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 formula for the compound crystallizing with a cubic closest packed array of sulfur ions, containing zinc ions in \(\frac{1}{8}\) of the tetrahedral holes and aluminum ions in \(\frac{1}{2}\) of the octahedral holes, is \(ZnAl_2S_4\).

Step-by-step solution

Calculate the number of sulfur ions in a ccp array

In a ccp structure, the number of atoms (in our case, sulfur ions) per unit cell is 4. This is because a ccp structure has 8 corners and 6 faces, and the atoms are shared accordingly: 1/8 atom per corner (8 x 1/8) = 1 atom, and 1/2 atom per face (6 x 1/2) = 3 atoms. Hence, there are a total of 4 sulfur ions in one unit cell.

Calculate the number of tetrahedral and octahedral holes

In a ccp array, the number of tetrahedral holes is twice the number of atoms, while the number of octahedral holes is equal to the number of atoms in a unit cell. So, for the 4 sulfur ions, there are 8 tetrahedral holes and 4 octahedral holes.

Calculate the number of zinc and aluminum ions in the unit cell

We know that zinc ions occupy 1/8 of the tetrahedral holes and aluminum ions occupy 1/2 of the octahedral holes. Hence, there are 8 x 1/8 = 1 zinc ion in one unit cell and 4 x 1/2 = 2 aluminum ions in one unit cell.

Determine the formula for the compound

Based on the number of ions in one unit cell, we can now derive the empirical formula. We have 4 sulfur ions, 1 zinc ion, and 2 aluminum ions in the unit cell. Hence, the formula for the compound is ZnAl2S4.