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Chapter 12: Problem 18

(a) Using the ionic radii in Table \(12.3, \mathrm{com}-\) pute the theoretical density of CsCl. (Hint: Use a modification of the result of Prob\(\operatorname{lem} 3.3 .)\) (b) The measured density is \(3.99 \mathrm{g} / \mathrm{cm}^{3} .\) How do you explain the slight discrepancy between your calculated value and the measured one?

Short Answer

Expert verified
Answer: The theoretical density of CsCl using the ionic radii from Table 12.3 is approximately \(3.95\ \mathrm{g/cm}^3\). It is slightly lower than the given measured density, which is \(3.99\ \mathrm{g/cm}^3\). The discrepancy could be due to differences in the actual ionic radii in the crystal lattice, various crystal defects, or slight changes in temperature or pressure during the measurements.

Step by step solution

01

Calculate the edge length of the unit cell

In a CsCl unit cell, chloride ions occupy the corners and face centers, while cesium ions are at the center of the cube. According to Table 12.3, the ionic radii of Cs+ and Cl- ions are given: - Cs+ (cesium ion): \(r_{Cs+} = 167\ \mathrm{pm}\) - Cl- (chloride ion): \(r_{Cl-} = 181\ \mathrm{pm}\) Since both ions touch along the cube edge, the edge length of the CsCl unit cell (a) is equal to the sum of the ionic radii of Cs+ and Cl-: \(a = r_{Cs+} + r_{Cl-} = 167\ \mathrm{pm} + 181\ \mathrm{pm} = 348\ \mathrm{pm}\)
02

Calculate the volume of the unit cell

We can now calculate the volume of the unit cell (V) using the edge length calculated in step 1. The formula for the volume of a cube is: \(V = a^3\) Substitute the value of a in the formula: \(V = (\ 348\ \mathrm{pm}\ )^3 = 42,153,664\ \mathrm{pm}^3\)
03

Calculate the mass of the ions present in the unit cell

To find the density, we need the mass of Cs+ and Cl- ions present in the unit cell. There is one Cs+ ion and one Cl- ion in each unit cell. The molecular weight of Cs and Cl are: - Cs: 132.90 \(\mathrm{g/mol}\) - Cl: 35.45 \(\mathrm{g/mol}\) The mass of the CsCl unit cell in grams is: \(\text{Mass} = \frac{(132.90\ \mathrm{g/mol} + 35.45\ \mathrm{g/mol})}{(6.022\times10^{23}\ \text{ions/mol})} = 2.79\times10^{-22}\ \mathrm{g}\)
04

Calculate the theoretical density of CsCl

The density of CsCl can be calculated using the mass and the volume of the unit cell: \(\text{Density} = \frac{\text{Mass}}{\text{Volume}}\) Convert the volume from 𝑝𝑚 \(^3\) to 𝑐𝑚 \(^3\) : \(\text{Volume} = 42,153,664\ \mathrm{pm}^3 \times (\ \frac{1\ \mathrm{cm}}{10^{12}\ \mathrm{pm}}\ )^3 = 4.22\times10^{-23}\ \mathrm{cm}^3\) Now, calculate the density: \(\text{Density} = \frac{2.79\times10^{-22}\ \mathrm{g}}{4.22\times10^{-23}\ \mathrm{cm}^3} \approx 3.95\ \mathrm{g/cm}^3\)
05

Compare the theoretical density with the measured density and explain the discrepancy

The calculated theoretical density of CsCl is approximately \(3.95\ \mathrm{g/cm}^3\), while the measured density given in the problem is \(3.99\ \mathrm{g/cm}^3\). The slight discrepancy between the calculated and measured densities could be due to differences in the actual ionic radii in the crystal lattice, various crystal defects, or slight changes in temperature or pressure during the measurements.

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Most popular questions from this chapter

A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius \(5.0 \mathrm{mm}\) \((0.20 \text { in. }) ;\) the specimen fractured at a load of \(3000 \mathrm{N}\left(675 \mathrm{lb}_{\mathrm{f}}\right)\) when the distance between the support points was \(40 \mathrm{mm}\) (1.6 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of \(15 \mathrm{mm}\) ( 0.6 in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is maintained at \(40 \mathrm{mm}(1.6 \mathrm{in.}) ?\)

Compute the atomic packing factor for the rock salt crystal structure in which \(r_{\mathrm{C}} / r_{\mathrm{A}}=0.414\).

Iron titanate, \(\mathrm{Fe} \mathrm{Ti} \mathrm{O}_{3},\) forms in the ilmenite crystal structure that consists of an HCP arrangement of \(\mathrm{O}^{2-}\) ions. (a) Which type of interstitial site will the \(\mathrm{Fe}^{2+}\) ions occupy? Why? (b) Which type of interstitial site will the \(\mathrm{Ti}^{4+}\) ions occupy? Why? (c) What fraction of the total tetrahedral sites will be occupied? (d) What fraction of the total octahedral sites will be occupied?

Show that the minimum cation-to-anion radius ratio for a coordination number of 6 is \(0.414 .[\) Hint: Use the NaCl crystal structure (Figure 12.2 ), and assume that anions and cations are just touching along cube edges and across face diagonals.

The unit cell for \(\mathrm{Fe}_{3} \mathrm{O}_{4}\left(\mathrm{FeO}-\mathrm{Fe}_{2} \mathrm{O}_{3}\right)\) has cubic symmetry with a unit cell edge length of \(0.839 \mathrm{nm} .\) If the density of this material is \(5.24 \mathrm{g} / \mathrm{cm}^{3},\) compute its atomic packing factor. For this computation, you will need to use ionic radii listed in Table 12.3.
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