Question

The $\mathbf{CsCl}$structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 61). Given that the density of cesium chloride is $\mathbf{3}\mathbf{.}\mathbf{97}\mathbf{}\mathbf{gm}\mathbf{/}{\mathbf{cm}}^{\mathbf{3}}$, 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 ${\mathbf{Cs}}^{\mathbf{+}}$and ${\mathbf{Cl}}^{\mathbf{-}}$ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of ${\mathbf{Cs}}^{\mathbf{+}}$is 169 pm, and the ionic radius of${\mathbf{Cl}}^{\mathbf{-}}$ is 181 pm.

Short answer

The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array. That means CsCl has a BCC structure.

Step-by-step solution

Given Information

The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array.

The density of cesium chloride is $3.97\mathrm{gm}/{\mathrm{cm}}^3$.

The chloride and cesium ions touch along the body diagonal of the cubic unit cell.

Concept Introduction

The bond distance mainly depends upon the bond strength and coordination number, and the ionic mainly depends upon the oxidation state.

Step 3: Calculates the mass of the unit cell

The total number of chloride ions present at 8 corners of the cube,$\left(\frac{1}{8}\times 8\right)$

The net ions of chloride in the unit cell =1

The net Cs⁺ ions in unit cell = 1

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{unit}\mathrm{cell}=\mathrm{Mass}\mathrm{of}{\mathrm{Cs}}^{+}\mathrm{ion}+\mathrm{Mass}\mathrm{of}{\mathrm{Cl}}^{-}\mathrm{ion} \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{unit}\mathrm{cell}=\frac{132.9\mathrm{gm}/\mathrm{mole}}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}}+\frac{35.45\mathrm{gm}/\mathrm{mole}}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{mole}} \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{unit}\mathrm{cell}=27.95\times {10}^{-23}\mathrm{gm} \end{gathered}$$

Step 4: The distance between the centers of adjacent Cs+ and Cl-ions in solids

The value of z = 1 because per unit cell, only one CsCl is present.

$$\begin{gathered} \mathrm{d}=\frac{\mathrm{z}\times \mathrm{M}}{{\mathrm{N}}_{\mathrm{A}}\times \mathrm{V}} \\ \mathrm{d}=\frac{1\times 168.36\mathrm{gm}/\mathrm{formula}\mathrm{unit}}{6.023\times {10}^{23}\mathrm{atoms}/\mathrm{formula}\mathrm{unit}\times {\mathrm{a}}^3} \\ {\mathrm{a}}^3=7.04\times {10}^{-23}{\mathrm{m}}^3 \\ \mathrm{a}=4.13\times {10}^{-8}\mathrm{m} \\ \sqrt{3}\mathrm{a}=2\left({\mathrm{r}}_{+}+{\mathrm{r}}_{-}\right) \\ \left({\mathrm{r}}_{+}+{\mathrm{r}}_{-}\right)=357\mathrm{pm} \end{gathered}$$

Step 5: Comparing this value with the expected distance based on the sizes of the ions

From the given value of the ionic radius of Cs* and CI ions, the expected value of the distance between the centers of adjacent Cs and CI ions would be as follows:

$$\begin{gathered} {\mathrm{Cs}}^{+}=169\mathrm{pm}\mathrm{and}{\mathrm{Cl}}^{-}=181\mathrm{pm} \\ {\mathrm{r}}_{+}+{\mathrm{r}}_{-}=169\mathrm{pm}+181\mathrm{pm} \\ {\mathrm{r}}_{+}+{\mathrm{r}}_{-}=350\mathrm{pm} \end{gathered}$$

Thus, this expected value is less than the distance calculated from the density by 7 pm.