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 \(\mathrm{pm} .\)

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 = \(\sqrt [3]{\text{Volume of one CsCl unit}}\)

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, \(d = \sqrt {a^2 + a^2 + a^2}\)

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.

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.