Hybridization Theory

Master sp, sp2, sp3, sp3d, sp3d2 hybridization, predict molecular geometries, understand orbital mixing for JEE Main & Advanced.

21 min read

Prerequisites

Before studying hybridization, review:

The Hook: The Carbon Chameleon

Connect: One Atom, Three Personalities

Carbon is the ultimate shape-shifter!

In diamond: Each carbon bonds to 4 neighbors → Tetrahedral → Hardest natural material In graphite: Each carbon bonds to 3 neighbors → Planar → Soft enough to write with In acetylene (welding torches): Each carbon bonds to 2 atoms → Linear → Burns at 3300°C!

Same atom. Same number of valence electrons. Completely different geometries!

How? The secret is hybridization — atoms can “remix” their orbitals to create bonds with different shapes and strengths!

In Oppenheimer (2023), graphite was used in nuclear reactors because of its planar sp² structure. Diamond wouldn’t work! Understanding hybridization isn’t just theory — it’s understanding why materials have the properties that make technology possible.

JEE Gold Mine: 3-4 questions every year worth 12-16 marks. Master the formula, and you’ll solve hybridization in 10 seconds flat!

The Core Concept

What is Hybridization?

Hybridization is the mixing of atomic orbitals of slightly different energies to form new hybrid orbitals of equal energy, suitable for bonding.

$$\boxed{\text{Atomic orbitals (different energies)} \xrightarrow{\text{mix}} \text{Hybrid orbitals (equal energy, new shapes)}}$$

In simple terms: Imagine you have different paint colors (s, p, d orbitals). Mix them together, and you get a completely new color (hybrid orbital) that’s uniform and perfect for painting (bonding)!

Why Do Atoms Hybridize?

Problem without hybridization:

Take carbon (ground state): 1s² 2s² 2p² — Carbon should form only 2 bonds (2 unpaired electrons in 2p)!

But we know CH₄ exists with 4 C-H bonds! How?

Solution: Hybridization!

  1. Promote one 2s electron to 2p (requires energy)
  2. Mix one 2s and three 2p orbitals → four sp³ hybrids
  3. Now carbon has 4 half-filled orbitals → can form 4 bonds!
  4. Energy released from forming 4 bonds > energy needed for promotion

Visualizing the Hybridization Process

Orbital Energy Diagram: Hybridization ProcessGround StateCarbon: 1s2 2s2 2p2Energy2pemptypxpypz2senergy gapOnly 2 unpaired e-Can form only 2 bonds!Excite(energy input)Excited StateCarbon: 1s2 2s1 2p32ppxpypz2spromotionNow 4 unpaired e-Can form 4 bonds!Hybridize(mix orbitals)Hybridized (sp3)4 equivalent sp3 orbitalssp3former 2pformer 2sAll orbitals equivalent!Intermediate energy25% s + 75% p characterKey Insight: Why Does Carbon Hybridize?Energy cost of promotion is LESS than energy gained from forming 2 extra bondsCH4 with 4 bonds is more stable than hypothetical CH2 with 2 bondsNet energy: Favorable!

The Magic Formula: Instant Hybridization

The Steric Number Formula

$$\boxed{H = \frac{1}{2}[V + M - C + A]}$$

Where H = Hybridization number (sum of orbitals)

  • V = Valence electrons of central atom
  • M = Number of monovalent atoms (H, Cl, Br, I, etc.)
  • C = Charge if cation (positive charge)
  • A = Charge if anion (negative charge)

Then:

H ValueHybridizationGeometry
2spLinear
3sp²Trigonal planar
4sp³Tetrahedral
5sp³dTrigonal bipyramidal
6sp³d²Octahedral
7sp³d³Pentagonal bipyramidal
JEE Shortcut: 10-Second Hybridization

Don’t draw Lewis structures! Use the formula directly:

Example: NH₃

  • V = 5 (N has 5 valence e⁻)
  • M = 3 (three H atoms)
  • C = 0, A = 0
  • H = ½[5 + 3] = ½[8] = 4 → sp³

Example: SF₆

  • V = 6 (S has 6 valence e⁻)
  • M = 6 (six F atoms)
  • H = ½[6 + 6] = 6 → sp³d²

Takes 5 seconds! Much faster than drawing Lewis structures!

