Question

Based on VSEPR theory, the number of $ 90 $ degree $ {\text{F - Br - F}} $ angles in $ Br{F_5} $ are:

Answer 1

Hint: We know that in accordance with VSEPR theory or Valence Shell Electron Pair Repulsion theory, the geometry of a compound is determined by the repulsions that its valence electron pairs undergo and the repulsion undergone by a molecule is such that the molecule may get distorted while attaining a state that minimizes repulsion and maximizes stability of the molecule.

Complete answer:
 $ Br{F_5} $ is Bromine pentafluoride. It is a fluoride of Bromine and is an interhalogen compound. Bromine has five valence electrons which form covalent bonds which are $ \sigma - bonds, $ with five Fluorine atoms. There is one lone pair that exists. The hybridization of $ Br{F_5} $ is $ s{p^3}{d^2}. $ This implies that there are six hybrid orbitals which are equivalent. One $ 4s, $ three $ 4p, $ and two $ 4d $ orbitals of Bromine atoms are involved in the hybridization. These hybrid orbitals are directed towards the corners of an octahedron. Four of them lie on one plane, one lies above and one lies below the plane.
 $ Br{F_5} $ has $ 5 $ bond pairs and $ 1 $ lone pair. Four of the bond pairs are equatorially aligned on the plane of the central atom, while one is axially aligned. The lone pair is on the opposite side of the axial bond pair. The shape that $ Br{F_5} $ exists in is square pyramidal. Since we already know that lone pair-bond pair repulsion is higher than bond pair-bond pair repulsion, the bond pairs adjacent to the lone pair, which are the four equatorial bond pairs, get distorted to attain a shape of minimum repulsion and maximum stability. The distortion occurs from $ 90^\circ $ to lower angle $ 84.8^\circ $ after the repulsion that they undergo. Hence, based on VSEPR theory, the number of 90 degree $ {\text{F - Br - F}} $ angles in $ Br{F_5} $ are zero.

Note:
The Lewis structure of the electron pairs in $ Br{F_5} $ adopts Octahedral geometry, as it has $ s{p^3}{d^2} $ hybridization. The bond angles between $ {\text{F - Br - F}} $ are $ 90^\circ $ without assuming VSEPR theory. However, it is very important to consider the valence electron repulsions in molecules in accordance with VSEPR as the shape and angles can be different from what was originally proposed.