QCC Notes
CLASS 11 · CHEMISTRYJEE MAIN × NEETहिंदी
§ 4.6NCERT Class 11 · Chemistry · Chapter 4

Hybridisation: sp, sp², sp³ & Molecular Shapes

Carbon’s ground state offers two unpaired electrons — yet methane makes four identical bonds. The reconciliation is hybridisation: mix the orbitals first, bond afterwards. Five types, one steric-number shortcut, and the s-character thread that ties angles, bond lengths and acidity together.

QCC Notes · Class 11 Chemistry NCERT-aligned Updated 26 Sep 2026 14 min read
01

What is Hybridisation? — Complete Theory

The paradox that forced the idea: carbon’s ground-state configuration is 1s² 2s² 2p² — only two unpaired electrons, so valence bond theory predicts CH₂. Methane has four identical C–H bonds. Promoting a 2s electron to 2p⁴ gives four unpaired electrons, but they would be unequal — three p-bonds and one s-bond of different energies. Neither picture survives contact with experiment. The fix (Pauling, 1931): before bonding, the atom’s valence orbitals intermix to form a new set of equivalent hybrid orbitals — identical in shape and energy, oriented for maximum separation, and better at overlap than any pure orbital.

Salient features. (1) The number of hybrid orbitals formed equals the number of atomic orbitals mixed. (2) Hybrid orbitals are identical in energy and shape. (3) They orient to maximise mutual repulsion distance — the same crowd logic as VSEPR (§ 4.4), which is why the two theories agree. (4) Hybrid orbitals overlap more effectively than pure orbitals → stronger, shorter bonds. (5) Hybridisation is a mathematical operation, not a physical event — no orbitals literally collide; it is a bookkeeping of wave functions that better matches the bonding.

Conditions. The orbitals hybridising must (i) belong to the valence shell of the same atom, and (ii) have comparable energy — 2s mixes with 2p but not with 3d while 2p seats exist. Crucially, promotion is not essential: filled orbitals can hybridise, which is exactly how NH₃’s nitrogen (2s² 2p³, nothing promoted) forms four sp³ hybrids with one holding the lone pair.

Table 1 — The five hybridisations: mix, geometry, angle, examples
TypeMixingHybrid orbitalsGeometryAngleExamples
sp1 s + 1 p2Linear180°BeCl₂, C₂H₂, CO₂
sp²1 s + 2 p3Trigonal planar120°BF₃, C₂H₄, SO₂* , NO₃⁻*
sp³1 s + 3 p4Tetrahedral109.5°CH₄, NH₃*, H₂O*, NH₄⁺
sp³d1 s + 3 p + 1 d5Trigonal bipyramidal120°, 90°PCl₅, SF₄*, ClF₃*
sp³d²1 s + 3 p + 2 d6Octahedral90°SF₆, XeF₄*, IF₅*

* starred examples carry lone pairs — hybridisation follows the steric number, and lone pairs occupy hybrid orbitals too.

The shortcut — steric number. For any central atom: σ-bonds + lone pairs = hybridisation index (multiple bonds count one σ each). NH₃: 3 σ + 1 lp = 4 → sp³. H₂O: 2 + 2 = 4 → sp³. BF₃: 3 + 0 = 3 → sp². PCl₅: 5 + 0 = 5 → sp³d. This is the same crowd count as VSEPR’s steric number — VSEPR tells you the shape, hybridisation tells you which orbitals make it possible; they are two languages for one geometry.

s-character — the hidden dial. Mixing an s orbital (spherical, close to nucleus) with p orbitals distributes that s-content across the hybrids: sp = 50% s, sp² = 33.3%, sp³ = 25%. More s-character pulls the hybrid orbital’s electrons closer to the nucleus, so the orbital is shorter, more electronegative, and overlaps harder. Consequences you can measure: C–H bonds shorten and strengthen along ethane (sp³) → ethene (sp²) → ethyne (sp); bond angles widen along the same line (109.5° → 120° → 180°); and terminal alkyne hydrogens turn acidic because the sp-carbon’s anion holds its lone pair in a 50%-s orbital. One dial, three exam topics.

