How to Find the Hybridization and Molecular Geometry of SF6?
Understanding the hybridization of SF6 is crucial for JEE Main aspirants aiming to master chemical bonding and molecular geometry. Sulfur hexafluoride (SF6) is a classic example of an expanded octet molecule, where the central sulfur atom accommodates more than eight electrons, leading to interesting questions about orbital mixing, shape, and bonding rationale. Its structure, symmetry, and bond angles all feature regularly across competitive and school-level Chemistry exams recognized by standard syllabi.
Hybridization is the process by which atomic orbitals mix to form new, equivalent hybrid orbitals suitable for bonding. To determine hybridization, consider the total number of electron domains (bond pairs + lone pairs) around the central atom. When this number exceeds four, involvement of d orbitals is possible—a key concept for molecules like SF6 that show an expanded octet beyond the usual s and p mixing.
- Hybridization helps predict the molecular shape and geometry accurately.
- The electron domain number (steric number) guides the hybridization type.
- Expanded octet compounds may utilize d orbitals (3d for period 3 and below).
- For main-group elements: 2 domains → sp, 3 → sp2, 4 → sp3, 5 → sp3d, 6 → sp3d2.
- Use Lewis dot diagrams to count bonded and lone pair regions.
- Always check for exceptions to the octet rule in period 3 and heavier elements.
Hybridization of SF6: Stepwise Approach
To find the hybridization of SF6, follow this sequence using electron domain analysis and Lewis structure:
- Write the Lewis structure: Place S at the center, connect six F atoms with single bonds (each S–F bond is a sigma bond).
- Count electron domains: 6 bond pairs, 0 lone pairs on S → total = 6.
- Apply the hybridization formula: domains = number of orbitals mixed → Need six hybrid orbitals.
- For six domains: s, three p, and two d orbitals combine → sp3d2 hybridization.
- Sulfur uses 3s, 3px, 3py, 3pz, 3dz2, 3dx2-y2.
- No lone pairs distort geometry—each hybrid orbital forms a bond with one fluorine atom.
The hybridisation of sulphur in SF6 is thus sp3d2, ensuring six equivalent S–F bonds and symmetrical electron distribution in three dimensions.
Structure, Shape, and Bond Angle of SF6
With sp3d2 hybridization, SF6 has distinct molecular geometry:
- Shape: **Octahedral**
- Bond angle: 90° between adjacent S–F bonds, 180° across.
- All S–F bonds are equivalent in length and strength.
- Highly symmetrical—no net dipole moment (non-polar molecule).
| Parameter | SF6 |
|---|---|
| Central atom | Sulfur (S) |
| Hybridization | sp3d2 |
| Shape/Geometry | Octahedral |
| Bond angle | 90° |
| Number of S–F bonds | 6 (all equivalent) |
| Polarity | Non-polar |
Remember: The perfect symmetry means that vector sum of all S–F bond dipoles cancels, making SF6 a non-polar compound—a frequent MCQ trap in JEE Main.
Comparison Table: SF6, SF4, and PCl5
Hybridization, structure, and bond angles vary with electron domain count. Compare these typical compounds for clear differentiation:
| Compound | Hybridization | Electron Domains | Geometry | Bond Angle |
|---|---|---|---|---|
| SF6 | sp3d2 | 6 (all bonds) | Octahedral | 90° |
| SF4 | sp3d | 5 (4 bonds + 1 lone pair) | Seesaw | <100°/>90° |
| PCl5 | sp3d | 5 (all bonds) | Trigonal bipyramidal | 120°, 90° |
For a deeper understanding, see Hybridization of SF4 and Hybridization in PCl5.
Exam Tricks and Shortcuts
- Count central atom bonds + lone pairs to determine electron domains (steric number).
- Steric number 6 → always sp3d2 (octahedral) for main-group compounds.
- For period 3 and beyond, sulfur, phosphorus, and similar can expand octet—hence d orbital involvement.
- All bond pairs & no lone pairs → regular geometry, all angles equal.
- Octahedral → 6 equivalent bonds, 90° angles; non-polar if all outer atoms are identical.
- Quick mnemonic: 2 (sp), 3 (sp2), 4 (sp3), 5 (sp3d), 6 (sp3d2).
- Practice with isoelectronic and similar compounds for robust exam prep.
Practice Problems for JEE
- Determine the hybridization of central atom in SF6. Justify your answer.
- State the shape and bond angles of SF6. Draw a neat diagram.
- Compare hybridizations of SF6 and SF4. What causes the difference?
