Question

In a compound $ {\text{X}}{{\text{Y}}_{\text{2}}}{{\text{O}}_{\text{4}}} $ , oxide ions are arranged in CCP and cations $ {\text{X}} $ are present in octahedral voids. Cations $ {\text{Y}} $ are equally distributed between octahedral and tetrahedral voids. Cations $ {\text{Y}} $ are equally distributed between octahedral and tetrahedral voids. The fraction of the octahedral voids occupied is:
(A) $ 1/2 $
(B) $ 1/4 $
(C) $ 1/8 $
(D) $ 1/6 $

Answer 1

Hint: To answer this question, you must recall the concept of cubic close packing in a solid ionic crystal. Voids are the gaps that are present between the constituent atoms, molecules or ions. The number of octahedral voids are equal to the number of particles present in the lattice and the number of tetrahedral voids are double that of the octahedral voids or the number of particles present in the lattice.

Complete step by step solution:
We know that a CCP lattice has a face centred cubic unit cell. So the number of particles present in the unit cell are 4. Or we can say that the number of oxide ions in the given crystal packing is $ n = 4 $
The number of tetrahedral voids are given by $ = 2 \times n = 8 $
The number of octahedral voids are given by $ n = 4 $
According to the formula of the compound, $ {\text{X}}{{\text{Y}}_{\text{2}}}{{\text{O}}_{\text{4}}} $ , the number of cations X in one unit cell $ = 1 $ and the number of cations Y in one unit cell $ = 2 $
It is also given in the question, that the cation Y is distributed equally in octahedral and tetrahedral voids, thus, we can say that one Y is in octahedral void and one in tetrahedral void. Also, X occupies the octahedral voids. So the number of octahedral voids occupied $ $ $ = 1 + 1 = 2 $
So the fraction of octahedral voids occupied is $ = \dfrac{2}{4} = 1/2 $
The correct answer is A.

Note:
It should be known that in a face centred cubic unit cell, atoms are present at the corners and the centres of all the faces of the cube. So, we can write that,
 $ {n_c} = $ number of atoms present at the corners of the fcc unit cell $ = 8 $
 $ {n_f} = $ number of atoms present on the six faces of the fcc unit cell $ = 6 $
 $ {n_i} = $ number of atoms present at the body center of the fcc unit cell $ = 0 $
 $ {n_e} = $ number of atoms present at the edge centres of the fcc unit cell $ = 0 $
Thus, the total number of particles in a face centred unit cell is
 $ n = \dfrac{{{n_c}}}{8} + \dfrac{{{n_f}}}{2} + \dfrac{{{n_i}}}{1} + \dfrac{{{n_e}}}{4} $
Substituting the values, we get
 $ n = \dfrac{8}{8} + \dfrac{6}{2} + \dfrac{0}{1} + \dfrac{0}{4}{\text{ }} $
 $ n = 1 + 3 + 0 + 0{\text{ }} $
Thus, $ n = 4 $ .