Why does boron trifluoride accept an electron pair from ammonia even though both BF₃ and NH₃ are covalent molecules? BF₃ is electron-deficient at boron and can act as a Lewis acid, accepting an electron pair into an available orbital. NH₃ has a lone pair on nitrogen and can act as a Lewis base, donating that pair to boron. Together they can form the Lewis acid-base adduct often drawn as F₃B←NH₃ or NH₃→BF₃, with the electron-pair arrow starting at nitrogen and pointing to boron. No proton needs to be transferred: this illustrates Lewis acid-base chemistry and coordinate covalent bonding rather than the ordinary Brønsted–Lowry proton-transfer definition.
For Hougang parents searching for JC H2 Chemistry tuition, a tutor for 9476 Chemical Bonding, BF₃ trigonal-planar structure, NH₃ lone pairs, Lewis acids and bases or coordinate bonding, the first useful check is whether your child can draw both molecules before the interaction and explain where the donated electron pair originates. The 2027 H2 Chemistry syllabus explicitly includes BF₃ and NH₃ among VSEPR shape examples and requires understanding coordinate covalent bonding. The particular BF₃–NH₃ adduct is a useful linked teaching example, not a separately named compulsory reaction in the bonding outcome. A diagnosis-led three-pax tutorial should repair the electron-counting misconception before asking for more diagrams.
Picture two molecules that look completely ordinary on a page. One has an available place for an electron pair; the other has an electron pair that is not currently used to bond to hydrogen. That structural complementarity is the beginning of an entire family of chemical interactions.
The central question is not whether the molecules already have bonds
BF₃ and NH₃ are both covalent molecules. That does not mean all atoms within them are equally able to donate or accept electrons in a new interaction.
Boron in BF₃ is bonded to three fluorine atoms. In the simplest electron-dot model, it has three bonding pairs around it and only six electrons counted within its local valence-shell bonding environment. This makes it electron-deficient compared with the ordinary octet.
Nitrogen in NH₃ has three N–H bonding pairs and one lone pair. That lone pair can be used to form an additional bond with a suitable electron-pair acceptor.
Why BF₃ has an incomplete octet
Boron belongs to Group 13 and has three valence electrons as a neutral atom. In the familiar BF₃ Lewis structure, it shares an electron pair with each of the three fluorine atoms, creating three B–F bonds.
The six bonding electrons around the central boron correspond to three pairs. The fluorine atoms have their own lone pairs, but those are not simply counted as boron’s complete octet in the elementary Lewis picture.
A student who says that every atom in every stable covalent molecule must have exactly eight surrounding electrons has turned a useful rule into an absolute one. Boron compounds are a familiar exception.
Why NH₃ has a lone pair available
Nitrogen belongs to Group 15 and has five valence electrons. In NH₃ it makes three single covalent bonds to hydrogen, leaving one nonbonding electron pair on nitrogen in the simple Lewis model.
The molecule’s available electron pair is why ammonia can act as an electron-pair donor in reactions with appropriate acceptors.
This does not mean every electron pair in every molecule is automatically available for bonding in the same way; orbital overlap, chemical environment and energetics affect real interactions.
Draw the two molecules before drawing the adduct
- BF₃: B is central with three B–F bonds and no boron lone pair in the simple Lewis diagram.
- NH₃: N is central with three N–H bonds and one lone pair on nitrogen.
- Arrow source: begin at nitrogen’s lone pair.
- Arrow destination: point towards electron-deficient boron.
- Result: a new N–B interaction is represented as a coordinate covalent bond.
A carefully labelled electron-pair arrow is not a decoration. It explains the donor and acceptor roles. Starting the arrow at BF₃ and pointing to ammonia reverses the chemical meaning.
Lewis acid and Lewis base definitions
A Lewis acid accepts an electron pair. A Lewis base donates an electron pair. These definitions focus on electron-pair sharing and can describe reactions that do not involve a hydrogen-ion transfer.
BF₃ is a Lewis acid because boron can accept a donated electron pair; NH₃ is a Lewis base because nitrogen can donate its lone pair.
Neither definition requires the acid to contain hydrogen or the base to contain OH⁻. That is why this example helps expand the student’s acid-base model beyond simple aqueous pH questions.
