VSEPR Theory and Molecular Shapes

Master VSEPR theory, predict molecular geometries, understand electron pair repulsions, and solve JEE problems on molecular shapes.

Prerequisites

Before diving into VSEPR theory, review:


The Hook: Why Water Isn’t Linear

Connect: The 104.5° That Changed Everything

Imagine if water was linear (H-O-H at 180°) instead of bent (104.5°)…

  • Ice wouldn’t float → oceans would freeze from bottom up → no marine life
  • Water wouldn’t be such a good solvent → no biochemistry, no life
  • Snowflakes wouldn’t have their beautiful hexagonal shapes

One tiny angle difference = The difference between a living planet and a dead rock!

In Interstellar (2014), they search for planets with liquid water. But it’s not just H₂O that matters — it’s the bent shape that gives water its unique properties! The reason? VSEPR Theory — Valence Shell Electron Pair Repulsion.

JEE Reality: This is a free marks topic! 2-3 guaranteed questions every year worth 8-12 marks. Master the patterns, and you’ll solve these in under 30 seconds each!


The Core Concept

What is VSEPR Theory?

VSEPR Theory predicts the 3D shapes of molecules based on one simple idea:

$$\boxed{\text{Electron pairs repel each other and arrange to minimize repulsion}}$$

In simple terms: Imagine balloons tied together at a central point. The balloons naturally spread out as far as possible to minimize bumping into each other. That’s exactly what electron pairs do around an atom!

The Fundamental Principle

Electron pairs (bonding + lone pairs) around a central atom arrange themselves to be as far apart as possible to minimize repulsion.

graph TD
    A[Count electron pairs
around central atom] --> B{Total electron pairs} B -->|2 pairs| C[Linear
180°] B -->|3 pairs| D[Trigonal Planar
120°] B -->|4 pairs| E[Tetrahedral
109.5°] B -->|5 pairs| F[Trigonal Bipyramidal
90°, 120°] B -->|6 pairs| G[Octahedral
90°] style A fill:#e1f5fe style C fill:#fff9c4 style D fill:#f0f4c3 style E fill:#dcedc8 style F fill:#c5e1a5 style G fill:#aed581

VSEPR Methodology: The 4-Step Process

Step 1: Draw Lewis Structure

Identify central atom, count valence electrons, distribute as bonds and lone pairs.

Step 2: Count Electron Pairs

Steric Number (SN) = Number of bonding pairs + Number of lone pairs

Important: Count multiple bonds as ONE bonding region (double = 1, triple = 1)

Step 3: Determine Electron Geometry

Based on total electron pairs (SN), determine how electron pairs arrange in space.

Step 4: Determine Molecular Geometry

Consider only atoms (ignore lone pairs) to determine molecular shape.

Golden Rule

Electron geometry (includes lone pairs) determines electron arrangement

Molecular geometry (only atoms) determines molecular shape

Example: H₂O

  • Electron geometry: Tetrahedral (4 electron pairs)
  • Molecular geometry: Bent (only 2 H atoms visible)

JEE Trap: Questions ask for “shape” → they want molecular geometry, not electron geometry!


Repulsion Order: Who Pushes Harder?

Not all electron pairs repel equally! Lone pairs occupy more space than bonding pairs.

$$\boxed{\text{LP-LP} > \text{LP-BP} > \text{BP-BP}}$$

Where:

  • LP = Lone Pair
  • BP = Bonding Pair

Why? Lone pairs are held by only one nucleus (the central atom), so they spread out more. Bonding pairs are held by two nuclei, so they’re more compressed.

Effect on Bond Angles

More lone pairs → Smaller bond angles

Example: CH₄ vs NH₃ vs H₂O

MoleculeLPBPBond AngleReason
CH₄04109.5°Standard tetrahedral
NH₃13107°1 LP pushes BP’s closer
H₂O22104.5°2 LP’s push BP’s even closer

Memory trick: “More Lone, Less Angle!” → Each lone pair reduces angle by ~2.5°


The Complete VSEPR Guide

2 Electron Pairs (SN = 2)

Linear Geometry (180°)

Electron Geometry: Linear Molecular Geometries: 1 shape possible

Type: AX₂ (2 BP, 0 LP) — LINEAR

Example: BeCl₂, CO₂, HCN

Cl-Be-Cl    or    O=C=O
    180°           180°

Visual Description:

  • Linear: Like a straight rod
  • Both atoms on opposite sides of central atom
  • Bond angle = 180°
CO₂ Visualization

Imagine a barbell with weights on both ends. The central carbon is the bar, and the oxygen atoms are the weights on opposite ends. Perfectly straight, perfectly balanced.

