HSC Exam Practice

Sample response. A covalent molecular substance consists of discrete, separate molecules held together by weak intermolecular forces. The covalent bonds exist within individual molecules, not between them. A covalent network solid has covalent bonds extending continuously throughout the entire crystal in a giant 3D lattice; no discrete molecules exist. The distinguishing structural feature is whether discrete molecules exist (molecular) or whether covalent bonds span the entire crystal (network).

Marking notes. 1 mark for a correct definition of covalent molecular substance (discrete molecules + IMFs between them); 1 mark for a correct definition of covalent network solid (continuous covalent bond network; no discrete molecules); 1 mark for explicitly stating the structural distinguishing feature (discrete molecules vs continuous network).

Sample response. The three types of intermolecular force are: (1) dispersion forces (London dispersion / van der Waals forces) — present in all molecules, temporary induced dipoles; (2) dipole–dipole forces — between polar molecules that have permanent dipoles; (3) hydrogen bonding — between a hydrogen atom bonded to a highly electronegative atom (F, O, or N) and a lone pair on a neighbouring electronegative atom. Hydrogen bonding is the strongest; dispersion forces are the weakest (for molecules of the same size).

Marking notes. 1 mark per correctly named type of IMF (3 marks total). Allow minor variation in naming dispersion forces (van der Waals, London forces). The strength order is implicit in identifying hydrogen bonding as strongest and dispersion as weakest.

Sample response. CO₂ is a covalent molecular substance. It exists as discrete O=C=O molecules held together by weak dispersion forces between molecules. Sublimation of CO₂ only requires overcoming these weak IMFs — the strong C=O covalent bonds within each molecule are not broken (BP −78°C). SiO₂, by contrast, is a covalent network solid. Each Si atom is covalently bonded to four oxygen atoms, and each oxygen bridges two silicon atoms, forming a continuous 3D Si–O–Si–O network throughout the crystal. To melt SiO₂, enormous numbers of strong Si–O covalent bonds throughout the entire lattice must be broken, requiring very large amounts of energy — hence the extremely high melting point of 1713°C. The difference is structural type, not the type of element or bond type.

Marking notes. 1 mark for correctly classifying CO₂ as covalent molecular and stating it has discrete molecules; 1 mark for stating that melting/subliming CO₂ breaks only weak IMFs (NOT covalent bonds); 1 mark for correctly classifying SiO₂ as a covalent network solid with a continuous Si–O bond network; 1 mark for stating that melting SiO₂ requires breaking covalent bonds throughout the lattice.

Sample response. Both diamond and graphite are covalent network solids made entirely of carbon atoms. In diamond, each carbon is bonded to four others in a rigid 3D tetrahedral network — there are no delocalised electrons and all bonds are directional. In graphite, each carbon is bonded to three others in flat hexagonal layers; the fourth electron per carbon is delocalised within each layer. Diamond is extremely hard (Mohs 10) because deforming it requires breaking strong C–C covalent bonds in all directions through the 3D lattice. Graphite is very soft (Mohs 1–2) because the layers are held together only by weak dispersion forces, allowing them to slide easily. Diamond does not conduct electricity because all electrons are localised in covalent bonds. Graphite conducts electricity within its layers because the delocalised electrons are free to carry charge.

Marking notes. 1 mark for correct structural distinction (diamond: 4-bonded, 3D; graphite: 3-bonded, layered with delocalised electrons); 1 mark for explaining diamond’s hardness (all bonds directional, must break C–C bonds in 3D); 1 mark for explaining graphite’s softness (layers slide, weak forces between layers); 1 mark for explaining conductivity difference (diamond: no free electrons; graphite: delocalised electrons within layers).

Sample response. H₂O has a bent (V-shaped) molecular geometry with a bond angle of approximately 104.5°. NH₃ has a trigonal pyramidal geometry with a bond angle of approximately 107°. Both have four electron domains around the central atom: H₂O has 2 bonding pairs + 2 lone pairs; NH₃ has 3 bonding pairs + 1 lone pair. VSEPR theory predicts that lone pairs repel other electron pairs more strongly than bonding pairs do, because the lone pair electron cloud is more diffuse and occupies more angular space. Water has two lone pairs, compressing the H–O–H bond angle to 104.5° (below the tetrahedral ideal of 109.5°). Ammonia has one lone pair, providing less compression and a slightly larger bond angle of 107°. Each additional lone pair on the central atom reduces the bond angle further.

Marking notes. 1 mark for correctly naming and stating the bond angle for H₂O (bent, 104.5°); 1 mark for correctly naming and stating the bond angle for NH₃ (trigonal pyramidal, 107°); 1 mark for applying VSEPR theory correctly (lone pairs repel more than bonding pairs, compress bond angle); 1 mark for explaining why H₂O’s angle is smaller than NH₃’s (H₂O has 2 lone pairs; NH₃ has 1 lone pair; more lone pairs → more compression).

Sample response. The student is incorrect. Neither I₂ nor F₂ can form hydrogen bonds, because hydrogen bonding requires a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) and a lone pair on a neighbouring electronegative atom. I₂ and F₂ do not contain hydrogen atoms at all. The correct explanation is that I₂ has a higher boiling point than F₂ because of stronger dispersion forces. I₂ has 106 electrons per molecule compared to F₂’s 18. More electrons create larger temporary dipoles, generating stronger dispersion forces that require more energy to overcome on boiling. The intramolecular I–I or F–F covalent bond strength is irrelevant to the boiling point — only the intermolecular forces between molecules matter.

