Chapter 18, Problem 18.94QA

Interpretation Introduction

To calculate:

a) The radius of the $C{s}^{+}$ ions in unit cell of $CsCl$.

b) The density of simple cubic $CsCl$.

c) The percent empty space in a unit cell of $CsCl$.

Answer to Problem 18.94QA

Solution:

a) The radius of the $C{s}^{+}$ ions  is $r= 175.8 pm$

b) $Density of \mathrm{s}\mathrm{i}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{ }\mathrm{c}\mathrm{u}\mathrm{b}\mathrm{i}\mathrm{c}\mathrm{ }CsCl=4.00\frac{g}{c{m}^3}$

c) Total volume of empty space $=32 \%$

Explanation of Solution

1) Concept:

a)  $CsCl$ has bcc unit cell crystal structure. Using the relation between radius and edge length, we can get the ionic radius of $C{s}^{+}$ ion.

b) Volume of the unit cell is the cube of the edge length. Using the given value of edge length, we can calculate its volume. Simple cubic has one atom per unit cell. So, using molar mass of $CsCl$, Avogadro’s number, and number of atoms per unit cell gives the mass of a unit cell. Values of mass and volume of unit cell gives the density of a unit cell of simple cubic.

c) The packing efficiency is calculated using volumes of one $Cs$ sphere and one $Cl$ sphere and volume of unit cell. Empty space available can be calculated using the total space of a unit cell and the packing efficiency.

2) Formula:

i) bcc: $r=\frac{l\sqrt{3}}{4},$$$ where $r$ is ionic radius and $l$ is the edge length.

ii) Simple cubic:$r=\frac{l}{2}=0.5\times l$, where $r$ is ionic radius and $l$ is the edge length

iii)  Volume,$V=l^3 c{m}^3$

iv)  $Density =\frac{mass}{volume}$

v) $V of sphere=\frac{4}{3}\pi r^3$

vi) $Packing efficiency= =\frac{V_{atoms}}{V_{unit cell}}\times 100\%$

3) Given:

i) $r of C{l}^{-}ions=181 pm$

ii) $l=412 pm$

4) Calculations:

a) Calculate the radius of $C{s}^{+}$ ions:

$4r= l\sqrt{3}$

$2C{s}^{+}+2C{l}^{-}= l\sqrt{3}$

$2C{s}^{+}+2 \times 181 pm= 412 pm \times \sqrt{3}$

$2C{s}^{+}+362 pm= 713.584 pm$

$2C{s}^{+}= 351.584 pm$

$C{s}^{+}= 175.8 pm$

b) Calculate the edge length in cm from $412 pm$:

$412 pm \times \frac{1 \times {10}^{-10} cm}{1 pm}=4.12 \times {10}^{-8} cm$

Calculate the volume of simple cubic $CsCl:$

$V=l^3={\left(4.12 \times {10}^{-8} cm\right)}^3=6.99\times {10}^{-23}c{m}^3$

Mass of $CsCl$ per unit cell:

$1 atom\times \frac{1 mol}{6.022\times {10}^{23}atom}\times 168.358 \frac{g}{mol}=2.7957\times {10}^{-22}g$

$Density =\frac{mass}{volume}= \frac{2.7957\times {10}^{-22} g}{6.99\times {10}^{-23 }c{m}^3}=4.00\frac{g}{c{m}^3}$

c) Volume of $Cs$ sphere $=\frac{4}{3}\pi r^3$:

$=\frac{4}{3}\times 3.14\times {\left(175.8 pm\times \frac{1 \times {10}^{-10} cm}{1 pm}\right)}^3$

$=\frac{4}{3}\times 3.14\times {\left(1.758\times {10}^{-8}cm\right)}^3$

$=\frac{4}{3}\times 3.14\times 5.4332\times {10}^{-24}c{m}^3$

$=2.2747\times {10}^{-23}c{m}^3$

Volume of $Cl$ sphere $=\frac{4}{3}\pi r^3$:

$=\frac{4}{3}\times 3.14\times {\left(181 pm\times \frac{1 \times {10}^{-10} cm}{1 pm}\right)}^3$

$=\frac{4}{3}\times 3.14\times {\left(1.81\times {10}^{-8}cm\right)}^3$

$=\frac{4}{3}\times 3.14\times 5.9297\times {10}^{-24}c{m}^3$

$=2.483\times {10}^{-23}c{m}^3$

So, the total volume of $CsCl$ is

$\left(2.2747\times {10}^{-23}c{m}^3\right)+\left(2.483\times {10}^{-23}c{m}^3\right)=4.7577\times {10}^{-23}c{m}^3$

Calculate the volume of simple cubic $CsCl$:

$V_{unit cell}=l^3={\left(4.12 \times {10}^{-8} cm\right)}^3=6.99\times {10}^{-23}c{m}^3$

$Packing efficiency= =\frac{V_{atoms}}{V_{unit cell}}\times 100\%$

$=\frac{4.7577\times {10}^{-23}c{m}^3}{6.99\times {10}^{-23}c{m}^3}\times 100\%=68 \%$

$The total volume of the empty space=100\%-68\%=32 \%$

Conclusion:

We calculate the radius of the cation from the relation between the edge length and radius. The density of the simple cube is calculated from the edge length. Packing efficiency is calculated from the volume of atoms and volume of unit cell.