Question

What is the formula for the compound that crystallizes with a cubic closest packed array of sulfur ions, and that contains zinc ions in $\frac{1}{8}$ of the tetrahedral holes and aluminum ions in $\frac{1}{2}$ of the octahedral holes?

Short answer

The compound crystallizes with a cubic closest packed array of sulfur ions, and contains zinc ions in $\frac{1}{8}$ of the tetrahedral holes and aluminum ions in $\frac{1}{2}$ of the octahedral holes. Upon analyzing the unit cell, it contains 4 Sulfur ions, 1 Zinc ion, and 2 Aluminum ions. The ratio of S:Zn:Al is 4:1:2, and therefore, the formula for the compound is ZnAl2S4.

Step-by-step solution

Determine the number of sulfur ions

In a cubic closest packed array, there are 4 sulfur ions (S) per unit cell.

Determine the number of tetrahedral and octahedral holes

In a cubic closest packed array, the number of octahedral holes is equal to the number of ions present, and there are twice as many tetrahedral holes as there are ions. Since we have 4 sulfur ions in the unit cell, there are: - 4 octahedral holes and - 8 tetrahedral holes

Calculate the number of occupied sites by Zinc and Aluminum ions

Given that zinc ions occupy $\frac{1}{8}$ of the tetrahedral holes and aluminum ions occupy $\frac{1}{2}$ of the octahedral, we can calculate the number of Zinc and Aluminum ions in the unit cell: - Number of Zinc ions (Zn) = $\frac{1}{8}\times 8$ tetrahedral holes = 1 Zn - Number of Aluminum ions (Al) = $\frac{1}{2}\times 4$ octahedral holes = 2 Al

Determine the formula of the compound

Now we have the number of each type of ion in the unit cell: - 4 Sulfur ions - 1 Zinc ion - 2 Aluminum ions The ratio of S:Zn:Al is 4:1:2. Therefore, the formula for the compound is ZnAl2S4.