9701 · 3.6

Intermolecular forces, electronegativity and bond properties — practice questions

Practice and worked examples for 9701 Intermolecular forces, electronegativity and bond properties. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Propanone ( $CH_3COCH_3$ , $M_r = 58.0$ ) boils at 56 °C, whereas butane ( $C_4H_{10}$ , $M_r = 58.0$ ) boils at -1 °C. Explain this difference.

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Identify molecules and IMFs: Both molecules have similar relative molecular masses and therefore a similar number of electrons. This means the strength of their van der Waals forces will be comparable.
Analyse polarity: Butane is a non-polar molecule, so it only has van der Waals forces between its molecules. Propanone is a polar molecule due to the polar C=O bond and its shape. Therefore, it has both van der Waals forces and stronger permanent dipole-dipole forces.
Relate IMFs to boiling point: Since the total intermolecular forces in propanone (vdW + pd-pd) are stronger than those in butane (vdW only), more energy is required to overcome these forces and separate the molecules. Consequently, propanone has a significantly higher boiling point.

Worked example 2

Explain why ammonia ( $NH_3$ ) is very soluble in water ( $H_2O$ ), but methane ( $CH_4$ ) is not.

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Analyse solute and solvent IMFs: Water is a polar solvent whose molecules are held together by strong hydrogen bonds (in addition to van der Waals forces).
Analyse ammonia ( $NH_3$ ): Ammonia is a polar molecule. The hydrogen atoms are bonded to a highly electronegative nitrogen atom. This means ammonia molecules can form hydrogen bonds with each other, and crucially, they can also form hydrogen bonds with water molecules. The energy released when these new hydrogen bonds form is sufficient to overcome the hydrogen bonds between water molecules and between ammonia molecules.
Analyse methane ( $CH_4$ ): Methane is a non-polar molecule. It only has weak van der Waals forces between its molecules. It cannot form hydrogen bonds with water. To dissolve, the strong hydrogen bonds between water molecules would need to be broken, but only weak van der Waals forces would form between methane and water. This process is not energetically favourable, so methane is insoluble.
Conclusion: Solubility follows the 'like dissolves like' principle. Substances that can form similar types of intermolecular forces to the solvent tend to be soluble. Ammonia can hydrogen bond with water, so it dissolves. Methane cannot, so it does not.

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