Chemistry · JEE Main & Advanced 2026 · Chapter 04

Chemical Bonding notes, MOT, VSEPR & PYQs for JEE 2026

Complete Chemical Bonding and Molecular Structure — Lewis structures, Formal Charge, VSEPR Theory, Valence Bond Theory (VBT), Hybridization, Molecular Orbital Theory (MOT), Bond Parameters, and Hydrogen Bonding. Shortcuts, 150+ worked examples, and JEE PYQs included.

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Chemical Bonding JEE Notes, MOT, VSEPR, PYQs
Chemical Bonding JEE Notes, MOT, VSEPR, PYQs
1

Kössel-Lewis Approach & Formal Charge

Chemical bonding deals with how atoms combine to form molecules. The Lewis approach explains this via the octet rule, while formal charge helps determine the most stable structure when multiple Lewis structures are possible.

1.1 Lewis Dot Structures & Octet Rule

Atoms combine to achieve a stable noble gas configuration (8 valence electrons, or 2 for H/He). Lewis structures represent valence electrons as dots around the element symbol.

Exceptions to Octet Rule (High PYQ Frequency)

  • Incomplete Octet: BeCl₂ (4 e⁻), BF₃ (6 e⁻), AlCl₃ (6 e⁻)
  • Odd Electron: NO, NO₂, ClO₂
  • Expanded Octet: PCl₅ (10 e⁻), SF₆ (12 e⁻), IF₇ (14 e⁻), XeF₄ (12 e⁻)

1.2 Formal Charge Calculation

Formal Charge Formula

\[F.C. = V - (L + \frac{S}{2})\]

Where: V = Valence electrons, L = Lone pair electrons, S = Shared (bonding) electrons

Rule: The structure with formal charges closest to zero (and negative charge on more electronegative atom) is the most stable.

PYQ — JEE Main 2022

Question: What is the formal charge on the central oxygen atom in the ozone (O₃) molecule?

Solution:

Structure of Ozone (O=O–O): Central O forms 1 double bond and 1 single bond, with 1 lone pair.

Valence electrons of O (V) = 6

Lone pair electrons on central O (L) = 2

Shared electrons (S) = 6 (2 bonds = 4 bonding pairs = 8 electrons? Wait, S = 6 means 3 pairs. Central O has 1 double bond (4e) + 1 single bond (2e) = 6 shared electrons).

\[F.C. = 6 - \left(2 + \frac{6}{2}\right) = 6 - (2 + 3) = +1\]
Formal Charge on Central O = +1

Answer: +1

2

Ionic Bond & Lattice Enthalpy

Ionic bonds are formed by the complete transfer of electrons from a metal to a non-metal. The strength of an ionic bond is measured by lattice enthalpy, which is a key concept for JEE.

2.1 Born-Haber Cycle

Lattice enthalpy cannot be measured directly. It is calculated using the Born-Haber cycle, which applies Hess's Law to the formation of an ionic compound from its elements.

Born-Haber Cycle for NaCl

\[\Delta H_f = \Delta H_{sub} + IE + \Delta H_{diss} + EA + U\]

Where:

  • ΔH_f = Enthalpy of formation
  • ΔH_sub = Sublimation energy of metal
  • IE = Ionization enthalpy of metal
  • ΔH_diss = Dissociation energy of non-metal
  • EA = Electron gain enthalpy of non-metal
  • U = Lattice enthalpy (usually negative)

2.2 Factors Affecting Lattice Enthalpy

Charge on Ions

Higher charge → stronger electrostatic force → higher lattice enthalpy

\[|U| \propto q^+ \times q^-\]

Example: MgO (q=+2,-2) has much higher lattice energy than NaCl (q=+1,-1)

Size of Ions

Smaller size → ions closer → stronger attraction → higher lattice enthalpy

\[|U| \propto \frac{1}{r^+ + r^-}\]

Example: LiF > LiCl > LiBr > LiI (as halide size increases)

PYQ — JEE Main 2024

Question: Arrange the following in order of increasing lattice enthalpy: NaCl, MgO, CaO, KCl.

Solution:

Lattice enthalpy depends on charge and size.

Step 1: Charge

NaCl and KCl have +1/-1 charges. MgO and CaO have +2/-2 charges.

Step 2: Size

Among +1/-1: K⁺ > Na⁺, so KCl has lower lattice energy than NaCl.