02

Visualising sp³ & Solving Any Atom

Ek mixing diagram, ek solver — watch carbon’s four orbitals merge first, then solve any steric number live.

sp³ hybridisation — from ground-state carbon to four equivalent hybrids FIG. 1 — BUILDING sp³: 1 s + 3 p → 4 EQUIVALENT HYBRIDS GROUND STATE 1s² 2s² 2p² 1s ↑↓ 2s ↑↓ 2p ↑ 2p ↑ 2p (empty) 2 unpaired → CH₂? promote PROMOTED STATE 2s¹ 2p³ — 4 unpaired 1s ↑↓ 2s ↑ 2p ↑ 2p ↑ 2p (empty) 2p (empty) 4 unpaired — but unequal mix C 4 sp³ hybrids identical · 25% s · 109.5° hybridisation is bookkeeping of wave functions — not a physical event
FIG. 1 — Carbon’s rescue in three frames: ground state offers 2 unpaired electrons; promotion offers 4 but unequal; hybridisation merges the four into equivalent sp³ hybrids — identical energy, tetrahedral 109.5°, each 25% s-character. The promotion costs energy; four strong bonds repay it with interest.
Try it live

Hybridisation Solver

Hybridisationsp³
GeometryTetrahedral
Ideal angle109.5°
s-character25%
50%sp
33.3%sp²
25%sp³

Solver accepts a steric number directly or a formula — σ-bonds + lone pairs are computed for you (multiple bonds count one σ; ionic charges handled). The s-character strip highlights the active row and tracks the electronegativity/angle consequences.

03

Solved Examples (Step-by-Step)

σ-count + lp → steric number → hybridisation. Jo chain yahan chalti hai, wahi solver me live chalti hai.

EXAMPLE 01Foundation · NEET

The tetrahedral quartet: CH₄, NH₃, H₂O, NH₄⁺

Determine the hybridisation of the central atom in CH₄, NH₃, H₂O and NH₄⁺, and relate each to its shape.

  1. CH₄4 σ + 0 lp = 4 → sp³, tetrahedral 109.5°.
  2. NH₃3 σ + 1 lp = 4 → sp³ — one hybrid holds the lp → trigonal pyramidal 107°.
  3. H₂O2 σ + 2 lp = 4 → sp³ — two lp hybrids → bent 104.5°.
  4. NH₄⁺4 σ (dative counts) + 0 lp = 4 → sp³, perfect tetrahedral — the lone pair became a bond, the angle springs back ✓

All four sp³ — hybridisation fixed, shape varies by lp count

EXAMPLE 02JEE Main · s-character chain

Ethane → ethene → ethyne: the C–H bond shrinks

Explain why the C–H bond length decreases and bond strength increases from C₂H₆ (sp³) to C₂H₄ (sp²) to C₂H₂ (sp).

  1. s-contentssp³ = 25% s · sp² = 33.3% s · sp = 50% s.
  2. EffectMore s-character → orbital closer to nucleus → shorter, more electronegative, harder-overlapping hybrid.
  3. ResultBond length: C–H(sp³) > C–H(sp²) > C–H(sp); strength exactly reversed.
  4. BonusSame dial explains why terminal alkyne H is acidic — the sp-anion’s lone pair sits in a 50%-s orbital, stabilised.

25% → 33.3% → 50% s-char — bond shortens and strengthens

EXAMPLE 03JEE Main · Hypervalent seats

PCl₅ and SF₆: where the extra orbitals come from

Phosphorus has three unpaired electrons, yet forms five bonds; sulfur forms six. Determine the hybridisation in PCl₅ and SF₆ and explain how the extra orbitals appear.

  1. PCl₅5 σ + 0 lp = 5 → sp³d — one empty 3d orbital (same shell, comparable energy) joins 3s + 3p. No promotion needed.
  2. SF₆6 σ + 0 lp = 6 → sp³d² — two 3d seats join. Octahedral, 90°.
  3. ShapesPCl₅: TBP — 3 equatorial P–Cl (120°), 2 axial (90°); axial bonds longer (more 90° repulsion).
  4. Trap checkNo electron promotion is invoked — empty comparable-energy orbitals participate directly ✓

PCl₅ sp³d · SF₆ sp³d² — empty same-shell d seats join

04

Practice Questions (With Solutions)

Attempt first — options lock after one shot, exactly like the real exam. Then read the working, chahe galti ho ya na ho.

Attempted 0/4 · Correct 0

Q1NEET · Classify the hybrid

The hybridisation of sulphur in SF₄ is:

Solution

  1. S in SF₄: 4 σ + 1 lp = 5 → sp³d (see-saw shape, from § 4.4).
  2. Option A is the four-bond reflex — lone pairs count as hybrid seats too. (B and C as printed: the intended discriminator is lone-pair counting.)

sp³d — steric number 5

Q2JEE Main · s-character link

Which C–H bond is the strongest?

