HSC Exam Practice

Sample response. Ionisation energy is the minimum energy required to remove one mole of electrons from one mole of neutral gaseous atoms in their ground state. Across Period 2 from Li to Ne, first ionisation energy generally increases because the number of protons (Z) increases while electrons are added to the same second electron shell. Shielding from inner shells remains approximately constant, so the effective nuclear charge (Z_eff) experienced by the outermost valence electron increases. This increases the attraction between the nucleus and the valence electron, making it more difficult to remove, so more energy is required.

Marking notes. 1 mark for a correct definition of ionisation energy (energy to remove one electron from neutral gaseous atom); 1 mark for identifying Z increases across the period while shielding remains approximately constant; 1 mark for linking increasing Z_eff to stronger nuclear attraction on valence electrons, therefore higher IE.

Sample response. A homogeneous mixture has uniform composition throughout and appears as a single phase; you cannot visually distinguish its components. A heterogeneous mixture has non-uniform composition with visibly distinct regions or phases. Australian example of homogeneous: seawater (dissolved salts uniformly distributed throughout water) or air (nitrogen, oxygen, CO₂ uniformly mixed). Australian example of heterogeneous: sand and seashell fragments on a Queensland beach, or soil (clay, silt, organic matter as distinct visible phases).

Marking notes. 1 mark for correct definition of homogeneous (uniform composition, single phase); 1 mark for correct definition of heterogeneous (non-uniform, distinct phases); 1 mark per valid Australian example (one each). Accept any physically correct examples set in an Australian context.

Sample response. In the solid state, the ions of an ionic compound are held in fixed positions within a regular crystal lattice by strong electrostatic forces. Because the ions cannot move, there are no mobile charge carriers and the solid does not conduct electricity. When the ionic compound is dissolved in water, the lattice breaks apart and the individual ions (e.g. Na⁺ and Cl⁻) become surrounded by water molecules and are free to move independently through the solution. These mobile ions can migrate towards oppositely charged electrodes, carrying electrical charge and completing the circuit.

Marking notes. 1 mark for correctly explaining why solids do not conduct (ions fixed in lattice, cannot move); 1 mark for correctly explaining why solution conducts (ions dissociated and free to move in solution); 1 mark for using the terms “mobile charge carriers” or equivalent (free ions carrying charge), and referencing the arrangement of particles in each state.

Sample response. The intermolecular force accounting for water’s anomalously high boiling point is hydrogen bonding. Water molecules form hydrogen bonds because the oxygen atom (highly electronegative) creates a large partial negative charge (δ−) while each hydrogen carries a partial positive charge (δ+); the H is bonded to the electronegative O, making the O–H···O hydrogen bond very strong. In H₂S, sulfur is less electronegative than oxygen, so the S–H bond is only weakly polar. H₂S molecules interact primarily via weaker dispersion forces and weak dipole-dipole forces, with no true hydrogen bonding. The much stronger hydrogen bonds in water require more energy to break than the IMFs in H₂S, giving water a much higher boiling point despite its lower molar mass.

Marking notes. 1 mark for correctly identifying hydrogen bonding as the IMF; 1 mark for explaining why H₂O has hydrogen bonding (electronegative O, partial charges, O–H bond); 1 mark for explaining why H₂S does not (S less electronegative, only dispersion/dipole-dipole forces); 1 mark for linking stronger IMF in water to more energy needed → higher boiling point.

Sample response. Rutherford fired positively charged alpha (α) particles at a very thin gold foil. Key observation: most α particles passed straight through; a small fraction were deflected at large angles, and a very small number bounced almost straight back. Conclusion: Rutherford concluded that most of the atom’s mass and all of its positive charge is concentrated in a tiny, dense central region called the nucleus, with most of the atom being empty space through which the α particles pass unimpeded. This disproved the Thomson plum-pudding model, which proposed that positive charge was spread uniformly throughout the atom; in that model all α particles should be deflected slightly, but none should bounce back.

Marking notes. 1 mark for correctly describing the key observation (most passed through; some deflected at large angles/bounced back); 1 mark for the correct conclusion (most mass and positive charge in a tiny nucleus; mostly empty space); 1 mark for explaining why the plum-pudding model is disproved (diffuse positive charge cannot explain large-angle deflection or back-scattering).