Why linear? 2 electron regions want to be as far apart as possible → 180° is maximum separation!


3 Electron Pairs (SN = 3)

Electron Geometry: Trigonal Planar (120°)

Type 1: AX₃ (3 BP, 0 LP) — TRIGONAL PLANAR

Example: BF₃, BCl₃, CO₃²⁻, NO₃⁻

     F
     |
 F—B—F
   120°

Visual Description:

  • Trigonal Planar: Like a Mercedes-Benz logo
  • All atoms in the same plane
  • Three arms at 120° angles
  • Flat, symmetrical

Example: BF₃ (boron trifluoride)

  • B at center, 3 F atoms around it
  • All in one plane, 120° apart
  • Looks like three spokes of a wheel

Type 2: AX₂E (2 BP, 1 LP) — BENT/V-SHAPED

Example: SO₂, O₃, NO₂⁻

   O
  //
 S
  \
   O
  ~120° (actually ~119°)

Visual Description:

  • Bent: Like a boomerang
  • Two atoms bonded, one lone pair (invisible)
  • Lone pair pushes atoms closer
  • Bond angle < 120° (usually ~119° for SO₂)
Quick Comparison

CO₂ (linear) vs SO₂ (bent) — Why different?

CO₂: C has 4 valence e⁻, forms 2 double bonds → 2 bonding regions, 0 LP → linear

SO₂: S has 6 valence e⁻, forms 2 double bonds → 2 bonding regions, 1 LP → bent

JEE Trick: Even though both have O=X=O formula, count lone pairs!


4 Electron Pairs (SN = 4)

Electron Geometry: Tetrahedral (109.5°)

Type 1: AX₄ (4 BP, 0 LP) — TETRAHEDRAL

Example: CH₄, CCl₄, NH₄⁺, SO₄²⁻

       H
       |
   H—C—H
       |
       H
   109.5°

Visual Description:

  • Tetrahedral: Like a triangular pyramid
  • 4 atoms arranged at corners of a tetrahedron
  • Central atom at the center
  • All bond angles = 109.5°
  • 3D structure (NOT flat!)

Imagine: A tripod with one leg pointing up and three pointing down (but more symmetrical). Or a soccer ball’s stitching at a vertex.

Type 2: AX₃E (3 BP, 1 LP) — TRIGONAL PYRAMIDAL

Example: NH₃, PCl₃, H₃O⁺

      N
     /|\
    H H H
    ~107°

Visual Description:

  • Trigonal Pyramidal: Like a tripod or a stool
  • 3 atoms form a triangular base
  • Central atom above the base (not in the plane!)
  • Lone pair at the top (invisible)
  • Bond angle ≈ 107° (less than 109.5° due to LP repulsion)

Imagine: A three-legged camera tripod. The camera (central atom) sits above the three legs (bonded atoms). There’s an invisible fourth leg (lone pair) pointing up.

Type 3: AX₂E₂ (2 BP, 2 LP) — BENT/V-SHAPED

Example: H₂O, H₂S, SCl₂

    O:  (2 lone pairs)
   / \
  H   H
 104.5°

Visual Description:

  • Bent: Like a boomerang or an open book
  • 2 atoms bonded at an angle
  • 2 lone pairs (invisible) push from opposite sides
  • Bond angle ≈ 104.5° (H₂O), smaller than 109.5°

Imagine: Take a tetrahedral structure, remove two opposite corners (atoms), replace with lone pairs. The remaining two atoms form a bent shape.

Common Mistake: Bent from 3 EP vs 4 EP

Two types of bent shapes — DON’T confuse!

