Water boils at 100°C. Diamond — made of a single element bonded identically to water's oxygen in terms of bond type — doesn't melt below 3550°C. Both are covalent. The difference isn't in the bonds themselves, but in whether those bonds form tiny isolated molecules or one giant interconnected structure spanning the entire crystal.
Core Content
This is the single most important concept in covalent chemistry for IQ2:
| Feature | Covalent molecular | Covalent network |
|---|---|---|
| Structure | Discrete molecules with IMFs between them | Continuous covalent bond network throughout the crystal |
| What breaks on melting? | Intermolecular forces (IMFs) | Covalent bonds |
| Melting point | Low to moderate (<300°C typical) | Very high (>1000°C typical) |
| Hardness | Soft, easily deformed | Extremely hard (all covalent bonds) |
| Conductivity | None in any state (except some with delocalised π electrons) | None (except graphite — delocalised electrons in layers) |
| Solubility | Polar molecules → dissolve in water; non-polar → don't | Insoluble in all common solvents |
| Examples | H₂O, CO₂, CH₄, C₆H₁₂O₆, I₂ | Diamond, graphite, SiO₂, SiC |
Within molecular substances, BP depends on the strength of IMFs. Stronger IMFs → higher BP.
Insert two side-by-side diagrams: (left) molecular — show 4–5 discrete water molecules with H-bonds drawn as dashed lines between them and covalent O-H bonds shown as solid lines within each molecule. Label: covalent bond (solid, strong), hydrogen bond/IMF (dashed, weak), molecules. (right) network — show a section of diamond-type structure with all carbon atoms connected by solid covalent bonds in all directions, no gaps, no discrete units. Label: covalent bond, no discrete molecules, continuous network.
| Substance | Structure | MP (°C) | Hardness | Conductivity | Notable property |
|---|---|---|---|---|---|
| Diamond | Each C bonded to 4 others in 3D tetrahedral network | 3550 | Hardest natural substance (10 Mohs) | None | Transparent; used in cutting tools |
| Graphite | Each C bonded to 3 others in layers; 1 delocalised e⁻ per C within layers | ~3650 | Soft (layers slide — Mohs 1–2) | Yes (within layers) | Lubricant, electrode material, pencil lead |
| Silicon dioxide (SiO₂) | Each Si bonded to 4 O; each O bridges two Si — 3D network | 1713 | Very hard | None | Sand, quartz, glass (when amorphous) |
| Silicon carbide (SiC) | Similar to diamond — Si and C alternate in tetrahedral network | 2730 | Extremely hard (9.5 Mohs) | None (slightly semiconducting) | Abrasives, cutting discs |
Worked Examples
Activities
1 Substance A: MP = −85°C, does not conduct in any state, dissolves slightly in water, soft, exists as a gas at room temperature.
2 Substance B: MP = 2730°C, extremely hard, does not conduct in any state, insoluble in all common solvents.
3 Substance C: BP = 100°C, does not conduct in any state, dissolves well in water, forms hydrogen bonds.
Question: "Why does water boil at 100°C while most covalent molecules boil much lower?"
Question: "Why does silicon dioxide (SiO₂) have a much higher melting point than carbon dioxide (CO₂)?"
Multiple Choice
Click to check. One attempt only.
1. When liquid water boils, what type of interactions are broken?
2. Which pair of substances most clearly illustrates the difference between covalent molecular and covalent network structures?
3. A covalent molecular substance has a boiling point of −33°C. A student claims this is because its covalent bonds are weak. This claim is:
4. The boiling points of the halogens are: F₂ (−188°C), Cl₂ (−35°C), Br₂ (59°C), I₂ (184°C). Which explanation correctly accounts for this trend?
5. Which statement about silicon dioxide (SiO₂) is correct?
Short Answer
6. Explain why iodine (I₂, BP 184°C) has a much higher boiling point than fluorine (F₂, BP −188°C), even though both are non-polar covalent molecular substances of the same type. 3 MARKS
7. A student is given data on two unknown substances: Substance X (MP −22°C, no conductivity in any state, dissolves in water) and Substance Y (MP 1713°C, no conductivity in any state, insoluble in all solvents). Classify each substance and explain all of its listed properties in terms of structure and bonding. 5 MARKS
8. Using your knowledge of intermolecular forces, explain why water (H₂O, MW = 18) has a boiling point of 100°C, which is dramatically higher than propane (C₃H₈, MW = 44, BP −42°C), even though propane is a larger molecule. 4 MARKS
1. Covalent molecular compound. The very low MP of −85°C indicates only weak IMFs between discrete molecules (not covalent bonds, which are strong). Gaseous state at room temperature confirms very weak IMFs. No conductivity is consistent with no free electrons or ions. Slight water solubility suggests mild polarity. This substance could be HCl (BP −85°C).
