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

Hint: 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.