In BrF3 molecule, the lone pairs occupy equatorial positions to minimize
- lone pair - bond pair repulsion only
- bond pair - bond pair repulsion only
- lone pair - lone pair repulsion and lone pair - bond pair repulsion
- lone pair - lone pair repulsion only.
Answer
(4) lone pair - lone pair repulsion only.
Solution
In BrF3, Br has 3 bond pairs + 2 lone pairs = 5 electron pairs ⇒ sp3d, trigonal bipyramidal arrangement.
As per NCERT (VSEPR theory), the repulsion order is lone pair–lone pair > lone pair–bond pair > bond pair–bond pair, and the strongest repulsion, lone pair–lone pair at 90 , must be avoided first.
⇒ With both lone pairs in equatorial positions they are 120 apart, so no lone pair–lone pair repulsion at 90 remains.
If one lone pair were axial, the two lone pairs would be at 90 to each other (strong lone pair–lone pair repulsion).
∴ The lone pairs occupy equatorial positions to minimize lone pair–lone pair repulsion, giving a bent T-shaped molecule.
Answer: (4)