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Chapter 10: Problem 75

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

Expert verified
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 is ZnAl鈧係鈧�.

Step by step solution

01

Determine the number of sulfur ions in the unit cell

In a cubic closest packed array, the sulfur ions will occupy the lattice points. In a cubic unit cell, there is 1 lattice point at each corner (8 corners) and 1 lattice point at the center of each face (6 faces). Each corner contributes 1/8 of an ion to the unit cell and each face contributes 1/2 of an ion to the unit cell. So, the total number of sulfur ions in the unit cell is: \[ N_\text{S} = 8\left(\frac{1}{8}\right) + 6\left(\frac{1}{2}\right) = 1+3 = 4 \]
02

Determine the number of tetrahedral and octahedral holes in the unit cell

In a ccp structure, the number of tetrahedral holes is twice the number of lattice points, and the number of octahedral holes is equal to the number of lattice points. Since the total number of sulfur ions in the unit cell is 4 (from step 1), there are 8 tetrahedral holes and 4 octahedral holes in the unit cell.
03

Determine the number of zinc ions in the unit cell

Zinc ions occupy 1/8 of the tetrahedral holes. Since there are 8 tetrahedral holes in the unit cell, the number of zinc ions in the unit cell can be calculated as: \[ N_\text{Zn} = \frac{1}{8}(8) = 1 \]
04

Determine the number of aluminum ions in the unit cell

Aluminum ions occupy 1/2 of the octahedral holes. Since there are 4 octahedral holes in the unit cell, the number of aluminum ions in the unit cell can be calculated as: \[ N_\text{Al} = \frac{1}{2}(4) = 2 \]
05

Write the formula of the compound

Now we can write the formula of the compound using the number of ions of each type in the unit cell. The formula is: \[ \text{Zn}_1\text{Al}_2\text{S}_4 \] Since all the coefficients are in the simplest whole-number ratio, the final formula is: \[ \text{ZnAl}_2\text{S}_4 \] The formula for the compound is ZnAl鈧係鈧�.

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