Chemical Bonding (Class 11): VSEPR, Hybridisation & Molecular Shapes Explained
How to predict the shape of a molecule using VSEPR theory and hybridisation — with a quick reference table for common geometries.
Step 1 — Count electron pairs
Find the number of bond pairs (from bonded atoms) and lone pairs on the central atom. Their total is the steric number, which decides the basic geometry.
Common geometries
| Steric no. | Hybridisation | Shape (no lone pairs) | Example |
|---|---|---|---|
| 2 | sp | Linear | BeCl₂ |
| 3 | sp² | Trigonal planar | BF₃ |
| 4 | sp³ | Tetrahedral | CH₄ |
| 5 | sp³d | Trigonal bipyramidal | PCl₅ |
| 6 | sp³d² | Octahedral | SF₆ |
Step 2 — Adjust for lone pairs
Lone pairs occupy more space than bond pairs, so they bend the shape. For example, H₂O has 4 electron pairs (sp³) but two lone pairs, giving a bent shape (~104.5°), and NH₃ is trigonal pyramidal (~107°).
Common mistakes
- Forgetting to count lone pairs when deciding shape.
- Confusing electron geometry with molecular shape.
- Assuming bond angles are always ideal — lone pairs reduce them.
FAQs
Is VSEPR enough for Class 11 boards?
Yes — VSEPR plus basic hybridisation answers most shape and bond-angle questions.
How do I quickly find hybridisation?
Use steric number: 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d².
Struggling with bonding & shapes?
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