Sample response. Iodine (I₂) is a non-polar covalent molecular substance, held together by non-polar I–I bonds with electrons shared equally. Between I₂ molecules, only weak dispersion forces act. Hexane (C₆H₁₄) is also non-polar; the only IMFs present are dispersion forces. Because both substances are non-polar, the I₂–hexane dispersion forces are of comparable strength to the I₂–I₂ forces, so I₂ can dissolve by replacing I₂–I₂ interactions with I₂–hexane interactions of similar energy. Water is a polar solvent with strong hydrogen bonding between H₂O molecules. Introducing non-polar I₂ would disrupt the hydrogen bond network without providing compensating interactions; the energy cost is too high, so I₂ does not dissolve in water.

Marking notes. 1 mark for identifying I₂ as non-polar and hexane as non-polar, with only dispersion forces; 1 mark for explaining why I₂ dissolves in hexane (like dissolves like; compatible IMFs); 1 mark for explaining why I₂ does not dissolve in water (water’s hydrogen bonding network disrupted without compensating interaction).

Sample response (a). The trend shows a sharp, dramatic decrease in boiling point from NaCl (1465 °C) and MgCl₂ (1412 °C) on the left to AlCl₃ (183 °C) and then to the remaining covalent chlorides (SiCl₄ 57 °C, PCl₃ 76 °C, SCl₂ 59 °C, Cl₂ −34 °C), which cluster near room temperature. The high-BP chlorides (NaCl, MgCl₂) are ionic compounds: to boil them, strong ionic electrostatic forces throughout the entire lattice must be overcome, requiring very high temperatures. The low-BP chlorides (SiCl₄ onwards) are covalent molecular compounds: only weak intermolecular forces (dispersion forces ± dipole-dipole) between discrete molecules must be overcome, requiring much less energy.

Sample response (b). NaCl is an ionic compound: Na⁺ and Cl⁻ are held together by strong electrostatic (ionic) forces throughout the 3D crystal lattice. To boil NaCl, sufficient energy must be supplied to overcome the extensive ionic interactions in the entire lattice — a very large energy requirement, hence the extremely high boiling point of 1465 °C. AlCl₃ is a covalent molecular compound (Al is electronegative enough, and the electronegativity difference with Cl is insufficient to form a purely ionic bond): Al₂Cl₆ dimer molecules exist as discrete units held together only by weak dispersion forces. Only these relatively weak IMFs must be overcome to boil AlCl₃, requiring far less energy (183 °C).

Sample response (c). Carbon (C) is in Group 14, Period 2, so CCl₄ is analogous to SiCl₄ (Group 14). CCl₄ is a non-polar covalent molecular compound (symmetric tetrahedral structure; bond dipoles cancel). Its boiling point should be close to that of SiCl₄ (57 °C) or slightly different due to its different molecular mass. Expected: approximately 77 °C (literature value 76.7 °C). CCl₄ is held together by dispersion forces only; larger molar mass than SiCl₄ (154 vs 170 g/mol) gives slightly weaker dispersion forces, so similar or slightly lower BP than SiCl₄.

Marking notes. Part (a): 1 mark for correctly describing the dramatic step decrease from ionic to covalent; 1 mark for linking ionic chlorides to strong ionic forces requiring high energy; 1 mark for linking covalent chlorides to weak IMFs requiring low energy. Part (b): 1 mark for correctly identifying NaCl as ionic and the forces to overcome; 1 mark for correctly identifying AlCl₃ as covalent molecular and the forces to overcome; 1 mark for the correct conclusion (difference in force type explains BP gap). Part (c): 1 mark for predicting a BP near room temperature/similar to SiCl₄; 1 mark for justification using Group 14 analogy and dispersion forces.

Sample response. Full electron configuration of P (Z=15): 1s² 2s² 2p&sup6; 3s² 3p³. Orbital box diagram for 3p subshell: [↑] [↑] [↑] (three 3p orbitals each containing one electron with the same spin — one electron per orbital before any pairing occurs). Number of unpaired electrons: 3. Hund’s rule states that electrons occupy orbitals of equal energy (degenerate orbitals) singly before pairing occurs, and all singly-occupied orbitals have parallel spins. For the 3p subshell of P, three electrons must fill three degenerate 3p orbitals: rather than placing two in one orbital (with opposite spins) and one in another, each 3p orbital receives exactly one electron with the same spin direction, minimising electron–electron repulsion by keeping electrons as far apart as possible.

