Question

In $Br{F}_3$ molecule, the lone pairs occupy equatorial positions to reduce

• A. Bond pair-lone pair repulsion only
• B. Bond pair-bond pair repulsion only
• C. Lone pair-lone pair repulsion and bond pair- lone pair repulsion
• D. Lone pair-lone pair repulsion only

Answer 1

The correct option is C Lone pair-lone pair repulsion and bond pair- lone pair repulsion

$Br{F}_3$ has $s{p}^{3}d$ hybridisation because there are three bonding $Br-F$ pairs and two lone pairs on $Br$ and it adopts trigonal bipyramidal geometry. In this geometry, lone pairs are always kept at equatorial positions as these are minimum repulsion sites. Having both the lone pair at the axial positions will create 6 lone pair-bond pair repulsions. Having one of the lone pairs at the axial position will create one lone pair-lone pair repulsion and three lone pair-bond pair repulsions. But lone pair at equatorial position will have only four lone pair- bond repulsions and hence minimises both types of repulsion i.e. lp-lp repulsion as well as bp-lp repulsion.