sp Hybridization: The Linear Specialist

sp Hybridization Diagram

sp Hybridization - Linear GeometryBond angle 180 degrees - Examples: BeCl2, C2H2 (Acetylene), CO2Orbital Mixings1 x 2s+-+1 x 2pmix2 x sp (180 degrees apart)++-pz2 x 2p (unhybridized)perpendicular to sp axissp Properties50% s-characterShortest, strongest bondsBeCl2 - Linear Molecule(Only sigma bonds, no lone pairs)ClsigmaBesigmaCl180 degreesH = 1/2[V + M] = 1/2[2 + 2] = 2H = 2 means spAcetylene (C2H2) - Triple Bond(1 sigma + 2 pi bonds between carbons)HCsppi bonds(p-p overlap)sigma(sp-sp)CspHAll atoms in a straight line (180 degrees)Triple Bond = 1 sigma (sp-sp) + 2 pi (p-p)Bond length: 120 pm (shortest C-C)2 unhybridized p orbitals make 2 pi bondssp Hybridization Key Points2 hybrid orbitals, 180 degrees apart (linear)50% s-character (highest among hybrids)Shortest bonds, most acidic C-HExamples: BeCl2, CO2, HCN, C2H2 (alkynes)2 unhybridized p orbitals available for pi bonds

Formation

One s orbital + One p orbital → Two sp hybrid orbitals

Characteristics

  • Number of orbitals: 2
  • Geometry: Linear
  • Bond angle: 180°
  • s-character: 50% (most s-character!)
  • Shape: Two lobes pointing in opposite directions

Examples

BeCl₂ (Beryllium Chloride)

Hybridization:

  • Be: 1s² 2s² → Excite → 1s² 2s¹ 2p¹
  • Mix 2s + 2p → two sp orbitals
  • Each sp orbital overlaps with Cl 3p
  • Result: Linear molecule (Cl-Be-Cl at 180°)

Formula check:

  • V = 2, M = 2, C = 0, A = 0
  • H = ½[2+2] = 2 → sp ✓

C₂H₂ (Acetylene/Ethyne)

Structure: H-C≡C-H

For each carbon:

  • sp hybridization (2 sp orbitals)
  • 1 sp overlaps with H (σ bond)
  • 1 sp overlaps with other C (σ bond)
  • 2 unhybridized p orbitals form 2 π bonds with other C

Result:

  • C-C has 1 σ (sp-sp) + 2 π (p-p) = triple bond
  • Linear geometry (H-C-C-H all in a line)
sp Bonding Pattern

sp hybridization means:

  • 2 σ bonds (from 2 sp orbitals)
  • Remaining unhybridized p orbitals → π bonds
  • Triple bond = sp + 2π bonds
  • Linear geometry (180°)

Key molecules:

  • CO₂: sp on C (O=C=O linear)
  • HCN: sp on C (H-C≡N linear)
  • BeF₂: sp on Be

sp² Hybridization: The Planar Pro

sp2 Hybridization Diagram

sp2 Hybridization - Trigonal Planar GeometryBond angle 120 degrees - Example: BF3, Ethene (C2H4)Orbital Mixing1 x 2s+2 x 2p3 x sp2+1 x 2p (unhybridized)perpendicularto plane33.3% s-character, 66.7% p-characterBF3 - Trigonal Planar(All atoms in same plane)BFFF120 deg120 deg120 degempty pz(perpendicular)Ethene (C2H4) - Double Bond(All 6 atoms in same plane)CCsigma (sp2-sp2)pi bond(p-p overlap)HHHH~120 degPi Bond Formation from Unhybridized p Orbitals+-pz (C1)++-pz (C2)sideways overlapCCpi bond (above and below sigma)Key Points:- Pi bond restricts rotation- Double bond = 1 sigma + 1 pi- Planar molecule (all in same plane)

Formation

One s orbital + Two p orbitals → Three sp² hybrid orbitals

Characteristics

  • Number of orbitals: 3
  • Geometry: Trigonal planar
  • Bond angle: 120°
  • s-character: 33.3%
  • Shape: Three lobes in a plane, 120° apart

Examples

BF₃ (Boron Trifluoride)

Hybridization:

  • B: 1s² 2s² 2p¹ → Excite → 1s² 2s¹ 2p²
  • Mix 2s + 2p + 2p → three sp² orbitals
  • Each sp² overlaps with F 2p
  • Result: Planar molecule (all in same plane, 120°)

Formula check:

