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Chapter 10: Problem 77

The \(\mathrm{CsCl}\) structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 67 ). Given that the density of cesium chloride is \(3.97 \mathrm{~g} / \mathrm{cm}^{3}\), and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent \(\mathrm{Cs}^{+}\) and \(\mathrm{Cl}^{-}\) ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of \(\mathrm{Cs}^{+}\) is \(169 \mathrm{pm}\), and the ionic radius of \(\mathrm{Cl}^{-}\) is \(181 \mathrm{pm}\).

Short Answer

Expert verified
The distance between adjacent Cs₊ and Cl₋ ions in the solid is found to be 357.25Å, which is very close to the sum of the ionic radii of Cs₊ and Cl₋ (350 pm or 3.50 Å). This agreement validates the assumptions made for the CsCl structure and indicates that the ionic sizes are an essential factor in determining inter-ionic distances in the solid.

Step by step solution

01

Determine the molecular weight of CsCl

In order to compute the molecular weight of CsCl, we need to find the atomic weights of Cs and Cl. The atomic weights are as follows: - Cesium (Cs): 132.9 g/mol - Chlorine (Cl): 35.45 g/mol Now, add the atomic weights to find the molecular weight of CsCl: \( Molecular\ weight\ of\ CsCl = 132.9 + 35.45 = 168.35\ g/mol \)
02

Calculate the number of moles per unit volume

Now we need to find the moles per unit volume. To do this, we can use the given density of CsCl and the molecular weight we found in step 1. The formula to calculate the moles per unit volume is: \( Moles\ per\ unit\ volume = \frac{Density}{Molecular\ weight} \) Substitute the given values and calculate the number of moles per unit volume: \( Moles\ per\ unit\ volume = \frac{3.97\ g/cm^3}{168.35\ g/mol} = 0.0236\ mol/cm^3 \)
03

Find the edge length of the cubic unit cell

Each unit cell in the CsCl structure contains one CsCl molecule since there's one Cs₊ ion at the center of each cubic array of Cl₋ ions. We know that there are 0.0236 moles of CsCl per cm³. Therefore, there are 0.0236 moles of cubic unit cells per cm³. Now, let's find the side length (a) of the cubic unit cell. We can use the following formula: \( \frac{1\ molecule}{1\ unit\ cell} \times (\frac{0.0236\ mol}{cm^3}) \times (\frac{6.022 \times 10^{23}\ cells}{mol}) = \frac{1}{a^{3}}\) Solving for edge length a, we get: \( a = (6.022 \times 10^{23}\times 0.0236)^{-1/3} = 4.126 \times 10^{-8} cm = 412.6\ \mathring{A} \)
04

Find the length of the body diagonal

As it is given that chloride and cesium ions touch along the body diagonal of the cubic unit cell, we will find the body diagonal using the edge length (a), which we found in step 3. The length of the body diagonal can be calculated using the Pythagorean theorem in 3D: \( Body\ diagonal = \sqrt{a^2 + a^2 + a^2} = \sqrt{3a^2} \) Using the edge length found in step 3, we get: \( Body\ diagonal = \sqrt{3(412.6)^2} = 714.5\ \mathring{A} \)
05

Compare the body diagonal with the sum of ionic radii

The sum of ionic radii for Cs₊ and Cl₋ are given as: Cs₊ ionic radius: 169 pm Cl₋ ionic radius: 181 pm Sum of ionic radii: 169 + 181 = 350 pm = 3.50 Angstroms Now, the distance between adjacent Cs₊ and Cl₋ ions in the solid is half of the body diagonal, which is 714.5/2 = 357.25 Angstroms. This value is very close to the sum of the ionic radii of Cs₊ and Cl₋ (3.50 Angstroms). Therefore, the distance calculated agrees well with the expected value based on the sizes of the ions.

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Most popular questions from this chapter

Like most substances, bromine exists in one of the three typical phases. \(\mathrm{Br}_{2}\) has a normal melting point of \(-7.2{ }^{\circ} \mathrm{C}\) and a normal boiling point of \(59^{\circ} \mathrm{C}\). The triple point for \(\mathrm{Br}_{2}\) is \(-7.3^{\circ} \mathrm{C}\) and 40 torr, and the critical point is \(320^{\circ} \mathrm{C}\) and \(100 \mathrm{~atm}\). Using this information, sketch a phase diagram for bromine indicating the points described above. Based on your phase diagram, order the three phases from least dense to most dense. What is the stable phase of \(\mathrm{Br}_{2}\) at room temperature and \(1 \mathrm{~atm} ?\) Under what temperature conditions can liquid bromine never exist? What phase changes occur as the temperature of a sample of bromine at \(0.10\) atm is increased from \(-50^{\circ} \mathrm{C}\) to \(200^{\circ} \mathrm{C}\) ?

Water in an open beaker evaporates over time. As the water is evaporating, is the vapor pressure increasing, decreasing, or staying the same? Why?

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