Question

Select the substance in each pair that should have the higher boiling point. In each case identify the principal intermolecular forces involved and account briefly for your choice.

(a) K2S or (CH3)3N,

(b) Br2 or CH3CH2CH2CH3

Answer 1

(a)
`K_(2)S` (b)
`Br_(2)`

Step-by-step explanation:

(a)
`K_(2)S` is an ionic compound in which
`K^(+)` and
`S^(2-)` ions are tightly bound together through ion-ion attraction force (principal intermolecular force).

`(CH_(3))_(3)N` is a polar covalent molecule. So dipole-dipole attraction force (principal intermolecular force) exists in this molecule.

We know, ion-ion attraction force is stronger than dipole-dipole attraction force. Therefore more energy is required to boil liquid
`K_(2)S` by disrupting stronger ion-ion attraction force as compared to
`(CH_(3))_(3)N`.

So,
`K_(2)S` has higher boiling point.

(b) Both
`Br_(2)` and
`CH_(3)CH_(2)CH_(2)CH_(3)` are nonpolar covalent compounds. Hence only london dispersion force(principal intermolecular force) exists in both molecules.

London dispersion force is proportional to molas mass of a molecule.

`Br_(2)` has higher molar mass as compared to
`CH_(3)CH_(2)CH_(2)CH_(3)` . Therefore
`Br_(2)` has higher boiling point.