HSC Exam Practice

Sample response. Intermolecular forces (IMFs) are attractions that act between separate molecules. They are distinct from intramolecular bonds (covalent, ionic, metallic) which act within a molecule or lattice. When a molecular substance boils, intermolecular forces between molecules are overcome; the covalent bonds within each molecule remain intact, so the molecules themselves are unchanged.

Marking notes. 1 mark for correctly defining IMFs as forces between molecules (not within). 1 mark for correctly identifying intramolecular bonds as within the molecule. 1 mark for stating that boiling breaks intermolecular forces, not intramolecular/covalent bonds.

Sample response. (a) CH₄: dispersion forces only. No polar bonds, no N/O/F–H bonds. No hydrogen bonding. (b) CHCl₃: dispersion + dipole-dipole. Polar molecule (C–Cl bonds with electronegative Cl create a net dipole). No N/O/F–H bond, so no hydrogen bonding. (c) NH₃: dispersion + hydrogen bonding (N–H···N). N is electronegative enough (χ = 3.0) to create a large δ+ on H; adjacent N has a lone pair. (d) HF: dispersion + hydrogen bonding (F–H···F). F is the most electronegative element (χ = 4.0), creating extremely strong H-bonds.

Marking notes. 1 mark per molecule for correct IMF identification with justification (4 marks total). Accept “London dispersion forces” as equivalent to dispersion forces. For CH₄ and CHCl₃, mark is lost only if hydrogen bonding is incorrectly claimed.

Sample response. Straight-chain alkanes are non-polar, so the only IMF present is dispersion forces. As chain length increases, the number of electrons in the molecule increases, making the electron cloud larger and more polarisable. Larger electron clouds produce stronger instantaneous temporary dipoles, leading to stronger dispersion forces between adjacent molecules. More energy is therefore required to overcome these forces during boiling, resulting in a higher boiling point with each additional –CH₂– unit.

Marking notes. 1 mark for identifying dispersion forces as the only IMF in alkanes. 1 mark for correctly explaining that increasing chain length → more electrons → stronger dispersion forces. 1 mark for linking stronger dispersion forces to a higher boiling point (more energy required to vaporise).

Sample response. Dispersion forces are present in ALL molecules, including non-polar ones. They arise from instantaneous uneven distributions of electrons (temporary dipoles) that induce dipoles in neighbouring molecules, creating weak, fleeting attractions. Example: CH₄, O₂, noble gases. Dipole-dipole forces exist only in polar molecules that have a permanent charge separation (δ+/δ−) due to electronegativity differences between bonded atoms. The δ+ end of one molecule is attracted to the δ− end of an adjacent molecule. Example: HCl, SO₂, CHCl₃. Polar molecules therefore experience both dispersion AND dipole-dipole forces; dipole-dipole forces are stronger than dispersion forces for molecules of similar size.

Marking notes. 1 mark for correctly describing dispersion forces (temporary dipoles, present in ALL molecules, including non-polar). 1 mark for correct explanation of origin of dispersion forces (instantaneous electron movement). 1 mark for correctly describing dipole-dipole forces (polar molecules only, permanent charge separation). 1 mark for correct explanation of origin of dipole-dipole (electronegativity difference → δ+/δ− ends attract).

Sample response. The student is incorrect. For hydrogen bonding to occur, hydrogen must be bonded to an atom that is sufficiently electronegative: specifically F, O, or N (not just any polar molecule with H). Chlorine has a Pauling electronegativity of approximately 3.0, which is insufficient to create the large partial positive charge on hydrogen necessary for H-bonding to the lone pair of an adjacent molecule. HCl therefore has only dispersion forces and dipole-dipole forces — no hydrogen bonding. This is a common misconception; HF (not HCl) is the hydrogen halide that forms H-bonds because F (χ = 4.0) is electronegative enough.

Marking notes. 1 mark for identifying the flaw: H-bonding requires H bonded to F, O, or N specifically — polarity alone is insufficient. 1 mark for explaining why Cl is not electronegative enough to enable H-bonding (accept Pauling χ(Cl) ≈ 3.0 vs F = 4.0, O = 3.5). 1 mark for correctly stating the IMFs actually present in HCl: dispersion + dipole-dipole only.

Sample response. Both n-pentane and neopentane have the same molecular formula (C₅H₁₂), the same number of electrons, and the same IMF type (dispersion forces only, as both are non-polar). The difference in boiling point arises from molecular shape. n-Pentane is a straight chain that can align closely with neighbouring molecules along its entire length, maximising surface contact area. Neopentane has a compact, nearly spherical branched shape with minimal surface contact. Greater contact area in n-pentane leads to stronger total dispersion forces and a higher boiling point (36 °C vs 9.5 °C). The IMF type responsible is dispersion forces.

Marking notes. 1 mark for identifying that both molecules have identical IMF type (dispersion forces only) and same electron count, so the difference is not due to IMF type or electron number. 1 mark for explaining that n-pentane’s linear shape allows greater surface contact with neighbouring molecules than neopentane’s compact branched shape. 1 mark for concluding that greater surface contact → stronger total dispersion forces → higher BP in n-pentane.

