Chemistry • Year 11 • Module 1 • Lesson 20
Module 1 Synthesis and Review
Apply the IQ1–IQ3 framework to unfamiliar substances, real data and comparative reasoning tasks to build Band 4–5 exam technique.
1. Interpret experimental data — identifying unknown substances
The table below gives experimental data for five unknown solid substances A–E. Use the Module 1 IQ1–IQ3 framework to classify each substance. 10 marks
| Substance | Melting point | Conducts electricity (solid) | Conducts electricity (molten) | Hardness / other |
|---|---|---|---|---|
| A | 801 °C | No | Yes | Brittle; dissolves in water giving conducting solution |
| B | 1085 °C | Yes | Yes | Malleable and ductile; good thermal conductor |
| C | −95 °C | No | No | Liquid at room temperature; non-polar; soluble in hexane |
| D | >3500 °C | No | No | Hardest natural material; insoluble in all solvents |
| E | 100 °C (boiling point) | No | No | Polar; dissolves many ionic and polar covalent substances |
1.1 Classify each substance (A–E) as: ionic solid, metallic solid, covalent network solid, or covalent molecular substance. For each, give one piece of evidence from the data and one structural reason for that evidence. 10 marks (2 per substance)
| Substance | Classification | One piece of evidence | Structural reason |
|---|---|---|---|
| A | |||
| B | |||
| C | |||
| D | |||
| E |
2. Compare diamond and graphite across five features
Complete the two-column table. For each feature write a concise description contrasting the two allotropes of carbon, and include the structural reason for each difference. 10 marks (1 per cell)
| Feature | Diamond | Graphite |
|---|---|---|
| Structural arrangement of C atoms | ||
| Hardness | ||
| Electrical conductivity | ||
| Melting point (approximate) | ||
| Typical industrial use (justify from properties) |
3. Predict and justify — silicon carbide in Australian industry
Silicon carbide (SiC) is manufactured at Alcoa’s Wagerup alumina refinery in Western Australia as an abrasive and refractory material. It is formed by reacting silica (SiO2) with carbon at around 2000 °C. In SiC, each Si atom is covalently bonded to four C atoms in a 3D tetrahedral network identical in structure to diamond.
6 marks
3.1 Predict the melting point of SiC (high / moderate / low) and justify your prediction with reference to the bonding and structure. 2 marks
3.2 Predict whether SiC will conduct electricity. Justify from its electron structure. 2 marks
3.3 Explain why SiC is appropriate for use as an abrasive material, using evidence from your predictions above. 2 marks
4. Interpret graph — first ionisation energy across Period 3
The graph below shows the first ionisation energy (IE1) of elements from sodium (Na, Z=11) to argon (Ar, Z=18) across Period 3. 8 marks
Figure 4.1. First ionisation energies (IE1) for Period 3 elements Na–Ar (kJ mol−1). Data sourced from WebElements Periodic Table (Periodic Table, University of Sheffield). Illustrative representation.
4.1 Describe the general trend in IE1 across Period 3 and explain it in terms of effective nuclear charge (Zeff) and atomic radius. 3 marks
4.2 Identify the two anomalous dips in the graph (at Al and S). Explain why Al has a lower IE1 than Mg, even though Al has a higher Z. 3 marks
4.3 Predict the trend in IE1 going down Group 1 (Li→Na→K→Rb) and explain it in terms of atomic structure. 2 marks
5. Case study — materials selection for the Sydney Harbour Bridge
The Sydney Harbour Bridge, completed in 1932, is made primarily of steel (an iron alloy). The main arch spans 503 m. The bridge must simultaneously: (a) carry enormous loads, (b) conduct heat away from expansion joints, (c) resist corrosion, and (d) be readily shaped and welded during construction. 6 marks
5.1 Using Module 1 IQ2 principles, explain why a metallic material is appropriate for requirements (a)–(c). For each requirement, name the relevant metallic bonding property and explain why it satisfies the requirement. 4 marks
5.2 An architect suggests replacing the steel with quartz glass (SiO2) because glass has a higher compressive strength. Evaluate this suggestion using Module 1 reasoning. In your answer, identify at least two properties of SiO2 (from IQ2) that would make it unsuitable for this application. 2 marks
Q1.1 — Classifying substances A–E
A — Ionic solid (e.g. NaCl). Evidence: does not conduct as solid but does as a melt or solution. Reason: ions are fixed in a crystal lattice as a solid (cannot move), but when molten or dissolved they are free to move and carry charge.
B — Metallic solid (e.g. copper, Cu). Evidence: conducts electricity as a solid and is malleable. Reason: the metallic lattice of delocalised electrons can carry charge in all states, and the electron ‘sea’ allows layers of metal cations to slide without bond rupture.
C — Covalent molecular substance (e.g. hexane, C6H14). Evidence: very low melting point (liquid at room temperature), non-polar. Reason: discrete molecules held together only by weak dispersion forces; little energy required to separate them.
D — Covalent network solid (diamond). Evidence: extremely high melting point (>3500 °C), hardest natural material, non-conductor. Reason: entire structure is a continuous 3D lattice of strong covalent C–C bonds; melting requires breaking all these bonds; no free electrons.
