Chemistry • Year 11 • Module 1 • Lesson 9
Covalent Compounds: Molecular and Network
Build HSC Band 5–6 extended-response technique on classifying covalent structures, evaluating property data, and designing investigations into molecular behaviour.
1. Data + scenario: comparing four covalent substances (Band 5–6)
8 marks Band 5–6
Scenario. A materials scientist at the University of New South Wales is comparing four covalent substances for use in a high-temperature industrial cutting tool. The table below summarises measured properties.
| Substance | Melting point (°C) | Hardness (Mohs) | Electrical conductivity | Solubility in water |
|---|---|---|---|---|
| W | 184 | 1–2 (soft solid) | None | Slightly soluble |
| X | 3550 | 10 (hardest known) | None | Insoluble |
| Y | 2730 | 9–9.5 | Slight (semiconductor) | Insoluble |
| Z | 1713 | 7 | None | Insoluble |
Illustrative data based on standard reference values.
Q1. Analyse and evaluate the data in the table to identify each substance, classify each as covalent molecular or covalent network, and assess which substance (X, Y, or Z) is most suitable for the high-temperature cutting-tool application. In your response you must:
- Identify the most likely real substance for each of W, X, Y, and Z and classify each as molecular or network. Justify each classification with specific reference to the property data.
- For each covalent network solid, explain why it has such a high melting point in terms of what must be broken to melt it.
- Explain why substance W has a melting point of 184°C despite being a small molecular substance, referencing its intermolecular forces.
- Evaluate which single substance among X, Y, and Z is most suitable for a high-temperature cutting tool, using evidence from at least three property criteria and reaching an explicit judgement.
- State one limitation of using property data alone (without structural data) to classify covalent substances.
2. Experimental design — determining whether an unknown covalent solid is molecular or network (Band 5–6)
7 marks Band 5–6
Research question. A Year 11 student is given an unknown white solid, Substance K. The student knows it is covalent (formed only from non-metal atoms) but cannot determine from visual inspection alone whether it is a covalent molecular substance or a covalent network solid. Design a scientific investigation using Year 11 laboratory equipment to classify Substance K.
Constraints: You have access to standard Year 11 laboratory equipment: Bunsen burner (max temp ≈800°C), electronic balance (0.01 g precision), test tubes, small crucible, conductivity tester, distilled water, non-polar solvent (hexane), and litmus paper. The investigation must be completed in one 60-minute lesson.
Q2. Design the investigation and present it in the format below.
- State a testable hypothesis that predicts the outcomes for both possible classifications (molecular vs network).
- List the independent variable, dependent variable(s), and at least two controlled variables.
- Describe the procedure in at least five numbered steps. Include tests for melting behaviour, solubility in water and hexane, and electrical conductivity.
- Describe what results would support each classification (molecular vs network) for each test.
- Explain what result would falsify your hypothesis.
- State two limitations of your design and suggest one improvement.
Q1 — Sample Band 6 response (8 marks), annotated
Identification and classification (4 marks):
W (MP 184°C, soft, no conduct, slightly soluble): Iodine (I2) — covalent molecular. The low MP of 184°C is consistent with only weak IMFs (dispersion forces) between discrete I2 molecules; the intramolecular I–I covalent bonds are not broken on melting. Slight water solubility indicates some polarity or interaction with the polar solvent. No conductivity confirms no ions or free electrons [1 mark for correct identification + classification with data justification].
X (MP 3550°C, Mohs 10, no conduct, insoluble): Diamond — covalent network. The extreme MP of 3550°C means the substance cannot be melted without breaking an enormous number of strong covalent bonds throughout the entire 3D network. The highest Mohs hardness (10) confirms all bonds in any direction are strong directional covalent bonds. Non-conducting because no free electrons or ions [1 mark].
Y (MP 2730°C, Mohs 9–9.5, slight conductivity, insoluble): Silicon carbide (SiC) — covalent network. Like diamond, the very high MP and extreme hardness indicate a continuous covalent network (alternating Si and C in a tetrahedral arrangement). Slight semiconductor behaviour distinguishes it from diamond [1 mark].
Z (MP 1713°C, Mohs 7, no conduct, insoluble): Silicon dioxide, SiO2 (quartz) — covalent network. Every Si is bonded to 4 O, and every O bridges two Si atoms, forming a 3D Si–O–Si–O network. The MP is lower than X or Y but still very high — consistent with a network where the Si–O bonds (bond energy ≈450 kJ mol−1) must all be broken to melt [1 mark].
Why network solids have high MPs (1 mark): X (diamond) melts by breaking C–C covalent bonds throughout the tetrahedral lattice; Y (SiC) by breaking Si–C bonds in an analogous tetrahedral network; Z (SiO2) by breaking Si–O bonds throughout the 3D SiO4 network. In all cases, the covalent bonds extend continuously in three dimensions — melting requires energy proportional to the number of bonds throughout the whole crystal, not just at a surface [1 mark].
Why W has MP of 184°C despite being molecular (1 mark): I2 is a non-polar diatomic molecule, so only dispersion forces act between molecules. However, I2 has 106 electrons per molecule, making it a large, highly polarisable molecule that experiences relatively strong dispersion forces compared to smaller halogens (e.g. Cl2 BP −35°C, F2 BP −188°C). The large electron cloud generates substantial temporary dipoles, explaining the higher-than-expected BP for a non-polar molecular solid [1 mark].
