IB DP Chemistry Reactivity 3.4: Electron-pair sharing: Lewis acids, complexes and mechanisms – 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.

This set gives you marked practice on the additional higher level half of Reactivity 3.4, “Electron-pair sharing reactions”, in IB DP Chemistry. It is numbered against the Diploma Programme Chemistry guide, published by the IB in February 2023 for first assessment in 2025, and every understanding from Reactivity 3.4.6 to 3.4.13 is tested at least once. All of it is HL only, for examination sessions from 2025 onwards.

Stuck on a method? Go back to the electron-pair sharing study guide or the revision notes. Other units are on the DP Chemistry hub, the DP Chemistry checklist records what you have covered, and the free diagnostics suggest where to go next.

Where a question asks you to describe a mechanism in words, say where each curly arrow starts and ends.

Questions

1. For each reaction, identify the Lewis acid and the Lewis base.

(a) BF₃ + F⁻ → BF₄⁻ [1] (b) Ni²⁺ + 6NH₃ → [Ni(NH₃)₆]²⁺ [1] (c) CH₃NH₂ + H⁺ → CH₃NH₃⁺ [1]

2. Explain why every Brønsted–Lowry base is also a Lewis base, but not every Lewis acid is a Brønsted–Lowry acid. Give one example of a Lewis acid that is not a Brønsted–Lowry acid. [3]

3. Boron trifluoride reacts with methanol, CH₃OH, to form a single adduct.

(a) Draw the Lewis formula of the adduct, showing all lone pairs, and identify the coordination bond. [2] (b) Describe how the electron domain geometry around boron changes in this reaction. [2] (c) State whether the coordination bond in the adduct differs from a B–O bond formed in any other way, once it has formed. [1]

4. Deduce the charge on each complex.

(a) Co³⁺ with four NH₃ and two Cl⁻ ligands [1] (b) Fe²⁺ with five CN⁻ ligands and one H₂O ligand [1] (c) V³⁺ with four Cl⁻ and two H₂O ligands [1] (d) Deduce the oxidation state of chromium in [Cr(OH)₆]³⁻. [1]

5. Explain, in terms of Lewis theory, how a water molecule acts as a ligand towards a transition element cation, and identify the donor atom in the cyanide ligand. [3]

6. Two halogenoalkanes, A (primary) and B (tertiary), are each reacted with aqueous hydroxide ions. Doubling [OH⁻] doubles the rate for A but has no effect on the rate for B.

(a) Deduce the rate equation for each reaction. [2] (b) Under starting conditions the rate for A is 2.4 × 10⁻⁵ mol dm⁻³ s⁻¹ and the rate for B is 1.5 × 10⁻⁴ mol dm⁻³ s⁻¹. Calculate each new rate if both [halogenoalkane] and [OH⁻] are tripled. [2] (c) Describe how the two energy profiles differ. [1]

7. 1-iodohexane reacts with aqueous sodium hydroxide. Describe and explain the mechanism, including the transition state. [4]

8. A pure sample of one enantiomer of 2-iodooctane reacts with hydroxide ions by an SN2 mechanism.

(a) Describe the stereochemistry of the octan-2-ol formed and explain why. [3] (b) State what would be formed instead if the reaction went entirely by SN1, and explain. [2]

9. Use these values: average bond enthalpies C–Cl 324 kJ mol⁻¹, C–Br 285 kJ mol⁻¹, C–I 228 kJ mol⁻¹.

(a) Predict the order of rate of hydrolysis for chlorocyclohexane, bromocyclohexane and iodocyclohexane, fastest first. [1] (b) Explain your order, referring to the data and the leaving group. [2] (c) A student argues that chlorocyclohexane should react fastest because C–Cl is the most polar bond. Explain why this is wrong. [1]

10. Cycloheptene, C₇H₁₂, a ring of seven carbons with one C=C, is a symmetrical alkene.

(a) Name the product with bromine. [1] (b) Describe and explain the mechanism for its reaction with bromine. [4] (c) Name the product with hydrogen chloride. [1]

11. 1-ethylcyclohexene reacts with water in the presence of an acid catalyst.

