What will be the most stable and acceptable structure of BrF3

Question

In the structure of BrF3 as my teacher said the two lone electron pairs of Br are on the equatorial position an two Fluorine axial position and one Fluorine on equatorial position as the steric no is 5 hence structure should be like trigonal bi-pyramidal. But what if we replace the two lone pairs of electron with the two fluorines on the axial position, this will make the angle between the lone pairs highest and hence repulsion between them should be lowest, and as there are two lone pairs on opposite axial sides the repulsion by them to the three bond pair of Br and F in my model should get canceled- ( my teacher said that to explain the structure of XeF4). So what will the most stable structure of BrF3.

I am a HIGH SCHOOL STUDENT and I know VSEPR THEORY

Answer 1

The bromine trifluoride molecule has been identified as T-shaped, and chemistry is an experimental science, so there's that.

The T shape is quite consistent with VSEPR theory once the true distribution of the nonbonding electrons is recognized. When an atom, such as the bromine in this case, has multiple pairs of nonbonding electrons, the orbitals they occupy are not the lobes that are usually drawn. Rather, following the orthogonality rules that apply to all molecular orbitals (also to the orbitals on an isolated atom), the true orbital wavefunctions are orthogonal combinations of the lobes. With two pairs these combinations are just the sum and the difference between the lobe wavefunctions. The difference appears as a pair of oppositely phased components approximating a valence $p$ orbital. The sum wavefunction is a single lobe centered on the nodal plane of the $p$-like difference orbital.

With equatorial lobes on $\ce{BrF3}$, the sum orbital points outwards from the bromine atom and repels the bonds around that atom, distorting the triangular molecule to its observed T shape. Had the lobes been axial instead, the sum orbital would have been centered on the bromine atom rather than pointing outwards, producing a less favorable electron-repulsion configuration.

Answer 2

Have a good read on this topic Lone Pair Repulsion. From the linked article:

In the case that there are five electron groups around a central atom, there are two different types of positions around the central atom: equatorial positions and axial positions. The three equatorial ligands are 120° from one another and are 90° from each of the two axial ligands. The axial positions have three adjacent groups oriented 90° away in space. Axial groups are thus more crowded than the equatorial positions with only two adjacent groups at 90°. The crowding of axial positions results in slight differences in bond distances; crowded axial groups have longer bonds than the less crowded equatorial groups. Lone pairs of electrons generally prefer to occupy equatorial positions rather than axial positions because the lone electron pairs are more repulsive than bonding electron pairs, and thus the lone pairs prefer the less crowded equatorial positions.

In case of $\ce{BrF3}$, there are two lone pairs which according to VSEPR theory occupies the equatorial position. Hence, the molecule is T-Shaped. This is also verified experimentally:

The bromine trifluoride molecule can be similarly described from careful microwave study as a trigonal bipyramid with fluorine atoms at the two apical positions and one equatorial positions (and unshared pairs at the other two equatorial positions), the four atoms thus being coplanar. The two small $\ce{F-Br-F}$ bond angles have the value 86°12.6'; the $\ce{Br-F}$ bond lengths are 1.721 Å (to one F atom) and 1.810 Å (to the other two).

_{Ref.: The Nature of the Chemical Bond and the Structure of Molecules and Crystals: An Introduction to Modern Structural Chemistry, Linus Pauling, Cornell University Press, 1960}