On this page
Fluorine is the most electronegative element, so it should pull the most electron density off boron and make BF3 the hungriest electron-pair acceptor of the boron halides. Yet a standard JEE question asks you to arrange BF3, BCl3 and BBr3 by Lewis acidity, and the expected answer runs the other way. This post gives you the answer examiners expect, a worked electron count, and an honest note on what chemists have found since.
Is this topic in the syllabus and in NCERT?
The JEE (Main) 2026 syllabus has a p-block unit covering Group 13 to Group 18: "general trends in physical and chemical properties of elements across the periods and down the groups, unique behaviour of the first element in each group". The NEET (UG) 2026 syllabus has the same unit. The current NCERT Class 11 chemistry books (Reprint 2026-27), Part I and Part II, list Units 1 to 9 only, with no p-block unit. Boron halides still appear: the Equilibrium unit uses BF3 + :NH3 → BF3:NH3 as its example of a Lewis acid with no proton, and the Chemical Bonding unit describes BCl3 as sp² hybridised and trigonal planar. So the trend itself is a JEE Main and NEET-UG topic that you will study outside the current NCERT text. Check this year's bulletin.
Why is BX3 a Lewis acid at all?
Boron has three valence electrons. In BX3 it forms three bonds and is left with six electrons around it, so it has an empty 2p orbital at right angles to the molecular plane. A Lewis base such as NH3 can donate its lone pair into that orbital.
Worked example: what changes when BF3 meets NH3?
Use formal charge = valence electrons − non-bonding electrons − (bonding electrons ÷ 2).
In BF3: boron has 3 − 0 − 6/2 = 0. Each F has 7 − 6 − 2/2 = 0. Boron has 6 electrons around it.
In F3B–NH3: boron now has four bonds: 3 − 0 − 8/2 = −1. Nitrogen has four bonds and no lone pair: 5 − 0 − 8/2 = +1. Boron has an octet.
Shape: boron goes from trigonal planar (sp², bond angles 120°) to roughly tetrahedral (sp³, about 109.5°). The three B–F bonds must bend out of the plane.
The back-bond picture: draw one resonance form with a B=F double bond. Now that fluorine has 7 − 4 − 4/2 = +1 and boron 3 − 0 − 8/2 = −1. A fluorine lone pair has partly filled boron's empty orbital before any base arrives.
What answer do exams expect, and why?
Electronegativity alone predicts BF3 > BCl3 > BBr3. Plumley and Evanseck (2009) note that this prediction is "opposite to that inferred by experiment". So the order you should write towards bases like ammonia or pyridine is:
BF3 < BCl3 < BBr3
The textbook explanation is pπ-pπ back bonding. Fluorine's 2p orbital and boron's 2p orbital are similar in size, so a fluorine lone pair overlaps well with boron's empty orbital and partly fills it. Chlorine (3p) and bromine (4p) overlap less well. So BF3's boron is the least electron-deficient, and its back bond must also be lost when it turns tetrahedral. Rowsell, Gillespie and Heard (1999) call this "the most popular and most widely quoted" explanation.
What does research say about the back-bonding answer?
Here the story gets more interesting, and it is worth knowing if you go on to study chemistry.
Rowsell, Gillespie and Heard found that the halogens stay close-packed as the adduct forms, so the B–X bonds must stretch. Short, strong B–F bonds cost more energy to stretch than longer B–Cl bonds, and they argue that this is why BCl3 is the stronger acid towards NH3. Towards a weak base like CO, BX3 hardly bends, and BF3 is the stronger acid.
Bessac and Frenking (2003) calculated that bending BCl3 and BF3 into the pyramidal shape costs nearly the same energy. They link BCl3's stronger bond with NH3 to its lower-lying LUMO.
Thayyil and Parambil (2022) say the back-bonding explanation "is believed by most" despite opposing findings, and conclude that the acidity trend can change with the strength of the base.
The lesson: "stronger Lewis acid" only means something once you say towards which base. For exam answers, stay with back bonding. For understanding, remember that it is a model.
Common mistakes and how to avoid them
Ranking by electronegativity. That gives the reverse order. Write the back-bonding argument instead.
Saying back bonding makes boron sp³. BF3 stays planar and sp². It becomes sp³ only in the adduct.
Calling NH3 the acid. NH3 donates the pair, so it is the Lewis base. Our side-by-side guide to Arrhenius, Bronsted-Lowry and Lewis acids has more examples.
Wrong formal charges in the adduct. Boron is −1 and nitrogen +1, not the other way round.
When a question mixes hybridisation, formal charge and acidity, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can check each link in your argument.
A practice routine for this week
Draw BF3, its B=F resonance form and F3B–NH3, with formal charges on every atom.
Write the back-bonding argument in five lines, naming the orbitals.
Redo the formal-charge count for BCl3 + pyridine.
Explain in two sentences why the order can depend on the base.
Before your next chemistry session, write the three-compound order and the one-line reason on a flashcard.