Contrast with Brønsted–Lowry acids and bases
A Brønsted–Lowry acid donates a proton, and a Brønsted–Lowry base accepts a proton. In the familiar reaction NH₃ + H⁺ → NH₄⁺, ammonia accepts a proton and therefore acts as a Brønsted–Lowry base.
In the reaction of NH₃ with BF₃, the molecule BF₃ does not donate a proton. It accepts an electron pair. The Lewis description is therefore directly useful in a way that the conventional proton-transfer description is not.
A child who insists BF₃ cannot be an acid because it has no H has confused two related but different acid definitions.
What is a coordinate covalent bond?
A coordinate or dative covalent bond is one in which the two electrons used to form the bonding pair are initially supplied by the same donor species. Once the bond is formed, it is part of the molecule’s covalent bonding system.
For NH₃→BF₃, the nitrogen lone pair supplies both bonding electrons when the N–B bond forms. The arrow notation indicates this source, not that two electrons permanently travel as separate particles in one direction through a wire.
After formation, chemical bonding is not accurately explained by claiming one end owns the pair forever in an ordinary classical sense. Coordinate bonding is a way to describe bond formation and electron-pair donation.
How the Lewis adduct is represented
A simple representation is H₃N→BF₃. The donor is N on the left and the acceptor is B on the right. The adduct contains a new N–B bond.
It may also be written with a single N–B covalent bond and formal charges to preserve the electron-bookkeeping convention. In that representation nitrogen is often assigned formal +1 and boron formal −1.
Formal charge is a bookkeeping tool derived from a specified Lewis structure. It should not be mistaken for a claim that the bonded atoms are isolated N⁺ and B⁻ ions floating separately.
Why BF₃ begins as trigonal planar
Valence Shell Electron Pair Repulsion theory provides a useful first model for molecular shape. Boron in isolated BF₃ has three regions of bonding electron density arranged around the central boron and no lone pair on that atom in the elementary structure.
Those three regions minimise their repulsions in an approximately trigonal-planar arrangement with ideal bond angles near 120°. The molecule is approximately flat around boron.
This shape is among the explicitly named VSEPR examples in the H2 syllabus. Students need to identify electron regions, not guess the shape from the number of atoms alone.
Why NH₃ begins as trigonal pyramidal
Nitrogen in isolated ammonia has four electron-pair regions: three N–H bonding regions and one lone pair. The four electron regions have an approximately tetrahedral arrangement, but only the positions of atoms define the molecular shape.
The three hydrogen atoms form a trigonal-pyramidal geometry around N, with an H–N–H bond angle about 107°. The lone pair tends to repel more strongly than a bonding pair in the familiar VSEPR explanation, compressing the H–N–H bond angle below the ideal tetrahedral 109.5°.
A student who labels ammonia tetrahedral because there are four electron regions has confused electron-domain geometry with the molecular shape defined by atom positions.
What changes to boron’s geometry on accepting the lone pair?
When the new N–B bond forms in the simple adduct, boron is now bonded to four atoms: three fluorines and one nitrogen. The arrangement around boron is then approximately tetrahedral rather than trigonal planar.
This is a powerful structural prediction. The student can move from three bonding regions around boron to four without memorising an entirely new unrelated molecule.
Real bond angles in specific adduct structures need not be exactly 109.5° because different substituents and interactions affect the geometry. The school model predicts the broad shape.
Does the nitrogen keep its lone pair after donating it?
The nitrogen lone pair becomes involved in forming the new N–B bonding interaction. In a simple electron-dot structure of the adduct, nitrogen has four bonds and no remaining lone pair corresponding to the original donating pair.
A student who draws three N–H bonds, the N–B bond and an additional unchanged nitrogen lone pair has overcounted the original nitrogen’s valence electrons.
A quick formal-charge check makes the mistake visible. With four bonds and no lone pair, nitrogen’s usual formal charge in that Lewis structure is +1.
The paired formal charges in the adduct
Formal charge follows a consistent Lewis bookkeeping rule. Nitrogen has five valence electrons in the free atom and is assigned one electron from each of its four bonds in the adduct. The resulting nitrogen formal charge is +1.
Boron has three valence electrons as a free atom. With four simple covalent bonds in the adduct Lewis drawing and no lone pairs on boron, it is assigned four bonding electrons for formal-charge purposes, giving a formal charge of −1.
The charges sum to zero, consistent with a neutral adduct formed from neutral BF₃ and NH₃. This is an optional check when formal-charge calculations are part of the learner’s course.
