Hybridisation and Molecular Shapes
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Direct answer
Count regions of electron density around the central atom — sigma bonds plus lone pairs, with each multiple bond counting as one region — and both hybridisation and shape follow: two regions sp and linear 180°, three sp2 and trigonal planar 120°, four sp3 and tetrahedral 109.5°, five sp3d and trigonal bipyramidal, six sp3d2 and octahedral, seven sp3d3 as in IF7. Molecular shape ignores lone pairs while electron-pair geometry includes them: water's four sp3 pairs give a bent 104.5° molecule, ammonia's give pyramidal 107°, squeezed below 109.5° because lone pair-bond pair repulsion exceeds bond pair-bond pair. In trigonal bipyramids lone pairs always sit equatorial, producing the exam's beloved distorted shapes: see-saw SF4, T-shaped ClF3, linear XeF2.
What you must remember
- Steric number rule: hybridisation = number of sigma bonds + number of lone pairs on the central atom; pi bonds never add regions, so CO2's two double bonds still mean sp.
- Reference molecules: sp — BeCl2, CO2, C2H2; sp2 — BF3, SO3, C2H4, NO3−; sp3 — CH4, NH3 (pyramidal, 107°), H2O (bent, 104.5°), NH4+; sp3d — PCl5; sp3d2 — SF6; sp3d3 — IF7 (72° and 90° angles).
- Repulsion order: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair — the reason angles shrink from 109.5° (CH4) to 107° (NH3) to 104.5° (H2O).
- TBP lone-pair rule: lone pairs occupy equatorial positions (fewer 90° interactions), giving SF4 see-saw, ClF3 T-shaped (two equatorial lone pairs) and XeF2 linear (three equatorial lone pairs).
- Octahedral derivatives: XeF4 has two lone pairs trans to each other — square planar; BrF5 one lone pair — square pyramidal.
- PCl5 asymmetry: axial bonds (240 pm) are longer and weaker than equatorial (202 pm) in the solid and vapour — NCERT's quoted numbers.
- I3− geometry: two bond pairs plus three equatorial lone pairs on the central iodine — linear anion, a standard surprise question.
From count to shape: xenon tetrafluoride and friends
Take XeF4. Total valence electrons: 8 + 4 × 7 = 36. Four Xe–F sigma bonds consume 8, each fluorine holds three lone pairs (24), leaving 4 on xenon — two lone pairs. Steric number 4 + 2 = 6: sp3d2, octahedral electron-pair geometry. The two lone pairs take opposite (trans) positions to minimise their mutual repulsion, and the molecular shape is square planar, all F–Xe–F angles 90° in one plane. Contrast BrF5, one lone pair short: square pyramidal, with the lone pair crowding the apex.
Now SF4. Six plus four sevens is 34 electrons; four bonds and one lone pair make steric number 5 — trigonal bipyramidal electron geometry. The lone pair claims an equatorial seat, and the molecule becomes a see-saw. ClF3 goes one further: three bonds plus two lone pairs, both equatorial, leaving a T-shape. Notice the pattern — every extra equatorial lone pair shaves a bond position off the equatorial plane while the two axial bonds persist. That single diagram, drawn once and remembered, answers half the shape questions NEET asks.
Counting slips that mislabel hybridisation
Carbon dioxide is the gateway error: two double bonds are two regions, sp, linear — candidates who count four bonds write sp3. Graphite's carbons are sp2 and diamond's sp3, a materials-flavoured favourite. The second slip is reporting electron-pair geometry when the question asks molecular shape: water is bent, not tetrahedral; the tetrahedron belongs to its electron pairs. Third, the axial-equatorial distinction in PCl5 — axial bonds are the longer, weaker pair, and interconversion through pseudo-rotation is beyond the syllabus but the length difference is asked. Fourth, hybridisation belongs to the central atom's orbitals, so species like NH4+ (sp3) and I3− (sp3d on the central iodine) need the charge folded into the electron count before any counting begins.
Frequently asked questions
What are the hybridisation and shape of SF4?
sp3d, see-saw — the lone pair occupies an equatorial position of the trigonal bipyramid, distorting the shape.
Why is XeF4 square planar rather than tetrahedral?
Its two lone pairs sit trans in an octahedral sp3d2 arrangement, leaving four bonds in one square plane.
How many regions of electron density does each carbon of ethyne have?
Two — one sigma bond to hydrogen and one triple bond to the other carbon — so each is sp hybridised and linear.
Why do lone pairs take equatorial positions in a trigonal bipyramid?
Equatorial placement keeps lone pairs at 90° to only two bonds instead of three, minimising repulsion.
Which bonds of PCl5 are longer and why?
The axial bonds (240 pm against 202 pm equatorial), because each axial pair suffers more 90° repulsions than an equatorial bond.