Problem 126
Question
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?
Step-by-Step Solution
Verified Answer
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 \(\frac{2}{2 \times 1} \times 100 = 100\%\).
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
4Step 4: 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) × 100
Percentage of occupied tetrahedral holes = (2 / (2 × 1)) × 100 = 100%
Therefore, 100% of the tetrahedral holes are occupied by potassium ions in the solid K2O.
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