Evaluation for cutting tool (1 mark): Substance X (diamond) is most suitable. Criterion 1 — melting point: X has the highest MP (3550°C) of all four, ensuring it will not melt or soften under high operating temperatures. Criterion 2 — hardness: Mohs 10, the highest possible, means it will scratch and cut virtually any material. Criterion 3 — conductivity: no electrical conductivity means it can be used safely without creating circuit shorts in industrial machinery. Although Y (SiC) has a very high MP (2730°C) and hardness (9–9.5) and is used in abrasive cutting discs as a cheaper alternative, X has superior performance on all three criteria. W is eliminated immediately (too soft, too low MP). Z (SiO2) has a lower MP (1713°C) and lower hardness than X or Y [1 mark for explicit data-based judgement nominating X].
Limitation (1 mark): Property data alone cannot distinguish between two network solids with similar MPs or between a molecular substance with strong IMFs and a network solid with moderate covalent bond density. For example, a very large molecular substance could have a similar MP to a low-density network solid. Structural data (X-ray crystallography, electron microscopy) would be needed to definitively confirm the presence or absence of discrete molecules [1 mark].
Marking criteria summary (8 marks): 1 = correctly identifies and classifies W (I2, molecular) with data evidence; 1 = correctly identifies and classifies X (diamond, network) with data evidence; 1 = correctly identifies and classifies Y (SiC, network) with data evidence; 1 = correctly identifies and classifies Z (SiO2, network) with data evidence; 1 = explains why network solids have high MPs in terms of covalent bond breaking throughout the lattice; 1 = explains W’s moderate BP using dispersion force strength (large molecule, many electrons); 1 = explicit evaluative judgement for cutting tool using three criteria (MP, hardness, conductivity); 1 = states one valid limitation of using property data alone to classify covalent substances.
Q2 — Sample Band 6 response (7 marks), annotated
Hypothesis (1 mark): If Substance K is a covalent molecular substance, it will melt or show significant softening below 800°C in the Bunsen flame, will dissolve (at least partially) in distilled water or hexane depending on its polarity, and will not conduct electricity. If Substance K is a covalent network solid, it will not melt below 800°C, will be insoluble in both water and hexane, and will not conduct electricity [1 — testable hypothesis predicting outcomes for both classifications, naming IV and DV].
Variables: Independent variable: the classification category tested (molecular vs network conditions). Dependent variables: melting temperature (or evidence of melting vs no melting), solubility in water and hexane, electrical conductivity of the solid and solution. Controlled variables: mass of Substance K used in each test (0.5 g), volume of solvent (10 mL), temperature of solvent (room temperature 25°C).
Procedure (1 mark): (1) Place ≈0.5 g of Substance K in a small crucible and heat strongly with a Bunsen burner for 2 minutes. Record whether the solid melts, softens, or shows no change. (2) Add ≈0.5 g of Substance K to 10 mL distilled water in a test tube. Stir for 1 minute. Note whether it dissolves (clear solution), partially dissolves (turbid solution), or remains as a solid (insoluble). Test any resulting solution with a conductivity meter and litmus paper. (3) Add ≈0.5 g of Substance K to 10 mL hexane (non-polar solvent) in a fume cupboard. Stir for 1 minute. Note solubility. (4) Test the electrical conductivity of the dry solid K using a conductivity meter with probes directly on the solid surface. (5) Record all observations in a results table and compare outcomes against the predicted results for covalent molecular and covalent network classifications [1 mark for 5+ steps including melting, solubility in 2 solvents, conductivity].
Expected results vs classification (1 mark): Molecular: melts or softens below 800°C; soluble in water (if polar) or hexane (if non-polar); no conductivity. Network: does NOT melt below 800°C; insoluble in both solvents; no conductivity [1 mark for clearly describing results supporting each classification for each test].
Falsification (1 mark): If the hypothesis predicts Substance K is molecular (hypothesis: melts below 800°C), but it shows no melting, remains solid, and is insoluble in all solvents, the hypothesis would be falsified — the evidence would instead support covalent network classification [1 mark].
Limitations (2 marks): (1) The Bunsen burner only reaches ≈800°C. If Substance K is a network solid with a MP above 800°C (e.g. diamond 3550°C), its failure to melt would be consistent with the network hypothesis but could not definitively prove it — a molecular substance with IMFs too strong to break below 800°C would give identical results [1]. (2) If Substance K is a non-polar molecular solid, it may be insoluble in water and only slightly soluble in hexane, which might be mistaken for network behaviour without the solubility differential being large enough to detect in 60 minutes [1].
Improvement: Use a muffle furnace capable of reaching 1500–2000°C to positively test for very high melting points characteristic of network solids; this would make the melting test definitive rather than absence-of-evidence only.
Marking criteria summary (7 marks): 1 = testable hypothesis predicting outcomes for both molecular and network scenarios (names IV, DV); 1 = five or more clear procedure steps including melting, water solubility, non-polar solvent solubility, and conductivity tests; 1 = explicit prediction of results for each test under each classification; 1 = states what result would falsify the hypothesis; 1 = first valid limitation; 1 = second valid limitation; 1 = one specific, feasible improvement to the design.