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 repulson only

Answer 1

lone pair lone pair repulsions...for better stability

Answer 2

Explanation:

The geometry of $BrF_{3}$ molecule has T-shaped geometry. And, hybridization of $BrF_{3}$ molecule is calculated as follows.

                Hybridization = $\frac{1}{2}$[Valence electrons in Br + No. of atoms attached to central atom]

                                        = $\frac{1}{2}[7 + 3]$

                                        = 5

Thus, hybridization of $BrF_{3}$ is $sp^{3}d$. As it contains two lone pair of electrons and it is known that the increasing order of repulsion in lone pairs, bond pairs and both of them is as follows.

                              l.p-l.p > l.p - b.p > b.p - b.p

So, in order to minimize the repulsion between both the lone pairs in $BrF_{3}$, they move away from each other. This means they occupy the equatorial position and due to this molecule becomes stable.

Thus, we can conclude that in $BrF_{3}$, the lone pairs occupy equatorial positions to minimize lone pair-lone pair repulson only.