Question
Predict the shape of SF₆ (sulphur hexafluoride) using VSEPR theory. What is the hybridisation of sulphur in SF₆?
Solution — Step by Step
VSEPR (Valence Shell Electron Pair Repulsion) theory states: electron pairs in the valence shell of the central atom repel each other and arrange themselves to minimise repulsion. The molecular shape is determined by the positions of the bonding pairs and lone pairs.
The shape we describe is the arrangement of the atoms, not the electron pairs (lone pairs are invisible in a molecular shape description).
Sulphur is the central atom. Fluorine is highly electronegative and forms single bonds with S.
Valence electrons of S = 6. Each F contributes 1 electron to form a bond with S. 6 F atoms → 6 bonding pairs with S.
Total electron pairs around S = 6 bonding pairs + 0 lone pairs on S.
(Sulphur uses all 6 of its valence electrons in bonding: electrons left for lone pairs.)
6 electron pairs around the central atom. To minimise repulsion, 6 pairs arrange themselves in an octahedral geometry — each pair at 90° from its four nearest neighbours and 180° from the pair directly opposite.
Octahedral arrangement gives maximum separation for 6 groups.
Since all 6 electron pairs are bonding pairs (no lone pairs), the molecular shape is the same as the electron pair geometry:
All F–S–F bond angles = 90° (for adjacent F atoms) and 180° (for opposite F atoms).
6 bonding pairs require 6 hybrid orbitals on S.
Hybridisation = (1 s orbital + 3 p orbitals + 2 d orbitals from the 3d subshell of S).
This is possible because sulphur is in Period 3 and has vacant 3d orbitals available — it can expand its octet beyond 8 electrons.
Why This Works
SF₆ is the textbook example of an “expanded octet” compound. Second-period elements (like N, O, F) cannot expand their octet because they have no available d orbitals. Sulphur, being in the third period, has low-energy 3d orbitals that can participate in bonding, allowing it to form 6 bonds.
The regular octahedral geometry of SF₆ (with all bond angles exactly 90° or 180°) arises because all 6 surrounding atoms are identical (all F) and there are no lone pairs to distort the geometry.
JEE Main and CBSE Class 11 regularly test VSEPR shapes. Key shapes from the 6-pair family: octahedral (0 LP), square pyramidal (1 LP, like IF₅), and square planar (2 LP, like XeF₄). The 2 lone pairs in XeF₄ go to axial positions, leaving 4 F in the equatorial plane.
Alternative Method
Using hybridisation directly:
- S has electronic configuration: [Ne] 3s² 3p⁴ 3d⁰
- To form 6 bonds, S promotes electrons to form 6 half-filled orbitals: one 3s, three 3p, and two 3d
- These 6 orbitals hybridise → 6 equivalent orbitals
- Each bonds with one F → regular octahedral geometry
Both approaches give the same answer. VSEPR is easier for predicting shape; hybridisation gives the orbital picture.
Common Mistake
Students sometimes write SF₆ shape as “trigonal bipyramidal” or “square pyramidal” by confusing it with PCl₅ (5 bond pairs, trigonal bipyramidal) or IF₅ (5 bonds + 1 lone pair, square pyramidal). SF₆ has 6 bonding pairs and 0 lone pairs → pure octahedral. Count the bond pairs carefully before selecting the shape.