AS Chemistry: Shapes of Molecules and Intermolecular Forces — Revision Notes

Condensed for the final weeks. For the full explanation, use the Shapes and Intermolecular Forces study guide.

Orbital overlap: sigma and pi bonds

A σ (sigma) bond forms by direct, head-on overlap of orbitals between two bonding atoms. A π (pi) bond forms by sideways overlap of adjacent p orbitals, above and below the σ bond.

A single bond is always one σ bond alone. A double bond is one σ bond plus one π bond — this is why the second bond in a C=C double bond is weaker and more reactive than the first: sideways overlap is less effective than head-on overlap.

Hybridisation describes how an atom’s orbitals mix before bonding:

Carbon type	Hybrid orbitals	Geometry	Example
sp³	Four equivalent hybrids	Tetrahedral	Methane, ethane
sp²	Three hybrids in a plane, one unhybridised p orbital left for a π bond	Trigonal planar	Ethene
sp	Two hybrids arranged linearly, two unhybridised p orbitals left for two π bonds	Linear	Ethyne

VSEPR — the one rule

Electron pairs repel and arrange themselves as far apart as possible. Lone pairs repel more strongly than bonding pairs, so each lone pair reduces the bond angle by roughly 2.5°.

repulsion:   lone-lone  >  lone-bond  >  bond-bond

Bond pairs	Lone pairs	Shape	Angle
2	0	Linear	180°
3	0	Trigonal planar	120°
4	0	Tetrahedral	109.5°
3	1	Trigonal pyramidal	107°
2	2	Bent / V-shaped	104.5°
5	0	Trigonal bipyramidal	120° and 90°
6	0	Octahedral	90°
4	2	Square planar	90°

The CH₄ → NH₃ → H₂O sequence (109.5 → 107 → 104.5) comes up often, and the explanation — one then two lone pairs, each repelling more than a bonding pair — is what earns the marks, not the numbers.

Polarity

A bond is polar if the two atoms differ in electronegativity. A molecule is polar only if the bond dipoles do not cancel.

• CO₂ — polar bonds, linear, dipoles cancel → non-polar.
• H₂O — polar bonds, bent, dipoles do not cancel → polar.
• CCl₄ — polar bonds, tetrahedral and symmetrical → non-polar.
• CHCl₃ — tetrahedral but asymmetric → polar.

Symmetry decides it. This is the standard trap.

The three intermolecular forces

Force	Present in	Relative strength
Induced dipole–induced dipole (van der Waals / London)	All molecules	Weakest, but grows with size
Permanent dipole–permanent dipole	Polar molecules	Intermediate
Hydrogen bonding	H bonded to N or O (the Cambridge 9701 assessed condition), plus a lone pair on N/O — F also qualifies scientifically but isn’t a required 9701 example	Strongest

Hydrogen bonding requires both: hydrogen directly bonded to N or O (the condition Cambridge 9701 assesses — H–F also hydrogen bonds by the same underlying rule, but 9701 only requires N–H/O–H examples), and a lone pair on an N or O of a neighbouring molecule to accept it. HCl does not hydrogen bond — chlorine is not electronegative enough to qualify, despite being an electronegative element.

Induced dipole forces increase with the number of electrons, which is why boiling points rise down the alkanes and down Group 17. Branching lowers boiling point because it reduces the surface contact area.

Where hydrogen bonding shows up

• Anomalous boiling points of H₂O and NH₃ against the trend in their groups (HF shows the same anomaly through H–F hydrogen bonding — scientifically valid, but beyond the 9701 N–H/O–H scope, included here for completeness).
• Ice is less dense than water — each H₂O forms four hydrogen bonds in a tetrahedral arrangement, creating an open lattice with holes. On melting the lattice partly collapses, so the liquid is denser.
• Solubility of alcohols, sugars and carboxylic acids in water.
• High surface tension and high specific heat capacity of water.

The ice-density explanation must mention the open lattice and the tetrahedral arrangement — “hydrogen bonds hold the molecules apart” alone is not enough.

Comparing boiling points — the method

1. Identify the strongest intermolecular force in each substance.
2. If they differ, the stronger force gives the higher boiling point.
3. If the same, compare number of electrons (bigger → stronger induced dipole forces).
4. If still tied, compare branching (more branching → lower).

Crucially: intermolecular forces break on boiling, not covalent bonds. Saying “the covalent bonds break” is a common error worth watching for — the covalent bonds within each molecule stay intact.

Exam traps

• Giving the shape without accounting for lone pairs.
• Saying a molecule is polar because its bonds are polar, ignoring symmetry.
• Claiming HCl or H₂S hydrogen bonds.
• Saying covalent bonds break during boiling.
• Forgetting that hydrogen bonding also needs an acceptor lone pair.
• Explaining ice’s density without mentioning the open tetrahedral lattice.

Self-test

1. State the VSEPR rule and the repulsion order.
2. Give the shapes and angles of CH₄, NH₃ and H₂O, and explain the trend.
3. Why is CO₂ non-polar but H₂O polar?
4. State the two requirements for hydrogen bonding.
5. Why is ice less dense than liquid water?

Answers: 1. Electron pairs repel and arrange themselves as far apart as possible; lone–lone > lone–bond > bond–bond. 2. Tetrahedral 109.5°, trigonal pyramidal 107°, bent 104.5°; each lone pair repels more strongly than a bonding pair, compressing the bond angle by about 2.5°. 3. Both have polar bonds, but CO₂ is linear so the dipoles cancel, while H₂O is bent so they do not. 4. Hydrogen bonded directly to N or O (the Cambridge 9701 assessed condition), and a lone pair on an N or O of a neighbouring molecule. 5. Each molecule forms four hydrogen bonds in a tetrahedral arrangement, producing an open lattice containing holes; melting partly collapses this, so liquid water is denser.