Chemistry • Year 11 • Module 1 • Lesson 9
Covalent Compounds: Molecular and Network
Lock in the core vocabulary, the molecular vs network distinction, and the role of intermolecular forces before tackling harder questions.
1. Term–definition match
The definitions below are shuffled. In the right-hand column write the matching term from this list: covalent bond, covalent molecular substance, covalent network solid, intermolecular forces (IMFs), dispersion forces, hydrogen bonding, VSEPR theory, tetrahedral, trigonal pyramidal, bent. 10 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 1.1 | A bond formed by sharing of electron pairs between two non-metal atoms; bond energies typically 150–1000 kJ mol−1. | |
| 1.2 | A substance where covalent bonds hold atoms together within separate, discrete molecules; has low melting and boiling points. | |
| 1.3 | A substance where covalent bonds extend continuously throughout the entire crystal in a giant 3D network; no discrete molecules; very high melting point. | |
| 1.4 | Attractions between separate molecules, including dispersion forces, dipole-dipole forces, and hydrogen bonding; much weaker than covalent bonds. | |
| 1.5 | Weak, temporary forces present in all molecules; increase in strength with molecular size (more electrons → larger temporary dipoles). | |
| 1.6 | A relatively strong intermolecular force between a hydrogen atom bonded to a highly electronegative atom (F, O, or N) and a lone pair on a neighbouring electronegative atom. | |
| 1.7 | A theory that predicts molecular geometry by assuming electron pairs arrange themselves to minimise electrostatic repulsion around the central atom. | |
| 1.8 | Molecular shape with 4 bonding domains and 0 lone pairs around the central atom; bond angle 109.5°. Example: CH4. | |
| 1.9 | Molecular shape with 4 electron domains and 1 lone pair; bond angle ≈107°. Example: NH3. | |
| 1.10 | Molecular shape with 4 electron domains and 2 lone pairs; bond angle ≈104.5°. Example: H2O. |
2. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
2.1 When water boils, the O–H covalent bonds within each water molecule are broken, releasing separate hydrogen and oxygen atoms. T / F
2.2 All covalent substances have low melting points because covalent bonds are weak. T / F
2.3 Carbon dioxide (CO2) and silicon dioxide (SiO2) are both covalent, but CO2 is molecular while SiO2 is a network solid. T / F
2.4 Iodine (I2) has a higher boiling point than fluorine (F2) because iodine forms stronger hydrogen bonds. T / F
2.5 Graphite conducts electricity within its layers because it has one delocalised electron per carbon atom available in the layer. T / F
2.6 Covalent network solids are insoluble in all common solvents because the strong covalent bonds throughout the lattice cannot be disrupted by solvent molecules. T / F
3. Fill-in-the-blank paragraph
Use the word bank to complete the passage. Each word is used once. 9 marks (1 per blank)
Word bank:
covalent bonds · dispersion · discrete · hydrogen bonding · intermolecular forces · molecular · network · low · very high
There are two types of covalent substances. A covalent ___________ substance consists of ___________ molecules held together by weak ___________; these substances have ___________ melting and boiling points because only the IMFs need to be broken on melting, not the covalent bonds. In contrast, a covalent ___________ solid has ___________ extending throughout the entire crystal, giving it a ___________ melting point. The strength of IMFs in molecular substances varies: non-polar molecules experience only ___________ forces, while molecules with N–H, O–H, or F–H bonds can form the much stronger ___________.
4. Function recall
Answer each question in 1–2 sentences using precise terms from the lesson. 8 marks (2 each)
4.1 What determines whether a covalent substance has a low or very high melting point? Name the two categories.
4.2 Why does CO2 have a very low boiling point (−78°C) despite having strong C=O covalent bonds?
4.3 Why does diamond have a melting point of 3550°C?
4.4 Why does water (H2O, MW = 18) have a much higher boiling point than hydrogen sulfide (H2S, MW = 34), which is a larger molecule?