2. Covalent network solid. MP 2730°C is extreme — only substances where strong covalent bonds extend throughout the crystal can reach this temperature. Extreme hardness confirms a continuous covalent network (all bonds must break to deform). No conductivity eliminates metals and graphite. Insolubility in all solvents is characteristic of network covalent solids. This substance is silicon carbide (SiC).
3. Covalent molecular compound (water, H₂O). A BP of 100°C is high for a small covalent molecule — this is explained by strong hydrogen bonding between H₂O molecules (O is highly electronegative → O–H bonds are strongly polar → O–H···O hydrogen bonds form). Classification is covalent molecular because it consists of discrete H₂O molecules with hydrogen bonds between them. Conducts only when ionised (pure water doesn't conduct well).
Response 1 — Error: Student A is incorrect. When water boils, the H–O–H covalent bonds do NOT break — the water molecules remain intact as individual H₂O molecules in the gas phase. What breaks is the intermolecular hydrogen bonds between adjacent water molecules. Water has a higher boiling point than many other molecules (such as H₂S) because it forms strong hydrogen bonds (O–H···O), not because of stronger intramolecular bonds. The strength of the H–O covalent bond is not relevant to the boiling point — it would only matter if you were breaking water molecules apart chemically.
Response 2 — Error: Student B incorrectly treated SiO₂ as a molecular substance. SiO₂ does NOT consist of discrete molecules — it is a covalent network solid where Si–O covalent bonds extend throughout the entire crystal. CO₂, by contrast, forms discrete O=C=O molecules with only weak dispersion forces between them. The higher MP of SiO₂ is because melting requires breaking strong Si–O covalent bonds (rather than just IMFs as in CO₂). The reason has nothing to do with molecular mass or IMF strength — it is entirely about structural type (network vs molecular).
1. B — Boiling breaks intermolecular hydrogen bonds. Covalent O–H bonds remain intact. Gas-phase water is still H₂O molecules.
2. C — CO₂ (molecular, BP −78°C) vs SiO₂ (network, MP 1713°C) is the canonical example of this contrast. A = both ionic; B = both metallic; D = both molecular (comparing IMF type, not structural category).
3. D — Low BP in molecular substances reflects weak IMFs between molecules. The covalent bonds within molecules are not broken on boiling and their strength is irrelevant to the BP.
4. A — Dispersion forces increase with molecular size (more electrons → stronger temporary dipoles → stronger dispersion). These are the only IMFs in non-polar diatomic halogens. No hydrogen bonding; covalent bond strength is irrelevant to BP.
5. B — SiO₂ is a covalent network solid with a continuous Si–O bond network. It does not consist of discrete molecules; it does not dissolve in water; it does not conduct when molten.
Q6 (3 marks): Both F₂ and I₂ are non-polar diatomic molecules — the only IMF acting between their molecules is dispersion forces (1 mark). Dispersion forces increase in strength with the number of electrons in a molecule. I₂ has 106 electrons compared to F₂'s 18 electrons — I₂ has many more electrons and a much larger, more easily polarised electron cloud (1 mark). The stronger dispersion forces between I₂ molecules require significantly more energy to overcome during boiling → much higher BP (184°C vs −188°C). No covalent bonds are broken in either case — only IMFs (1 mark).
Q7 (5 marks): Substance X is a covalent molecular compound (1 mark). Its low MP of −22°C indicates only weak IMFs between discrete molecules, which are easily overcome — covalent bonds within molecules are not broken on melting (1 mark). No conductivity in any state confirms no free electrons or mobile ions (1 mark). Water solubility indicates the substance is polar — like-dissolves-like; polar molecular substances can interact with polar water molecules and dissolve. Substance Y is a covalent network solid (1 mark). MP 1713°C requires breaking enormous numbers of strong covalent bonds extending throughout the crystal — consistent with a network solid (this is SiO₂, quartz). No conductivity confirms no free electrons or mobile ions — the strong, directional covalent bonds hold all electrons in place. Insolubility: the Si–O network is too stable and strongly bonded to be disrupted by water molecules (1 mark across Y properties).
Q8 (4 marks): Propane (C₃H₈) is a non-polar hydrocarbon molecule — it cannot form hydrogen bonds or dipole-dipole interactions, so only weak dispersion forces act between propane molecules (1 mark). Despite its larger size (MW 44 vs 18), propane's BP is −42°C because these dispersion forces are still relatively weak (1 mark). Water (H₂O) contains two polar O–H bonds where oxygen's high electronegativity creates a large partial negative charge on O and partial positive charge on H. This allows water molecules to form strong intermolecular hydrogen bonds (O–H···O) with each other (1 mark). Hydrogen bonds are much stronger than the dispersion forces in propane — approximately 5–10× stronger — so far more energy is required to separate water molecules, resulting in a boiling point 142°C higher than propane despite water being the smaller molecule (1 mark).
Tick when you've finished all activities and checked your answers.