2022 H2 Chem P1 Q1 — Chemical bonding Solution

The answer is C.

Boiling involves overcoming the forces of attraction between the particles:
• \(\text{CH}_3\text{CH}_2\text{CH}_3\): instantaneous dipole-induced dipole (id-id) attractions only, the weakest, so the lowest boiling point.
• \(\text{CH}_3\text{CH}_2\text{NH}_2\) and \(\text{CH}_3\text{CH}_2\text{OH}\): hydrogen bonding between molecules. Both form an average of one intermolecular hydrogen bond per molecule, but the O-H bond is more polar than the N-H bond because O is more electronegative than N, so the hydrogen bonds between \(\text{CH}_3\text{CH}_2\text{OH}\) molecules are stronger and \(\text{CH}_3\text{CH}_2\text{OH}\) has the higher boiling point of the two.
• NaF: strong ionic bonds in a giant ionic lattice, which need the most energy to overcome, so the highest boiling point.

Order: \(\text{CH}_3\text{CH}_2\text{CH}_3\) < \(\text{CH}_3\text{CH}_2\text{NH}_2\) < \(\text{CH}_3\text{CH}_2\text{OH}\) < NaF.

• A is wrong: it places \(\text{CH}_3\text{CH}_2\text{OH}\) below \(\text{CH}_3\text{CH}_2\text{NH}_2\), but its hydrogen bonds are stronger.
• B is wrong: NaF, with ionic bonding, must have the highest boiling point, above \(\text{CH}_3\text{CH}_2\text{NH}_2\).
• D is wrong: \(\text{CH}_3\text{CH}_2\text{NH}_2\) forms hydrogen bonds, so it boils higher than \(\text{CH}_3\text{CH}_2\text{CH}_3\), which has only id-id attractions.