AS Chemistry: Shapes of Molecules and Intermolecular Forces — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Shapes and Intermolecular Forces revision notes

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Section A

1. State the principle used to predict the shape of a simple molecule. [2]

2. Give the shape and bond angle of (a) BF₃, (b) NH₃, (c) H₂O. [3]

3. Explain why the bond angle in NH₃ is smaller than in CH₄. [2]

4. State the two conditions required for hydrogen bonding. [2]

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Section B

5. Consider CO₂ and H₂O.

(a) Both contain polar bonds. Explain why CO₂ is non-polar but H₂O is polar. [3]

(b) H₂O boils at 100 °C; H₂S boils at −60 °C. Explain this difference. [3]

(c) Explain, in terms of structure, why ice is less dense than liquid water. [3]

6. The boiling points of four substances are shown.

Substance	Boiling point / °C
CH₄	−162
C₄H₁₀	−0.5
CH₃OH	65
NaCl	1465

(a) Explain the difference between CH₄ and C₄H₁₀. [2]

(b) Explain why CH₃OH boils higher than C₄H₁₀, despite similar relative molecular masses. [2]

(c) Explain why NaCl boils far higher than all three. [2]

(d) A student writes: “When methane boils, the covalent bonds break.” Explain what is wrong with this statement. [2]

7. Distinguish between a σ bond and a π bond, and state how many of each are present in a C=C double bond. [3]

8. Describe the hybridisation and bond angles around the carbon atoms in ethene, C₂H₄. [3]

9. A student writes “hydrogen bonding is a separate force from van der Waals’ forces.” Explain what is wrong with this statement. [2]

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Answers

1. Electron pairs repel and arrange themselves as far apart as possible [1]; lone pairs repel more than bonding pairs [1].

2. (a) Trigonal planar, 120° [1]. (b) Trigonal pyramidal, 107° [1]. (c) Bent/V-shaped, 104.5° [1].

3. NH₃ has one lone pair whereas CH₄ has none [1]; the lone pair repels more strongly than a bonding pair, compressing the bond angle [1].

4. Hydrogen bonded directly to N or O (the Cambridge 9701 assessed condition — N–H and O–H groups) [1]; a lone pair on an N or O of a neighbouring molecule to accept it [1]. (H–F also hydrogen bonds by the same N/O/F rule taught more broadly in chemistry, but 9701 only requires N–H/O–H examples.)

5. (a) CO₂ is linear and symmetrical, so the two bond dipoles are equal and opposite and cancel [1]. H₂O is bent [1], so the dipoles do not cancel and there is a net dipole [1].

(b) H₂O has hydrogen bonding between molecules [1]; H₂S has only permanent dipole–dipole forces and (weaker) induced dipole–induced dipole forces, but no hydrogen bonding (sulfur is not electronegative enough) [1]. More energy is required to overcome the stronger intermolecular forces in water [1].

(c) Each water molecule forms four hydrogen bonds in a tetrahedral arrangement [1], producing an open lattice containing holes [1]. On melting this partly collapses, so molecules pack closer and the liquid is denser [1].

6. (a) C₄H₁₀ has more electrons [1], so the induced dipole–induced dipole forces between molecules are stronger and more energy is needed to separate them [1].

(b) CH₃OH has an O–H group so forms hydrogen bonds [1], which are stronger than the induced dipole forces in C₄H₁₀ [1].

(c) NaCl is a giant ionic lattice [1]; strong electrostatic attractions between oppositely charged ions throughout the structure must be overcome [1].

(d) Boiling overcomes the intermolecular forces between molecules, not the covalent bonds within them [1]. The covalent C–H bonds remain intact — methane gas is still CH₄ [1]. This is the single most heavily penalised sentence in the topic.

7. A σ bond forms by direct, head-on overlap of orbitals [1]; a π bond forms by sideways overlap of adjacent p orbitals, above and below the σ bond [1]. A C=C double bond has one σ bond and one π bond [1]. A single bond is always one σ bond alone, since sideways overlap needs a second, already-present σ bond to overlap alongside.

8. Each carbon is sp² hybridised [1], with three hybrid orbitals arranged in a plane at 120° [1], leaving one unhybridised p orbital on each carbon to overlap sideways and form the π bond of the C=C double bond [1]. This contrasts with an sp³ carbon (as in ethane), which has four equivalent hybrid orbitals arranged tetrahedrally with no unhybridised p orbital left over.

9. The syllabus uses van der Waals’ forces as the umbrella term that includes hydrogen bonding, permanent dipole–dipole forces and induced dipole–induced dipole (dispersion) forces [1] — it is not a separate, competing category from hydrogen bonding [1]. Calling every intermolecular force “van der Waals’” as if it meant only the weak dispersion type is the more common version of this mistake.

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Where marks are usually lost

• Saying a molecule is polar because its bonds are polar, ignoring symmetry.
• Claiming H₂S hydrogen bonds.
• Explaining ice’s density without “open lattice” and “tetrahedral”.
• Saying covalent bonds break on boiling.
• Comparing molecule size rather than number of electrons for induced dipole strength.
• Saying a double bond is “two identical bonds” rather than one σ and one weaker, more reactive π bond.
• Forgetting that the unhybridised p orbital, not the hybrid orbitals, is what forms the π bond.
• Treating “van der Waals’ forces” as excluding hydrogen bonding, rather than as the umbrella term covering it.