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Lewis Adducts and Exam Phrasing

Show how a Lewis adduct forms with a correctly drawn arrow, naming the donor and acceptor.

  • GCE A-Level H2 Chemistry 9476-2027

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Lewis questions are usually “spot the donor/acceptor” questions. If you can identify the lone pair donor, the electron-pair acceptor, and the correct product charge, you can answer most prompts in one clean sentence. Review the three acid–base definitions when a question asks you to choose the appropriate model.

Definitions (Must Know)

A. Lewis acid and Lewis base

  • Lewis acid: electron-pair acceptor.
  • Lewis base: electron-pair donor.

B. Lewis adduct

A Lewis adduct is the product formed when a Lewis base donates a lone pair to a Lewis acid to form a coordinate (dative) bond.

Example: BF₃ + :NH₃ → F₃B < -NH₃

Lewis adduct diagramDiagram showing lone pair donor, electron pair acceptor, coordinate bond arrow direction, and final adduct charge check.Donor · Lewis baseNH₃N supplies the bonding pairelectron pairAcceptor · Lewis acidBF₃Electron-deficient B accepts the pairadduct formsProduct · Lewis adductF₃B ← NH₃Bond arrow: donor → acceptoroverall charge: 0 → 0
BF3-NH3 adduct: point the coordinate arrow from lone-pair donor NH3 to acceptor BF3, and identify donor as base and acceptor as acid.

C. Coordinate (dative) bond

A coordinate (dative) bond is a covalent bond where both electrons in the shared pair come from the same atom.

See also: Dative Bonding and Common Examples.

Key Ideas (What Earns Marks)

  • A Lewis base donates an electron pair. In the required examples, this is a lone pair on species such as NH₃, H₂O, OH⁻ or Cl⁻; a π bond can also supply an electron pair in later organic chemistry.

  • A Lewis acid can accept an electron pair. It may be electron-deficient or have an empty orbital, as in BF₃ and AlCl₃; some cations, including H⁺, are also electron-pair acceptors.

  • The arrow/dative bond direction is donor → acceptor.

  • A reaction with electron-pair transfer but no proton transfer is Lewis acid–base chemistry, not Brønsted–Lowry chemistry.

  • Exam phrasing mapping:

  • “Electron-pair donor” means Lewis base.

  • “Electron-pair acceptor” means Lewis acid.

Quick Recall (Spotting Lewis Acids/Bases)
  • Lewis base = has a lone pair (often negative / neutral with lone pairs). - Lewis acid = can accept a lone pair (positive, or electron-deficient / empty orbital). - Arrow direction in an adduct = donor → acceptor.

Detailed Explanations

A. How to identify Lewis acids and bases (workflow)

  1. Find the electron-pair donor (a lone pair in these examples; negative charge is a useful clue, not a definition).
  2. Find the electron-pair acceptor (electron-deficient or positive).
  3. Write the product/adduct and ensure the overall charge is conserved.
  4. State: “donates/accepts an electron pair” explicitly.

Because the Lewis base provides both electrons in the new bond, the coordinate-bond arrow must start at the base (lone pair) and point to the acid (empty orbital / electron-deficient centre).

Mini example: AlCl₃ + Cl⁻ → AlCl₄⁻

  • Total charge on the left is -1, so the product must be -1.
  • Cl⁻ donates a lone pair (Lewis base) to electron-deficient AlCl₃ (Lewis acid).
Lewis adduct diagramDiagram showing lone pair donor, electron pair acceptor, coordinate bond arrow direction, and final adduct charge check.Donor · Lewis baseCl⁻Cl− supplies the bonding pairelectron pairAcceptor · Lewis acidAlCl₃Electron-deficient Al accepts the pairadduct formsProduct · Lewis adductAlCl₄⁻Bond arrow: donor → acceptoroverall charge: −1 → −1
AlCl3 + Cl- adduct check: arrow from Cl- (donor) to AlCl3 (acceptor), then confirm total charge is conserved from -1 to -1.

B. Common adduct patterns

Ion adduct: AlCl₃ + Cl⁻ → AlCl₄⁻

  • Cl⁻ donates a lone pair to Al → Lewis base.
  • AlCl₃ accepts an electron pair → Lewis acid.

Proton as a Lewis acid: NH₃ + H⁺ → NH₄⁺

  • H⁺ accepts an electron pair to form the N–H bond → Lewis acid.

Worked Examples

Modelled example 1

Form a chloride adduct

Core

Problem

For AlCl₃ + Cl⁻ → AlCl₄⁻, identify the Lewis acid and Lewis base and justify both roles.
Study the worked solution

  1. Find the acceptor

    Method

    Identify electron-deficient AlCl₃.

    Reason

    Aluminium accepts a lone pair to complete the adduct.

    Working

    AlCl₃ is the Lewis acid.

  2. Find the donor

    Method

    Identify the lone pair on Cl⁻.

    Reason

    A Lewis base donates an electron pair.

    Working

    Cl⁻ is the Lewis base.

  3. Audit the product

    Method

    Retain the total charge in AlCl₄⁻.

    Reason

    The neutral acid plus a 1- base gives a 1- adduct.

    Working

    Total charge: 0 + (-1) = -1.

