Rationalize the difference in boiling points for each of the following pairs of substances: $$\begin{array}{rr}{\text { a. Ar }} & {-186^{\circ} \mathrm{C}} \\\ {\mathrm{HCl}} & {-85^{\circ} \mathrm{C}}\end{array}$$ $$\begin{array}{rr}{\text { b. } \mathrm{HF}} & {20^{\circ} \mathrm{C}} \\\ {\mathrm{HCl}} & {-85^{\circ} \mathrm{C}}\end{array}$$ $$\begin{array}{cc}{\text { c. } \mathrm{HCl}} & {-85^{\circ} \mathrm{C}} \\\ {\mathrm{LiCl}} & {1360^{\circ} \mathrm{C}}\end{array}$$ $$\begin{array}{ccc}{\text { d. } n \text { -pentane }} & {\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}} & {36.2^{\circ} \mathrm{C}} \\ {n \text { -hexane }} & {\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}} & {69^{\circ} \mathrm{C}}\end{array}$$

In summary, the differences in boiling points for the given pairs of substances can be explained by the type and strength of the intermolecular forces present in these molecules: a. Ar has weak London dispersion forces, while HCl has stronger dipole-dipole interactions and hydrogen bonding, leading to a higher boiling point for HCl. b. Both HF and HCl exhibit dipole-dipole interactions, but the hydrogen bonding in HF is much stronger due to fluorine's higher electronegativity, resulting in a higher boiling point for HF. c. HCl has dipole-dipole interactions and hydrogen bonding, while LiCl has strong ionic bonds, which require much more energy to break, causing a higher boiling point for LiCl. d. n-pentane and n-hexane have London dispersion forces, but n-hexane's longer carbon chain and higher molecular weight result in stronger forces and a higher boiling point compared to n-pentane.