Question

Question: Repeat the calculation of problem 39 for CsCl, taking the Madelung constant from Table 21.5 and taking the radii of ${\mathbf{\text{Cs}}}^{\mathbf{+}}$and${\mathbf{\text{Cl}}}^{\mathbf{-}}$to be role="math" localid="1661442448149" $\mathbf{\text{1.67}}\stackrel{\mathbf{\circ }}{\mathbf{A}}$and $\mathbf{\text{1.81}}\stackrel{\mathbf{\circ }}{\mathbf{A}}$.

Short answer

${\text{633.24 kJ mol}}^{\text{- 1}}$

Step-by-step solution

The internuclear separation between cesium and chlorine ions.

As the radii of${\text{Cs}}^{\text{+}}$ and${\text{Cl}}^{-}$ are $\begin{array}{rcl}& & 1.67\stackrel{\circ }{A}\end{array}$and $\begin{array}{rcl}& & 1.81\stackrel{\circ }{A}\end{array}$, respectively.

Therefore;

role="math" localid="1661442608622" $\begin{array}{rcl}{\text{R}}_{\text{o}}& =& 1.67\stackrel{\circ }{A}+1.81\stackrel{\circ }{A}\\ & =& 13.48\stackrel{\circ }{A}\\ & =& 3.48\times {10}^{-10}\text{m}\end{array}$

The energy needed to dissociate 1.00 mol of crystalline CsCl into its gaseous ions.

Using the Madelung constant of 1.7627 for the structure CsCl;

$\text{lattice energy =}\frac{{\text{N}}_{\text{A}}{e}^2\text{M}}{4\pi {\epsilon }_0{R}_0}$

$$\begin{gathered} \text{lattice energy =}\frac{\left(6.02\times {10}^{23}mo{l}^{-1}\right){\left(1.602\times {10}^{-19}\right)}^2\left(1.7627\right)}{4\left(3.14\right)(8.854\times {10}^{-12}{\text{C}}^2{\text{J}}^{-1}{\text{m}}^{-1})(3.48\times {10}^{-10}\text{m})} \\ =7.036\times {10}^5{\text{J mol}}^{-1} \\ =703.6{\text{kJ mol}}^{-1} \end{gathered}$$

The10% of the lattice energyis;

$\frac{\text{703.6}}{100}\times \text{10\% = 70.36}$

Assuming that the repulsive energy reduces the lattice energy by 10% from the pure Coulomb energy.

$703.6{\text{- 70.36 = 633.24 kJ mol}}^{-1}$