IB DP Chemistry Structure 2.2: The covalent model: bonds, Lewis formulas, shapes and intermolecular forces – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

Every question here targets Structure 2.2.1 to 2.2.10, the first part of the covalent model, as laid out in the IB’s February 2023 Chemistry guide for Diploma Programme courses first assessed in 2025. All of it is SL and HL content except one part on transition element complexes, marked “HL only”.

Use the Structure 2.2 study guide and the Structure 2.2 revision notes first. The Diploma Chemistry overview links every unit, the checklist of Diploma Chemistry understandings records progress, and a free diagnostic test helps you spot gaps in other units.

Questions

1. Deduce the Lewis formula of hypochlorous acid, HOCl, in which O is the central atom. [2]

2. Boron trifluoride reacts with a fluoride ion to form the tetrafluoroborate ion, BF₄⁻.

(a) Deduce the Lewis formula of BF₄⁻. [2] (b) Explain how a coordination bond forms in this reaction. [1] (c) Predict the shape of BF₄⁻ and its F–B–F bond angle. [2] (d) (HL only) The complex ion [Fe(H₂O)₆]³⁺ forms in aqueous solution. State the number of coordination bonds in the ion and which atom donates the electron pair in each. [2]

3. Hydrazine (N₂H₄), diazene (N₂H₂) and nitrogen (N₂) contain nitrogen–nitrogen single, double and triple bonds respectively. Arrange the three nitrogen–nitrogen bonds in order of increasing bond length and explain the order. [3]

4. For each species, state the electron domain geometry, the molecular geometry and the predicted bond angle.

(a) SiCl₄ [2] (b) H₃O⁺ [2] (c) SCl₂ [2]

5. In carbonyl chloride, COCl₂, carbon is the central atom. Explain why the Cl–C–Cl bond angle is less than 120°. [2]

6. Use these electronegativity values: H 2.2, B 2.0, C 2.6, N 3.0, F 4.0.

(a) Show the bond polarity of a C–F bond using partial charges. [1] (b) Deduce whether each of CF₄, CHF₃, BF₃ and NF₃ has a net dipole moment. Give a reason for each. [4]

7. This question is about network structures and allotropes.

(a) Explain why graphite conducts electricity but diamond does not. [2] (b) Explain why C₆₀ fullerene melts at a much lower temperature than diamond. [2] (c) Describe the structure of silicon dioxide. [2] (d) Use these bond enthalpies: C–C 346 kJ mol⁻¹, Si–Si 226 kJ mol⁻¹. Explain the difference in strength. [2]

8. State the strongest type of intermolecular force acting between:

(a) molecules of fluoromethane, CH₃F [1] (b) molecules of methanol, CH₃OH [1] (c) a chloromethane molecule and an iodine molecule (other than London forces) [1] (d) molecules of hexane, C₆H₁₄ [1]

9. Use these data.

Substance	Molar mass / g mol⁻¹	Boiling point / °C
oxygen, O₂	32.00	−183
methanol, CH₃OH	32.05	65
fluoromethane, CH₃F	34.04	−78

Explain the order of boiling points in terms of intermolecular forces. [5]

10. Covalent substances and their physical properties.

(a) Explain why methanol is completely miscible with water but hexan-1-ol is only slightly soluble. [2] (b) Hydrogen chloride dissolved in water conducts electricity, but hydrogen chloride dissolved in methylbenzene does not. Explain this difference. [2]

11. A mixture of three dyes, P, Q and R, is separated by thin layer chromatography on a silica plate using a non-polar solvent. The solvent front moves 8.0 cm from the base line. Spot P moves 1.6 cm, Q moves 4.4 cm and R moves 6.8 cm.

(a) Calculate the RF value of each dye. [2] (b) Deduce which dye is the most polar. Explain your answer. [2] (c) A pure reference dye run on the same plate has RF 0.55. Identify which dye it may be. [1] (d) State why the reference must be run on the same plate with the same solvent. [1] (e) On a second plate, with a different solvent, dye P has RF 0.36 and the solvent front moves 7.5 cm. Calculate the distance moved by P. [2]

12. This question is about ethanenitrile, CH₃CN.

(a) Deduce its Lewis formula. [2] (b) Predict the bond angle around each carbon atom. [2] (c) Deduce whether the molecule is polar. [1] (d) State the strongest intermolecular force between ethanenitrile molecules. [1] (e) Suggest why ethanenitrile mixes with water even though it cannot form hydrogen bonds with itself. [1]

Answers

1. H–O–Cl with single bonds [1]; two lone pairs on O and three lone pairs on Cl (14 valence electrons in total) [1]. Examiner insight: a Lewis formula with missing lone pairs on the outer atoms is incomplete; show every valence electron, not just the bonds.

2. (a) Four B–F single bonds with three lone pairs on each F (32 electrons) [1]; square brackets with a 1− charge outside [1]. (b) The fluoride ion donates a lone pair into the vacant orbital of the electron-deficient boron atom, so both shared electrons come from F⁻ [1]. (c) Tetrahedral [1]; 109.5° [1]. (d) Six coordination bonds [1]; in each, the O atom of a water molecule donates the lone pair to Fe³⁺ [1]. Examiner insight: in part (b), “boron shares electrons with fluorine” is unlikely to earn credit; state that both electrons of the pair come from one atom.

3. N≡N < N=N < N–N (shortest to longest) [1]. More shared pairs mean more electron density between the nuclei [1], so the nuclei are attracted more strongly and pulled closer together [1]. Examiner insight: an order alone is unlikely to earn more than the first point; link shared pairs to the pull of the nuclei.