  • V = 3, M = 3
  • H = ½[3+3] = 3 → sp² ✓

C₂H₄ (Ethylene/Ethene)

Structure: H₂C=CH₂

For each carbon:

  • sp² hybridization (3 sp² orbitals)
  • 2 sp² overlap with H (σ bonds)
  • 1 sp² overlaps with other C (σ bond)
  • 1 unhybridized p orbital forms π bond with other C

Result:

  • C=C has 1 σ (sp²-sp²) + 1 π (p-p) = double bond
  • Planar geometry (all 6 atoms in same plane)

Benzene (C₆H₆)

Each carbon:

  • sp² hybridization
  • Forms 3 σ bonds (2 with C, 1 with H)
  • 1 unhybridized p orbital → delocalized π system (aromatic ring)

Result: Flat hexagonal ring with delocalized electrons above and below plane

sp² Bonding Pattern

sp² hybridization means:

  • 3 σ bonds (from 3 sp² orbitals)
  • 1 unhybridized p orbital → π bond
  • Double bond = sp² σ + 1π bond
  • Trigonal planar geometry (120°)

Key molecules:

  • C₂H₄: sp² on C (ethylene, planar)
  • CO₃²⁻: sp² on C (carbonate, planar)
  • NO₃⁻: sp² on N (nitrate, planar)
  • Graphite: sp² on C (planar sheets)

sp³ Hybridization: The Tetrahedral Champion

sp3 Hybridization Diagram

sp3 Hybridization - Tetrahedral GeometryExample: Methane (CH4) - Bond angle 109.5 degreesOrbital Mixing1 x 2s+3 x 2p4 x sp325% s-character75% p-characterMethane (CH4) - 3D StructureCHHwedgeHwedgeHdash (away)109.5 degrees109.5 degreesElectronic ConfigurationGround state:2s22p2up-downup upexciteExcited:2s12p3hybridizesp3:upupupupBond Formationsp3 (C)overlap1s (H)=sigmaC-H sigma bondsp3 Key PropertiesBond angle: 109.5 degrees (tetrahedral)s-character: 25% (1 s + 3 p orbitals)Geometry: TetrahedralExamples: CH4, NH3, H2O, CCl4, diamond

Formation

One s orbital + Three p orbitals → Four sp³ hybrid orbitals

Characteristics

  • Number of orbitals: 4
  • Geometry: Tetrahedral
  • Bond angle: 109.5°
  • s-character: 25%
  • Shape: Four lobes pointing to corners of a tetrahedron

Examples

CH₄ (Methane)

Hybridization:

  • C: 1s² 2s² 2p² → Excite → 1s² 2s¹ 2p³
  • Mix 2s + 2p + 2p + 2p → four sp³ orbitals
  • Each sp³ overlaps with H 1s
  • Result: Tetrahedral molecule (all angles 109.5°)

Formula check:

  • V = 4, M = 4
  • H = ½[4+4] = 4 → sp³ ✓

NH₃ (Ammonia)

Hybridization:

  • N: 1s² 2s² 2p³ (already has 3 unpaired + 1 pair)
  • Mix 2s + 2p + 2p + 2p → four sp³ orbitals
  • 3 sp³ overlap with H (σ bonds)
  • 1 sp³ holds lone pair

Result:

  • Electron geometry: Tetrahedral
  • Molecular geometry: Trigonal pyramidal (1 LP)
  • Bond angle: ~107° (LP-BP repulsion reduces from 109.5°)

H₂O (Water)

Hybridization:

  • O: 1s² 2s² 2p⁴ (2 unpaired + 2 pairs)
  • Mix 2s + 2p + 2p + 2p → four sp³ orbitals
  • 2 sp³ overlap with H (σ bonds)
  • 2 sp³ hold lone pairs

Result:

  • Electron geometry: Tetrahedral
  • Molecular geometry: Bent (2 LP)
  • Bond angle: ~104.5° (2 LP-BP repulsions reduce angle)
sp³ with Lone Pairs

sp³ hybridization doesn’t mean tetrahedral shape!

Molecular shape depends on lone pairs:

  • 4 BP, 0 LP: Tetrahedral (CH₄) — 109.5°
  • 3 BP, 1 LP: Trigonal pyramidal (NH₃) — ~107°
  • 2 BP, 2 LP: Bent (H₂O) — ~104.5°

Lone pairs occupy sp³ orbitals too! They just don’t show up in molecular shape.