Among +2/-2: Ca²⁺ > Mg²⁺, so CaO has lower lattice energy than MgO.

\[\text{KCl} < \text{NaCl} < \text{CaO} < \text{MgO}\]

Answer: KCl < NaCl < CaO < MgO

3

Covalent Character & Fajan's Rule

No bond is 100% ionic or 100% covalent. Fajan's rule predicts the covalent character of ionic bonds based on polarization.

Fajan's Rule — Most Repeated JEE Concept

A compound shows more covalent character if:

  • Small cation: High polarizing power
  • Large anion: High polarizability
  • High charge: Increases polarization
\[\text{Covalent Character} \propto \frac{\text{Charge}}{\text{Size of Cation} + \text{Size of Anion}}\]

JEE Shortcut — Polarization

Covalent character order for alkali metal halides:

\[\text{LiCl} > \text{NaCl} > \text{KCl} > \text{RbCl} > \text{CsCl}\]

Covalent character order for silver halides:

\[\text{AgF} < \text{AgCl} < \text{AgBr} < \text{AgI}\]

PYQ — JEE Main 2021

Question: Which of the following is most covalent?

(a) SnCl₂    (b) SnCl₄    (c) FeCl₂    (d) FeCl₃

Solution:

According to Fajan's rule, higher the charge on cation, greater is its polarizing power, hence more covalent character.

Sn in SnCl₂ has +2 charge; Sn in SnCl₄ has +4 charge.

Fe in FeCl₂ has +2 charge; Fe in FeCl₃ has +3 charge.

Since +4 > +3 > +2, SnCl₄ has the highest polarizing power.

\[\text{SnCl}_4 \text{ is most covalent}\]

Answer: (b) SnCl₄

4

VSEPR Theory — Molecular Shapes

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the geometry of molecules based on the repulsion between electron pairs. This is a highly tested topic in JEE.

Order of Repulsion (Must Memorize)

\[LP-LP > LP-BP > BP-BP\]

Reason: Lone pairs are attracted to only one nucleus, so they spread out more and occupy more space.

4.1 VSEPR Chart (Steric Number vs Shape)

Steric No.BPLPHybridizationShapeExamples
220spLinearBeCl₂, CO₂
330sp²Trigonal PlanarBF₃, SO₃
321sp²BentSO₂, O₃
440sp³TetrahedralCH₄, NH₄⁺
431sp³PyramidalNH₃, PCl₃
422sp³Bent / V-shapeH₂O, OF₂
550sp³dTrigonal BipyramidalPCl₅
541sp³dSee-sawSF₄, TeCl₄
532sp³dT-shapeClF₃, BrF₃
523sp³dLinearXeF₂, I₃⁻
660sp³d²OctahedralSF₆
651sp³d²Square PyramidalBrF₅, XeOF₄
642sp³d²Square PlanarXeF₄
770sp³d³Pentagonal BipyramidalIF₇

PYQ — JEE Main 2023

Question: Which of the following molecules has a see-saw shape?

(a) XeF₄    (b) SF₄    (c) BF₃    (d) ClF₃

Solution:

Steric Number = (Valence e⁻ + Monovalent atoms)/2

SF₄: S has 6 valence e⁻, 4 F atoms. SN = (6+4)/2 = 5.

Steric Number 5 means sp³d hybridization.

Lone Pairs = 5 - 4 = 1. (4 Bond Pairs, 1 Lone Pair).

Shape with 4 BP and 1 LP = See-saw.

\[\text{SF}_4 \text{ has see-saw shape}\]

Answer: (b) SF₄

5

Valence Bond Theory & Hybridization

VBT explains covalent bond formation by the overlap of half-filled atomic orbitals. Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for bonding.

Hybridization Shortcut Formula

\[H = \frac{1}{2}(V + M - C + A)\]
  • V = Valence electrons of central atom
  • M = Number of monovalent atoms (H, F, Cl, etc.)
  • C = Charge on cation
  • A = Charge on anion

5.1 Types of Overlap

Sigma (σ) Bond

  • Formed by axial (head-on) overlap
  • Stronger, free rotation possible
  • s-s, s-p, p-p overlaps

Pi (π) Bond

  • Formed by lateral (sideways) overlap
  • Weaker, restricts rotation
  • Only p-p overlap (needs unhybridized p orbitals)

PYQ — JEE Main 2024

Question: The hybridization and geometry of XeOF₄ are:

(a) sp³d, trigonal bipyramidal

(b) sp³d², square pyramidal

(c) sp³d², octahedral

(d) sp³d, see-saw

Solution:

Xe in XeOF₄: Valence e⁻ (V) = 8

Monovalent atoms (M) = 4 (F atoms). O is divalent, so it doesn't count in M.