Bent from 3 electron pairs (AX₂E):

  • Example: SO₂
  • Electron geometry: Trigonal planar
  • Bond angle ≈ 119-120°

Bent from 4 electron pairs (AX₂E₂):

  • Example: H₂O
  • Electron geometry: Tetrahedral
  • Bond angle ≈ 104-105°

JEE Trap: Both are “bent,” but different angles! Always count total electron pairs first!


5 Electron Pairs (SN = 5)

Electron Geometry: Trigonal Bipyramidal (90° and 120°)

Key Feature: TWO types of positions:

  • Axial: 2 positions above and below (90° to equatorial)
  • Equatorial: 3 positions in the middle plane (120° to each other)

Important: Lone pairs ALWAYS occupy equatorial positions first (less repulsion!)

Type 1: AX₅ (5 BP, 0 LP) — TRIGONAL BIPYRAMIDAL

Example: PCl₅, PF₅

      Cl (axial)
       |
Cl—P—Cl  (equatorial, 120° apart)
       |
      Cl (axial)

Visual Description:

  • Trigonal Bipyramidal: Like two triangular pyramids joined at the base
  • 3 atoms in the middle (equatorial) at 120°
  • 2 atoms above and below (axial) at 90° to equatorial
  • NOT all bond angles equal!

Imagine: A bow tie or two Egyptian pyramids stuck together base-to-base.

Type 2: AX₄E (4 BP, 1 LP) — SEE-SAW/DISTORTED TETRAHEDRAL

Example: SF₄, XeO₂F₂

      F (axial)
       |
    F—S—F (equatorial)
       |
      F (axial)
  (LP at equatorial)

Visual Description:

  • See-Saw: Like a playground see-saw or teeter-totter
  • Lone pair occupies one equatorial position
  • 2 axial atoms (above/below)
  • 2 equatorial atoms (one position empty due to LP)
  • Asymmetrical, distorted

Bond angles: ~90° (axial-equatorial), ~120° (equatorial-equatorial), but distorted by LP

Imagine: A see-saw at a playground — two kids on the ends (axial), one kid on one side (equatorial), empty seat on other side (lone pair).

Type 3: AX₃E₂ (3 BP, 2 LP) — T-SHAPED

Example: ClF₃, BrF₃

      F
      |
    F—Cl—F
   (2 LP at equatorial)

Visual Description:

  • T-Shaped: Like the letter “T”
  • 2 lone pairs occupy equatorial positions
  • 2 atoms at axial positions (up and down)
  • 1 atom at remaining equatorial position
  • Forms a T

Bond angles: ~90° (axial to equatorial), slightly less than 90° due to LP repulsion

Imagine: A capital letter T. The vertical part has atoms at top and bottom (axial), the horizontal part has one atom extending to the side (equatorial).

Type 4: AX₂E₃ (2 BP, 3 LP) — LINEAR

Example: XeF₂, I₃⁻

F—Xe—F
(3 LP at equatorial)

Visual Description:

  • Linear: Straight line (like AX₂)
  • 3 lone pairs occupy ALL equatorial positions
  • 2 atoms at axial positions (least repulsion)
  • Bond angle = 180°

Imagine: A barbell again, but with invisible balloons (lone pairs) in the middle plane pushing the atoms to opposite sides.

5 Electron Pairs Memory Trick

“Lone pairs love the equator!” — They always go to equatorial positions first.

Why? Equatorial positions have 2 neighbors at 90° (axial positions have 3 neighbors at 90°). Lone pairs need more space, so they choose the less crowded equatorial spots!

Patterns:

  • 0 LP: Trigonal bipyramidal (all atoms)
  • 1 LP: See-saw (LP at equatorial)
  • 2 LP: T-shaped (both LP at equatorial)
  • 3 LP: Linear (all LP at equatorial, atoms at axial)

6 Electron Pairs (SN = 6)

Electron Geometry: Octahedral (90°)

Key Feature: All 6 positions are equivalent (unlike TBP)

Type 1: AX₆ (6 BP, 0 LP) — OCTAHEDRAL

Example: SF₆, PF₆⁻, [Fe(CN)₆]³⁻

      F
      |
  F—S—F
      |
      F
(+ 1 F front, 1 F back)

Visual Description:

  • Octahedral: Like two square pyramids joined at the base
  • 6 atoms at corners of an octahedron
  • All bond angles = 90°
  • Highly symmetrical

Imagine: A die (6-sided cube) with atoms at each face’s center. Or two Egyptian pyramids (square base) stuck together.

