Answer
(a) $K_{2}S$ should have higher boiling point than $(CH_{3})_{3}N$ because the electrostatic force in $K_{2}S$ are stronger than dispersion forces in $(CH_{3})_{3}N$.
(b) $Br_{2}$ should have higher boiling point than $CH_{3}CH_{2}CH_{2}CH_{3}$. Both molecules have only dispersion forces. $Br_{2}$ has more electrons than $CH_{3}CH_{2}CH_{2}CH_{3}$; as the number of electrons in the molecules increases, the strength of the dispersion forces also increases.
Work Step by Step
(a) $K_{2}S$ should have higher boiling point than $(CH_{3})_{3}N$ because the electrostatic force in $K_{2}S$ are stronger than dispersion forces in $(CH_{3})_{3}N$.
(b) $Br_{2}$ should have higher boiling point than $CH_{3}CH_{2}CH_{2}CH_{3}$. Both molecules have only dispersion forces. $Br_{2}$ has more electrons than $CH_{3}CH_{2}CH_{2}CH_{3}$; as the number of electrons in the molecules increases, the strength of the dispersion forces also increases.