4. (a) Tetrahedral electron domain geometry and tetrahedral molecular geometry [1]; 109.5° [1]. (b) Tetrahedral electron domain geometry; trigonal pyramidal molecular geometry [1]; about 107° [1]. (c) Tetrahedral electron domain geometry; bent molecular geometry [1]; about 104.5° [1]. Examiner insight: give both geometries when both are asked for; writing only “tetrahedral” for H₃O⁺ does not answer the molecular geometry.

5. The C=O double bond holds more electron density than a C–Cl single bond, so it repels the other domains more strongly [1]; this pushes the two C–Cl bonds closer together, so the Cl–C–Cl angle closes below 120° [1]. Examiner insight: the first step needs a comparison of repulsion between domains, not just “double bonds are bigger”.

6. (a) δ+C–Fδ− [1]. (b) CF₄: tetrahedral, four identical polar bonds cancel, so no net dipole [1]. CHF₃: the C–H bond (0.4) differs from the C–F bonds (1.4), so the dipoles do not cancel; it is polar [1]. BF₃: trigonal planar, three identical dipoles at 120° cancel; no net dipole [1]. NF₃: trigonal pyramidal with a lone pair on N, so the N–F dipoles cannot cancel; it is polar [1]. Examiner insight: each verdict needs its reason (shape or unequal bonds); a bare list of verdicts is unlikely to earn full credit.

7. (a) Graphite: each carbon bonds with three electrons; the fourth is delocalised and carries charge within a layer [1]. Diamond: all outer electrons sit in localised bonds [1]. (b) C₆₀ consists of separate molecules held together only by London forces [1]; melting diamond requires breaking many strong covalent bonds throughout the network [1]. (c) A covalent network in which each Si atom is bonded tetrahedrally to four O atoms [1] and each O atom is bonded to two Si atoms [1]. (d) C–C is stronger [1] because carbon atoms are smaller, so the bond is shorter and the shared pair is closer to both nuclei [1]. Examiner insight: in part (b), “fullerene has weak bonds” is unlikely to earn credit; name the intermolecular force and contrast it with covalent bonds in diamond.

8. (a) Dipole–dipole forces [1]. (b) Hydrogen bonding [1]. (c) Dipole-induced dipole forces [1]. (d) London (dispersion) forces [1]. Examiner insight: fluoromethane is a common trap: its H atoms are bonded to C, not F, so it cannot hydrogen bond.

9. Molar masses are similar, so London forces are comparable [1]. Oxygen is non-polar and has London forces only, so it has the lowest boiling point [1]. Fluoromethane is polar, so it also has dipole–dipole forces and boils higher than oxygen [1]. Methanol has an H atom bonded to O, so it forms hydrogen bonds between molecules [1]; these are the strongest forces, so methanol boils highest [1]. Examiner insight: begin by noting that the molar masses are similar; without that step, the comparison of force types is not justified.

10. (a) Methanol’s O–H group forms hydrogen bonds with water molecules [1]; hexan-1-ol’s long non-polar chain cannot hydrogen bond with water, so it dissolves only slightly [1]. (b) In water, HCl reacts to form ions (H₃O⁺ and Cl⁻), which are mobile charged particles [1]; in methylbenzene it stays as covalent molecules, so there are no mobile ions [1]. Examiner insight: conductivity answers must name the mobile charged particles; “it dissolves” is not enough.

11. (a) RF = distance moved by dye ÷ distance moved by solvent front [1]; P = 1.6 ÷ 8.0 = 0.20, Q = 4.4 ÷ 8.0 = 0.55, R = 6.8 ÷ 8.0 = 0.85 [1]. (b) P is the most polar [1]: silica is a polar stationary phase, so the most polar dye is attracted most strongly to it and travels least far [1]. (c) Q (same RF) [1]. (d) RF depends on the solvent and stationary phase, so values compare only under identical conditions [1]. (e) Distance = RF × solvent front distance [1] = 0.36 × 7.5 = 2.7 cm [1]. Examiner insight: an RF above 1, or one quoted with a unit, signals a wrong method; measure both distances from the base line.

12. (a) H₃C–C≡N with three C–H bonds, one C–C bond and one C≡N triple bond [1]; one lone pair on N (16 valence electrons) [1]. (b) Around the CH₃ carbon: 109.5° (four domains) [1]; around the nitrile carbon: 180° (two domains, the triple bond counting as one) [1]. (c) Polar: the C≡N bond is polar and the linear C–C≡N end has no opposing dipole to cancel it [1]. (d) Dipole–dipole forces [1]. (e) The lone pair on N can accept a hydrogen bond from an O–H of water [1]. Examiner insight: for part (b), counting the triple bond as three domains gives a wrong shape; state the domain count for each carbon separately.

Where marks are usually lost

• Drawing Lewis formulas with bonds only, missing the lone pairs on the outer atoms.
• Forgetting the brackets and charge on an ion such as BF₄⁻.
• Describing a coordination bond without saying both electrons come from one atom.
• Treating a multiple bond as more than one electron domain.
• Giving electron domain geometry instead of molecular geometry.
• Stating that a molecule is polar without referring to its shape.
• Claiming hydrogen bonding in CH₃F or CH₃Cl, where H is bonded to carbon.
• Saying “bonds break” when a simple molecular substance boils.
• Missing the step that molar masses are comparable before ranking forces.
• Measuring RF to the end of the plate or paper instead of the solvent front.

Next steps

• Recap the key facts: Structure 2.2 revision notes
• Book a free trial class
• Relearn a section: Structure 2.2 study guide
• Try all free 10-minute diagnostics
• Tick off progress on the checklist of Diploma Chemistry understandings
• See every unit on the Diploma Chemistry overview

Official syllabus

Source document: the International Baccalaureate Organization’s Chemistry guide for the Diploma Programme, first assessment 2025, published in February 2023. Sections used: Structure 2.2.1 to 2.2.10.