Chemistry • Year 11 • Module 1 • Lesson 6
Chemical Bonding Overview
Build HSC Band 5–6 extended-response technique on structure-property relationships, the graphite exception, and multi-substance classification with justification.
1. Data + scenario: classifying materials used in the Sydney Harbour Bridge (Band 5–6)
8 marks Band 5–6
Scenario. The Sydney Harbour Bridge uses several distinct materials for different structural purposes. A materials engineer records the following properties for four unmarked samples (W, X, Y, Z) recovered from the bridge and its surrounding infrastructure during maintenance.
| Sample | MP (°C) | Conducts (solid) | Conducts (molten or aq) | Mechanical behaviour | Water solubility |
|---|---|---|---|---|---|
| W | 1538 | Yes (high) | Yes | Malleable, ductile | Insoluble (reacts with acids) |
| X | 801 | No | Yes (dissolved) | Hard, brittle | Soluble |
| Y | −117 | No | No | Soft, waxy | Insoluble |
| Z | 1713 | No | No | Extremely hard, brittle | Insoluble |
Illustrative data. MP values are representative of real substances; conductivity and mechanical data are measured under standard laboratory conditions.
Q1. Analyse and evaluate the property data above to classify each sample, identify a likely substance for each, and evaluate which sample presents the greatest challenge to classification. In your response you must:
- Classify each sample (W–Z) as ionic compound, covalent molecular, covalent network, or metallic element, and justify each classification using at least two properties per sample.
- Name a likely specific substance for each sample based on the MP and other data.
- Identify which sample (if any) is most likely to be misclassified from a single-property analysis, and explain the risk with reference to a specific property overlap.
- Explain why sample Z cannot be classified as ionic even though it has a high melting point and is hard, using conductivity as your key evidence.
- State one additional property measurement that would further confirm your classification of sample W, and explain how it would confirm this.
2. Experimental design — distinguishing diamond from silicon dioxide (Band 5–6)
7 marks Band 5–6
Research question. A geologist has two unmarked powdered samples. Both have melting points above 1700 °C, neither conducts electricity as solids, and both are extremely hard. One is diamond (C), the other is silicon dioxide (SiO2, quartz). Design a scientific investigation to distinguish between them.
Constraints: You have standard Year 11 lab equipment plus a Mohs hardness testing kit, dilute hydrofluoric acid (HF, with appropriate safety supervision), a simple conductivity tester, and access to a Bunsen burner. You cannot melt either sample.
Q2. Design the investigation and present it in the format below.
- State your hypothesis — predict which observable difference you expect and why, using your knowledge of the bonding in each substance.
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe the procedure in at least four numbered steps, including at least two different property tests to distinguish the substances.
- Explain what result would falsify your hypothesis for one of the tests.
- State two limitations of your design, and suggest one improvement.
Q1 — Sample Band 6 response (8 marks), annotated
Classification with justification:
Sample W (iron, Fe) — Metallic element [1]. High MP (1538°C) is consistent with metallic bonding; conducts as a solid (delocalised electrons, always mobile); malleable and ductile (non-directional metallic bonding allows ion layers to slide with electron sea intact); insoluble in water but reacts with acids (typical metallic behaviour). Likely identity: iron (Fe), MP = 1538°C exactly matches published value.
Sample X (sodium chloride, NaCl) — Ionic compound [1]. MP 801°C is high but not extreme; does not conduct as a solid (ions fixed in lattice); conducts when dissolved in water (mobile ions); hard but brittle; soluble in water. Likely identity: NaCl, MP matches exactly.
Sample Y (ethanol C2H5OH or similar) — Covalent molecular [1]. MP −117°C is very low (weak intermolecular forces only); no conductivity in any state (no free electrons or ions); soft and waxy (weak dispersion forces); insoluble in water (non-polar). Likely identity: ethanol (C2H5OH), MP = −114°C is close; or a hydrocarbon wax.
Sample Z (silicon dioxide, SiO2) — Covalent network solid [1]. MP 1713°C is very high (strong covalent bonds throughout); no conductivity in solid or molten state; extremely hard (all bonds must be broken to deform). Likely identity: SiO2 (quartz/silica), MP = 1713°C exact match.
