By Kokil Jain, Chemistry faculty at Pramana Institute (IIT Kanpur, 14 years teaching)
Chemical bonding is one of the most frequent Chemistry chapters in JEE Main, and NEET asks a shape or hybridisation question almost every year. Drawing Lewis structures for XeF₄ or I₃⁻ in the exam takes time and invites mistakes. One formula gives you the hybridisation, the number of lone pairs and the shape, without drawing anything.
The formula
Steric number (SN) = ½ [V + M − C + A]
- V = valence electrons of the central atom
- M = number of monovalent atoms attached (H, F, Cl, Br, I)
- C = charge on a cation, A = charge on an anion
- Divalent atoms such as O and S add nothing, so leave them out.
Then lone pairs on the central atom = SN − number of atoms bonded to it.
| SN | Hybridisation | 0 lone pairs | 1 lone pair | 2 lone pairs | 3 lone pairs |
|---|---|---|---|---|---|
| 2 | sp | Linear | |||
| 3 | sp² | Trigonal planar | Bent | ||
| 4 | sp³ | Tetrahedral | Pyramidal | Bent | |
| 5 | sp³d | Trigonal bipyramidal | See-saw | T-shaped | Linear |
| 6 | sp³d² | Octahedral | Square pyramidal | Square planar | |
| 7 | sp³d³ | Pentagonal bipyramidal (IF₇) | Distorted octahedral (XeF₆) |
Worked examples
| Species | Working | SN | Hybridisation | Lone pairs | Shape |
|---|---|---|---|---|---|
| NH₃ | (5 + 3)/2 | 4 | sp³ | 1 | Pyramidal |
| NH₄⁺ | (5 + 4 − 1)/2 | 4 | sp³ | 0 | Tetrahedral |
| CO₂ | (4 + 0)/2 | 2 | sp | 0 | Linear |
| SO₄²⁻ | (6 + 0 + 2)/2 | 4 | sp³ | 0 | Tetrahedral |
| PCl₅ | (5 + 5)/2 | 5 | sp³d | 0 | Trigonal bipyramidal |
| SF₄ | (6 + 4)/2 | 5 | sp³d | 1 | See-saw |
| ClF₃ | (7 + 3)/2 | 5 | sp³d | 2 | T-shaped |
| I₃⁻ | (7 + 2 + 1)/2 | 5 | sp³d | 3 | Linear |
| XeF₄ | (8 + 4)/2 | 6 | sp³d² | 2 | Square planar |
| XeF₆ | (8 + 6)/2 | 7 | sp³d³ | 1 | Distorted octahedral |
In I₃⁻, the central I is bonded to two I atoms, which count as monovalent, and the anion charge adds 1.
When the trick does not apply
- Coordination compounds: use valence bond theory or crystal field theory, not this formula.
- Atoms that aren't the central atom: the formula gives the hybridisation of the central atom only. For carbon chains in organic molecules, count σ bonds and lone pairs on each carbon instead.
FAQs
Why don't oxygen atoms count in the formula? The two electrons the central atom shares with a doubly bonded O are already counted in V, and together they make one electron domain (the σ part of the double bond). So O needs no extra term.
Is this method accepted in board exams? Board answers expect the hybridisation and shape, which this method gives. Show the Lewis structure if the question asks for it.
Which shapes are asked most? See-saw (SF₄), T-shaped (ClF₃), square planar (XeF₄) and linear I₃⁻ come up repeatedly.
Practise bonding with weekly tests in a batch of at most 40 at Pramana, Geeta Bhawan Square, Indore. See programs and fees or book free counselling.
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