IB Chemistry HLTopic 6 — Electron Pair SharingPaper 1 & 2Core idea~9 min read
Lewis Acid and Base Reactions
Last page gave you the definitions. This page shows them doing real work — a molecule with a hole in it, a molecule with a spare pair, and the bond that forms when the two meet.
📚 What you need to know
A coordinate bond forms when a Lewis base donates a lone pair to a Lewis acid.
The classic example is NH3 + BF3. Ammonia is the base, boron trifluoride is the acid.
Boron in BF3 forms three sp2 hybrid orbitals, which leaves an empty 2p orbital.
That empty orbital is the hole the lone pair drops into.
No proton moves, so neither species is a Brønsted–Lowry acid or base here.
An electrophile is a Lewis acid; a nucleophile is a Lewis base.
The same idea builds complex ions such as [Cu(H2O)6]2+.
Ammonia meets boron trifluoride
Boron sits in group 13, so it only brings three outer electrons to the party. After bonding to three fluorines it has just six electrons around it, not eight. It is two short, and it would very much like to fix that.
Ammonia has the opposite problem. Nitrogen has a full octet plus a lone pair that is not doing anything useful.
Put them together and the nitrogen lone pair slots into boron’s empty orbital. One new bond, both electrons from nitrogen.
The arrow inside the product is the coordinate bond. It points away from the atom that supplied both electrons.
Where the empty orbital comes from
Boron has three outer electrons: two in 2s and one in 2p. To make three identical bonds it promotes one 2s electron into a 2p orbital, then mixes one s with two p orbitals to give three sp2 hybrid orbitals.
Three orbitals, three electrons, three bonds to fluorine. But boron started with four orbitals available, so one 2p orbital is left over with nothing in it. That leftover is the hole.
The red box is the vacant 2p orbital. Nothing lives there, which is exactly why ammonia can move in.
You do not need to redraw this in an exam very often, but you do need the sentence: “boron forms three sp2 hybrid orbitals, leaving a vacant 2p orbital that can accept a lone pair.” That one line earns marks on its own.
Only electron pairs move
Look carefully at the BF3 and NH3 reaction and check for a hydrogen ion. There isn’t one. No proton is donated, no proton is accepted.
That means Brønsted–Lowry theory has nothing to say about this reaction at all. It is not that the theory gives a wrong answer — it simply cannot be applied. Lewis theory can.
Quick test. If you can point at an H+ that has moved, both theories apply. If you cannot, only Lewis does. That single check answers most “explain why” questions on this topic.
The organic version of exactly the same thing
Take a carbocation such as (CH3)3C+. The positive carbon has only six electrons around it, so it has a gap in just the same way boron does. Now bring in a bromide ion with four lone pairs.
Carbocation and bromide
(CH3)3C+ + :Br– → (CH3)3C–Br
An organic chemist calls that carbocation an electrophile and the bromide a nucleophile. A physical chemist calls them a Lewis acid and a Lewis base. Both are describing the same electron pair moving into the same gap.
Reaction
Lewis acid
Lewis base
Proton moved?
NH3 + BF3
BF3
NH3
No
H+ + OH–
H+
OH–
Yes
(CH3)3C+ + Br–
(CH3)3C+
Br–
No
Cu2+ + 6H2O
Cu2+
H2O
No
Building a complex ion
Transition metal ions are small, highly charged and have empty orbitals available. That makes them excellent Lewis acids. Drop a copper(II) ion into water and six water molecules line themselves up around it, each donating an oxygen lone pair.
Six coordinate bonds, all pointing inwards towards the metal. The copper ion is the Lewis acid and every water molecule is a Lewis base. Nothing is deprotonated and no charge changes — the 2+ is still 2+ on the other side.
Worked examples
WORKED EXAMPLE
Explain, in terms of orbitals, why BF3 reacts with NH3 but CF4 does not.
Count the electrons round the central atom
In BF3, boron has three bonds and no lone pair, so only six electrons.
What is left over
Boron uses three sp2 orbitals for bonding, so one 2p orbital is empty and can take a pair.
Now do the same for CF4
Carbon has four bonds, so eight electrons and no empty low-energy orbital.
BF3 is a Lewis acid; CF4 has nowhere to put a lone pair“electron-deficient” is the key phrase — get it into the answer
WORKED EXAMPLE
HCOO– reacts with H2O to give HCOOH and OH–. State which species is amphoteric, and describe the two roles it can play.
Look at what water does here
One of its δ+ hydrogens accepts a lone pair from the methanoate oxygen, so here water is the Lewis acid.
Now think of a different partner
With Cu2+, water donates its oxygen lone pair, so there it is the Lewis base.
Name the property
Behaving as either acid or base depending on the partner is called being amphoteric.
Water is amphoteric: Lewis acid here, Lewis base with a metal ionalways tie the answer to the specific reaction in front of you
💡 Exam tip
Draw the coordinate bond as an arrow pointing from the donor atom to the acceptor. Direction is marked.
Show the lone pair on the base in your diagram. If it is not drawn, the examiner cannot see where the electrons came from.
For “explain why this is not a Brønsted–Lowry reaction”, say plainly that no proton is transferred.
Learn the sp2 line for boron. It comes up almost every time BF3 appears.
In complex ions the metal is always the Lewis acid and the ligands are always the Lewis bases.
Curly arrows here are double-headed — a whole pair moves.
⚠️ Common mix-up
Calling BF3 an acid because of the F. Fluorine has nothing to do with it. Boron is the acidic centre.
Drawing the coordinate arrow from acid to base. It goes the other way: donor to acceptor.
Saying nitrogen “loses” its lone pair. The pair is now shared, not given away.
Forgetting the charge on a complex ion. Neutral ligands do not change it, so [Cu(H2O)6] is still 2+.
Thinking hybridisation only matters for carbon. Boron, nitrogen and oxygen all hybridise too.
Mixing up the empty orbital with a lone pair. Empty means zero electrons. A lone pair means two.
Up next: Coordination Bonds — what happens when a metal ion collects a whole ring of Lewis bases, and how to count them properly.
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