Type 2: AX₅E (5 BP, 1 LP) — SQUARE PYRAMIDAL

Example: BrF₅, IF₅, [NiCN₅]³⁻

      F
      |
  F—Br—F
   | |
    F—F
(LP opposite to apex)

Visual Description:

  • Square Pyramidal: Like the Egyptian pyramids
  • 4 atoms form a square base
  • 1 atom at the apex (top)
  • Lone pair at the bottom (opposite to apex)

Bond angles: ~90° (mostly), slightly distorted by LP

Imagine: The Great Pyramid of Giza — square base, apex at top, lone pair hidden underground.

Type 3: AX₄E₂ (4 BP, 2 LP) — SQUARE PLANAR

Example: XeF₄, ICl₄⁻, [PtCl₄]²⁻

    F
    |
F—Xe—F
    |
    F
(2 LP above and below plane)

Visual Description:

  • Square Planar: Like a flat square
  • 4 atoms at corners of a square
  • 2 lone pairs above and below the plane
  • All atoms in one plane
  • Bond angles = 90°

Imagine: A perfectly flat piece of paper with atoms at the four corners. Lone pairs are ghosts floating above and below the paper.


Summary Table: All VSEPR Shapes

SNFormulaBPLPElectron GeometryMolecular GeometryBond AngleExample
2AX₂20LinearLinear180°CO₂, BeCl₂
3AX₃30Trigonal planarTrigonal planar120°BF₃, CO₃²⁻
3AX₂E21Trigonal planarBent~120°SO₂, NO₂⁻
4AX₄40TetrahedralTetrahedral109.5°CH₄, NH₄⁺
4AX₃E31TetrahedralTrigonal pyramidal~107°NH₃, PCl₃
4AX₂E₂22TetrahedralBent~104.5°H₂O, H₂S
5AX₅50TBPTrigonal bipyramidal90°, 120°PCl₅
5AX₄E41TBPSee-saw<90°, <120°SF₄
5AX₃E₂32TBPT-shaped~90°ClF₃
5AX₂E₃23TBPLinear180°XeF₂, I₃⁻
6AX₆60OctahedralOctahedral90°SF₆
6AX₅E51OctahedralSquare pyramidal~90°BrF₅
6AX₄E₂42OctahedralSquare planar90°XeF₄

Interactive Demo: Explore VSEPR Molecular Geometries

Use this interactive 3D visualization to explore all VSEPR geometries. Select preset molecules like CH4, NH3, H2O, or customize electron pair counts. Drag to rotate the model and observe how lone pairs affect bond angles. Watch the difference between electron geometry and molecular geometry in real-time!

How to Use This Visualization
  1. Select a Preset: Choose from common molecules like CH4, NH3, H2O, SF6, XeF4, etc.
  2. Custom Mode: Use sliders to set total electron pairs (2-6) and lone pairs
  3. Rotate: Drag on the molecule to rotate it in 3D space
  4. Observe: Watch how lone pairs (purple clouds) push bonding pairs closer together
  5. Compare: Notice the difference between electron geometry and molecular geometry!

Key Insight: The visualization shows LP-LP > LP-BP > BP-BP repulsion - lone pairs compress bond angles!


Memory Tricks & Patterns

The Lone Pair Effect

“Each lonely pair steals 2.5° from the party!”

