Step-by-Step Guide to Determining the Hybridization and Shape of SF4
The Hybridization Of SF4 is a classic example favored in JEE Main Chemistry due to its non-standard geometry and the influence of lone pairs on molecular shape. Understanding why SF4 adopts a see-saw geometry (not tetrahedral or square planar) requires a firm grasp of sp3d hybridization, the VSEPR principle, and the behavior of hypervalent molecules like sulfur fluorides. In this page, we cover the calculation, diagrams, and application-based examples for SF4 hybridization, in sync with JEE Main syllabus and exam trends.
Hybridization Of SF4: Basic Overview and Significance
Sulphur tetrafluoride (SF4) is a hypervalent molecule where sulfur exceeds the octet by using d-orbitals for bonding. The unique properties, reactivity, and geometry of SF4 feature regularly in JEE Chemistry, especially in questions from p-block elements and chemical bonding. Recognising its molecular structure, hybridization, and bond angles is essential for scoring well in objective and subjective questions.
Electronic Configuration & Lewis Structure of SF4
Start by identifying the number of valence electrons and then drawing the Lewis structure for SF4, as these steps reveal the distribution of bonds and lone pairs around sulfur:
- Sulfur (S) valence electrons: 6 (from 3s23p4)
- Each Fluorine (F) atom: 7 valence electrons (five F atoms × 7 = 35)
- But only four F atoms are present, so total = 6 (S) + 4×7 (F) = 34 electrons
- In Lewis structure, sulfur forms four single S–F bonds (8 electrons) and retains one lone pair (2 electrons)
The central S atom has four bonding pairs and one lone pair. This will impact both hybridization and geometry, a point commonly tested in JEE Main theoretical MCQs.
Hybridization Calculation in SF4: Step-wise Derivation
To determine the hybridization for SF4, use the following logical approach, which matches the formula-based questions found in JEE:
- Step 1: Calculate total electron pairs (regions of electron density) around S.
- Step 2: Number of bonded atoms (F): 4, Number of lone pairs on central S: 1
- Step 3 (Formula): Hybridization number = ½ [V + M – C + A], where V = group number for S (6), M = monovalent atoms (4), C = cationic charge (0), A = anionic charge (0)
- So, Hybridization number = ½ [6 + 4] = 5
- Step 4: Number 5 → sp3d hybridization
- This corresponds to five orbitals mixed: one s, three p, and one d (sp3d)
Therefore, in SF4 the sulfur atom is sp3d hybridized. The five orbitals arrange to minimize repulsion, setting the stage for its geometry. For more on hybrid orbital details, see hybridization concepts.
Shape & Geometry of SF4: VSEPR Theory Explained
Although the electronic geometry is trigonal bipyramidal (by VSEPR theory), the actual molecular shape is see-saw due to the arrangement of bonding and lone pairs. Here's the critical insight for JEE:
- The five electron domains (4 bonds + 1 lone pair) arrange trigonal bipyramidal geometry
- Lone pair occupies an equatorial position to minimize repulsion
- Molecular shape becomes see-saw; NOT tetrahedral (due to 5, not 4, regions)
- Bond angles are distorted: axial F–S–F ≈ 173°, equatorial F–S–F ≈ 102°
- Presence of lone pair compresses angles below ideal values
| Molecule | Hybridization | Shape | Lone Pairs | Bond Angles |
|---|---|---|---|---|
| SF4 | sp3d | see-saw | 1 | ~102°, ~173° |
| SF6 | sp3d2 | octahedral | 0 | 90°, 180° |
| SF2 | sp3 | bent | 2 | ~98° |
| XeF4 | sp3d2 | square planar | 2 | 90°, 180° |
This approach leverages VSEPR theory to refine the basic hybridization result, a must for accurate reasoning in exam settings. For more geometry illustrations and principles, refer to chemical bonding MCQ practice and group 16 elements overview.
Comparison with Related Molecules: SF2, SF6, XeF4, ClF3
To deepen your understanding of SF4 hybridization, compare it with other hypervalent and related molecules, as commonly requested in JEE reasoning or assertion-reason type questions:
- SF6: sp3d2, octahedral (all positions bonded, no lone pairs)
- SF2: sp3, bent (2 bonding pairs, 2 lone pairs)
- XeF4: sp3d2, square planar (4 bonds, 2 lone pairs)
- ClF3: sp3d, T-shaped (3 bonds, 2 lone pairs), same hybridization but different shape due to lone pairs
Practising these contrasts is a proven route to clarity for JEE Main conceptual MCQs. Use the above comparisons and tables for quick mnemonic support.
JEE Main Sample Problems on Hybridization Of SF4
Typical JEE Main problems involve direct hybridization calculation, comparison, or shape identification. Practice with the following examples:
- Calculate the hybridization of the central atom in SF4. Show steps.
- Among SF4, SF2, and SF6, which has a see-saw shape?
- Select the molecule matching square planar geometry: SF4, SF6, XeF4, or ClF3?
- Which hybridization corresponds to five sigma bonds and zero lone pairs?
- In SF4, what positions do the lone pair and F atoms occupy?