(a) Draw or describe the two carbocations that could form when H⁺ adds. [2] (b) Predict the major product and explain your choice. [2] (c) Describe the remaining steps of the mechanism from the carbocation to the alcohol, and state why the acid is a catalyst. [3]

12. Benzene reacts with the nitronium ion, NO₂⁺.

(a) Explain why benzene undergoes substitution rather than addition, despite being unsaturated. [2] (b) Describe and explain the mechanism by which benzene reacts with NO₂⁺. [4] (c) In the mixture of concentrated nitric and sulfuric acids that produces NO₂⁺, state the role of HNO₃ in Brønsted–Lowry terms. [1]

Answers

1. (a) Acid BF₃, base F⁻ [1]. [1] (b) Acid Ni²⁺, base NH₃ [1]. [1] (c) Acid H⁺, base CH₃NH₂ [1]. [1] Examiner insight: Name both roles in each part; identifying only the acid leaves the answer incomplete.

2. A Brønsted–Lowry base accepts H⁺ by donating a lone pair to it [1], so it is an electron-pair donor, which is the Lewis definition of a base [1]. A Lewis acid need not contain transferable hydrogen, for example BF₃ or Fe³⁺ [1]. [3] Examiner insight: The example must contain no H⁺ to donate; quoting HCl as a Lewis-only acid gains nothing.

3. (a) Oxygen keeps one lone pair and bonds to CH₃, H and B; boron bonds to three F atoms (each with three lone pairs) and to O [1]. The O–B bond is the coordination bond, both electrons from oxygen [1]. [2] (b) BF₃ has three electron domains: trigonal planar [1]. The adduct has four around boron: tetrahedral [1]. [2] (c) No difference; once formed, it is an ordinary covalent bond [1]. [1] Examiner insight: Lone pairs on every fluorine and the remaining pair on oxygen are part of a Lewis formula; omitting them leaves the structure incomplete.

4. (a) +3 + 2(−1) = +1, [Co(NH₃)₄Cl₂]⁺ [1]. [1] (b) +2 + 5(−1) = −3, [Fe(CN)₅(H₂O)]³⁻ [1]. [1] (c) +3 + 4(−1) = −1, [VCl₄(H₂O)₂]⁻ [1]. [1] (d) x + 6(−1) = −3, so +3 [1]. [1] Examiner insight: The answer is the overall charge of the complex, written outside the square bracket; repeating the metal ion’s own charge misses the ligand contribution.

5. The oxygen of water has a lone pair that it donates to the cation [1], forming a coordination bond; water is the Lewis base and the cation the Lewis acid [1]. In CN⁻ the donor atom is carbon [1]. [3] Examiner insight: “Water bonds to the metal” is not enough; the answer needs donation of a lone pair.

6. (a) A: rate = k[RX][OH⁻] [1]. B: rate = k[RX] [1]. [2] (b) A: 2.4 × 10⁻⁵ × 3 × 3 = 2.2 × 10⁻⁴ mol dm⁻³ s⁻¹ [1]. B: 1.5 × 10⁻⁴ × 3 = 4.5 × 10⁻⁴ mol dm⁻³ s⁻¹ [1]. [2] (c) A (SN2) has one hump; B (SN1) has two humps with a carbocation intermediate between them [1]. [1] Examiner insight: Treat each reaction separately in (b); tripling only the halogenoalkane for A ignores the hydroxide term.

7. Curly arrow from an oxygen lone pair of the hydroxide ion to C1, attacking from the side opposite iodine [1]. Curly arrow from the C–I bond to I [1]. Transition state: [HO···C···I]⁻ in brackets with ‡, other three groups in one plane, overall negative charge [1]. Products hexan-1-ol and I⁻; one step because C1 is primary, uncrowded, and a primary carbocation would be unstable [1]. [4] Examiner insight: An arrow must start at a lone pair or a bond, not at the minus sign, and must end on the δ+ carbon.

