Chapter 9, Problem 146CP

Rubidium chloride has the sodium chloride structure at normal pressures but assumes the cesium chloride structure at high pressures. (See Exercise 69.) What ratio of densities is expected for these two forms? Does this change in structure make sense on the basis of simple models? The ionic radius is 148 pm for Rb⁺ and 181 pm for CI⁻.

Interpretation Introduction

Interpretation:

Rubidium chloride has two structures at different pressures. The ratio of the density of these two forms has to be determined.

Concept introduction:

In packing of atoms or molecules of a solid, the atoms/molecules are imagined as spheres. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing. Cubic close packing structure has face-centered cubic (FCC) unit cell. FCC unit cell has 4 units of atom/molecule per unit cell. In FCC unit cell the components touch along the edge of the cubic unit cell. A simple cubic unit cell has 1 unit of atom/molecule per unit cell. In this unit cell, the components touch along the body diagonal of the unit cell.

Answer to Problem 146CP

Answer

The ratio of the density of the two forms of Rubidium chloride is 1.30.

Explanation of Solution

Explanation

Calculate the volume of unit cell of $\text{RbCl}$ in its $\text{NaCl}$ structure.

$\begin{array}{l}{\text{given data: ionic radius of Rb}}^{\text{+}}\text{=}\text{148}\text{pm}\\ \text{ionic}\text{radius}\text{of}{\text{Cl}}^{\text{-}}\text{=}\text{181}\text{pm}\end{array}$

                  $\begin{array}{l}\text{edge}\text{length,}\text{l}\text{=}{\text{2r}}_{{\text{Rb}}^{\text{+}}}\text{+}{\text{2r}}_{{\text{Cl}}^{\text{-}}}\\ \text{l}\text{=}\text{2}\left(\text{148}\text{+}\text{181}\right)\text{pm}\text{=}\text{658}\text{pm}\end{array}$

                    $\text{volume,}{\text{l}}^{\text{3}}\text{=}{\left(\text{658}\text{×}{\text{10}}^{\text{-12}}\text{m}\right)}^{\text{3}}\text{=}\text{2}\text{.85}\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}$

              At normal pressure structure of Rubidium chloride is similar to that of Sodium chloride. The ionic radius of the ${\text{Rb}}^{\text{+}}$ ion and ${\text{Cl}}^{\text{-}}$ ion are given. The formula for the edge length of the FCC cubic unit cell containing a molecule is known as $\text{l}\text{=}{\text{2r}}_{\text{c}}\text{+}{\text{2r}}_{\text{a}}$ where ${\text{r}}_{\text{c}}\text{and}{\text{r}}_{\text{a}}$ are ionic radius of cation and anion respectively. The value is substituted and the volume of the unit cell containing ${\text{Rb}}^{\text{+}}$ ions and ${\text{Cl}}^{\text{-}}$ ions is calculated.

Calculate the mass and density of unit cell of $\text{RbCl}$ in its $\text{NaCl}$ structure.

$\begin{array}{l}\text{Average mass of one RbCl}\text{molecule}\text{=}\frac{\text{molecular}\text{mass}\text{of}\text{RbCl}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{\text{120}\text{.9}\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}\text{20}\text{.08}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

$\text{Each}\text{FCC}\text{unit}\text{cell}\text{has}\text{4}\text{RbCl}\text{molecules}\text{.}\text{Therefore,}$

$\begin{array}{l}\text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}\text{4}\text{×}\text{average}\text{mass}\text{of}\text{RbCl}\text{molecule}\\ \text{=}\text{4}\text{×}\text{20}\text{.08}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}\text{80}\text{.32}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

$\begin{array}{l}\text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{=}\frac{80.32\text{×}{\text{10}}^{\text{-23}}\text{g}}{2.85\text{×}{\text{10}}^{\text{-22}}{\text{cm}}^{\text{3}}}\\ \text{=}2.82{\text{g/cm}}^{\text{3}}\end{array}$

           Each FCC unit cell contains 4 $\text{RbCl}$ molecules like in the case of Sodium chloride. Therefore four times the average mass of one $\text{RbCl}$ molecule gives mass of a unit cell. Volume of the unit cell is calculated in the previous step and the mass of the unit cell are substituted in the formula, $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$.

Calculate the volume of unit cell of $\text{RbCl}$ in its $\text{CsCl}$ structure.

