VSEPR Theory — Shape of SF₆ Explained

Question

Using VSEPR theory, predict the shape of SF₆. Determine the hybridization of sulfur, the number of bond pairs and lone pairs, and explain why all six S–F bonds are equivalent. Also compare SF₆ with SF₄ and SCl₂ to illustrate how lone pairs affect shape.

Solution — Step by Step

Step 1: Count Valence Electrons Around Sulfur

Sulfur is the central atom in SF₆.

S is in Group 16: 6 valence electrons Each F is in Group 17: 7 valence electrons, but F only forms 1 bond (it will complete its octet with 3 lone pairs)

Bonds formed by S: We need to find out how many F atoms S bonds to.

Since the formula is SF₆, S forms 6 bonds with 6 F atoms.

Step 2: Check Valence Electrons on S After Bonding

Normal valence electron count for S = 6 After forming 6 bonds: 6 − 6 = 0 lone pairs on S

Wait — but S only has 6 valence electrons and forms 6 bonds (using 6 electrons for bonding, shared). In a Lewis structure:

• S contributes 1 electron to each of the 6 bonds
• S uses all 6 valence electrons for bonding → 0 lone pairs on S

This is an expanded octet — S has 12 electrons around it (6 bonding pairs × 2). This is allowed for Period 3 elements because S has available 3d orbitals.

Step 3: Apply VSEPR

Electron pairs around S:

• Bonding pairs: 6
• Lone pairs: 0
• Total electron pairs: 6

From the VSEPR table:

Electron geometry = Octahedral Molecular geometry = Octahedral (no lone pairs to distort)

Step 4: Hybridization

Total electron pairs around S = 6 → sp³d² hybridization

S uses: 1 × 3s + 3 × 3p + 2 × 3d = six sp³d² hybrid orbitals

Each sp³d² orbital overlaps with a 2p orbital of F to form a σ bond.

Step 5: Why Are All Six S–F Bonds Identical?

In the octahedral arrangement, all six positions are geometrically equivalent — each sp³d² orbital points toward one corner of a regular octahedron. There are no "axial" or "equatorial" distinctions (unlike in trigonal bipyramidal geometry). All six S–F bonds point to corners of a perfect octahedron at 90° to each other.

This contrasts with SF₄ (next section below), where one lone pair breaks the symmetry.

Why This Works — Expanded Octets in Period 3+

SF₆ has 12 electrons around S — violating the octet rule. This is possible because:

1. Sulfur is in Period 3 (n=3 shell) → has 3s, 3p, AND 3d orbitals available
2. The 3d orbitals are close enough in energy to 3s and 3p to participate in hybridization (forming sp³d²)
3. Period 2 elements (C, N, O, F) cannot exceed the octet — they have no available d orbitals in the n=2 shell

Elements that can have expanded octets: P, S, Cl, Br, I, Xe (Period 3 and beyond)

Examples of expanded octets with S:

• SF₂: 2 bonds + 2 lone pairs = 4 pairs → sp³, bent (normal octet)
• SF₄: 4 bonds + 1 lone pair = 5 pairs → sp³d, see-saw (expanded)
• SF₆: 6 bonds + 0 lone pairs = 6 pairs → sp³d², octahedral (expanded)

Alternative Method — Comparing SF₆, SF₄, SCl₂

This comparison is a JEE favourite. Let's see how adding lone pairs changes everything.

SCl₂ (S with 2 bonds): S in SCl₂: 2 bonds + 2 lone pairs = 4 electron pairs → sp³ → bent shape, ~103° bond angle

SF₄ (S with 4 bonds): S in SF₄: 4 bonds + 1 lone pair = 5 electron pairs → sp³d → see-saw shape The lone pair occupies an equatorial position (equatorial positions have less repulsion from adjacent orbitals). The axial F–S–F angle is 173° (compressed from 180° by the equatorial lone pair); equatorial F–S–F angle is 101° (compressed from 120°).

SF₆ (S with 6 bonds): S in SF₆: 6 bonds + 0 lone pairs = 6 electron pairs → sp³d² → perfect octahedral No lone pairs → no distortion → all angles exactly 90°.

The progressive removal of lone pairs (as we go from SCl₂ to SF₄ to SF₆) "upgrades" both the hybridization and the symmetry of the molecule.

Common Mistake