IB DP Chemistry - S2.2.9 Intermolecular forces and physical properties- Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S2.2.9 Intermolecular forces and physical properties- Study Notes – New Syllabus
IITian Academy excellent Study Notes and effective strategies will help you prepare for your IB DP Chemistry exam.
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
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Structure 2.2.9 — Intermolecular Forces and Physical Properties of Covalent Substances
Structure 2.2.9 — Intermolecular Forces and Physical Properties of Covalent Substances
Covalent compounds consist of molecules held together by intermolecular forces (IMFs), not ionic or metallic bonds. These forces are weaker than intramolecular covalent bonds, but determine many physical properties.
Types of Intermolecular Forces (in increasing strength)
- London (Dispersion) Forces – temporary dipoles between all molecules; only type in non-polar substances.
- Dipole–Dipole Interactions – occur between polar molecules with permanent dipoles.
- Hydrogen Bonding – strong dipole–dipole interaction when H is bonded to N, O, or F.
General Rule: For substances of similar molar mass, the strength follows:
\( \text{London Forces} < \text{Dipole–Dipole} < \text{Hydrogen Bonding} \)
Physical Properties of Covalent Substances (explained via structure and IMFs)
Volatility:
Covalent substances with weak intermolecular forces (e.g. London forces) are volatile (evaporate easily), e.g. hexane. Stronger IMFs (e.g. hydrogen bonds in water) decrease volatility.
Melting/Boiling Point:
Affected by IMF strength. Stronger IMF = higher boiling point.
For example:
- \( \text{CH}_4 \) (non-polar, London forces) → very low boiling point
- \( \text{H}_2\text{O} \) (hydrogen bonding) → much higher boiling point
Electrical Conductivity:
Most simple covalent molecules do not conduct electricity as they have no free ions or electrons.
Solubility:
Depends on polarity:
- Polar covalent substances (e.g. ethanol) dissolve in polar solvents (e.g. water)
- Non-polar molecules (e.g. iodine) dissolve in non-polar solvents (e.g. hexane)
- “Like dissolves like” principle applies
Example
Compare the boiling points of \( \text{CH}_4 \) and \( \text{NH}_3 \). Explain the difference in terms of intermolecular forces.
▶️Answer/Explanation
\( \text{CH}_4 \): Non-polar → only London dispersion forces.
\( \text{NH}_3 \): Polar → hydrogen bonding due to N-H bonds.
Hydrogen bonds in \( \text{NH}_3 \) are much stronger than London forces in \( \text{CH}_4 \), so more energy is needed to boil \( \text{NH}_3 \).
→ Therefore, \( \text{NH}_3 \) has a higher boiling point than \( \text{CH}_4 \).
Example
Explain why iodine (\( \text{I}_2 \)) is insoluble in water but soluble in hexane.
▶️Answer/Explanation
\( \text{I}_2 \) is a non-polar covalent molecule.
Water is polar and forms hydrogen bonds.
→ “Like dissolves like”: non-polar substances dissolve in non-polar solvents.
Hexane is non-polar, so \( \text{I}_2 \) dissolves in hexane due to London dispersion interactions.
\( \text{I}_2 \) does not dissolve in water because it cannot form hydrogen bonds or dipole interactions with water.
Example
Explain why ethanol (\( \text{C}_2\text{H}_5\text{OH} \)) has a much higher boiling point than ethane (\( \text{C}_2\text{H}_6 \)) even though both have similar molar mass.
▶️Answer/Explanation
Both molecules have similar molar mass (~46 g/mol), so London forces are similar.
However, ethanol has an -OH group → it can form hydrogen bonds.
Ethane has only London dispersion forces.
→ Hydrogen bonding is stronger than London forces, so ethanol requires more energy to boil.
Hence, ethanol has a much higher boiling point than ethane.