X is the FCC packing
Y is the octahedral voids and Z is the tetrahedral voids.
So, the number of octahedral voids in FCC lattice = \[Y = 4\]
Thus, the total number of atoms left in \[Y = 4 - 1 = 3\]
Number of tetrahedral voids = \[Z = 8\]
Thus, the total number of atoms left in \[Z = 8 - 2 = 6\]
To derive the formula of the crystal lattice:
No. of atoms of left in X lattice = \[6 \times \;\left[ {Face - centered{\text{ }}atoms} \right] + \left( {8 - 2} \right) \times \;\left[ {Corner{\text{ }}atoms} \right]\]
= $3 + \dfrac{3}{4}$ = $\dfrac{{15}}{4}$
so, the formula of compound becomes${X_5}{Y_4}{Z_8}$.

$\therefore $The option D is correct answer.

Note:
From the above question, we need to remember that there are 4 octahedral voids and 8 tetrahedral voids in an FCC unit cell. And we must know that the face Centered Cubic (fcc) or Cubic Close Packed (ccp) These are two different names for the same lattice.