The compounds \(\mathrm{Na}_{2} \mathrm{O}, \mathrm{CdS},\) and \(\mathrm{ZrL}_{4}\) all can be described as cubic closest packed anions with the cations in tetrahedral holes. What fraction of the tetrahedral holes is occupied for each case?

A cubic closest packed structure is a type of crystal structure where anions are packed as closely together as possible in a cubic arrangement. In this arrangement, there are two tetrahedral holes per anion.

For each compound, we should calculate the number of anions and tetrahedral holes based on their formulas. We have: 1. For \(\mathrm{Na}_{2} \mathrm{O}\), there are 2 Na (cation) for each O (anion). With ccp structure, each O anion gives 2 tetrahedral holes. 2. For \(\mathrm{CdS}\), there is 1 Cd (cation) for each S (anion). Each S anion gives 2 tetrahedral holes. 3. For \(\mathrm{ZrL}_{4}\), there is 1 Zr (cation) for each 4 L (anions). Each L anion gives 2 tetrahedral holes.

Now that we have the number of cations and tetrahedral holes for each compound, we can calculate the fraction of tetrahedral holes occupied. 1. For \(\mathrm{Na}_{2} \mathrm{O}\): Number of cations (Na) = 2 Number of tetrahedral holes (O) = 1 * 2 = 2 Fraction of occupied tetrahedral holes = \(\frac{2}{2}\) = 1 2. For \(\mathrm{CdS}\): Number of cations (Cd) = 1 Number of tetrahedral holes (S) = 1 * 2 = 2 Fraction of occupied tetrahedral holes = \(\frac{1}{2}\) 3. For \(\mathrm{ZrL}_{4}\): Number of cations (Zr) = 1 Number of tetrahedral holes (L) = 4 * 2 = 8 Fraction of occupied tetrahedral holes = \(\frac{1}{8}\) So the fractions of tetrahedral holes occupied for the compounds are: 1. \(\mathrm{Na}_{2} \mathrm{O}\): 1 2. \(\mathrm{CdS}\): \(\frac{1}{2}\) 3. \(\mathrm{ZrL}_{4}\): \(\frac{1}{8}\)