             $\begin{array}{l}\text{body}\text{diagonal}\text{=}{\text{2r}}_{{\text{Rb}}^{\text{+}}}\text{+}{\text{2r}}_{{\text{Cl}}^{\text{-}}}\text{=}\text{l}\sqrt{\text{3}}\\ \text{l}\sqrt{\text{3}}\text{=}\text{2}\left(\text{148}\text{+}\text{181}\right)\text{pm}\text{=}\text{658}\text{pm}\end{array}$

                        $\begin{array}{l}\\ \text{edge}\text{length,}\text{l}\text{=}\frac{\text{658}\text{×}{\text{10}}^{\text{-12}}\text{m}}{\sqrt{\text{3}}}=\text{380}\text{pm}\\ \text{volume,}{\text{l}}^3={\left(380\times {10}^{-12}\text{m}\right)}^3\\ =\text{5}\text{.49}\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\end{array}$

At high pressure structure of Rubidium chloride is similar to that of Cesium chloride. The ionic radius of the ${\text{Cs}}^{+}$ ion and ${\text{Cl}}^{\text{-}}$ ion are given. The formula for the edge length of the simple cubic unit cell containing a molecule is known as $\text{l}\text{=}\frac{{\text{2r}}_{{\text{Rb}}^{\text{+}}}\text{+}{\text{2r}}_{{\text{Cl}}^{\text{-}}}}{\sqrt{\text{3}}}$ where ${\text{r}}_{\text{c}}\text{and}{\text{r}}_{\text{a}}$ are ionic radius of cation and anion respectively. The value is substituted and the volume of the unit cell containing ions ${\text{Cs}}^{+}$ and ${\text{Cl}}^{\text{-}}$ ions is calculated.

Calculate the mass and density of unit cell of $\text{RbCl}$ in its $\text{CsCl}$ structure.

$\begin{array}{l}\text{Average mass of one RbCl}\text{molecule}\text{=}\frac{\text{molecular}\text{mass}\text{of}\text{RbCl}}{\text{Avogadro}\text{number}}\\ \text{=}\frac{\text{120}\text{.9}\text{g}}{\text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}}\\ \text{=}\text{20}\text{.08}\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

$\text{Each}\text{simple}\text{cubic}\text{unit}\text{cell}\text{has}\text{one}\text{RbCl}\text{molecule}\text{.}\text{Therefore,}$

         $\begin{array}{l}\text{mass}\text{of}\text{a}\text{unit}\text{cell}\text{=}1\text{×}\text{average}\text{mass}\text{of}\text{RbCl}\text{molecule}\\ \text{=}1\text{×}\text{20}\text{.08}\text{×}{\text{10}}^{\text{-23}}\text{g}\\ \text{=}20.08\text{×}{\text{10}}^{\text{-23}}\text{g}\end{array}$

$\begin{array}{l}\text{density}\text{=}\frac{\text{mass}}{\text{volume}}\\ \text{=}\frac{20.08\text{×}{\text{10}}^{\text{-23}}\text{g}}{5.49\text{×}{\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}}\\ \text{=}3.66{\text{g/cm}}^{\text{3}}\end{array}$

Each simple cubic unit cell contains one $\text{RbCl}$ molecule like in the case of Cesium chloride. Therefore the average mass of one $\text{RbCl}$ molecule gives mass of a unit cell. Volume of the unit cell is calculated in the previous step and the mass of the unit cell are substituted in the formula, $\text{density}\text{=}\frac{\text{mass}}{\text{volume}}$.

Compare the density of two forms of Rubidium chloride.

Let density of $\text{RbCl}$ structure similar to $\text{NaCl}$ structure be ${\text{d}}_{\text{1}}$.

Let density of $\text{RbCl}$ structure similar to $\text{CsCl}$ structure be ${\text{d}}_{2.}$.

$\begin{array}{l}\text{Then,}\text{ratio}\text{of}\text{densities}\text{of}\text{the}\text{two}\text{forms}\text{of}\text{RbCl}\text{is,}\\ \frac{{\text{d}}_{\text{2}}}{{\text{d}}_1}=\frac{\text{3}\text{.66}{\text{g/cm}}^{\text{3}}}{\text{2}\text{.82}{\text{g/cm}}^{\text{3}}}=1.30\end{array}$

The ratio of densities of the two forms of $\text{RbCl}$ is calculated.

Conclusion

Conclusion

The structure of $\text{RbCl}$ in its high pressure form has more density due to the closest packing of the ions as a result of high pressure.