Guided practice 2

Explain why boron trifluoride is a Lewis acid

About 6 min

Problem

Explain why BF₃ is a Lewis acid in BF₃ + :NH₃ → F₃B < -NH₃.

Try this before viewing the solution

Hints

Hint 1: boron feature
Boron is electron-deficient and has an incomplete octet in BF₃.
Hint 2: follow the pair
The coordinate arrow starts at the nitrogen lone pair and points toward boron.
View solution step by step

  1. Name the acceptor feature

    Method

    State that boron is electron-deficient or has an empty orbital.

    Reason

    This allows BF₃ to receive a lone pair.

    Working

    BF₃ has an incomplete octet at boron.

  2. Follow electron-pair movement

    Method

    State that BF₃ accepts the lone pair donated by NH₃.

    Reason

    An electron-pair acceptor is a Lewis acid.

    Working

    BF₃ is the Lewis acid; NH₃ is the Lewis base.

Common misconception 3

Correct a coordinate-arrow direction

Find and correct the mistake

Learner diagram

A learner draws the coordinate arrow for BF₃ + :NH₃ from boron in BF₃ toward nitrogen in NH₃. Diagnose the arrow.

Choose the arrow origin

The arrow begins at

View solution step by step

  1. Locate the electrons

    Method

    Start at the lone pair on nitrogen.

    Reason

    NH₃ supplies both electrons in the new bond.

    Working

    Donor: :NH₃.

  2. Point to the acceptor

    Method

    Direct the arrow toward boron in BF₃.

    Reason

    BF₃ accepts that electron pair.

    Working

    Coordinate arrow: donor → acceptor.

Examiner practice 4

Treat a proton as a Lewis acid

4 marks

Problem

For NH₃ + H⁺ → NH₄⁺, identify the Lewis acid and Lewis base and justify each role using electron-pair language. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Identify the donor

    1 mark

    Method

    Name ammonia as the Lewis base.

    Reason

    Nitrogen has a lone pair available for bonding.

    Working

    NH₃ is the base.

  2. State donation

    1 mark

    Method

    State that ammonia donates an electron pair.

    Reason

    Both electrons in the new N–H bond originate from ammonia.

    Working

    Electron-pair donor ⇒ Lewis base.

  3. Identify the acceptor

    1 mark

    Method

    Name H⁺ as the Lewis acid.

    Reason

    The proton receives the pair forming the bond.

    Working

    H⁺ is the acid.

  4. State acceptance

    1 mark

    Method

    State that the proton accepts an electron pair.

    Reason

    Electron-pair acceptance is the Lewis acid definition.

    Working

    Electron-pair acceptor ⇒ Lewis acid.

Challenge 5

Transfer Lewis reasoning to a metal complex

Minimal support

Problem

For Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺, identify the Lewis acid and base, state the electron-pair direction, and justify the product charge.

Try this before viewing the solution

Hints

Hint 1: ligand pairs
Each neutral ammonia ligand donates a nitrogen lone pair to the metal ion.
Hint 2: charge audit
Four neutral ligands do not change the 2 + total charge supplied by copper.
View solution step by step

  1. Assign Lewis roles

    Method

    Name Cu²⁺ as acid and NH₃ as base.

    Reason

    The metal ion accepts electron pairs donated by ammonia lone pairs.

    Working

    Acceptor: Cu²⁺; donor: NH₃.

  2. Audit the adduct charge

    Method

    Retain the 2 + charge in the complex ion.

    Reason

    Cu²⁺ contributes + 2 and all four NH₃ ligands are neutral.

    Working

    + 2 + 4(0) = +2, so the product is [Cu(NH₃)₄]²⁺.

Common Mistakes

  • Arrow drawn in the wrong direction (acceptor → donor).
  • Using “proton donor/acceptor” language when there is no proton transfer.
  • Forgetting the charge on the adduct (e.g. AlCl₄⁻).
  • Writing “electron donor/acceptor” instead of electron-pair donor/acceptor.
  • Calling NH₃ an acid “because it contains H”.

Exam Tips

  • In Lewis questions, always use the phrase “donates/accepts an electron pair”.
  • If asked to “state”, give one line: “X is a Lewis acid because it accepts an electron pair.”
  • If asked to “explain”, add the feature: lone pair on donor; empty orbital/electron deficiency on acceptor.
  • When writing adduct equations, check charges are balanced.

Mind Stretchers

Mind stretcher 1Extension

Explain why BF₃ acts as a Lewis acid in its reaction with NH₃ but does not act as a Brønsted–Lowry acid.

Show Answer

Mark scheme:

  • Lewis acid: BF₃ accepts an electron pair (electron-deficient).
  • Brønsted–Lowry acid requires proton donation.
  • In this reaction BF₃ does not donate a proton, so it does not act as a Brønsted–Lowry acid.

Mind stretcher 2Extension

Consider the reaction:

BF₃ + F⁻ → BF₄⁻

(a) Identify the Lewis acid and base. (b) Explain why the product has a 1− charge.

Show Answer

Mark scheme:

  • (a) Lewis acid: BF₃ (electron-pair acceptor; electron-deficient). Lewis base: F⁻ (lone pair donor).
  • (b) Total charge on the left is -1 (because of F⁻), so the adduct must also have charge -1; therefore the product is BF₄⁻.
Syllabus and review details

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