For tetrahedral base (109.5°):

  • 0 LP: CH₄ = 109.5°
  • 1 LP: NH₃ = 107° (−2.5°)
  • 2 LP: H₂O = 104.5° (−5°)

Shape Family Tree

2 EP: Linear only
3 EP: Planar (flat) shapes only
4 EP: 3D shapes start
5 EP: Complex shapes, LP prefer equatorial
6 EP: 3D symmetrical shapes

Visual Mnemonics

BeCl₂: Becomes Linear (180°) BF₃: Boron Forms Trigonal planar (120°) CH₄: Carbon Has Tetrahedral (109.5°) NH₃: Nitrogen’s Pyramid (107°) H₂O: Highly Bent (104.5°) PCl₅: Phosphor BP shape (Bipyramidal) SF₆: Sulfur Octahedral (6 faces)


Special Cases and Exceptions

Multiple Bonds Count as ONE Region

Critical VSEPR Rule

Double and triple bonds count as ONE electron region for VSEPR!

Example: CO₂

  • Lewis structure: O=C=O
  • You might think “4 bonds = 4 regions”
  • WRONG! Each C=O is one bonding region
  • Total: 2 regions → Linear

Example: HCN

  • Lewis structure: H-C≡N
  • H-C is 1 region, C≡N is 1 region
  • Total: 2 regions → Linear

Expanded Octets (Period 3 and Beyond)

Elements in Period 3 and higher can accommodate more than 8 electrons using vacant d-orbitals.

Examples:

  • PCl₅: P has 10 electrons (5 bonds)
  • SF₆: S has 12 electrons (6 bonds)
  • IF₇: I has 14 electrons (7 bonds)

Why? They have accessible 3d, 4d, 5d orbitals for electron accommodation.

Resonance Doesn’t Change Shape

Example: CO₃²⁻

  • Has 3 resonance structures (double bond rotates)
  • But VSEPR shape remains trigonal planar
  • Why? Average bond order doesn’t change electron regions

Common Mistakes to Avoid

Trap #1: Confusing Electron and Molecular Geometry

Mistake: Saying H₂O is “tetrahedral”

Correct:

  • H₂O has tetrahedral electron geometry (4 electron pairs)
  • But bent molecular geometry (only 2 H atoms visible)

JEE questions ask for “shape of molecule” → Answer = Bent, not tetrahedral!

Rule: Unless specifically asked for “electron geometry,” always give molecular geometry!

Trap #2: Counting Multiple Bonds Wrong

Mistake: Counting C=O double bond as 2 bonding regions

Correct: C=O is ONE bonding region (even though it has 2 bonds)

Example: CO₂

  • Wrong count: 4 bonds → 4 regions → tetrahedral
  • Correct count: 2 bonding regions (each C=O) → Linear ✓
Trap #3: Forgetting Lone Pairs on Central Atom

Mistake: Drawing Lewis structure but not identifying lone pairs on central atom

Example: SO₂

  • S has 6 valence electrons
  • Forms 2 double bonds (uses 4 electrons)
  • Remaining 2 electrons = 1 lone pair ← Don’t forget this!
  • Result: 2 BP + 1 LP → Bent, not linear

JEE Tip: After drawing Lewis structure, ALWAYS count lone pairs on central atom!


Practice Problems

Level 1: Foundation (NCERT Style)

Problem 1

Question: Predict the shape of the following molecules: (a) BeCl₂ (b) BF₃ (c) CH₄

Solution:

(a) BeCl₂:

  • Be has 2 valence electrons
  • Forms 2 bonds with Cl
  • SN = 2 (2 BP, 0 LP)
  • Shape: Linear (180°)

(b) BF₃:

  • B has 3 valence electrons
  • Forms 3 bonds with F
  • SN = 3 (3 BP, 0 LP)
  • Shape: Trigonal planar (120°)

(c) CH₄:

  • C has 4 valence electrons
  • Forms 4 bonds with H
  • SN = 4 (4 BP, 0 LP)
  • Shape: Tetrahedral (109.5°)
Problem 2

Question: Why is H₂O bent and not linear?

Solution:

Lewis structure of H₂O:

  • O has 6 valence electrons
  • Forms 2 bonds with H (uses 2 electrons)
  • Remaining 4 electrons = 2 lone pairs

VSEPR Analysis:

  • Total electron pairs = 2 BP + 2 LP = 4
  • Electron geometry: Tetrahedral
  • Molecular geometry: Bent (only 2 H atoms)
  • Bond angle ≈ 104.5° (less than 109.5° due to LP-LP repulsion)

Answer: H₂O is bent because oxygen has 2 lone pairs that repel the bonding pairs, creating a bent shape instead of linear.


