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Chapter 10: Problem 136

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 structure of \(\mathrm{K}_{2} \mathrm{O}\), the ratio of potassium ions to tetrahedral holes is 1:1. Therefore, 100% of the tetrahedral holes are occupied by potassium ions.

Step by step solution

01

Determine the ratio of the ions in the compound

The given compound is \(\mathrm{K}_{2} \mathrm{O}\). According to its formula, it has 2 potassium ions (K) for every 1 oxide ion (O).
02

Find the number of tetrahedral holes in a cubic closest packed structure

In a cubic closest packed structure (ccp), the number of tetrahedral holes is twice the number of atoms or ions. In this case, for each oxide ion (O), there are two tetrahedral holes.
03

Calculate the ratio of potassium ions to tetrahedral holes

Since there are 2 potassium ions for every oxide ion, and for each oxide ion, there are two tetrahedral holes, the ratio of potassium ions to tetrahedral holes is as follows: Ratio = Number of potassium ions / Number of tetrahedral holes = 2 potassium ions / (1 oxide ion x 2 tetrahedral holes) Ratio = 2 potassium ions / 2 tetrahedral holes The ratio simplifies to: Ratio = 1 potassium ion / 1 tetrahedral hole
04

Determine the percentage of occupied tetrahedral holes

Now that we have the ratio of potassium ions to tetrahedral holes, we can determine the percentage of occupied tetrahedral holes by multiplying the ratio by 100: Percentage of occupied tetrahedral holes = Ratio × 100% = (1 potassium ion / 1 tetrahedral hole) × 100% Percentage of occupied tetrahedral holes = 100% Thus, in the cubic closest packed structure of \(\mathrm{K}_{2} \mathrm{O}\), 100% of the tetrahedral holes are occupied by potassium ions.

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