Deduce the structure of BrF5 on the basis of VSEPR Theory?
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Solution
Part 1: VSEPR theory (Valence shell electron pair repulsion):
This theory helps us to predict the shape of molecules from the number of electron pairs and lone pairs of electrons that surrounds the central atoms.
According to this theory, all the valence shell electrons surrounding the central atom arrange themselves in such a way so as to be as far away from each other as possible to avoid repulsion.
Part 2: Structure of BrF5
The number of valence electrons in BrF5 molecule is:
7 + 5 × 7 = 42
The distribution of these electrons is given below in the diagram:
In the BrF5 molecule, bromine is the central atom, and its atomic number is 35.
It has seven valence electrons in its outermost shell.
The ground state electronic configuration of bromine is: [Ar] 4s2 3d10 4p5.
It also has an empty d orbital which can increase its covalency to form five bonds.
The two electrons from its valence shell are excited to the empty d orbital, due to this bromine undergoes a hybridization of sp3d2.
In BrF5, one 4s, three 4p and two 4d orbitals take part in hybridization.
The central atom bromine forms 5 sigma bonds with fluorine atoms.
There are six electron pairs around the central atom, out of which five are bond pairs and one is lone pair.
According to VSEPR theory, the geometry of the molecule is octahedral, but the shape is square pyramidal.
