Noble gas compounds — structure and bonding of XeF₂, XeF₄, XeF₆

Question

Describe the structure and bonding of $\text{XeF}_2$ , $\text{XeF}_4$ , and $\text{XeF}_6$ . Predict their shapes using VSEPR theory and state their hybridisation.

(JEE Main 2023, similar pattern — also important for CBSE boards)

Solution — Step by Step

Xe has 8 valence electrons. In $\text{XeF}_2$ , 2 are used for bonding with F, leaving 3 lone pairs on Xe.

Bond pairs: 2
Lone pairs: 3
Electron geometry: Trigonal bipyramidal ( $sp^3d$ )
Molecular shape: Linear (F atoms at the two axial positions, 3 lone pairs in equatorial plane)
Bond angle: 180°

Lone pairs occupy equatorial positions because they experience less repulsion there (only 2 close neighbours at 90° vs 3 in axial positions).

In $\text{XeF}_4$ , 4 electrons bond with F, leaving 2 lone pairs on Xe.

Bond pairs: 4
Lone pairs: 2
Electron geometry: Octahedral ( $sp^3d^2$ )
Molecular shape: Square planar (4 F atoms in a plane, 2 lone pairs on opposite sides — trans to each other)
Bond angle: 90°

The two lone pairs are trans (180° apart) to minimise lp-lp repulsion.

In $\text{XeF}_6$ , 6 electrons bond with F, leaving 1 lone pair.

Bond pairs: 6
Lone pairs: 1
Electron geometry: Pentagonal bipyramidal or capped octahedral ( $sp^3d^3$ )
Molecular shape: Distorted octahedral (not perfectly symmetric due to the lone pair)

The lone pair has no fixed position — it is stereochemically active and distorts the regular octahedral shape.

Summary Table

Compound	Bond pairs	Lone pairs	Hybridisation	Shape
$\text{XeF}_2$	2	3	$sp^3d$	Linear
$\text{XeF}_4$	4	2	$sp^3d^2$	Square planar
$\text{XeF}_6$	6	1	$sp^3d^3$	Distorted octahedral

Why This Works

Xenon can form compounds because it has a large enough atomic size and low enough ionisation energy (especially among heavier noble gases) to share electrons with highly electronegative fluorine. The expanded octet is possible because Xe has vacant $5d$ orbitals that participate in bonding.

The shapes follow directly from VSEPR theory: lone pairs always occupy positions that minimise repulsion. In $\text{XeF}_2$ , the 3 lone pairs go to the equatorial plane of a trigonal bipyramid. In $\text{XeF}_4$ , the 2 lone pairs go trans in an octahedron.

Alternative Method

Using the 3-centre 4-electron (3c-4e) bond model, $\text{XeF}_2$ can be explained without $d$ -orbital participation. Each F-Xe-F unit uses one $p$ orbital of Xe and one $p$ orbital from each F, creating a bonding, non-bonding, and antibonding MO set. This explains the linear geometry naturally.

For JEE, the three Xe fluorides are always compared in a table. Memorise: $\text{XeF}_2$ (linear), $\text{XeF}_4$ (square planar), $\text{XeF}_6$ (distorted octahedral). Also remember the Xe oxide: $\text{XeO}_3$ is pyramidal ( $sp^3$ , 1 lone pair) — a common follow-up question.

Common Mistake

Students often predict $\text{XeF}_4$ as tetrahedral (like $\text{CH}_4$ ). This would be correct only if there were no lone pairs on Xe. With 2 lone pairs, the shape is square planar, not tetrahedral. Always count lone pairs on the central atom before predicting shape — this is the step most students skip.