- Predict the polarity of SF6 based on its shape. Explain with reference to bond dipoles.
- A MCQ: "Which species shows sp3d2 hybridization?". Choices: (A) PCl5 (B) SF6 (C) BF3 (D) SO2. Correct answer: (B) SF6.
Applying these principles systematically, as practiced in Vedantu’s JEE Chemistry courses, ensures full marks for hybridization, geometry, and comparative structure questions related to SF6 and other p-block compounds. For related theories, revise VSEPR Theory, Hybridization basics, and Lewis Structures to visualize orbital arrangement confidently.
The hybridization of SF6 underlines key exam concepts—expanded octet, sp3d2 orbital mix, octahedral shape, and bond angle symmetry. Mastery of these features, alongside thoughtful application of exam shortcuts, builds a strong foundation for all advanced bonding and geometry topics in JEE Main Chemistry.
FAQs on Hybridization of SF6 Explained: Stepwise Method, Shape & Bond Angle
1. What is the hybridization in SF6?
Sulfur hexafluoride (SF6) exhibits sp3d2 hybridization. In SF6, the sulfur atom utilizes one s-orbital, three p-orbitals, and two d-orbitals to form six equivalent sp3d2 hybrid orbitals. These arrange in an octahedral geometry to bond with six fluorine atoms.
2. How do you find the hybridization of SF6 step by step?
The hybridization of SF6 can be found using these steps:
- Find the central atom: Sulfur (S).
- Calculate valence electrons of S (Group 16): 6.
- Count bonds with fluorine atoms: 6 sigma bonds
- No lone pairs on S in SF6
- Hybridization = number of sigma bonds + lone pairs = 6 (6+0)
- Corresponds to sp3d2 hybridization, giving an octahedral shape.
3. What is the molecular shape and bond angle of SF6?
SF6 has an octahedral molecular shape. All six S–F bonds are arranged at 90° to each other.
- Shape: Octahedral
- Bond angle: 90°
- Each fluorine atom is placed at the corners of an octahedron, making SF6 highly symmetrical.
4. Why are d orbitals involved in SF6 hybridization?
In SF6, sulfur uses two of its 3d orbitals to accommodate more than eight electrons, allowing for the formation of six bonds (expanded octet).
- Sulfur has an atomic number of 16; its 3d orbitals are available after the 3s and 3p orbitals
- In SF6, two 3d orbitals hybridize with one 3s and three 3p to form six sp3d2 hybrid orbitals
- This enables bonding with six fluorine atoms
5. How is the hybridization of SF4 different from SF6?
SF4 and SF6 have different central atom hybridizations due to a different number of bonded atoms and lone pairs.
- SF4: Sulfur forms 4 sigma bonds and has 1 lone pair; hybridization is sp3d (trigonal bipyramidal, seesaw shape).
- SF6: Sulfur forms 6 sigma bonds and no lone pairs; hybridization is sp3d2 (octahedral shape).
6. What is the similarity and difference between the hybridization of PCl5 and SF6?
Both PCl5 and SF6 exhibit expanded octet hybridizations, but with different geometries.
- PCl5: sp3d hybridization (trigonal bipyramidal shape, 5 bonds)
- SF6: sp3d2 hybridization (octahedral shape, 6 bonds)
- Both use d orbitals for bonding, allowing more than 8 electrons around the central atom
7. Can SF6 have a different shape if lone pairs are present?
In SF6, there are no lone pairs on the central sulfur atom, so the shape remains octahedral. If lone pairs were present, the geometry would deviate from octahedral, but for SF6 the structure is always octahedral due to six bonding pairs and zero lone pairs.
8. Is SF6 an exception to the octet rule?
SF6 is a classic example of an expanded octet exception to the octet rule.
- Sulfur in SF6 shares 12 valence electrons (6 bonds) instead of the usual 8.
- Elements in period 3 and below (like sulfur) can expand their valence shell using available d orbitals.
9. Why is SF6 non-polar despite polar S–F bonds?
SF6 is non-polar because its symmetric octahedral geometry causes all bond dipoles to cancel each other out, even though each S–F bond is polar.
- The molecular symmetry leads to zero net dipole moment
10. Is there a shortcut to find the hybridization of SF6?
A quick trick to determine hybridization is to count the number of sigma bonds plus lone pairs on the central atom.
- If the sum is 6 (like in SF6), hybridization = sp3d2 (octahedral).
- This method works well for molecules with no multiple bonds or resonance.