Level 2: JEE Main Type

Problem 3

Question: Arrange the following in order of increasing bond angle: H₂O, NH₃, CH₄, BF₃

Solution:

Step 1: Identify shapes and electron pairs

  • H₂O: 2 BP, 2 LP → Bent (tetrahedral base) → ~104.5°
  • NH₃: 3 BP, 1 LP → Trigonal pyramidal → ~107°
  • CH₄: 4 BP, 0 LP → Tetrahedral → 109.5°
  • BF₃: 3 BP, 0 LP → Trigonal planar → 120°

Step 2: Order by bond angle

H₂O (104.5°) < NH₃ (107°) < CH₄ (109.5°) < BF₃ (120°)

Answer: H₂O < NH₃ < CH₄ < BF₃

JEE Tip: More lone pairs → smaller bond angle (for same electron geometry)

Problem 4

Question: Which of the following has a square planar shape? (A) CCl₄ (B) XeF₄ (C) NH₄⁺ (D) SF₄

Solution:

(A) CCl₄:

  • C has 4 valence e⁻, 4 Cl atoms
  • SN = 4 (4 BP, 0 LP) → Tetrahedral

(B) XeF₄:

  • Xe has 8 valence e⁻, 4 F atoms
  • 4 bonds use 4 electrons, remaining 4 = 2 LP
  • SN = 6 (4 BP, 2 LP) → Square planar

(C) NH₄⁺:

  • N has 5 valence e⁻, but +1 charge → 4 electrons
  • 4 bonds, 0 LP
  • SN = 4 (4 BP, 0 LP) → Tetrahedral

(D) SF₄:

  • S has 6 valence e⁻, 4 F atoms
  • 4 bonds, 1 LP
  • SN = 5 (4 BP, 1 LP) → See-saw

Answer: (B) XeF₄


Level 3: JEE Advanced Type

Problem 5: Multi-Molecule Comparison

Question: Among NH₃, NH₄⁺, and NH₂⁻, which has: (a) The largest bond angle? (b) The smallest bond angle? Explain with reasoning.

Solution:

NH₃ (Ammonia):

  • N has 5 valence e⁻, 3 H atoms
  • 3 BP, 1 LP
  • Shape: Trigonal pyramidal
  • Bond angle ≈ 107°

NH₄⁺ (Ammonium ion):

  • N has 5 valence e⁻, +1 charge → 4 electrons available for bonding
  • 4 BP, 0 LP
  • Shape: Tetrahedral
  • Bond angle = 109.5°

NH₂⁻ (Amide ion):

  • N has 5 valence e⁻, −1 charge → 6 electrons total
  • 2 H atoms → 2 BP, 2 LP
  • Shape: Bent
  • Bond angle ≈ 104.5°

Answers: (a) Largest bond angle: NH₄⁺ (109.5°) — no lone pairs! (b) Smallest bond angle: NH₂⁻ (104.5°) — 2 lone pairs compress the most!

Pattern: As LP increases, bond angle decreases

  • 0 LP: 109.5°
  • 1 LP: ~107°
  • 2 LP: ~104.5°
Problem 6: Complex Shape Prediction

Question: Predict the shape of ClF₃ and explain why all bond angles are NOT equal.

Solution:

Step 1: Lewis structure

  • Cl has 7 valence e⁻, 3 F atoms
  • 3 bonds use 3 electrons, remaining 4 = 2 LP
  • Total: 3 BP + 2 LP = 5 electron pairs

Step 2: Electron geometry

  • 5 electron pairs → Trigonal bipyramidal

Step 3: Place lone pairs

  • Lone pairs prefer equatorial positions (less repulsion)
  • 2 LP occupy 2 equatorial spots
  • 3 F atoms: 2 at axial, 1 at remaining equatorial

Step 4: Molecular shape

  • T-shaped (3 atoms, 2 LP)

Step 5: Bond angles

  • Axial F—Cl—Equatorial F: ~90° (but slightly less due to LP repulsion)
  • Axial F—Cl—Axial F: ~180° (slightly less)

Why not equal? The 2 lone pairs at equatorial positions repel the bonding pairs unevenly:

  • LP-BP repulsion > BP-BP repulsion
  • Axial bonds are pushed slightly toward each other
  • Equatorial bond is pushed away from LP’s

Answer: ClF₃ is T-shaped with bond angles ≈ 87-88° (F-Cl-F, axial to equatorial) and ~175° (axial to axial), not exactly 90° and 180° due to lone pair repulsions.


