Understanding Hybridisation of BrF₅: Bromine Pentafluoride

Bromine Pentafluoride (BrF₅) is a colourless and highly reactive liquid. It is used in uranium processing. It also finds application in rocket propellants. Though not a compound students encounter often, it is an excellent example of sp³d² hybridisation. It also helps understand molecular geometry distortion due to lone pairs.

Let’s explore how hybridisation in BrF₅ takes place and why it leads to a square pyramidal shape.

What is the Hybridisation of BrF₅?

The process of hybridisation involves the mixing of atomic orbitals to form new hybrid orbitals that explain the molecule’s bonding and geometry. In BrF₅, bromine forms 5 bonds with fluorine atoms and holds 1 lone pair. There are a total of 6 electron domains around the central atom.

This corresponds to sp³d² hybridisation: a mix of 1 s, 3 p, and 2 d orbitals.

Using the Hybridisation Formula

We can determine the hybridisation using the simple formula:

formula

Step-by-step calculation:

Valence electrons of central atom (Bromine) = 7
Number of monovalent atoms (Fluorine atoms) = 5
Negative charge = 0
Positive charge = 0

formula

Interpretation:

A hybridisation number of 6 corresponds to sp³d² hybridisation.

Breakdown of BrF₅ Hybridisation

Here is a complete breakdown of the hybridisation:

Electronic Configuration of Bromine (Br)

The atomic number of Bromine is 35. Its ground state configuration is as follows:

[Ar] 4s² 3d¹⁰ 4p⁵

Bromine has 7 valence electrons in its ground state (4s² 4p⁵). To form 5 bonds, it must excite one electron to a 4d orbital.

So, the excited state configuration will be as follows:

4s¹ 4p⁴ 4d²

Ground state vs excited state orbital configuration of bromine showing 6 available orbitals

Step 2: Formation of Hybrid Orbitals

Now, six orbitals are ready to hybridise:

1 from 4s
3 from 4p
2 from 4d

Result: 6 equivalent sp³d² hybrid orbitals

Usage:

5 orbitals → Form sigma bonds with 5 fluorine atoms
1 orbital → Contains a lone pair

Diagram showing hybrid orbitals with 5 bonding domains and 1 lone pair

Molecular Geometry of BrF₅

Although 6 regions of electron density suggest an octahedral electron geometry, the presence of one lone pair distorts the shape.

According to VSEPR theory, the lone pair occupies one position, and the remaining five fluorine atoms take the rest.
The resulting molecular geometry becomes square pyramidal, not regular octahedral.

Property	Description
Hybridisation	sp³d²
Electron Geometry	Octahedral
Molecular Shape	Square Pyramidal
Bond Angle	~90°
Central Atom	Bromine
Peripheral Atoms	5 Fluorine atoms
Lone Pairs on Br	1
Bond Type	σ (sigma) bonds
Polarity	Polar molecule

BrF₅ molecule showing square pyramidal geometry with a lone pair at the top

Formal Charge in BrF₅

Let’s confirm its stability by calculating the formal charge:

Bromine (Br):

Valence electrons = 7
Bonding electrons = 10 (5 bonds)
Lone pair electrons = 2

FC = VE - (NBE + 1/2 BE)

Formal Charge = 7 – (2 + ½×10) = 7 – (2 + 5) = 0

Each Fluorine:

Valence electrons = 7
Bonding = 2
Lone pairs = 6

Formal Charge = 7 – (6 + 1) = 0

So, the molecule is neutral and stable.

Summing Up

BrF₅ exhibits sp³d² hybridisation involving six orbitals from the bromine atom. It forms five sigma bonds and contains one lone pair. Due to lone pair-bond pair repulsion, the molecule adopts a square pyramidal shape, not a perfect octahedron.