Sample response (a). Main IMF column: CH₄: dispersion forces only. CH₃F: dispersion + dipole-dipole. CH₃OH: dispersion + hydrogen bonding. C₂H₆: dispersion forces only. H₂O: dispersion + hydrogen bonding. NH₃: dispersion + hydrogen bonding. Explanation for CH₃F: fluoromethane is polar (C–F bond; F is highly electronegative), so it has dipole-dipole forces in addition to dispersion forces. However, the hydrogen atoms in CH₃F are bonded to carbon, not to fluorine; there is no F–H bond, so the requirement for H-bonding (H directly bonded to F, O, or N) is not met. Therefore CH₃F does not form hydrogen bonds despite containing F.

Sample response (b). Ranking CH₄, CH₃F, CH₃OH: Lowest to highest: CH₄ (−161 °C) < CH₃F (−78 °C) < CH₃OH (65 °C). CH₄ has the lowest BP because it has only weak dispersion forces (non-polar, small, 10 electrons) — the weakest IMFs. CH₃F has a polar C–F bond, so dispersion forces plus dipole-dipole forces act; this extra attraction elevates the BP above CH₄ despite both molecules being small. CH₃OH has the highest BP because the O–H bond enables hydrogen bonding (O–H···O), which is far stronger than dipole-dipole forces alone; this large additional attraction requires the most energy to overcome on vaporisation.

Sample response (c). H₂O vs NH₃: Both form hydrogen bonds, but water (BP 100 °C) boils higher than ammonia (BP −33 °C) despite very similar molecular masses (18 vs 17 g mol⁻¹). Water can form up to four hydrogen bonds per molecule (two as donor via O–H bonds; two as acceptor via the two lone pairs on O). Ammonia can form fewer H-bonds per molecule (one N–H bond donor; one lone pair acceptor), and the N–H···N bonds are somewhat weaker than O–H···O bonds because oxygen is more electronegative than nitrogen (χ(O) = 3.5 vs χ(N) = 3.0). The more extensive H-bonding network in water requires more energy to disrupt, giving the higher BP.

Marking notes. Part (a): 1 mark for correctly identifying 6 IMF entries; 1 mark for explaining CH₃F has no H-bonding (H bonded to C, not F); 1 mark for explaining polarity of C–F as the source of dipole-dipole forces. Part (b): 1 mark for correct ranking; 1 mark for explaining CH₄ lowest (dispersion only); 1 mark for explaining CH₃OH highest (H-bonding). Part (c): 1 mark for noting water forms more H-bonds per molecule (up to 4) vs NH₃; 1 mark for linking to higher BP (more energy required to disrupt the more extensive network).

Sample response. The claim contains a useful approximation but is not universally true, and a careful analysis shows when it holds and when it fails. The claim correctly captures the essential IMF hierarchy: for molecules of similar size and electron count, the type of IMF dominates — hydrogen bonding is far stronger than dipole-dipole, which is stronger than dispersion forces alone. Evidence supporting the claim: water (MW 18, hydrogen bonding, BP 100 °C) boils far higher than methane (MW 16, dispersion only, BP −161 °C); ethanol (MW 46, H-bonding, BP 78 °C) boils far higher than propane (MW 44, dispersion only, BP −42 °C). These examples show that for roughly comparable sizes, the presence of H-bonding reliably predicts a higher BP than dispersion-only counterparts. However, the claim breaks down when molecules differ greatly in size. Dispersion forces increase with electron count and molecular size; for very large non-polar molecules, the cumulative effect of many weak dispersion interactions can exceed the hydrogen bonding in a small molecule. A clear counterexample: hexadecane (C₁₆H₃₄, MW 226, dispersion only, BP 287 °C) has a far higher boiling point than water (BP 100 °C) despite water’s strong H-bonding. The 130+ electrons in hexadecane generate dispersion forces cumulatively stronger than water’s H-bonds. A second factor that the claim ignores is molecular shape: n-pentane (BP 36 °C) and neopentane (BP 9.5 °C) have identical molecular formulae and both have only dispersion forces — yet different BPs due to shape-dependent contact area. This shows that even within one IMF type, factors beyond IMF type (molecular geometry) matter. In summary, the statement is a useful heuristic for comparing molecules of similar size, but it must be qualified: both the type and the cumulative strength (which depends on molecular size, electron count, and molecular shape) of all IMFs determine boiling point. The claim holds when molecules being compared are of similar size; it breaks down when comparing small H-bonding molecules with large non-polar ones.

Marking criteria (7 marks). 1 = identifies at least one specific example (with named molecule and BP) where the claim is supported (H-bonding molecule has higher BP than similar-sized dispersion-only molecule). 1 = identifies a specific counterexample (named molecule + BP) where a dispersion-only molecule has a higher BP than a H-bonding molecule (e.g. hexadecane vs water). 1 = correctly explains the counterexample mechanistically (large electron count → strong cumulative dispersion forces). 1 = discusses the role of molecular size and electron count in determining dispersion force strength. 1 = discusses at least one additional factor beyond IMF type (e.g. molecular shape, number of H-bonds per molecule, electronegativity of H-bond donor). 1 = reaches an explicit evidence-based judgement (the claim holds for similar-sized molecules; breaks down for very different sizes). 1 = uses precise scientific terminology throughout (H-bonding, dispersion forces, electron count, polarisability, electronegativity, BP trend) with at least three named molecular examples with BPs.