5. Build a concept map
Draw labelled arrows between the six terms below to show how they connect. Each arrow must carry a linking phrase (e.g. “are broken when”, “determine”, “only present in”). Aim for at least 6 labelled arrows. 6 marks (1 per valid labelled arrow)
Supplied terms: covalent molecular substance · intermolecular forces · melting point · covalent network solid · covalent bonds · hydrogen bonding.
6. Label the VSEPR molecular shapes
The diagram below shows five molecular structures around a central atom. Write the correct molecular shape name, the number of lone pairs, and the approximate bond angle for each label A–E. 10 marks (1 shape + 1 bond angle each)
| Label | Molecular shape name | Lone pairs on central atom | Approximate bond angle |
|---|---|---|---|
| A | |||
| B | |||
| C | |||
| D | |||
| E |
Q1 — Term–definition match
1.1 covalent bond • 1.2 covalent molecular substance • 1.3 covalent network solid • 1.4 intermolecular forces (IMFs) • 1.5 dispersion forces • 1.6 hydrogen bonding • 1.7 VSEPR theory • 1.8 tetrahedral • 1.9 trigonal pyramidal • 1.10 bent.
Q2 — True / false with correction
2.1 False. When water boils, the intermolecular hydrogen bonds between water molecules are broken. The O–H covalent bonds within each water molecule remain intact — the gaseous phase still consists of H2O molecules.
2.2 False. Only covalent molecular substances have low melting points (because only weak IMFs between molecules are broken on melting). Covalent network solids such as diamond (MP 3550°C) and SiO2 (MP 1713°C) have extremely high melting points because the entire covalent network must be broken.
2.3 True.
2.4 False. I2 has a higher boiling point than F2 because I2 is a much larger molecule with more electrons, producing much stronger dispersion forces. Neither I2 nor F2 forms hydrogen bonds (hydrogen bonding requires an N–H, O–H, or F–H bond, not an F–F or I–I bond).
2.5 True.
2.6 True.
Q3 — Cloze paragraph
In order: molecular / discrete / intermolecular forces / low / network / covalent bonds / very high / dispersion / hydrogen bonding.
Q4.1 — What determines melting point category
Whether the substance is a covalent molecular substance or a covalent network solid. Molecular substances have low melting points (only weak IMFs are broken); network solids have very high melting points (strong covalent bonds throughout the lattice must be broken).
Q4.2 — Why CO2 has a very low BP despite strong C=O bonds
CO2 is a covalent molecular substance — it consists of discrete, separate O=C=O molecules. When CO2 boils, only the weak dispersion forces between molecules are overcome. The strong C=O covalent bonds within each molecule are not broken at −78°C.
Q4.3 — Why diamond has an extremely high MP
Diamond is a covalent network solid in which every carbon atom is covalently bonded to four others in a continuous 3D tetrahedral lattice. To melt diamond, an enormous number of strong C–C covalent bonds throughout the entire crystal must be broken, requiring an extremely large amount of energy.
Q4.4 — Why H2O has a higher BP than H2S despite being smaller
Water molecules form strong hydrogen bonds (O–H···O) because oxygen is highly electronegative. H2S cannot form hydrogen bonds because sulfur is not electronegative enough. The hydrogen bonds in water are significantly stronger than the weak dispersion and dipole-dipole forces in H2S, so much more energy is needed to boil water despite it being the smaller molecule.
Q5 — Sample concept map
Correct maps should include arrows such as:
- covalent molecular substance — has molecules held together by → intermolecular forces
- intermolecular forces — determine → melting point
- hydrogen bonding — is a type of → intermolecular forces
- covalent network solid — melts by breaking → covalent bonds
- covalent bonds — also determine → melting point
- covalent molecular substance — has lower → melting point — than → covalent network solid
Award 1 mark per valid labelled arrow (minimum 6, maximum 6 marked).
Q6 — VSEPR molecular shapes
A: Linear — 0 lone pairs — 180°. B: Trigonal planar — 0 lone pairs — 120°. C: Tetrahedral — 0 lone pairs — 109.5°. D: Trigonal pyramidal — 1 lone pair — ≈107°. E: Bent/V-shaped — 2 lone pairs — ≈104.5°.
Award 1 mark for the correct shape name and 1 mark for the correct bond angle per label.