Diamond (Carbon Network)

Each carbon:

  • sp³ hybridization
  • Forms 4 σ bonds with 4 other C atoms
  • 3D network structure

Result: Extremely strong, rigid structure (hardest natural material!)

sp³d Hybridization: The Five-Fold Master

sp³d Hybridization Diagram

sp3d Hybridization - Trigonal Bipyramidal Geometry5 orbitals - Bond angles: 90 degrees (axial-equatorial), 120 degrees (equatorial) - Example: PCl5Orbital Mixing for sp3ds1 x 3s+3 x 3p+1 x 3d5 x sp3dsp3d Properties20% s-characterRequires Period 3+(needs d-orbitals)PCl5 - Trigonal Bipyramidal(3 equatorial + 2 axial positions)PClaxialClaxial90 degClequatorialClequatorialClequatorial120 deg120 degequatorial planeLone Pairs Go Equatorial First!Why equatorial position?Equatorial: only 2 neighbors at 90 degAxial: 3 neighbors at 90 deg (more repulsion!)Molecular Shapes from sp3d:5 BP, 0 LP:Trigonal Bipyramidal (PCl5)4 BP, 1 LP:See-saw (SF4)3 BP, 2 LP:T-shaped (ClF3)2 BP, 3 LP:Linear (XeF2)SF4 - See-saw Shape(1 LP at equatorial)SFFFFLP (equatorial)Angles distorted from idealsp3d Key Points for JEE5 hybrid orbitals from 1s + 3p + 1d (Period 3+ only)Axial bonds are longer/weaker than equatorial bondsFormula: H = 5 means sp3d hybridizationLP always in equatorial (less 90 deg repulsion)

Formation

One s + Three p + One d orbital → Five sp³d hybrid orbitals

Requirement: Atom must be in Period 3 or higher (needs accessible d-orbitals)

Characteristics

  • Number of orbitals: 5
  • Geometry: Trigonal bipyramidal
  • Bond angles: 90° (axial-equatorial), 120° (equatorial-equatorial)
  • s-character: 20%
  • Shape: 3 orbitals in plane (equatorial), 2 perpendicular (axial)

Examples

PCl₅ (Phosphorus Pentachloride)

Hybridization:

  • P: [Ne] 3s² 3p³ → Excite → 3s¹ 3p³ 3d¹
  • Mix 3s + three 3p + one 3d → five sp³d orbitals
  • Each sp³d overlaps with Cl 3p
  • Result: Trigonal bipyramidal (TBP) structure

Formula check:

  • V = 5, M = 5
  • H = ½[5+5] = 5 → sp³d ✓

Bond lengths:

  • Axial P-Cl bonds: longer (weaker, more repulsion)
  • Equatorial P-Cl bonds: shorter (stronger)

SF₄ (Sulfur Tetrafluoride)

Hybridization:

  • S: 3s² 3p⁴ → Excite → 3s¹ 3p³ 3d¹
  • Five sp³d orbitals
  • 4 sp³d overlap with F (σ bonds)
  • 1 sp³d holds lone pair (at equatorial position!)

Result:

  • Electron geometry: Trigonal bipyramidal
  • Molecular geometry: See-saw (1 LP at equatorial)
sp³d Lone Pair Rule

In sp³d, lone pairs ALWAYS go to equatorial positions first!

Why? Equatorial has only 2 neighbors at 90° (less repulsion), while axial has 3 neighbors at 90° (more repulsion).

Shapes from sp³d:

  • 5 BP, 0 LP: Trigonal bipyramidal (PCl₅)
  • 4 BP, 1 LP: See-saw (SF₄) — LP at equatorial
  • 3 BP, 2 LP: T-shaped (ClF₃) — 2 LP at equatorial
  • 2 BP, 3 LP: Linear (XeF₂) — 3 LP at all equatorial