Solution

  1. sp = 50% s-character → orbital closest to nucleus → best overlap → strongest, shortest bond.
  2. D is the “hybridisation is only geometry” misconception — s-character has measurable bond consequences.

(C) sp C–H of ethyne

Q3NEET · Count carefully

The hybridisation of the central atom in SO₄²⁻ is:

Solution

  1. S bonds four O atoms — four σ directions (each S–O linkage counts once for geometry) → steric number 4 → sp³, tetrahedral.
  2. Option C tempts via “S has 6 valence electrons, draws double bonds” — π counting never enters hybridisation; only σ-directions + lone pairs do.

(B) sp³

Q4JEE Main · Conditions

Which of the following is NOT a necessary condition for hybridisation?

Solution

  1. Half-filling is not required — filled orbitals hybridise (NH₃’s 2s² lone-pair hybrid) and even empty ones (PCl₅’s 3d).
  2. A, B and D are genuine conditions/features from NCERT’s list.

(C) Half-filled requirement — false

05

Key Rules & Takeaways

Type card

Eight lines that solve this topic

Hybrids formed = orbitals mixed1s+3p → 4 sp³ — nothing created, nothing destroyed.
Conditions: same atom · valence shell · comparable EPromotion NOT required — filled/empty orbitals can join.
Steric no. = σ + lp = hybrid index2 sp · 3 sp² · 4 sp³ · 5 sp³d · 6 sp³d² — the one-line solver.
sp 180° · sp² 120° · sp³ 109.5°sp³d TBP 120/90 · sp³d² octahedral 90°.
Lone pairs occupy hybrid orbitalsH₂O sp³ with 2 lp hybrids — seats counted, not bonds.
s-character: sp 50% > sp² 33.3% > sp³ 25%More s → shorter/stronger bonds, wider angles, acidier H.
Unhybridised p → π bondssp² carbon keeps 1 p (C₂H₄); sp keeps 2 (C₂H₂’s two π).
sp³d/sp³d² use same-shell empty dPCl₅ (3d¹) · SF₆ (3d²) — no promotion, no next-shell d.

Anchor hybrids: BeCl₂ sp · BF₃ sp² · CH₄/NH₃/H₂O/NH₄⁺ sp³ · PCl₅ sp³d · SF₆ sp³d²  ·  s-characters: 50 · 33.3 · 25%  ·  Angles: 180 · 120 · 109.5°

  1. Hybridisation solves the equivalence paradox — unequal ground-state orbitals cannot make four identical bonds; equivalent hybrids can.
  2. One shortcut runs the topic — σ-bonds + lone pairs = hybridisation index; the same steric number as VSEPR, so the two theories never disagree.
  3. s-character is the topic’s hidden thread — it links bond length, bond strength, bond angle and acidity across sp/sp²/sp³ in one gradient.
  4. π bonds come from leftovers — hybrids build the σ framework; unhybridised p orbitals form every π bond — the double-bond misconception dies here.
06

FAQs

What is hybridisation in chemistry?

Hybridisation is the intermixing of valence atomic orbitals of an atom — of slightly different energies but comparable overall energy — to form a new set of equivalent hybrid orbitals, identical in shape and energy. The number of hybrid orbitals formed equals the number of atomic orbitals mixed, and hybrid orbitals overlap more effectively, forming stronger bonds. Hybridisation is a mathematical rearrangement, not a physical event.

What are the conditions required for hybridisation?

The orbitals being hybridised must belong to the valence shell of the same atom; they should have comparable energy — an ns orbital can mix with np but not with 3d while 2p is available; and promotion of electrons is not essential — hybridisation of an empty orbital, a half-filled or even a fully-filled orbital can participate, as in NH3 where a filled orbital hybridises.

How do you find the hybridisation of a central atom quickly?

Count the steric number: sigma bonds plus lone pairs on the central atom (a multiple bond counts as one sigma). Steric number 2 gives sp, 3 gives sp2, 4 gives sp3, 5 gives sp3d and 6 gives sp3d2. For example, NH3 has three sigma bonds and one lone pair — steric number 4 — so nitrogen is sp3 hybridised.

Why is the C–H bond in ethyne (sp carbon) stronger and shorter than in ethane (sp3 carbon)?

sp hybrid orbitals contain 50% s-character, sp2 have 33% and sp3 only 25%. Greater s-character pulls electrons closer to the nucleus, making the hybrid orbital shorter, more electronegative and more effective at overlap. Hence the sp C–H bond in ethyne is stronger and shorter than the sp2 bond in ethene, which is stronger and shorter than the sp3 bond in ethane — and the acidity of terminal alkynes follows the same s-character logic.

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