Question

The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 71 ). 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 expected distance between adjacent Cs+ and Cl- ions based on their ionic radii is 350 pm. By calculating the edge length of the unit cell and the body diagonal length, we can find the distance between the centers of adjacent ions using the formula \(\frac{d}{2}\). Comparing these two distances will help us validate the assumption that the Cs+ and Cl- ions touch along the body diagonal of the cubic unit cell.

Step-by-step solution

Find the edge length of the unit cell

From the density of cesium chloride (3.97 g/cm³), we can find the edge length of the unit cell, since we know the mass and size of both Cs+ and Cl- ions. The formula for density is: Density = \(\frac{\text{mass}}{\text{volume}}\) \[3.97 \frac{\text{g}}{\text{cm}^3} = \frac{\text{mass of one CsCl unit}}{\text{volume of one CsCl unit}}\] Since there is one Cs+ and one Cl- ion per unit cell: mass of Cs = 132.91 g/mol (from periodic table), mass of Cl = 35.45 g/mol (from periodic table), mass of one CsCl unit (g/mol) = mass of Cs + mass of Cl Volume of one CsCl unit = \(\frac{\text{mass of one CsCl unit}}{3.97 \frac{\text{g}}{\text{cm}^3}}\) The volume of a cube is given by V = a³. Therefore, the edge length of the unit cell (a) is given by: a = \(\sqrt [3]{\text{Volume of one CsCl unit}}\)

Calculate the body diagonal length

We know that Cs+ and Cl- touch along the body diagonal of the cubic unit cell. We can find the length of the body diagonal (d) using the Pythagorean theorem in three dimensions. We have, \(d = \sqrt {a^2 + a^2 + a^2}\)

Calculate the distance between adjacent Cs+ and Cl- ions

Since the Cs+ and Cl- ions touch along the body diagonal of the cubic unit cell, we can conclude that the sum of their ionic radii is equal to the distance between the centers of adjacent ions, which we designate as D. D = ionic radius of Cs+ + ionic radius of Cl- D = 169 pm + 181 pm = 350 pm Now that we have found the expected distance based on the ionic radii, we should compare this value with the one obtained using the edge length and body diagonal of the unit cell.

Compare the two distances

From the calculations above, we have the following: Expected distance based on ionic radii = 350 pm Distance based on unit cell edge length and body diagonal = \(\frac{d}{2}\) Calculate the value of \(\frac{d}{2}\), compare it to the expected distance, and discuss if they are close enough to validate the assumption that the Cs+ and Cl- ions touch along the body diagonal.

Key concepts

Ionic Radii

Understanding ionic radii helps us figure out how ions fit together in structures like CsCl. Ionic radii refer to the size of an ion in a crystal lattice. - **CsCl Structure**: In the CsCl structure, we have cesium ions (\(\text{Cs}^{+}\)) and chloride ions (\(\text{Cl}^{-}\)).- **Cesium Ion Radius**: The ionic radius of \(\text{Cs}^{+}\) is 169 pm.- **Chloride Ion Radius**: The ionic radius of \(\text{Cl}^{-}\) is 181 pm.These radii are significant because they determine how the ions pack together in the crystal lattice. In the case of CsCl, the ions touch along the body diagonal, meaning their radii sum up to give us the distance between centers of adjacent ions. This sum happens to be 350 pm.

Cubic Unit Cell

The CsCl structure forms a cubic unit cell, which is crucial for understanding its physical properties. A unit cell is the smallest repeating unit of a crystal lattice.- **Simple Cubic Geometry**: The CsCl unit cell is a simple cubic array, meaning each corner of the cube is occupied by a chloride ion and a cesium ion sits in the center.- **Edge Length**: The cube's edge length, represented as \(a\), can be derived from the density of cesium chloride using the formula: \[ a = \sqrt[3]{\text{Volume of one CsCl unit}} \]- **Uniform Packing**: Each unit cell contains one cesium ion and one chloride ion, maintaining charge neutrality and structural stability.Understanding the cubic unit cell is critical because it helps in calculating other properties such as density and body diagonals.

Density Calculation

Calculating density provides insights into how tightly packed a structure is, affecting its physical characteristics.The density of a solid can be determined using the formula:\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]For CsCl, given a density of 3.97 g/cm³, we proceed as follows:- **Mass Calculation**: The mass of one CsCl unit is obtained from the sum of the atomic masses: \[ \text{mass of CsCl unit} = 132.91 + 35.45 = 168.36 \text{ g/mol} \]- Convert this to a single molecule's mass and then use the density formula to determine the unit cell's volume:Volume = \( \frac{\text{mass of one CsCl unit}}{3.97} \) cm³.Knowing both, we can extract the edge length of the unit cell, which is pivotal in further calculations.

Body Diagonal

The body diagonal of a cubic unit cell is the line that runs from one corner of the cube, through its center, to the opposite corner. In the context of the CsCl structure, this is where ions touch each other. - **Three-Dimensional Pythagorean Theorem**: We use this theorem to calculate the body diagonal, \(d\), as follows: \[ d = \sqrt{a^2 + a^2 + a^2} = a\sqrt{3} \]- **Distance Calculation**: Since Cs+ and Cl- ions touch along this diagonal, the length of the body diagonal represents the sum of the ionic radii of the ions multiplied by two. Mathematically: \[ \text{Distance between ions} = \frac{d}{2} \]- By comparing this calculated value to the expected ionic distance (350 pm), we can confirm the structure's accuracy and determine how closely packed the ions are in the crystal.