Solutions:
- SF4 hybridization is sp3d (total 5 electron domains: 4 bonds + 1 lone pair)
- Only SF4 is see-saw
- XeF4 is square planar
- sp3d: five sigma bonds, zero lone pairs (example: PF5)
- In SF4, the lone pair occupies an equatorial position to reduce repulsion
Summary Table: Quick Revision for Hybridization Of SF4
| Attribute | SF4 |
|---|---|
| Central atom | Sulfur (S) |
| Hybridization type | sp3d |
| Lone pairs on S | 1 |
| Bonded atoms | 4 (F) |
| Geometry (VSEPR) | Trigonal bipyramidal (electron), See-saw (molecular) |
| Why not tetrahedral? | 5 domains (vs. 4 for tetrahedral) |
| Key comparison | ClF3 (T-shaped), SF6 (octahedral), SF2 (bent) |
Refer to this quick revision table before JEE Main MCQ or assertion-reason hybrids involving the Hybridization Of SF4.
Further Practice & Related JEE Main Links
Explore related problems, structural questions, and JEE-relevant summaries at these trusted resources:
- Hybridization Of SF4 – Practice Page
- Types of Hybridization in Chemistry
- Hybridization of ClF3 Explained
- Chemical Bonding & Structure Test Practice
- P-block Elements Practice Questions
- Group 16 Elements – Sulfur Compounds
- Hybridization of XeF4 Compared
- P-Block Elements Mock Test
- Hybridization of SF6 – Full Details
- Hybridization of BF3
- JEE Main Chemistry – Topic Practice
- P-block Elements – Revision Notes
- D and F Block Elements – Practice
- Hybridisation of PH3 Analysed
- Group 17 Elements: Halogens
- Organic Compound Characterisation
Mastering the Hybridization Of SF4 builds logic for p-block and coordination chemistry, and Vedantu’s resources offer more targeted preparation in line with JEE Main demands.
FAQs on Hybridization of SF4: Concept, Diagrams & Examples
1. What is the hybridization of SF4?
SF4 exhibits sp³d hybridization due to the arrangement of five electron pairs (four bonding pairs and one lone pair) around the central sulphur atom. This hybridization results in a trigonal bipyramidal geometry. Key points include:
- Sulphur forms four sigma bonds with fluorine and retains one lone pair.
- Total steric number = 5 —> sp³d hybridization.
- The shape becomes see-saw due to one lone pair's influence on molecular geometry.
2. Why is SF4 not tetrahedral in shape?
SF4 is not tetrahedral because it has five electron pairs (four bonding, one lone pair) which require a trigonal bipyramidal electron geometry. Due to the lone pair, the actual molecular shape is see-saw rather than tetrahedral. Summary:
- A tetrahedral shape requires only four bonding pairs and no lone pairs.
- Lone pair repulsion alters the ideal shape.
- VSEPR theory predicts see-saw geometry for SF4.
3. How do you find the shape of SF4?
To find the shape of SF4, count the valence electron pairs around the sulphur atom (steric number = 5), then apply VSEPR theory:
- Draw Lewis structure: 4 bonding pairs (S-F) and 1 lone pair.
- Steric number 5 —> Trigonal bipyramidal electron geometry.
- One position is a lone pair, so the molecular geometry is see-saw shaped.
4. What is sp3d hybridization?
sp³d hybridization is a mixing of one s, three p, and one d orbital of the central atom to form five equivalent hybrid orbitals. These arrange in a trigonal bipyramidal geometry.
- Example: SF4, PCl5
- Bond angles — 90°, 120° (ideal case)
5. What is the lone pair arrangement in SF4?
In SF4, the lone pair occupies an equatorial position in the trigonal bipyramidal geometry to minimize repulsion. This causes:
- Four S–F bonds remain around sulphur.
- Lone pair is in equatorial site → less crowded, less repulsion with bonding pairs.
6. How is SF4 different from SF6 in hybridization?
SF4 has sp³d hybridization with a see-saw shape, while SF6 has sp³d² hybridization and forms an octahedral structure. Comparison:
- SF4: 4 bonding pairs, 1 lone pair (sp³d, see-saw)
- SF6: 6 bonding pairs, 0 lone pairs (sp³d², octahedral)
7. Can SF4 exist with two lone pairs on the central atom?
No, SF4 cannot exist with two lone pairs on sulphur because sulphur in SF4 forms four S–F bonds, using up eight of its valence electrons, leaving only one lone pair. Two lone pairs would reduce bonding to three fluorines, which is not SF4.
8. Is sp4 a valid type of hybridization?
No, sp4 is not a valid or recognized hybridization. The accepted types include sp, sp2, sp3, sp3d, and sp3d2. Each type corresponds to the number and nature of atomic orbitals mixed.
9. Does SF4 have equal bond angles?
No, SF4 does not have equal bond angles. Due to its see-saw shape and the presence of a lone pair, SF4 has two sets of S–F bond angles:
- Axial–equatorial bonds: ∼89.5°
- Equatorial–equatorial bonds: ∼101°
- This distortion is caused by lone pair–bond pair repulsion.
10. Is the hybridization of SF4 the same as XeF4?
No, the hybridization of SF4 is sp³d, while XeF4 has sp³d² hybridization. SF4 forms a see-saw shape; XeF4 forms a square planar geometry due to two lone pairs on xenon.
