JEE Main 2021 · 27 August · morning · Single correct · +4 / −1
In which one of the following molecules strongest back donation of an electron pair from halide to boron is expected?
- (a) BCl₃
- (b) BF₃
- (c) BBr₃
- (d) BI₃
Answer and solution
Answer (b) BF₃
Solution
Chemistry ma’am · 3 steps
Back donation is a 2p to 2p overlap, and only fluorine's orbitals match boron's in size.
- Empty orbital on boron. In BX₃, boron is sp² hybridised and has only 6 electrons around it. Its unhybridised 2p orbital is empty and stands perpendicular to the plane of the molecule.
- Back donation. A filled p orbital on the halogen can push a lone pair sideways into that empty 2p orbital. This forms a pπ–pπ bond. How strong it is depends on how well the two orbitals overlap.
Molecule Halogen orbital Overlap with B 2p Back donation BF₃ 2p same size, effective strongest BCl₃ 3p larger, weaker weaker BBr₃ 4p much larger, poor weak BI₃ 5p very diffuse, poorest weakest - So. Back donation is strongest in BF₃, option (b). This is also why BF₃ is the weakest Lewis acid of the four: its empty orbital is already partly filled by fluorine.
The idea to keep. In back bonding, size match beats electronegativity. Same-period 2p–2p overlap is the best, so back donation falls BF₃ > BCl₃ > BBr₃ > BI₃, and Lewis acidity rises in that same order.
Answer from the NTA final answer key. Solution by PrepWiser.