Quick Revision Box

Total EP0 LP1 LP2 LP3 LP
2Linear (180°)
3Trig. planar (120°)Bent (~120°)
4Tetrahedral (109.5°)Trig. pyramidal (~107°)Bent (~104.5°)
5TBP (90°, 120°)See-sawT-shaped (~90°)Linear (180°)
6Octahedral (90°)Sq. pyramidalSq. planar (90°)

Most Tested Molecules in JEE

MoleculeShapeBond AngleKey Feature
CO₂Linear180°Double bonds count as 1 region
H₂OBent104.5°2 LP compress angle
NH₃Trig. pyramidal107°1 LP reduces angle
CH₄Tetrahedral109.5°Perfect tetrahedral
PCl₅Trig. bipyramidal90°, 120°2 different angles
SF₆Octahedral90°All positions equivalent
XeF₂Linear180°3 LP at equatorial
XeF₄Square planar90°2 LP above/below plane

Real-World Applications

Why Molecular Shapes Matter

1. Drug Design: Drugs fit into enzyme active sites like a key in a lock. The shape of the molecule determines if it fits!

  • Example: Inhibitor drugs are designed to be exact shapes to block enzyme activity

2. Smell and Taste: Your nose and tongue detect molecules based on shape, not just composition!

  • Limonene (lemon scent) vs Carvone (mint scent) — same formula, different shapes!

3. DNA Structure: The double helix exists because of the bent shape of water that stabilizes it through hydrogen bonding. Linear water → No DNA → No life!

4. Ozone Layer: O₃ is bent (not linear like O₂), which makes it absorb UV radiation differently. The bent shape is crucial for protecting Earth from harmful UV!

5. Catalysts: Transition metal complexes have specific shapes (octahedral, square planar) that determine their catalytic activity. Wrong shape → No catalysis!


Teacher’s Summary

Key Takeaways

1. VSEPR = Electron pairs minimize repulsion

  • Count total electron pairs (bonding + lone)
  • Arrange to maximize separation
  • Simple concept, powerful predictions!

2. Electron geometry ≠ Molecular geometry

  • Electron geometry: includes lone pairs
  • Molecular geometry: only atoms (what you “see”)
  • JEE asks for molecular geometry!

3. Repulsion order determines angles

  • LP-LP > LP-BP > BP-BP
  • More lone pairs → Smaller bond angles
  • Each LP reduces angle by ~2.5° (tetrahedral base)

4. Multiple bonds = ONE bonding region

  • C=O, C≡N count as 1 region each
  • Don’t count individual bonds in multiple bonds!

5. For 5 EP: Lone pairs love equatorial

  • Equatorial positions have less repulsion
  • Fill equatorial first with LP’s
  • Determines see-saw, T-shaped, linear shapes

6. Patterns for quick solving:

  • 2 EP → Linear (only 1 option)
  • 3 EP → Planar/Bent (2 options)
  • 4 EP → Tetrahedral/Pyramidal/Bent (3 options)
  • 5 EP → TBP/See-saw/T/Linear (4 options!)
  • 6 EP → Octahedral/Sq.Pyramidal/Sq.Planar (3 options)

7. JEE Strategy:

  • Draw Lewis structure (10 sec)
  • Count BP and LP (5 sec)
  • Look up table/remember pattern (5 sec)
  • Total: 20 seconds per question!

“Lone pairs are like invisible bullies — they take up more space and push everyone else closer together!”


Within Chemical Bonding

Atomic Structure Foundation

Organic Chemistry Applications

Coordination Chemistry

Inorganic Applications

Physics Connection

Mathematics