Question

The compounds ${\mathbf{Na}}_{\mathbf{2}}\mathbf{O}\mathbf{,}\mathbf{}\mathbf{CdS}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{ZrI}}_{\mathbf{4}}$all can be describedas cubic closest packed anions with the cations in tetrahedral holes. What fraction of the tetrahedral holes are occupied for each case?

Short answer

The tetrahedral void fraction for ${\mathrm{Na}}_2\mathrm{O},\mathrm{CdS},\mathrm{and}{\mathrm{ZrI}}_4$is 100, 50, and 50 percent, respectively.

Step-by-step solution

Given Information

Compounds${\mathrm{Na}}_2\mathrm{O},\mathrm{CdS},\mathrm{and}{\mathrm{ZrI}}_4$ have cubic closest packed anions with the cations in tetrahedral holes

Concept Introduction

The packing fraction of any compound is defined as the space in the unit cell occupied by the atoms in the cell.

Explanation

The ${\mathrm{Na}}_2\mathrm{O}$compound has an anti-fluorite structure. In the anti-fluorite structure, tetrahedral voids are occupied by ${\mathrm{Na}}^{+}$ions completely.

Therefore, the tetrahedral void fraction is 100 percent in ${\mathrm{Na}}_2\mathrm{O}$.

The CdS has the same structure as the ZnS, i.e., ccp. In the crystal, the Cd²⁺ ions are situated in the tetrahedral voids.

Hence, the void fraction in this compound is 50 percent in CdS.

The ${\mathrm{ZrI}}_4$has the ccp arrangement. The ${\mathrm{Zr}}^{4+}$ions are located in the tetrahedral voids.

The tetrahedral void fraction is 50 percent.