Chapter 10, Problem 10B.1E

Interpretation Introduction

Interpretation:

The energy of the London interaction for two argon atoms separated by $\text{1}\text{.0}\text{nm}$ has to be calculated.

Concept introduction:

Polar molecules are the molecules having a positive and negative end and so there will be a charge separation.

A polar molecule is a molecule where the polar bonds are asymmetrically arranged (the dipoles do not cancel)

A nonpolar molecule is a molecule with no polar bonds or a molecule where the polar bonds are symmetrically arranged.

In polar molecule the charge separation occurred with respect to the difference in electronegativity of atoms in the molecule.

Calculation of the dispersion energy:

  $\begin{array}{l}\text{V}\left(\text{r}\right)\text{=}\text{-}\frac{\text{3}}{\text{2}}\text{×}\frac{{\text{α}}_{\text{1}}{}^{\text{'}}{\text{α}}_{\text{2}}{}^{\text{'}}}{{\text{r}}^{\text{6}}}\text{×}\frac{{\text{I}}_{\text{1}}{\text{I}}_{\text{2}}}{{\text{I}}_{\text{1}}\text{+}{\text{I}}_{\text{2}}}\\ \\ {\text{α}}_{\text{1}}{}^{\text{'}}{\text{α}}_{\text{2}}{}^{\text{'}}\text{-Polarizability}\text{volumes}\\ {\text{I}}_{\text{1}}{\text{I}}_{\text{2}}\text{-}\text{Ionization}\text{energies}\\ \text{r}\text{-}\text{separation}\text{between atoms}\\ \end{array}$

If ${\text{I}}_{\text{1}}\text{=}{\text{I}}_{\text{2}}$, the equation can be re-written as below:

  $\text{V}\left(\text{r}\right)\text{=}\text{-}\frac{\text{3}}{\text{2}}\text{×}\frac{{\text{α}}_{\text{1}}{}^{\text{'}}{\text{α}}_{\text{2}}{}^{\text{'}}}{{\text{r}}^{\text{6}}}\text{×}\frac{\text{1}}{\text{2}}\text{I}$