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Chapter 10: 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 solid K2O, the compound has a cubic closest packed array (face-centered cubic) of oxide ions, with 4 oxide ions in the unit cell. There are 8 tetrahedral holes available in the unit cell, and 8 potassium ions are present (as given by the K2O formula). Therefore, 100% of the tetrahedral holes are occupied by potassium ions in this solid.

Step by step solution

01

Determine the number of oxide ions in the unit cell of K2O

In a face-centered cubic (fcc) structure, the oxide ions occupy the corners and the center of each face of the unit cell. There are 8 corner ions and 6 face-centered ions. We need to determine the number of oxide ions in the unit cell, considering that each corner ion is shared by 8 adjacent cells and each face-centered ion is shared by 2 adjacent cells. Number of oxide ions in unit cell = (Number of corner ions × 1/8) + (Number of face-centered ions × 1/2) Number of oxide ions in unit cell = (8 × 1/8) + (6 × 1/2) = 1 + 3 = 4
02

Calculate the number of tetrahedral holes in the unit cell

In a face-centered cubic arrangement, the number of tetrahedral holes is twice the number of anions (oxide ions, in this case). Number of tetrahedral holes = 2 × Number of oxide ions in the unit cell Number of tetrahedral holes = 2 × 4 = 8
03

Determine the number of potassium ions in the unit cell

The formula for the compound K2O indicates that there are 2 potassium ions for each oxide ion in the structure. So, we can determine the number of potassium ions in the unit cell of K2O by: Number of potassium ions = 2 × Number of oxide ions in the unit cell Number of potassium ions = 2 × 4 = 8
04

Calculate the percentage of occupied tetrahedral holes

Now we can calculate the percentage of tetrahedral holes occupied by potassium ions in the compound K2O: Percentage of occupied tetrahedral holes = (Number of potassium ions in the unit cell / Number of tetrahedral holes) × 100 Percentage of occupied tetrahedral holes = (8 / 8) × 100 = 100% Therefore, 100% of the tetrahedral holes are occupied by potassium ions in the solid K2O.

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