Charge = 0. Hybridization H = 1/2(8 + 4) = 6.

Steric Number = 6 → sp³d² hybridization.

Shape: SN=6, 1 LP (due to Xe=O double bond consuming 2 e⁻, leaving 4 Xe-F bonds and 1 Xe=O bond. Total 5 BP, 1 LP).

Geometry = Square Pyramidal.

\[\text{sp}^3\text{d}^2 \text{, Square Pyramidal}\]

Answer: (b) sp³d², square pyramidal

6

Molecular Orbital Theory (MOT)

MOT is the most advanced theory of chemical bonding. Unlike VBT, it treats the entire molecule as a single unit. This is the highest weightage topic in Chemical Bonding for JEE.

Bond Order Formula (Most Repeated PYQ)

\[B.O. = \frac{N_b - N_a}{2}\]

Where Nb = electrons in bonding MOs, Na = electrons in antibonding MOs

Higher bond order → stronger bond, shorter bond length, greater stability.

6.1 Energy Level Diagrams

There are two energy level sequences depending on the molecule:

For Z ≤ 7 (N₂, C₂, B₂)

σ1s, σ*1s, σ2s, σ*2s, π2p, σ2p, π*2p, σ*2p

Note: π2p has lower energy than σ2p due to less s-p mixing.

For Z ≥ 8 (O₂, F₂, Ne₂)

σ1s, σ*1s, σ2s, σ*2s, σ2p, π2p, π*2p, σ*2p

Note: σ2p has lower energy than π2p.

Magnetic Nature Shortcut

If all MOs are fully paired → Diamagnetic. If any unpaired electron exists in MO → Paramagnetic.

\[\text{O}_2 \text{ (Paramagnetic)} \quad ; \quad \text{N}_2 \text{ (Diamagnetic)}\]

B₂ and C₂ are also paramagnetic due to unpaired electrons in π2p orbitals!

PYQ — JEE Main 2023

Question: Which of the following species is diamagnetic?

(a) O₂    (b) O₂⁺    (c) O₂⁻    (d) O₂²⁻

Solution:

O₂ has 12 valence e⁻. Sequence for Z≥8: σ2s, σ*2s, σ2p, π2p(4), π*2p(2), σ*2p

O₂: 2 unpaired e⁻ in π*2p (Paramagnetic)

O₂⁺ (11 e⁻): 1 unpaired e⁻ in π*2p (Paramagnetic)

O₂⁻ (13 e⁻): 1 unpaired e⁻ in π*2p (Paramagnetic)

O₂²⁻ (14 e⁻): π*2p is fully filled. No unpaired electrons!

\[\text{O}_2^{2-} \text{ (Peroxide) is Diamagnetic}\]

Answer: (d) O₂²⁻

7

Bond Parameters & Dipole Moment

Bond parameters like bond length, bond angle, and dipole moment determine the physical properties and reactivity of molecules.

7.1 Dipole Moment

Formula

\[\mu = q \times d\]

Where q = charge, d = distance. Units: Debye (D), 1D = 3.335 × 10⁻³⁰ C·m

NH₃ vs NF₃ Dipole Moment (Classic JEE PYQ)

NH₃ has a higher dipole moment (1.47 D) than NF₃ (0.23 D).

Reason: In NH₃, orbital dipole of lone pair and bond dipoles of N-H are in the same direction. In NF₃, the N-F bond dipoles oppose the lone pair dipole, reducing the net dipole moment.

PYQ — JEE Main 2021

Question: Which of the following molecules has zero dipole moment?

(a) NH₃    (b) H₂O    (c) BF₃    (d) SO₂

Solution:

NH₃ is pyramidal (net μ ≠ 0)

H₂O is bent (net μ ≠ 0)

SO₂ is bent (net μ ≠ 0)

BF₃ is trigonal planar and perfectly symmetrical. The three B-F bond dipoles cancel each other out.

\[\mu_{\text{BF}_3} = 0\]

Answer: (c) BF₃

8

Hydrogen Bonding

Hydrogen bonding is a special type of dipole-dipole interaction. It's responsible for many unique properties of water and biological molecules.

