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 $1.67\stackrel{°}{\text{A}}$ and $\mathbf{1}\mathbf{.81}\stackrel{\mathbf{°}}{\mathbf{\text{A}}}$

Short answer

${\text{633.24kJmol}}^{\text{-1}}$

Step-by-step solution

The internuclear separation between cesium and chlorine ions.

As the radii of ${\text{Cs}}^{\text{+}}$ and ${\text{Cl}}^{-}$ are $1.67\stackrel{\text{o}}{\text{A}}$ and$1.81\stackrel{\text{o}}{\text{A}}$ , respectively.

Therefore;

$\begin{array}{c}{\text{R}}_{\text{o}}=1.67\stackrel{\text{o}}{\text{A}}+1.81\stackrel{\text{o}}{\text{A}}\\ =3.48\stackrel{\text{o}}{\text{A}}\\ =3.48\times {10}^{-10}\text{m}\end{array}$

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

Using the Madelung constant of 1.7627for the structure CsCl;

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

$\begin{array}{c}\text{latticeenergy=}\frac{(6.02\times {10}^{23}mo{l}^{-1}){(1.602\times {10}^{-19})}^2\left(1.7627\right)}{4\left(3.14\right)(8.854\times {10}^{-12}{\text{C}}^2{\text{J}}^{-1}{\text{m}}^{-1})(3.48\times {10}^{-10}\text{m})}\\ =7.036\times {10}^5{\text{Jmol}}^{-1}\\ =703.6{\text{kJmol}}^{-1}\end{array}$

The 10% of the lattice energy is;

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

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

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