Question

In the spinel structure, oxide ions forms CCP whereas $\dfrac{1}{8}th$ tetrahedral voids are occupied by ${A^{2 + }}$ cation and $\dfrac{1}{2}$ of octahedral voids by ${B^{3 + }}$ cations. If oxide ion is replaced by ${X^{ - \dfrac{8}{3}}}$ ion, the number of atomic vacancy per unit cell is:

Answer 1

A CCP arrangement has a total of 4 spheres per unit cell and an HCP arrangement has 8 spheres per unit cell. However, both configurations have a coordination number of 12. The packing efficiency is the fraction of volume in a crystal structure that is occupied by constituent particles, rather than empty space.

Complete answer:
Now firstly let us calculate the charge in the unit cell so the total negative charge present on the unit cell is = $4 \times - 2 = - 8$ and
The number of ${A^{2 + }}$ cation present per unit cell will be: $\dfrac{1}{8} \times 8 = 1atom$
The number of ${B^{3 + }}$ cation present per unit cell will be: $\dfrac{1}{1} \times 4 = 2atom$
Now the total positive charge present on the unit cell will be : $1( + 2) + 2( + 3) = 8$ as the cation A has two positive charges and B has three positive charges.
Hence the unit cell is electrically neutral.
Now according to question when the oxide ions is replaced with ion having charge equal to $- \dfrac{8}{3}$ ,
The total negative charge on the unit cell becomes $4 \times - \dfrac{8}{3} = - \dfrac{{32}}{3}$ .
Now out of this the cations will balance positive eight charge so the remaining negative charge is $- \dfrac{8}{3}$ which remains unbalanced and this is charge of one new replaced ion so there will be one cation vacancy available.

Note:
The arrangement of the atoms in a crystalline solid affects atomic coordination numbers, interatomic distances, and the types and strengths of bonding that occur within the solid. An understanding of atomic packing in a unit cell and crystal lattice can give insight into the physical, chemical, electrical, and mechanical properties of a given crystalline material.