8. (a) A single enantiomer of octan-2-ol forms [1]. OH⁻ attacks from the side opposite iodine [1], so the configuration at C2 is inverted [1]. [3] (b) A racemic mixture [1], because the planar carbocation can be attacked equally from either face [1]. [2] Examiner insight: “Inversion” alone, with no link to back-side attack, explains nothing about why it happens.

9. (a) Iodocyclohexane > bromocyclohexane > chlorocyclohexane [1]. [1] (b) C–I (228) is weaker than C–Br (285) and C–Cl (324), so it breaks most easily [1]; I⁻ is the most stable, best leaving group [1]. [2] (c) The slow part of the reaction is C–X bond cleavage, so the weakest bond, not the most polar one, controls the rate [1]. [1] Examiner insight: Use the numbers given; “C–I is weakest” with the values quoted is stronger than a general statement.

10. (a) 1,2-dibromocycloheptane [1]. [1] (b) The high electron density of the π bond repels electrons in Br₂, making the nearer Br δ+ [1]. Curly arrow from the π bond to Brδ+ and from the Br–Br bond to the far Br [1]. A carbocation and Br⁻ form [1]. Curly arrow from a Br⁻ lone pair to C⁺ [1]. [4] (c) chlorocycloheptane [1]. [1] Examiner insight: The induced dipole on Br₂ must be shown or stated; a non-polar Br₂ with no δ+ cannot act as an electrophile.

11. (a) H⁺ on the CH of the ring gives a tertiary carbocation on the carbon bearing the ethyl group [1]; H⁺ on that ethyl-bearing carbon gives a secondary carbocation on the neighbouring ring carbon [1]. [2] (b) 1-ethylcyclohexan-1-ol [1], because the tertiary carbocation is more stable, stabilised by three alkyl groups releasing electron density [1]. [2] (c) A water molecule donates an oxygen lone pair to C⁺ [1]. The protonated alcohol loses H⁺ [1]. H⁺ is used in step 1 and regenerated, so it is a catalyst [1]. [3] Examiner insight: Justify the major product with carbocation stability; quoting the “more hydrogens” rule alone does not explain.

12. (a) Benzene’s six π electrons are delocalised over the ring, which gives extra stability [1]; addition would destroy this, but substitution keeps the ring intact [1]. [2] (b) Curly arrow from the delocalised ring to N of NO₂⁺ [1]. Intermediate: horseshoe over five carbons with +, sixth carbon bearing H and NO₂ [1]. Curly arrow from the C–H bonding pair back into the π system [1]; H⁺ lost, delocalisation restored, nitrobenzene formed [1]. [4] (c) HNO₃ acts as a base: it accepts H⁺ from H₂SO₄ [1]. [1] Examiner insight: The horseshoe must leave out the carbon carrying both H and NO₂; a full circle in the intermediate is a common way to lose credit.

Where marks are usually lost

• Swapping “donor” and “acceptor” in Lewis definitions.
• Choosing a hydrogen-containing acid as an example of a Lewis-only acid.
• Leaving lone pairs off Lewis formulas of adducts.
• Forgetting that cyanide donates through carbon.
• Including the nucleophile in an SN1 rate equation, or leaving it out of SN2.
• Drawing front-side attack in SN2, which hides the inversion.
• Explaining halogenoalkane reactivity by bond polarity.
• Missing the induced dipole on Br₂.
• Picking the alcohol from the less stable carbocation.
• Drawing a full circle in the benzene intermediate.

Next steps

• Electron-pair sharing revision notes
• Electron-pair sharing study guide
• All DP Chemistry units on the hub
• DP Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Questions written against the Diploma Programme Chemistry guide (International Baccalaureate Organization), first assessment 2025, published February 2023, sub-topic Reactivity 3.4, understandings 3.4.6 to 3.4.13 (additional higher level).