In BrF3 molecule, the lone pairs occupy equatorial positions to minimize

Question

In BrF3 molecule, the lone pairs occupy equatorial positions to minimize (A) lone pair-bond pair repulsion only (B) bond pair-bond pair repulsion only (C) lone pair-lone pair repulsion and lone pair-bond pair repulsion (D) lone pair-lone pair repulsion only

Tardigrade Admin · Accepted Answer

In BrF3 molecule, Br is sp3d hybrid, but geometry is T-shaped due to distortion of geometry from trigonal-bipyramidal to T-shaped by the involvement of lone pair-lone pair repulsion. Here Ip-Ip repulsion =0 Ip-bp repulsion =4 bp-bp repulsion =2

Q. In $B r F_{3}$ 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

In $B r F_{3}$ molecule, Br is $s p^{3} d$ hybrid, but geometry is T-shaped due to distortion of geometry from trigonal-bipyramidal to T-shaped by the involvement of lone pair-lone pair repulsion. Here
$I p - I p$ repulsion $= 0$
$I p - b p$ repulsion $= 4$
$b p - b p$ repulsion $= 2$

Q. In BrF3​ molecule, the lone pairs occupy equatorial positions to minimize