Formula of a Compound and Number of Voids Filled

When particles are close-packed resulting in either ccp or hcp structure, two types of voids are generated. While the number of octahedral voids present in a lattice is equal to the number of close packed particles, the number of tetrahedral voids generated is twice this number. In ionic solids, the bigger ions (usually anions) form the close packed structure and the smaller ions (usually cations) occupy the voids. If the latter ion is small enough then tetrahedral voids are occupied, if bigger, then octahedral voids. All octahedral or tetrahedral voids are not occupied. In a given compound, the fraction of octahedral or tetrahedral voids that are occupied, depends upon the chemical formula of the compound.

In a compound XY₂O₄, oxide ions are arranged in ccp structure. Cation X are present in octahedral voids and cation Y are equally distributed between octahedral and tetrahedral voids. What is the fraction of octahedral voids occupied?

\(\frac{1}{2}\)

The correct answer is option 2. \(\frac{1}{2}\).

In a compound XY₂O₄,

oxide ions are arranged in ccp structure (FCC unit cell)

No. of oxide ions \(= 4 = \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 1 + 3= 4\)

Cation X are present in octahedral voids

No. of X cations = 1

1 O-void is occupied by cation X

Cation Y are equally distributed between octahedral and tetrahedral voids

No. of Y cations = 2

1 O-void is occupied by Y cation and 1 T-void is occupied by Y cation

Total no. of O-voids occupied = 2

No. of O-voids in a unit cell = 4

Fraction of O-voids occupied = \(\frac{2}{4}\) = \(\frac{1}{2}\)