Question
Determine the hybridization and shape of the following molecules: SF₆, IF₅, and XeF₄. Also state the bond angles where applicable.
(JEE Main 2023 — Chemical Bonding)
Solution — Step by Step
For each molecule, we apply the formula: Total electron pairs = (valence electrons of central atom + electrons contributed by surrounding atoms) ÷ 2
| Molecule | Central atom | Valence e⁻ | Bonds | Lone pairs | Total e⁻ pairs |
|---|---|---|---|---|---|
| SF₆ | S | 6 | 6 (with F) | 0 | 6 |
| IF₅ | I | 7 | 5 (with F) | 1 | 6 |
| XeF₄ | Xe | 8 | 4 (with F) | 2 | 6 |
All three have 6 electron pairs — this is the key observation.
Six electron pairs always means sp³d² hybridization. The central atom uses one s, three p, and two d orbitals to form six equivalent hybrid orbitals arranged octahedrally.
This is why all three share the same parent geometry — the octahedron. The shape we observe, however, changes based on how many of those six positions are occupied by lone pairs.
Lone pairs occupy more space than bonding pairs — they push other pairs closer together and distort the geometry.
- SF₆: 6 bond pairs, 0 lone pairs → Octahedral shape, bond angle 90°
- IF₅: 5 bond pairs, 1 lone pair → Square pyramidal shape, bond angle slightly less than 90°
- XeF₄: 4 bond pairs, 2 lone pairs → Square planar shape, bond angle 90°
The two lone pairs in XeF₄ sit opposite each other (axial positions) to minimize repulsion — this is the critical reasoning step.
In an octahedral arrangement, two lone pairs can either be adjacent (90° apart) or opposite (180° apart). Adjacent placement means two strong lone-pair–lone-pair repulsions at 90°. Opposite placement means zero such repulsions.
So XeF₄’s lone pairs go to the two axial positions, leaving four F atoms in the equatorial plane → square planar.
Why This Works
VSEPR theory treats electron pairs — both bonding and non-bonding — as regions of negative charge that repel each other. They arrange themselves to be as far apart as possible, which gives us the parent geometry from the total electron pair count.
The observed shape only counts the atoms, not the lone pairs. Think of it as: hybridization + total pairs → framework; lone pairs → invisible occupants that distort that framework. SF₆ has no invisible occupants, so what you predict is exactly what you see.
For IF₅ and XeF₄, the lone pair(s) occupy specific positions dictated by the rule: lone pairs prefer positions where they experience least repulsion from other lone pairs. In IF₅ with one lone pair, it goes axial. In XeF₄ with two, both go axial (opposite each other), compressing the molecule flat.
Alternative Method
You can use the AXₙEₘ notation directly:
- SF₆ → AX₆E₀ → Octahedral
- IF₅ → AX₅E₁ → Square Pyramidal
- XeF₄ → AX₄E₂ → Square Planar
Here, X = bonding pairs, E = lone pairs. This notation maps directly to VSEPR shape tables and is faster in exams when you’ve memorised the table. Both methods give identical results — use whichever you’re faster at under time pressure.
Common Mistake
Confusing hybridization with shape for XeF₄.
Many students write the shape of XeF₄ as octahedral because sp³d² suggests octahedral geometry. The hybridization is sp³d², and the electron pair geometry is octahedral — but the molecular shape (what you name based on atom positions only) is square planar. In JEE and NEET, the question always asks for the shape of the molecule. Writing “octahedral” costs you a mark here.
Quick check for shape naming: count only the bonds (atoms attached), ignore lone pairs for the shape name. Then check if lone pairs distort the bond angles. For NEET MCQs, if two molecules share the same hybridization, look at lone pair count to distinguish their shapes.