Question

In a gaseous RbF molecule, the bond length is $\mathbf{2}\mathbf{.}\mathbf{274}\mathbf{\times }{\mathbf{10}}^{-10}\mathbf{m}$ m. Using the data from Appendix F and making the same oversimplified assumption as in the prior problem on the shape of the potential curve from ${\mathbf{Rb}}^{+}\mathbf{+}{\mathbf{F}}^{-}$* to an internuclear separation of $\mathbf{2}\mathbf{.}\mathbf{274}\mathbf{\times }{\mathbf{10}}^{-10}\mathbf{m}$, calculate the energy in $\mathbf{kJ}{\mathbf{mol}}^{-1}$ required to dissociate RbF to neutral atoms.

Short answer

The energy in kJ/mol required to dissociate RbF to neutral atoms is 535.94 kJ/mol.

Step-by-step solution

Given

Bond length=2.274 m

Ionization Energy of Rb= 403.0

Electron affinity of F= 328.0

Dissociation energy=?

Permmitiivity of vaccume= 8.854

Concept used

Ions which are bonded by an ionic bond are attracted to each other ( due to opposite charges) by an electrostatic force. Potential energy of such a system is given by Coulamb’s law:

$\begin{array}{c}{\mathbf{\Delta E}}_d\mathbf{=}\frac{{\mathbf{q}}_1{\mathbf{q}}_2{\mathbf{N}}_A}{\mathbf{4}{\mathbf{\pi \epsilon }}_o{\mathbf{R}}_e{\mathbf{10}}^3}\mathbf{-}{\mathbf{\Delta E}}_{\infty }\mathbf{kJ}\mathbf{/}\mathbf{mol}\\ \mathrm{Where}\\ {\mathbf{q}}_1{\mathbf{q}}_2\mathbf{=}\mathbf{Charges}\mathbf{on}\mathbf{the}\mathbf{ions}\\ {\mathbf{N}}_A\mathbf{=}\mathbf{Avogadro}\mathbf{\text{'}}\mathbf{s}\mathbf{number}\\ {\mathbf{\epsilon }}_o\mathbf{=}\mathbf{permmitivity}\mathbf{of}\mathbf{vaccume}\\ {\mathbf{R}}_e\mathbf{=}\mathbf{Distance}\mathbf{of}\mathbf{separation}\\ {\mathbf{\Delta E}}_{\infty }\mathbf{=}\mathbf{Ionization}\mathbf{energy}\mathbf{of}\mathbf{cation}\mathbf{-}\mathbf{Electron}\mathbf{affinity}\mathbf{ofanion}\end{array}$

Calculation

$\begin{array}{c}{\text{ΔE}}_{\mathrm{\infty }}\text{=\hspace{0.17em}IE\hspace{0.17em}of\hspace{0.17em}Rb-\hspace{0.17em}EA\hspace{0.17em}of\hspace{0.17em}F}\\ {\text{ΔE}}_{\mathrm{\infty }}=403-328\\ {\text{ΔE}}_{\mathrm{\infty }}=75\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔE}}_{\text{d}}\text{=\hspace{0.17em}}\frac{{\text{q}}_{\text{1}}{\text{q}}_{\text{2}}{\text{N}}_{\text{A}}}{{\text{4πε}}_{\text{o}}{\text{R}}_{\text{e}}{\text{10}}^{\text{3}}}{\text{-ΔE}}_{\mathrm{\infty }}\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔE}}_{\text{d}}=\frac{(-1.602\times {10}^{-19}{\text{C}}^{\text{-1}})\times 6.023\times {10}^{23}{\text{\hspace{0.17em}mol}}^{-}}{(4\times 3.1416)(8.854\times {10}^{-12}{\text{C}}^{\text{2}}{\text{J}}^{\text{-1}}{\text{m}}^{\text{-1}})(2.274\times {10}^{-10}\mathrm{m})\left({10}^3{\text{J\hspace{0.17em}kJ}}^{\text{-1}}\right)}-75\mathrm{k}\text{J/mol}\\ {\text{ΔE}}_{\text{d}}=0.0610937\times {10}^4\\ {\text{ΔE}}_{\text{d}}=610.94\text{\hspace{0.17em}kJ/mol-\hspace{0.17em}}75\text{\hspace{0.17em}kJ/mol}\\ {\text{ΔE}}_{\text{d}}=535.94\text{\hspace{0.17em}kJ/mol}\end{array}$