Condition for H-Bonding

H atom must be bonded to a highly electronegative atom (F, O, or N).

\[\text{Strength: F-H...F} > \text{O-H...O} > \text{N-H...N}\]

8.1 Types of Hydrogen Bonding

Intermolecular H-Bond

Between two or more molecules. Increases boiling point and solubility.

Examples: H₂O, HF, NH₃, Alcohol

Intramolecular H-Bond

Within the same molecule. Decreases boiling point and solubility.

Examples: o-Nitrophenol, o-Hydroxybenzaldehyde

PYQ — JEE Main 2024

Question: Which compound has the lowest boiling point?

(a) p-Nitrophenol    (b) o-Nitrophenol    (c) m-Nitrophenol    (d) Phenol

Solution:

o-Nitrophenol forms intramolecular hydrogen bonds (between -OH and -NO₂ groups within the same molecule).

p-Nitrophenol and m-Nitrophenol form intermolecular hydrogen bonds, which require more energy (higher temperature) to break during boiling.

Since intramolecular H-bonding isolates the molecule, it has the lowest boiling point.

\[\text{o-Nitrophenol has the lowest BP}\]

Answer: (b) o-Nitrophenol

Previous Year Questions Analysis

JEE Main (Last 10 Years)

  • ✓ Molecular Orbital Theory (MOT): 30%
  • ✓ Hybridization & VSEPR: 25%
  • ✓ Dipole Moment & Shapes: 15%
  • ✓ Formal Charge: 10%
  • ✓ Hydrogen Bonding: 10%
  • ✓ Fajan's Rule: 10%

JEE Advanced (Last 10 Years)

  • ✓ Bond Order & Magnetic Nature: 25%
  • ✓ VSEPR Shapes (Xe compounds): 20%
  • ✓ Dipole Moment of Substituted Benzenes: 20%
  • ✓ Intramolecular vs Intermolecular H-Bonds: 15%
  • ✓ Resonance & Formal Charge: 20%

Top 10 Most Repeated Question Types

  1. Bond order and magnetic nature of O₂, N₂, NO, CO species
  2. Identifying hybridization and geometry of Xe compounds (XeF₄, XeF₆, XeOF₄)
  3. Formal charge on atoms in Ozone (O₃) or Carbonate (CO₃²⁻)
  4. Order of dipole moment (e.g., NH₃ vs NF₃)
  5. Covalent character using Fajan's rule (e.g., SnCl₄ vs SnCl₂)
  6. Identifying shape of molecules with lone pairs (SF₄, ClF₃, I₃⁻)
  7. Boiling point comparison based on H-bonding (o vs p-nitrophenol)
  8. Number of sigma and pi bonds in a given structure
  9. Lattice enthalpy order based on ionic size/charge
  10. Paramagnetic vs Diamagnetic nature of B₂, C₂, N₂, O₂

Weightage Analysis

JEE Main: 8-12 marks (2-3 questions)
JEE Advanced: 10-15 marks (3-4 questions)
Difficulty Level: Easy to Hard
Time Required: 4-5 hours practice
Download Complete PYQ PDF (2015-2024)

Practice Problem Set

Level 1: Basic (JEE Main Standard)

  1. Draw the Lewis structure of CO₃²⁻ and calculate the formal charge on each oxygen.
  2. Predict the shape of I₃⁻ using VSEPR theory.
  3. Calculate the bond order of NO⁺ and predict its magnetic nature.
  4. Which has a higher dipole moment: NH₃ or NF₃? Explain.
  5. How many sigma and pi bonds are present in acetylene (C₂H₂)?

Level 2: Intermediate (JEE Main/Advanced)

  1. Arrange in order of increasing lattice energy: NaF, NaCl, MgO, MgCl₂.
  2. Explain why O₂ is paramagnetic while N₂ is diamagnetic using MOT.
  3. Why does o-nitrophenol have a lower boiling point than p-nitrophenol?
  4. Identify the hybridization of Xe in XeF₂, XeF₄, and XeF₆.
  5. Which is more covalent: AgCl or AgI? Justify using Fajan's rule.