sp³d² Hybridization: The Octahedral Giant

sp³d² Hybridization Diagram

sp3d2 Hybridization - Octahedral Geometry6 orbitals - All bond angles 90 degrees - Examples: SF6, [Fe(CN)6]4-Orbital Mixing for sp3d2s1 x 3s+3 x 3p+2 x 3d6 x sp3d2sp3d2 Properties16.7% s-characterAll 6 positions equivalentPeriod 3+ elements onlySF6 - Perfect Octahedral(All 6 F atoms equivalent, all angles 90 degrees)SFFFFFfrontFback90 deg90 degH = 1/2[V + M] = 1/2[6 + 6] = 6H = 6 means sp3d2Shapes from sp3d2 with Lone PairsLone pairs go opposite (180 degrees apart)Minimizes LP-LP repulsion6 BP, 0 LP:Octahedral (SF6)5 BP, 1 LP:Square Pyramidal (BrF5)4 BP, 2 LP:Square Planar (XeF4)Note: 2 LPs always opposite each other!(Trans position, 180 degrees apart)XeF4 - Square Planar(2 LPs opposite each other)XeFFFFLP (above)LP (below)90 degAll F-Xe-F angles = 90 deg(flat molecule)sp3d2 Key Points for JEE6 hybrid orbitals from 1s + 3p + 2dAll 6 positions equivalent in pure octahedral2 LPs in XeF4 go opposite (trans) for min repulsionCommon in coordination compounds

Formation

One s + Three p + Two d orbitals → Six sp³d² hybrid orbitals

Requirement: Period 3+ elements only

Characteristics

  • Number of orbitals: 6
  • Geometry: Octahedral
  • Bond angles: 90°
  • s-character: 16.7%
  • Shape: 6 orbitals pointing to corners of an octahedron

Examples

SF₆ (Sulfur Hexafluoride)

Hybridization:

  • S: 3s² 3p⁴ → Excite → 3s¹ 3p³ 3d²
  • Mix 3s + three 3p + two 3d → six sp³d² orbitals
  • Each sp³d² overlaps with F 2p
  • Result: Perfect octahedral structure

Formula check:

  • V = 6, M = 6
  • H = ½[6+6] = 6 → sp³d² ✓

Properties:

  • All bond angles = 90°
  • All S-F bonds equal in length and strength
  • Highly symmetrical, very stable

XeF₄ (Xenon Tetrafluoride)

Hybridization:

  • Xe: [Kr] 5s² 5p⁶ → Excite → 5s¹ 5p³ 5d²
  • Six sp³d² orbitals
  • 4 sp³d² overlap with F (σ bonds)
  • 2 sp³d² hold lone pairs (opposite positions!)

Result:

  • Electron geometry: Octahedral
  • Molecular geometry: Square planar (2 LP opposite)
sp³d² Lone Pair Rule

In sp³d², all 6 positions are equivalent, BUT:

Lone pairs prefer to be opposite each other (180° apart) to minimize repulsion.

Shapes from sp³d²:

  • 6 BP, 0 LP: Octahedral (SF₆) — all 90°
  • 5 BP, 1 LP: Square pyramidal (BrF₅) — LP at one position
  • 4 BP, 2 LP: Square planar (XeF₄) — 2 LP opposite

sp³d³ Hybridization: The Rare Seven

Formation

One s + Three p + Three d orbitals → Seven sp³d³ hybrid orbitals

Very rare! Only seen in large atoms like I, Xe under special conditions.

Characteristics

  • Number of orbitals: 7
  • Geometry: Pentagonal bipyramidal
  • Bond angles: 72° (in pentagon), 90° (axial to pentagon)

Example

IF₇ (Iodine Heptafluoride)

Hybridization:

  • I: 5s² 5p⁵ → Excite → 5s¹ 5p³ 5d³
  • Seven sp³d³ orbitals
  • All 7 overlap with F atoms

Result: Pentagonal bipyramidal (5 F in a pentagon, 2 F above/below)

Formula check:

  • V = 7, M = 7
  • H = ½[7+7] = 7 → sp³d³ ✓

Hybridization and Bond Properties

s-Character Effect

More s-character → Stronger, shorter bonds

Hybridizations-characterBond StrengthBond Length
sp50%StrongestShortest
sp²33.3%MediumMedium
sp³25%WeakestLongest
sp³d20%WeakerLonger
sp³d²16.7%WeakestLongest

Example: C-C bonds

  • C≡C (sp-sp): 120 pm (shortest, strongest)
  • C=C (sp²-sp²): 134 pm (medium)
  • C-C (sp³-sp³): 154 pm (longest, weakest)

Electronegativity Effect

More s-character → Higher electronegativity

Why? s-orbitals are closer to nucleus than p-orbitals.