E — Covalent molecular substance (water, H2O). Evidence: boiling point 100 °C, non-conductor, polar solvent. Reason: discrete H2O molecules held by hydrogen bonds (an intermolecular force), which are stronger than dispersion forces, accounting for the relatively elevated boiling point; the polar nature allows it to dissolve polar and ionic substances.
Q2 — Diamond vs graphite comparison
Structural arrangement: Diamond — each C bonded to 4 others in a 3D tetrahedral network. Graphite — each C bonded to 3 others in flat hexagonal layers; layers are held together by weak dispersion forces.
Hardness: Diamond — hardest natural material; every C–C bond must be broken to deform. Graphite — soft and slippery; layers can slide past each other because only weak inter-layer forces must be overcome.
Electrical conductivity: Diamond — non-conductor; all 4 valence electrons locked in covalent bonds, no free charge carriers. Graphite — good conductor (within layers); 1 delocalised electron per C atom free to move across the layer.
Melting point: Diamond — >3500 °C (sublimes); strong 3D network of C–C bonds throughout. Graphite — also very high (~3600 °C); covalent bonds within layers must be broken.
Industrial use: Diamond — cutting/drilling tools (e.g. diamond-tipped drill bits, Australian mining industry) because of hardness. Graphite — electrodes in electrolysis cells (e.g. aluminium smelting at Portland, VIC) because of in-plane conductivity and chemical inertness.
Q3.1 — SiC melting point
High melting point (above 2700 °C). SiC is a covalent network solid in which every Si and C atom is joined by a continuous 3D lattice of strong covalent Si–C bonds. To melt SiC, all of these covalent bonds throughout the lattice must be broken — an enormous energy input. This is the same reason diamond and SiO2 have very high melting points.
Q3.2 — SiC electrical conductivity
SiC is a poor electrical conductor (effectively an insulator at room temperature, though it is a semiconductor). All valence electrons of Si (4) and C (4) are used in covalent Si–C bonds throughout the network. There are no free or delocalised electrons to carry electrical charge. (Note: SiC is a semiconductor with a larger band gap than Si; a Band 6 student could mention this, but insulator is acceptable here.)
Q3.3 — SiC as an abrasive
SiC is appropriate as an abrasive because it is extremely hard (a consequence of the 3D covalent network resisting deformation in all directions) and has a very high melting point (can be used at high temperatures without degrading). Abrasives must be harder than the material they are grinding; SiC’s hardness (Mohs 9–9.5) means it can scratch and grind metals and ceramics used in industrial processing.
Q4.1 — IE1 trend across Period 3
The general trend is an increase in IE1 from Na to Ar [1]. Across Period 3, the number of protons (Z) increases while the electrons are added to the same third shell, so the shielding from inner shells remains approximately constant. This causes Zeff (experienced by the outermost electron) to increase progressively [1]. As Zeff increases, the atomic radius decreases (outer electrons are pulled closer to the nucleus) and the attraction between the nucleus and the outermost electron strengthens, requiring more energy to remove it [1].
Q4.2 — Anomalous dips at Al and S
Identified dips: Al (lower than Mg) and S (lower than P) [1]. For Al vs Mg: Mg has configuration [Ne]3s² and Al has [Ne]3s²3p¹. The 3p electron in Al is in a higher-energy sub-shell (3p) that is slightly further from the nucleus and is additionally shielded by the 3s electrons. This makes the 3p electron in Al easier to remove than the 3s electron in Mg, despite Al having a higher Z [1]. For S vs P: P has configuration [Ne]3s²3p³ with one electron in each 3p orbital (Hund’s rule); S has [Ne]3s²3p&sup4; with one 3p orbital doubly-occupied. The two electrons sharing one orbital experience extra electron–electron repulsion, making it easier to remove one electron from S than expected [1].
Q4.3 — IE1 trend down Group 1
IE1 decreases going down Group 1 (Li > Na > K > Rb) [1]. Each successive element has one additional electron shell, increasing the atomic radius and increasing shielding of the nuclear charge. The outermost valence electron is therefore further from the nucleus and more shielded, reducing the effective nuclear charge it experiences, so less energy is needed to remove it [1].
Q5.1 — Metallic bonding and bridge requirements
(a) Load-bearing (high tensile strength): metals have strong metallic bonding (electrostatic attraction between metal cations and the sea of delocalised electrons), giving high tensile strength and resistance to deformation under load [1]. (b) Thermal conductivity: delocalised electrons conduct heat by kinetic energy transfer throughout the metallic lattice, allowing expansion joints to dissipate heat energy [1]. (c) Malleability/weldability (requirement d): the delocalised electron sea allows layers of metal cations to slide over each other without breaking the metallic bond, so steel can be shaped and welded without cracking [1]. (The question groups a–c but award any three valid property–requirement links for 4 marks.)
Q5.2 — Why SiO2 is unsuitable
Any two of the following [1 per property with valid reason]:
- Brittle: SiO2 (covalent network) shatters under tensile stress rather than deforming; a bridge must flex under load (e.g. traffic, wind). Glass cannot be deformed and welded during construction.
- Poor thermal conductivity: SiO2 is a thermal insulator; expansion joints in a bridge must allow heat to conduct away to prevent local stress buildup.
- Cannot be welded or shaped: SiO2 cannot be hammered or welded at achievable temperatures; metallic steel can.