Why the molecule does not become a salt solution
A covalent acid–base adduct is not the same as dissolving sodium chloride to produce separate Na⁺ and Cl⁻ ions. The BF₃–NH₃ example forms a new coordinate covalent bond in the connected molecular species.
The bond polarity and charge distribution may be significant, but the elementary structure is still a molecular adduct rather than an ordinary aqueous collection of separated ions.
That difference helps students link electron-pair interactions to chemical bonding without confusing them with the movement of free ions in an electrolyte.
BF₃ is electron-deficient despite strong B–F bonds
The presence of three covalent B–F bonds does not automatically complete boron’s octet. Bond strength and the number of electrons surrounding the central atom are different properties.
Fluorine’s electronegativity also affects the electronic structure of BF₃, and more advanced treatments discuss fluorine-to-boron electron donation and the relative Lewis acidity of boron halides.
Those refinements need not be imported into every JC question. The first model is enough to explain why an available acceptor orbital and ammonia’s lone pair can make a Lewis adduct.
Is the reaction necessarily irreversible?
Lewis acid–base association can have a thermodynamic balance that depends on temperature, pressure, solvent and other chemical conditions. Writing one arrow to show the familiar adduct does not prove that every molecule remains bound permanently under all circumstances.
At the H2 teaching level, focus on the formation of the donor–acceptor interaction and its structures. Full quantitative adduct equilibrium treatment is a separate advanced topic.
A tutor should distinguish a useful static Lewis drawing from a claim about perfect conversion in a real reaction mixture.
Another coordinate-bond example: ammonium
In the formation of NH₄⁺ from NH₃ and H⁺, the nitrogen lone pair donates into a bond with the proton. After bonding, ammonium has four N–H bonds and an overall +1 ionic charge.
That example is explicitly familiar in the H2 chemical-bonding syllabus. The distinction from BF₃ is that the acceptor is H⁺ in one case and an electron-deficient neutral molecule in the other.
The shared principle is that a lone pair is donated to an electron-pair acceptor. The chemical identity and charge of the resulting species must still be calculated correctly.
Why the product shapes are not just a drawing exercise
The arrangement of electron domains influences molecule geometry, and geometry affects how bond dipoles combine, what orbitals can overlap and how molecules interact with each other.
BF₃’s trigonal-planar symmetry is one reason the net molecular bond dipoles cancel in the ideal isolated BF₃ molecule even though B–F bonds are polar. NH₃’s trigonal-pyramidal structure produces a net molecular dipole.
This is an additional connection between VSEPR, bond polarity and the molecular interactions covered in H2 Chemistry. The Lewis adduct changes both connectivity and the structural symmetry that students originally counted.
A changed structural problem
Suppose the student has just explained NH₃→BF₃. Now replace the electron-pair acceptor with a proton. Would ammonia still be acting as a donor? Yes: its nitrogen lone pair can bond to H⁺ to form NH₄⁺.
Now replace the donor with a molecule that has no suitable accessible lone pair for the given interaction. The student should not draw an arbitrary coordinate bond merely because two covalent molecules are close together.
The rule is about actual electron-pair availability and a compatible acceptor, not the mere fact that two molecules appear on the same exam page.
A three-part Lewis structure checking routine
- Count: the central atom’s valence electrons and the electron domains already present.
- Locate: the donor lone pair and the acceptor’s incomplete-octet or available-orbital situation.
- Draw: the donation arrow from the electron pair to the new bonding site.
- Recount: bonds, lone pairs and any formal charges in the product.
- Predict: how the number of electron regions changes the expected molecular shape.
- Interpret: distinguish Lewis acid/base behaviour from a proton-transfer reaction when appropriate.
This sequence reduces the common mistake of beginning with a memorised adduct shape while forgetting how the bond formed.
Five mistakes behind wrong two-mark answers
- ‘BF₃ cannot be an acid because it contains no hydrogen.’ It can accept an electron pair and therefore act as a Lewis acid.
- ‘Ammonia has no lone pair.’ Nitrogen has one in the usual isolated NH₃ structure.
- ‘The arrow points from boron to nitrogen.’ Nitrogen supplies the bonding pair in the discussed adduct.
- ‘BF₃ is initially tetrahedral.’ Isolated BF₃ is trigonal planar around boron.