Example:

  • sp carbon (50% s): More electronegative than sp³ carbon (25% s)
  • Acidity: HC≡CH (sp C-H, pKₐ ≈ 25) more acidic than CH₄ (sp³ C-H, pKₐ ≈ 50)
JEE Trick: s-Character Patterns

Higher s-character means:

  • ✓ Shorter bonds
  • ✓ Stronger bonds
  • ✓ Higher electronegativity
  • ✓ Smaller bond angles (wait, what?)

Bond angle paradox:

  • Pure p-p overlap → 90°
  • More s-character → angle moves toward ideal geometry
  • sp: 180° (linear)
  • sp²: 120° (trigonal)
  • sp³: 109.5° (tetrahedral)

For JEE: If they ask “which C-H bond is shortest?” → Look for sp carbon!

Memory Tricks & Patterns

The Hybridization Ladder

sp³d³  ← Rare, Period 5+
sp³d²  ← Octahedral starts here
sp³d   ← d-orbitals start here (Period 3+)
sp³    ← Most common in organic chemistry
sp²    ← Double bonds live here
sp     ← Triple bonds live here

Quick Recognition Table

If molecule has…Hybridization likely…
Triple bond (≡)sp on that atom
Double bond (=)sp² on that atom
Single bonds onlysp³ (or sp³d if >4 bonds)
5 bondssp³d
6 bondssp³d²

Geometry Mnemonics

“2-3-4-5-6 → Linear-Plane-Tetra-TBP-Octa”

  • 2 orbitals → Linear (sp)
  • 3 orbitals → Planar (sp²)
  • 4 orbitals → Tetrahedral (sp³)
  • 5 orbitals → TBP (sp³d)
  • 6 orbitals → Octahedral (sp³d²)

Common Mistakes to Avoid

Trap #1: Hybridization ≠ Shape

Mistake: “sp³ means tetrahedral shape”

Correct:

  • sp³ is the hybridization (orbital mixing)
  • Shape depends on lone pairs!
    • NH₃: sp³ hybridization, trigonal pyramidal shape
    • H₂O: sp³ hybridization, bent shape

JEE Rule: State hybridization as “sp³” but shape as the molecular geometry from VSEPR!

Trap #2: Forgetting Excited State

Mistake: “Carbon has 2s² 2p², so it can only form 2 bonds”

Correct:

  • Carbon promotes one 2s electron to 2p (excited state)
  • Now it has 4 half-filled orbitals
  • Hybridization creates 4 equivalent sp³ orbitals
  • Forms 4 bonds!

Energy justification: Energy gained from 4 bonds > energy needed for promotion

Trap #3: d-Orbitals in Period 2

Mistake: “NH₅ can form with sp³d hybridization”

Wrong! Nitrogen is in Period 2 — no accessible d-orbitals!

Correct:

  • sp³d requires Period 3+ (3d, 4d, 5d orbitals available)
  • Period 2 (C, N, O, F) can NEVER exceed sp³
  • Maximum bonds: 4 (sp³) for Period 2 elements

Why can’t Period 2 use d-orbitals?

  • 2d orbitals don’t exist (quantum mechanics)
  • Next available d is 3d (too high in energy)
  • So Period 2 is limited to 2s and 2p orbitals only!

Practice Problems

Level 1: Foundation (NCERT Style)

Problem 1

Question: Identify the hybridization of carbon in: (a) CH₄ (b) C₂H₄ (c) C₂H₂

Solution:

(a) CH₄:

  • V = 4, M = 4
  • H = ½[4+4] = 4 → sp³
  • Geometry: Tetrahedral

(b) C₂H₄:

  • For each C: V = 4, M = 2 (2 H) + 1 (other C) = 3 effective
  • Actually, simpler: count bonds
  • 3 σ bonds (2 C-H, 1 C-C) → sp²
  • Geometry: Trigonal planar

(c) C₂H₂:

  • For each C: V = 4, 1 H + 1 C
  • 2 σ bonds → sp
  • Geometry: Linear

Quick method: Count σ bonds!

  • 4 σ → sp³
  • 3 σ → sp²
  • 2 σ → sp
Problem 2

Question: Why is H₂O bent and not linear despite sp³ hybridization?