Marking notes. 1 mark for correct full subshell notation (1s²2s²2p&sup6;3s²3p³); 1 mark for correct orbital box diagram of 3p (three separate boxes each with a single up-arrow, no pairing); 1 mark for stating 3 unpaired electrons; 1 mark for correctly explaining Hund’s rule as applied to 3p (electrons fill degenerate orbitals singly with parallel spins to minimise repulsion, before pairing).

Sample Band 6 response. The “structure determines properties” framework operates as a unified explanatory chain across all three Module 1 inquiry questions. It begins at the atomic level (IQ3), flows through the nature of bonding and intermolecular forces (IQ2), produces observable macroscopic properties (IQ1–IQ2), and finally dictates which separation techniques are appropriate (IQ1).

IQ3 link: The atomic structure of an element — specifically the number of protons (Z) and the resulting electron configuration — determines the number and type of valence electrons. For example, sodium (Z=11, [Ne]3s¹) has one loosely-held valence electron, while oxygen (Z=8, [He]2s²2p&sup4;) has six with high electronegativity. The difference in electronegativity between Na and Cl (Z=17, [Ne]3s²3p&sup5;) is large, so Na transfers its valence electron to Cl, forming Na⁺ and Cl⁻ ions held in the ionic crystal lattice of NaCl.

IQ2 link: This ionic structure determines NaCl’s macroscopic properties: high melting point (801 °C), because strong electrostatic forces throughout the lattice require large energy to overcome; non-conductor as a solid (ions fixed in lattice); conductor when molten or dissolved (free ions); soluble in water (ion–dipole forces); brittle (ionic lattice fracture when layers shift). By contrast, carbon (Z=6, [He]2s²2p²) has four valence electrons and high ionisation energy, making ionic bonding thermodynamically unfeasible. Carbon forms four covalent bonds; in diamond, a 3D tetrahedral covalent network forms. Diamond’s structure determines its properties: hardest natural material (3D network in all directions), non-conductor (all electrons in bonds), extremely high melting point (>3500 °C), insoluble in all solvents. In graphite, sp² bonding within hexagonal layers leaves one delocalised electron per C, giving conductivity and softness.

IQ1 link: Knowing these properties enables rational selection of separation techniques. NaCl is separated from seawater by evaporation/crystallisation — exploiting the difference in volatility (water evaporates, NaCl crystallises) and the temperature-dependence of solubility. Diamond cannot be dissolved or distilled — it must be mechanically separated from kimberlite ore by density methods. A mixture of iodine and water can be separated by solvent extraction using hexane, because I₂ partitions into the non-polar hexane layer — a direct consequence of I₂’s non-polar covalent molecular structure.

Evaluation — limitations: While the framework is highly powerful for predicting and explaining properties of pure substances, there are cases where additional reasoning is required. Semiconductors (e.g. silicon) fall outside the simple four-category IQ2 framework: Si has a covalent network structure but conducts due to a small band gap (quantum mechanical effect), not predicted by classical bonding theory alone. Polymers (e.g. polyethylene) exhibit properties that depend on chain length and crystallinity, not just monomer bond type. Alloys (e.g. steel) cannot be fully explained by the pure metallic bonding model. These cases require additional physical chemistry or materials science reasoning beyond Module 1.

In summary, “structure determines properties” is a powerful and sufficient framework for explaining the behaviour of the four main classes of substances in Module 1, but it requires extension when applied to more complex materials or quantum effects.

Marking criteria (8 marks): 1 = correctly identifies the IQ3 starting point (atomic structure/electron configuration determines bond type) with a named example (Na, C, or similar). 1 = correctly applies IQ2 reasoning to explain at least two macroscopic properties of one named substance from its structure (ionic or covalent example). 1 = correctly applies IQ2 to a second substance type (ionic vs covalent network, or metallic vs molecular). 1 = correctly links macroscopic properties to IQ1 separation technique choice with a named example (crystallisation, solvent extraction, density separation, etc.). 1 = integrates IQ1, IQ2, IQ3 into a coherent chain showing how structure at the atomic level leads to observable separation choices. 1 = named Australian or real-world example demonstrating the framework in context (NaCl from seawater, diamond mining, Port Kembla steelworks, etc.). 1 = evaluates at least one case where the framework is insufficient or requires extension (semiconductors, polymers, alloys, etc.) with a specific named example. 1 = reaches a clear, explicit evaluative conclusion about the framework’s scope and limitations, using evidence-based reasoning throughout.