Question

In $\text{ Br}{{\text{F}}_{\text{3}}}$ molecule, the lone pairs of electron occupy an equatorial position 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

Answer 1

In$\text{ Br}{{\text{F}}_{\text{3}}}$, there are three bonding pairs $\text{Br-F}$ and two lone pairs on the bromine atom.to avoid the repulsion, the lone pairs accommodate the two equatorial position, and the bond pair occupies the three-position, such that it reduces the repulsion between the lone pair’s lone pair and well as between the lone pair and bond pair repulsion.

Complete step by step answer:
Let’s first determine the hybridization of$\text{ Br}{{\text{F}}_{\text{3}}}$. The bromine is the central atom and its electronic configuration is as follows;
$\text{1}{{\text{s}}^{\text{2}}}\text{ }\\!\\!~\\!\\!\text{ 2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{ }\\!\\!~\\!\\!\text{ 3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{\text{10}}}\text{4}{{\text{s}}^{\text{2}}}\text{4}{{\text{p}}^{\text{5}}}$
The bromine forms the interhalogen compound with the fluorine. Fluorine can form bonds with the bromine some of the electrons of the bromine are shifted into the $\text{4d}-$ orbit. Fluorine has a higher oxidative capacity and thus it forces the bromine to promote the electron to a higher level.
Now, the bromine uses its d-orbitals to undergo hybridization.
Now, bromine can use the d-orbitals for hybridization.
$\text{ Br }$ has a total of seven electrons in its outermost shell. On bond formation, three of the electrons are involved in $\text{Br}-\text{F}$ bonds and 2 lone pairs are left on the bromine. Here , the value of the hybrid orbital is 5, 3 bond pairs, and 2 lone pairs give rise to $\text{ s}{{\text{p}}^{\text{3}}}\text{d }$hybrid orbitals. The lone pairs also take part in hybridization.
the $\text{ Br}{{\text{F}}_{\text{3}}}$ have molecular geometry of T-shaped or trigonal bipyramidal and have a bond angle of ${{86.2}^{0}}$which is slightly less than the usual ${{90}^{0}}$ bond angle. The structure can be
BrF3 molecular geometry is said to be T-shaped or Trigonal Bipyramidal. The lone pairs can be placed on the axial position or equatorial position as shown below With a bond angle of 86.2o which is slightly smaller than the usual 90°.

Here, in bent T-shaped geometry, both lone pairs acquire the equatorial position. At the equatorial position, the lone pair experiences the least repulsion. The number of interactions is as follows;
Lone pair-lone pair repulsion is 0.
Lone pair-bond pair repulsion is 4.
Bond pair-bond pair repulsion is 2.
This, stablises the $\text{ Br}{{\text{F}}_{\text{3}}}$molecule.
Therefore, in $\text{ Br}{{\text{F}}_{\text{3}}}$ a molecule, the lone pairs of electron occupy an equatorial position to minimize lone pair-lone pair repulsion and lone pair-bond pair repulsion.

Hence, (C) is the correct option.

Note: At first glance, $\text{ Br}{{\text{F}}_{\text{3}}}$ may look like a trigonal planar structure. But the lone pair on the bromine makes its trigonal bipyramidal. Do not forget that lone pairs near to the atom thus have a higher repulsive nature and distort the geometry.