Question

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

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

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: \( Molecular\ weight\ of\ CsCl = 132.9 + 35.45 = 168.35\ g/mol \)

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: \( Moles\ per\ unit\ volume = \frac{3.97\ g/cm^3}{168.35\ g/mol} = 0.0236\ mol/cm^3 \)

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: \( a = (6.022 \times 10^{23}\times 0.0236)^{-1/3} = 4.126 \times 10^{-8} cm = 412.6\ \mathring{A} \)

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: \( Body\ diagonal = \sqrt{3(412.6)^2} = 714.5\ \mathring{A} \)

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.

Key concepts

Ionic Radii

When exploring the world of crystal structures, understanding ionic radii is essential. Ionic radii refer to the effective distance from the nucleus of an ion to its outermost electron shell. In our problem, the ionic radii are essential for calculating distances between ions in the CsCl crystal structure.
The ionic radius of a cesium ion (\(\text{Cs}^+\) ) is 169 picometers (pm), while that of a chloride ion (\(\text{Cl}^-\)) is 181 pm. These values help determine how ions are arranged and touch in a structure. By adding these radii, you can predict the distance between the ion centers based on their sizes, which is crucial for understanding how they pack in the solid state.
This total distance, called the sum of ionic radii, is found to be 350 pm or 3.5 Ångströms (Å). This gives us a theoretical basis to compare with actual measured values, offering insights into the accuracy of crystal models.

CsCl Structure

The CsCl structure is a classic example of a simple cubic crystal system. In this structure, chloride ions form a cubic lattice with a cesium ion sitting at the center of each cube. This arrangement is essential for understanding many properties of the compound, including its density and how ions interact.
In this structure, the cesium and chloride ions touch each other along the body diagonal. This unique arrangement allows for close interactions between the cations and anions, contributing to the compound's stability. The method of placing ions in a cubic array helps maximize contact between oppositely charged ions, minimizing potential energy and enhancing the structural integrity of the solid.
Understanding the CsCl structure helps in comprehending more complex ionic compounds and their properties.

Density Calculations

Calculating the density of a crystal structure like CsCl involves understanding the relationship between mass, volume, and the number of moles. The given density of cesium chloride is 3.97 g/cm³, which tells us how much mass is in a given volume of the solid.
To find the moles per unit volume, you use the formula:\[Moles\ per\ unit\ volume = \frac{\text{Density}}{\text{Molecular\ weight}} \]Using the molecular weight of CsCl (168.35 g/mol) and its density, you find that there are 0.0236 moles of CsCl per cm³.
This information allows you to work out how many unit cells, like tiny repeating blocks of the crystal, fit into a cubic centimeter of the material, aiding in determining the unit cell dimensions and confirming the crystal packing.

Unit Cell Dimensions

Unit cell dimensions define the size and shape of the smallest repeating unit in a crystal lattice. For CsCl, the unit cell is cubic, containing one Cs⁺ ion at the center and Cl⁻ ions at each corner.
To find the edge length (\(a\)) of the cubic unit cell, consider that it holds one CsCl molecule. Using Avogadro's number (\(6.022 \times 10^{23}\) molecules/mol), you solve:\[a = \left( \frac{6.022 \times 10^{23} \times 0.0236}{1} \right)^{-1/3}\]This gives an edge length of 412.6 pm or 4.126 Å. From this, you can calculate the body diagonal and use it to verify the close contact between Cs⁺ and Cl⁻ along this line.
The calculated distances between ions are checked against expected values, offering insights into crystal stability and confirming theoretical models with physical measurements.