Understanding Hybridisation of XeF₆: Xenon Hexafluoride

Xenon Hexafluoride (XeF₆) is a gas that is a highly reactive compound. It consists of one xenon atom covalently bonded to six fluorine atoms. It has a distorted structure. It’s an excellent example of sp³d³ hybridisation in inorganic chemistry.

Let us understand how hybridisation happens in XeF₆. Read on to learn how it leads to its bonding and molecular shape.

What is the Hybridisation of XeF₆?

Xenon Hexafluoride consists of one central xenon atom, which is bonded to fluorine atoms and has one lone pair left. The molecule undergoes sp³d³ hybridisation since xenon has 7 regions of electron density (6 bonding pairs + 1 lone pair).

Breakdown of XeF₆ Hybridisation

XeF₆ is a xenon fluoride, a compound of the noble gas xenon. The lone pair takes up some position in the molecule, leading to repulsion and a bit of distortion from the original octahedral structure.

Here is a complete understanding of its hybridisation.

Electronic Configuration of Xenon

The atomic number of xenon is 54.

The ground state of xenon :

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶

Does not have unpaired electron → insufficient to form six bonds

Excited state configuration:

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s¹ 5p³ 5d³

Now, 6 unpaired electrons → enough to form bonds

And now, space for lone pairs also

Ground state vs excited state orbital diagram

Formation of Hybrid Orbitals

sp³d³ hybridisation occurs when 1 s orbital, 3 p orbitals, and 3 d orbitals mix to form equivalent hybrid orbitals.
The result:
→ 7 sp³d³ hybrid orbitals are formed
→ These accommodate a lone pair and are used to form bonds between atoms.

Bond Formation in Xenon Hexafluoride

The xenon atom uses:

6 sp³d³ hybrid orbitals overlap with the p orbitals in the fluorine atom and form 6 σ bonds.
The remaining 1 sp³d³ hybrid orbital contains lone pairs of electrons.

Result:

6 Xe–F σ bonds
1 lone pair on the Xe atom
Hybridisation type: sp³d³
Bond angle: ~ 180° and ~ 90°
Geometry: Distorted Octahedral

Bonding of Xenon Hexafluoride

Details At A Glance

Property	Details
Molecule	Xenon Hexafluoride (XeF₆)
Hybridisation	sp³d³
Geometry	Distorted Octrahedral
Bond angle	~ 180°, ~ 90°
Bonding	6 σ bonds (Xe–F), 1 lone pair
Unhybridised Orbitals	Most valence orbitals participate in sp³d³ hybridisation; higher orbitals remain unhybridised.
Xenon valency satisfied?	Yes, by forming 6 bonds and holding 1 lone pair

Formal Charge in XeF₆

To determine if the Lewis structure of XeF₆ is stable, we calculate the formal charge on each atom using the formula:

Formal charge = Valence electrons – (Lone pair electrons + ½ × Bonding electrons)

Step-by-step for each atom:

Xenon (Xe) :

Valence electrons: 8
Lone pairs: 1 (2 electrons)
Bonding electrons: 12 (6 Xe–F bonds)

Formal charge = 8 – (2 + ½ × 12) = 8 – (2 + 6) = 8 – (8) = 0

Fluorine (F) – each

Valence electrons: 7
Lone pairs: 3 (6 electrons)
Bonding electrons: 2 (1 single bond with Xe)

Formal charge = 7 – (6 + ½ × 2) = 7 – (6 + 1) = 7 – (7) = 0

Thus, all atoms in XeF₆ carry zero formal charge, which confirms that the Lewis structure is stable and correct.

Summing Up

The xenon in XeF₆ forms: 6 σ bonds and has 1 lone pair. sp³d³ hybridisation leads to distorted octahedral molecular geometry and bond angles < 90°, < 180°. The change in angle is because of the repulsion one lone pair has with bonding pairs.