Most likely to be misclassified from single-property analysis [1]: Sample W (iron) and Sample Z (SiO2) both have high MPS and are hard, creating a risk that a student focusing only on melting point might classify Z as metallic or W as covalent network. The critical distinguishing property is conductivity: W conducts as a solid (metallic), while Z does not conduct in any state (covalent network). A student using only MP would misclassify the pair.
Why Z cannot be ionic despite high MP and hardness [1]: Ionic compounds conduct electricity when molten (mobile ions). Sample Z does not conduct in the molten state, ruling out ionic classification. In covalent network solids, no charged species are ever released on melting — there are no ions to carry charge. Both properties (high MP and hardness) are shared by ionic and covalent network solids, so conductivity when molten is the decisive discriminator.
Additional property to confirm W is metallic [1]: Measuring thermal conductivity — metals conduct heat as well as electricity, via free electron movement. High thermal conductivity would confirm the metallic character and distinguish W from covalent network solids (low thermal conductivity). Alternatively, testing malleability (hammering) and confirming W deforms rather than fractures would also confirm metallic identity.
Marking criteria summary (8 marks): 1 mark each for correct classification + at least two justifying properties per sample (W, X, Y, Z = 4 marks); 1 mark for correctly identifying the W/Z pair as highest misclassification risk with a specific property overlap; 1 mark for explaining why Z is not ionic using molten conductivity as evidence; 1 mark for a valid additional test with explanation for W.
Q2 — Sample Band 6 response (7 marks), annotated
Hypothesis: If one sample is diamond and the other is SiO2, then the diamond sample will react with oxygen when heated (combustion to CO2), while the SiO2 sample will not combust. Additionally, the SiO2 sample will react with dilute HF, while diamond will not. Independent variable: identity of sample (diamond vs SiO2). Dependent variable: observable reaction with HF and/or combustion behaviour. Controlled variables: mass of sample (0.05 g each), concentration of HF, temperature of combustion test. [1]
Procedure: (1) Label two test tubes A and B. Place 0.05 g of sample 1 into A and 0.05 g of sample 2 into B. Add 1 mL of dilute HF (with appropriate safety supervision) to each. Record any gas evolution, change in solution appearance, or dissolution of the solid after 10 minutes. SiO2 reacts: SiO2(s) + 4HF(aq) → SiF4(g) + 2H2O(l); diamond does not react. (2) Rinse both samples thoroughly. Heat a small amount of each in a stream of pure oxygen from a gas jar above a Bunsen flame. Record whether a gas is produced; test the gas with limewater to confirm CO2 (diamond combustion). (3) Use the Mohs hardness kit: scratch both samples with known hardness reference minerals (corundum = 9, topaz = 8). Diamond scratches all; SiO2 is scratched by anything harder than 7. (4) Record all results in a data table. [1]
Falsification: If neither sample reacts with HF, the hypothesis that one is SiO2 would be falsified; both would be consistent with diamond or another non-reactive covalent network solid. [1]
Limitations: (1) HF is extremely hazardous; careful safety protocols required, and minor contamination of one sample could produce false positives. [1] (2) The combustion test requires pure oxygen and high temperatures — a standard Bunsen burner may not generate enough heat to combust diamond, yielding an inconclusive result. [1]
Improvement: Repeat each chemical test three times on separate portions to improve reliability; use inductively coupled plasma (ICP) spectroscopy or elemental analysis to confirm the presence of Si in one sample (SiO2) and the absence of Si in the other (diamond = pure C). [1]
What results would show: The SiO2 sample reacts with HF (gas evolved, solid dissolves) and does not combust; the diamond sample does not react with HF but does burn in oxygen to produce CO2 (confirmed by limewater turning milky). These results together uniquely identify each sample. [1]
Marking criteria (7 marks): 1 = testable hypothesis with IV, DV and reason; 1 = four steps including at least two different chemical tests; 1 = falsification condition; 1 = first valid limitation; 1 = second valid limitation; 1 = specific improvement; 1 = correct chemistry (equations or descriptions) throughout.