Marking notes. 1 mark for identifying the error (neither I₂ nor F₂ can form hydrogen bonds — correct criterion stated); 1 mark for naming the correct IMF (dispersion forces); 1 mark for explaining the trend in terms of electron count / molecular size (I₂ has more electrons → stronger dispersion forces → higher BP).

Sample response (a). The boiling point increases steadily from F₂ (−188°C) to Cl₂ (−35°C) to Br₂ (+59°C) to I₂ (+184°C) as molecular mass and electron count increase [describe trend: 1 mark]. This trend is explained by dispersion forces (London dispersion forces): as the number of electrons in the molecule increases, the electron cloud becomes larger and more polarisable, generating stronger temporary dipoles and therefore stronger dispersion forces between molecules. More energy is required to overcome these stronger forces on boiling, giving a higher boiling point [name + explain IMF: 1 mark]. Hydrogen bonding is not involved: none of the halogens contain hydrogen atoms [1 mark].

Sample response (b). The error is that the student attributes the higher BP of Br₂ to stronger covalent (intramolecular) bonds [identify error: 1 mark]. Boiling a molecular substance does not break covalent bonds — only the intermolecular forces between molecules are overcome on boiling; the Br–Br and Cl–Cl covalent bonds remain intact in the gas phase [1 mark]. The correct explanation is that Br₂ (70 electrons) has significantly more electrons than Cl₂ (34 electrons), producing stronger dispersion forces between Br₂ molecules; more energy is needed to separate them, giving a higher boiling point [1 mark].

Sample response (c). At₂ would have a boiling point significantly above 184°C (I₂’s BP) [1 mark]. At₂ has 170 electrons per molecule — more than I₂’s 106. Following the trend (more electrons → stronger dispersion forces → higher BP), At₂ will have stronger dispersion forces than I₂ and will therefore boil at a higher temperature. The predicted value is approximately +300–340°C, consistent with the roughly linear increase seen from Cl₂ to I₂ [1 mark].

Marking notes. Part (a): 1 mark for correctly describing the trend (increasing BP with increasing MW/electron count); 1 mark for explaining via dispersion forces and electron count/polarisability; 1 mark for noting no hydrogen bonds are formed. Part (b): 1 mark for identifying the error (covalent bond strength does not determine BP); 1 mark for stating that only IMFs are broken on boiling; 1 mark for correct explanation using dispersion forces and electron count. Part (c): 1 mark for predicting BP above 184°C with justification (170 electrons > 106 electrons); 1 mark for correctly linking to dispersion force trend.

Sample response. The statement is partially correct but oversimplified: bond type alone is insufficient to predict melting point; structural type and intermolecular force strength are also essential. For covalent substances, knowing only that “bonding is covalent” is not enough to predict melting point. The critical distinction is whether the substance is covalent molecular or covalent network. A covalent molecular substance melts by overcoming only the weak intermolecular forces between discrete molecules; the strong covalent bonds within molecules are not broken. CO₂, for example, is covalent but is a gas at room temperature (BP −78°C) because only weak dispersion forces between O=C=O molecules need to be overcome. In contrast, SiO₂ (also covalent, also bonded to oxygen) melts at 1713°C because it is a network solid — melting requires breaking Si–O covalent bonds throughout an enormous 3D lattice. Even among covalent molecular substances, bond type alone does not determine melting point: molecular size and polarity determine IMF strength, which in turn determines the melting/boiling point. H₂O (18 g mol⁻¹) boils at 100°C, far above propane (C₃H₈, 44 g mol⁻¹, BP −42°C), because water forms strong hydrogen bonds (O–H···O) while propane has only weak dispersion forces. Both are covalent molecular; the bond type alone cannot explain this 142°C difference. Similarly, iodine (I₂, BP 184°C) has a far higher boiling point than fluorine (F₂, BP −188°C) despite both having non-polar covalent bonds — because I₂ is a much larger molecule with more electrons and stronger dispersion forces. In conclusion, the statement is incorrect as a generalisation. Predicting melting point for covalent substances requires knowing: (1) whether the substance is molecular or network (structural type), and (2) for molecular substances, the type and strength of IMFs (determined by polarity and molecular size), not just the bond type.

Marking criteria (6 marks). 1 = correctly identifies that structural type (molecular vs network) is the primary determinant for covalent substances, not bond type alone. 1 = uses CO₂ vs SiO₂ (or equivalent) as a named example where bond type is the same (covalent) but structural type determines the dramatic MP difference. 1 = explains that within molecular substances, IMF type and strength determine melting/boiling point, not intramolecular bond strength. 1 = uses a second named example comparing molecular substances with different IMF strengths (e.g. H₂O vs C₃H₈, or I₂ vs F₂) to show that even within “covalent molecular”, bond type alone cannot predict BP. 1 = explicitly evaluates the original statement, identifying what is partially correct (bond type matters: ionic vs covalent vs metallic sets a broad range) and what is incorrect (within covalent: structural type and IMF strength are also essential). 1 = reaches a coherent, evidence-based judgement that integrates structural type and IMF type as two additional factors beyond bond type.