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Chapter 9: Problem 126

The structure of the compound \(\mathrm{K}_{2} \mathrm{O}\) is best described as a cubic closest packed array of oxide ions with the potassium ions in tetrahedral holes. What percent of the tetrahedral holes are occupied in this solid?

Short Answer

Expert verified
In the cubic closest packed (ccp) structure of K2O, there are two tetrahedral holes per oxide ion. Since the ratio of potassium ions (K^+) to oxide ions (O^2-) is 2:1, both of the tetrahedral holes are occupied by the potassium ions. Therefore, the percentage of occupied tetrahedral holes is \(\frac{2}{2 \times 1} \times 100 = 100\%\).

Step by step solution

01

Identify the number of oxide ions and tetrahedral holes in a ccp structure

In a cubic closest packed structure, there is one oxide ion (O^2-) per lattice point. The number of tetrahedral holes in a ccp structure is twice the number of lattice points, meaning there are two tetrahedral holes per oxide ion.
02

Determine the chemical formula of the compound

We know that the compound's formula is K2O. Therefore, the ratio of potassium ions (K^+) to oxide ions (O^2-) is 2:1.
03

Calculate the number of occupied tetrahedral holes

Since the ratio of potassium ions to oxide ions is 2:1, we know that for every two potassium ions (K^+), there is one oxide ion (O^2-). In a ccp structure, each oxide ion is associated with two tetrahedral holes. Thus, the two potassium ions in the compound will occupy two of the tetrahedral holes.
04

Calculate the percentage of occupied tetrahedral holes

We know that there are two tetrahedral holes per oxide ion in a ccp structure. Since two of the tetrahedral holes are occupied by potassium ions, the percentage of occupied tetrahedral holes can be calculated as: Percentage of occupied tetrahedral holes = (Number of occupied tetrahedral holes / Total number of tetrahedral holes) 脳 100 Percentage of occupied tetrahedral holes = (2 / (2 脳 1)) 脳 100 = 100% Therefore, 100% of the tetrahedral holes are occupied by potassium ions in the solid K2O.

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