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Chapter 10: Problem 70

The compounds \(\mathrm{Na}_{2} \mathrm{O}, \mathrm{CdS}\), and \(\mathrm{Zr} \mathrm{I}_{4}\) all can be described as cubic closest packed anions with the cations in tetrahedral holes. What fraction of the tetrahedral holes is occupied for each case?

Short Answer

Expert verified
In cubic closest packed anions with cations in tetrahedral holes, the fraction of occupied tetrahedral holes for Na2O is 1, for CdS is 0.5, and for ZrI4 is 0.125.

Step by step solution

01

Find the ratio of cations to anions for each compound

For each compound, we need to find the ratio of the cations to the anions. 1. Na2O: 2 Na+ cations and 1 O2- anion 2. CdS: 1 Cd2+ cation and 1 S2- anion 3. ZrI4: 1 Zr4+ cation and 4 I- anions
02

Calculate the number of tetrahedral holes in a ccp structure

In a cubic closest packed structure, there are twice as many tetrahedral holes as there are anions. We will use this information to calculate the number of tetrahedral holes for each compound. 1. Na2O: Number of tetrahedral holes = 2 × 1 (anions) = 2 2. CdS: Number of tetrahedral holes = 2 × 1 (anions) = 2 3. ZrI4: Number of tetrahedral holes = 2 × 4 (anions) = 8
03

Calculate the fraction of occupied tetrahedral holes

Now, we can find the fraction of occupied tetrahedral holes for each compound by dividing the number of cations (found in step 1) by the number of tetrahedral holes (found in step 2). 1. Na2O: Fraction of occupied tetrahedral holes = 2 (cations) / 2 (tetrahedral holes) = 1 2. CdS: Fraction of occupied tetrahedral holes = 1 (cations) / 2 (tetrahedral holes) = 0.5 3. ZrI4: Fraction of occupied tetrahedral holes = 1 (cations) / 8 (tetrahedral holes) = 0.125 In conclusion, the fraction of occupied tetrahedral holes for each compound is: 1. Na2O: 1 2. CdS: 0.5 3. ZrI4: 0.125

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