Solution:

Hybridization analysis:

  • O has sp³ hybridization (H = ½[6+2] = 4)
  • 4 sp³ orbitals in tetrahedral arrangement
  • 2 sp³ orbitals form σ bonds with H
  • 2 sp³ orbitals hold lone pairs

Shape determination:

  • Electron geometry: Tetrahedral (4 electron pairs)
  • Molecular geometry: Bent (only 2 H atoms visible)
  • LP-LP repulsion pushes H atoms closer
  • Bond angle: 104.5° (not 109.5°)

Answer: sp³ hybridization creates tetrahedral electron geometry, but the molecular shape is bent due to 2 lone pairs occupying 2 of the 4 sp³ orbitals.

Level 2: JEE Main Type

Problem 3

Question: Which of the following has sp³d² hybridization? (A) PCl₅ (B) SF₆ (C) BF₃ (D) NH₃

Solution:

(A) PCl₅:

  • V = 5, M = 5
  • H = ½[5+5] = 5 → sp³d ✗

(B) SF₆:

  • V = 6, M = 6
  • H = ½[6+6] = 6 → sp³d² ✓

(C) BF₃:

  • V = 3, M = 3
  • H = ½[3+3] = 3 → sp² ✗

(D) NH₃:

  • V = 5, M = 3
  • H = ½[5+3] = 4 → sp³ ✗

Answer: (B) SF₆

Problem 4

Question: Arrange the following C-H bonds in order of decreasing bond length: C₂H₆ (ethane), C₂H₄ (ethene), C₂H₂ (ethyne)

Solution:

Identify hybridization:

  • C₂H₆: C is sp³ (single bonds only)
  • C₂H₄: C is sp² (C=C double bond)
  • C₂H₂: C is sp (C≡C triple bond)

s-character:

  • sp: 50% s
  • sp²: 33.3% s
  • sp³: 25% s

Bond length trend: More s-character → shorter bond (s-orbitals closer to nucleus)

Order:

  • sp (shortest) < sp² < sp³ (longest)
  • C₂H₂ < C₂H₄ < C₂H₆

Answer: C₂H₆ > C₂H₄ > C₂H₂ (decreasing bond length)

Actual values:

  • C₂H₆ (sp³ C-H): ~109 pm
  • C₂H₄ (sp² C-H): ~108 pm
  • C₂H₂ (sp C-H): ~106 pm

Level 3: JEE Advanced Type

Problem 5: Complex Ion Hybridization

Question: Determine the hybridization of the central atom in: (a) [Fe(CN)₆]⁴⁻ (b) [Ni(CO)₄] (c) [Cu(NH₃)₄]²⁺

Solution:

(a) [Fe(CN)₆]⁴⁻:

  • Central atom: Fe
  • 6 ligands (CN⁻) bonded
  • Coordination number = 6
  • Hybridization: sp³d² (or d²sp³ — same thing)
  • Geometry: Octahedral

(b) [Ni(CO)₄]:

  • Central atom: Ni
  • 4 ligands (CO) bonded
  • Coordination number = 4
  • Hybridization: sp³
  • Geometry: Tetrahedral

(c) [Cu(NH₃)₄]²⁺:

  • Central atom: Cu²⁺
  • 4 ligands (NH₃) bonded
  • Coordination number = 4
  • Hybridization: dsp² (or sp²d)
  • Geometry: Square planar (NOT tetrahedral!)

Note: For coordination complexes, geometry depends on the metal, oxidation state, and ligand field. dsp² gives square planar, sp³ gives tetrahedral.

Problem 6: Multi-Atom Hybridization

Question: In the molecule CH₂=CH-CN: (a) Identify hybridization of each carbon (b) Predict all C-C bond lengths qualitatively (c) Which C-H bond is shortest?

Solution:

Structure: H₂C=CH-C≡N

(a) Hybridization:

C₁ (left carbon, in CH₂):

  • Forms 3 σ bonds (2 C-H, 1 C=C)
  • Hybridization: sp²

C₂ (middle carbon):

  • Forms 3 σ bonds (1 C-H, 1 to C₁, 1 to C₃)
  • Hybridization: sp²

C₃ (right carbon, in CN):

  • Forms 2 σ bonds (1 to C₂, 1 C≡N)
  • Hybridization: sp

(b) C-C bond lengths:

  • C₁=C₂: Double bond (sp²-sp²) → ~134 pm
  • C₂-C₃: Single bond (sp²-sp) → ~146 pm (shorter than typical sp³-sp³ due to sp contribution)

Order: C₁=C₂ < C₂-C₃

(c) Shortest C-H bond:

C-H bond lengths depend on hybridization:

  • C₁-H: sp² C-H
  • C₂-H: sp² C-H

Wait, there are NO sp C-H bonds in this molecule! (C₃ is bonded to N, not H)

So all C-H bonds are sp² → all equal in length (~108 pm)

Answer: (a) C₁: sp², C₂: sp², C₃: sp (b) C₁=C₂ (shortest) < C₂-C₃ (c) All C-H bonds equal (all sp²)

Quick Revision Box

HybridizationOrbitals MixedNumberGeometryBond AngleExample
sps + p2Linear180°BeCl₂, C₂H₂
sp²s + 2p3Trigonal planar120°BF₃, C₂H₄
sp³s + 3p4Tetrahedral109.5°CH₄, NH₃
sp³ds + 3p + d5Trigonal bipyramidal90°, 120°PCl₅, SF₄
sp³d²s + 3p + 2d6Octahedral90°SF₆, XeF₄
sp³d³s + 3p + 3d7Pentagonal bipyramidal72°, 90°IF₇

Formula Recall

$$\boxed{H = \frac{1}{2}[V + M - C + A]}$$
  • H = 2 → sp
  • H = 3 → sp²
  • H = 4 → sp³
  • H = 5 → sp³d
  • H = 6 → sp³d²

s-Character Memory

Hybridizations-characterProperty
sp50%Shortest, strongest bonds
sp²33.3%Medium
sp³25%Longest, weakest bonds
sp³d20%Weaker
sp³d²16.7%Weakest

Real-World Applications

Where Hybridization Rules

1. Diamond vs Graphite:

  • Diamond: sp³ → 3D network → hardest material
  • Graphite: sp² → planar sheets → soft, lubricant, conducts electricity

2. Polymers:

  • Polyethylene: sp³ C-C chains → flexible plastics
  • Kevlar: sp² aromatic rings → bullet-proof vests (rigid, strong)

3. Semiconductors:

  • Silicon: sp³ → tetrahedral lattice → computer chips
  • Graphene: sp² carbon → strongest 2D material, future electronics

4. Pharmaceuticals:

  • Drug shape depends on hybridization
  • sp² and sp centers create rigid structures (important for enzyme binding)
  • sp³ creates flexible chains

5. Fuels:

  • Gasoline (sp³ C-C): Low energy per bond → burns efficiently
  • Acetylene (sp C≡C): High energy → welding torches at 3300°C!

6. Organic Synthesis:

  • Alkenes (sp²): Reactive double bonds → most reactions happen here
  • Alkynes (sp): Even more reactive → building blocks for complex molecules

Teacher’s Summary

Key Takeaways

1. Hybridization = orbital mixing for better bonding

  • Atoms mix s, p, (d) orbitals → equal energy hybrids
  • Allows formation of more bonds than expected
  • Energy payoff: 4 bonds (CH₄) releases more energy than promotion costs

2. Use the magic formula for instant answers

$$H = \frac{1}{2}[V + M - C + A]$$
  • Saves 30 seconds per question (no Lewis structures needed!)
  • H = 2, 3, 4, 5, 6 → sp, sp², sp³, sp³d, sp³d²

3. Hybridization determines geometry (electron geometry)

  • sp → Linear (180°)
  • sp² → Trigonal planar (120°)
  • sp³ → Tetrahedral (109.5°)
  • sp³d → TBP (90°, 120°)
  • sp³d² → Octahedral (90°)

4. Hybridization ≠ Molecular shape

  • Hybridization includes lone pairs
  • Molecular shape only considers atoms
  • NH₃: sp³ hybridization, trigonal pyramidal shape

5. More s-character = shorter, stronger bonds

  • sp (50% s) > sp² (33%) > sp³ (25%)
  • C≡C shortest, C-C longest
  • Explains why triple bonds are strongest!

6. d-orbitals only for Period 3+

  • Period 2 (C, N, O, F): Maximum sp³ (no d-orbitals!)
  • Period 3+ (P, S, Cl): Can use sp³d, sp³d²
  • Explains why SF₆ exists but OF₆ doesn’t

7. For JEE:

  • Memorize the formula (saves massive time!)
  • Know s-character trends (bond length questions!)
  • Recognize hybridization from structure (count σ bonds!)
  • Don’t confuse hybridization with shape

“Hybridization is nature’s way of remixing orbitals to create the perfect bonding playlist!”

Within Chemical Bonding

Cross-Subject: Physics

Cross-Subject: Math