- ‘The new adduct is two free ions in solution.’ It is represented as a covalently connected Lewis acid-base adduct.
- ‘The lone pair remains untouched after becoming the new bond.’ The donor’s electron bookkeeping must be updated.
A tutor should identify which electron pair or bond count was lost in the student’s drawing before assigning more VSEPR diagrams.
What the 2027 H2 Chemistry syllabus actually asks
The revised H2 Chemistry 9476 syllabus covers ionic, covalent, metallic and coordinate covalent bonding, VSEPR shapes and bond angles, and the polarity of bonds and molecules. BF₃ (trigonal planar) and NH₃ (trigonal pyramidal) are explicitly among the specified molecular-shape examples.
The formation of ammonium and the Al₂Cl₆ molecule are named coordinate-bonding examples in the syllabus. The BF₃–NH₃ Lewis adduct used here is a coherent extension connecting electron deficiency, donor lone pairs and molecular shape; it is not presented as an extra named compulsory reaction.
Use the official SEAB 2027 A-Level syllabus directory and H2 Chemistry 9476 bonding syllabus to set the required depth for the student’s examination year.
Where to go for a full VSEPR shape explanation
The present article focuses on electron-pair donation and the structural transformation from free BF₃ and NH₃ to an adduct. It is not intended to repeat every possible octahedral, bent and trigonal-pyramidal molecule.
The existing Hougang VSEPR Shapes and Bond Angles guide contains that wider shape reference. The Hougang Brønsted–Lowry and Lewis Acids tutorial owns the broader acid-base comparison.
How a three-pax JC tutorial should diagnose this topic
Imagine three students drawing the adduct. One knows the lone pair but incorrectly calls BF₃ a Lewis base. Another labels BF₃ correctly but draws the arrow from boron to nitrogen. The third draws the arrow correctly but insists the boron remains trigonal planar after acquiring a fourth bond.
A tutor can use one common explanation, then give each learner a different changed example and ask for independent electron counting. This makes the reasoning visible rather than leaving everyone with a correctly copied picture.
A three-student group is helpful when it allows those individual explanations to be examined and revisited, not simply because it is small.
A five-minute parent-and-student review
- Why is the boron atom in isolated BF₃ called electron-deficient?
- How many bonding pairs and lone pairs surround nitrogen in NH₃?
- Which direction should the donation arrow point in NH₃→BF₃?
- Why does the central boron’s expected shape change after adduct formation?
- How does a Lewis acid differ from a Brønsted–Lowry acid?
- Is the BF₃–NH₃ adduct itself an explicitly named mandatory bonding example in 9476?
- What remains useful about the ammonia-to-ammonium example?
If a student can draw and explain NH₄⁺ and then return to the BF₃–NH₃ adduct without memorised arrows, they have learned a transferable chemical-bonding principle.
Questions Hougang JC parents ask
Is BF₃ acidic even without an H atom?
It can act as a Lewis acid because boron accepts a donated electron pair. Lewis acidity does not require proton donation.
Does ammonia give away two electrons permanently?
The nitrogen lone pair becomes part of a coordinate covalent bond. It is not the same as the separate transfer of two electrons in a redox half-equation.
Is BF₃ trigonal planar or tetrahedral?
Isolated BF₃ is trigonal planar around boron. In the illustrated adduct, boron has four bonds and an approximately tetrahedral environment.
Do all H2 students need the BF₃–NH₃ adduct itself?
Treat it as a useful extension unless assigned by the school. The 2027 syllabus explicitly names BF₃ and NH₃ shapes and coordinate bonding using other specified examples.
When is a tutor more useful than a new diagram pack?
When a student can reproduce a structure but cannot say which atom donated the pair or justify the resulting electron count, a precise first-principles correction is more valuable than repeating the same drawing.
A bond can begin with a spare pair and an available orbital
BF₃ and NH₃ illustrate how a chemical structure can reveal reactivity. Boron can accept an electron pair; nitrogen can donate one. The resulting bond connects Lewis acid-base chemistry, coordinate covalent bonding and VSEPR geometry in one coherent explanation.
Continue with the Hougang JC Chemistry Tuition guide, the VSEPR reference and the Hougang Chemistry Hub. For a diagnosis-led three-pax H2 Chemistry consultation and confirmation of teaching venue and class availability, contact eduKate Singapore.