Level 3: Advanced (JEE Advanced/Olympiad)

  1. Out of O₂, O₂⁺, O₂⁻, and O₂²⁻, which has the longest bond length? Prove with MOT.
  2. Draw the molecular orbital energy level diagram for CO and calculate its bond order.
  3. Predict the shapes of ClF₃ and BrF₅ and state their hybridization.
  4. Calculate the dipole moment of a molecule with charge 1.6 × 10⁻¹⁹ C and bond length 1.5 Å.
  5. Why does B₂ show paramagnetism even though it is in period 2? Explain using MOT.
Get Solutions PDF with Step-by-Step Working

> FAQ

Frequently Asked Questions

What is the formula for calculating bond order in MOT?

Bond Order (B.O.) = 1/2 * (Number of electrons in bonding molecular orbitals - Number of electrons in antibonding molecular orbitals). A higher bond order means a stronger and shorter bond.

What is the VSEPR theory postulate?

Valence Shell Electron Pair Repulsion (VSEPR) theory states that electron pairs around the central atom arrange themselves to minimize repulsion. The order of repulsion is Lone Pair-Lone Pair > Lone Pair-Bond Pair > Bond Pair-Bond Pair.

How is formal charge calculated?

Formal Charge = (Valence electrons) - (Non-bonding electrons) - (1/2 * Bonding electrons). The most stable Lewis structure has the smallest formal charges on atoms.

Why is O2 paramagnetic?

According to Molecular Orbital Theory (MOT), O2 has two unpaired electrons in the degenerate π*2p antibonding molecular orbitals. This presence of unpaired electrons makes it paramagnetic, a fact that Valence Bond Theory failed to explain.

What is the order of strength of hydrogen bonds?

The strength of hydrogen bonding depends on the electronegativity of the atom bonded to hydrogen. The order is F-H...F > O-H...O > N-H...N. Intermolecular hydrogen bonding is generally stronger than intramolecular hydrogen bonding.

How many questions come from chemical bonding in JEE Mains?

On average 2-3 questions (8-12 marks) appear from chemical bonding in every JEE Main session. Most questions are from Hybridization, MOT bond order calculations, and VSEPR shapes.

What is the weightage of chemical bonding in JEE Advanced?

Chemical bonding carries roughly 6-8% weightage in JEE Advanced. It is a highly conceptual chapter, and questions often combine MOT with periodic properties or dipole moment with structural isomerism.

What are the limitations of the Octet Rule?

The octet rule fails for: 1) Odd electron molecules like NO, NO2, 2) Incomplete octet in BeCl2, BF3, 3) Expanded octet in PCl5, SF6, IF7 (where central atom has 10, 12, 14 electrons), and 4) It does not explain the shape of molecules or relative stability of species.

What is the difference between sigma and pi bonds?

Sigma (σ) bonds are formed by axial (head-on) overlap of orbitals and are stronger. Pi (π) bonds are formed by lateral (sideways) overlap of p-orbitals and are weaker. A single bond is a sigma bond; a double bond has 1σ + 1π; a triple bond has 1σ + 2π.

Which has more dipole moment: NH3 or NF3?

NH3 has a higher dipole moment (1.47 D) than NF3 (0.23 D). In NH3, the orbital dipole of the lone pair and the bond dipoles of N-H are in the same direction. In NF3, the N-F bond dipoles oppose the orbital dipole of the lone pair, reducing the net dipole moment.

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Complete Chemical Bonding Mastery Guide for JEE 2026

Why Chemical Bonding is THE Most Important Chapter

Chemical Bonding is the bridge between Physical and Inorganic Chemistry. Here's why it's crucial:

  • Highest Weightage: 2-3 questions in JEE Main, 3-4 in JEE Advanced
  • Foundation for Organic & Inorganic: Hybridization is used everywhere in reaction mechanisms; Lattice energy explains salt solubility.
  • High Scoring: MOT and VSEPR questions are purely logical and formula-based.
  • Rank Decider: Exceptional cases (like B₂ paramagnetism or NH₃ vs NF₃ dipole) distinguish top rankers.

Topic-wise Weightage Analysis

MOT (Bond Order, Magnetism) 30%
Hybridization & VSEPR 25%
Dipole Moment & Bond Parameters 15%
Formal Charge & Resonance 10%
Hydrogen Bonding & Fajan's Rule 20%

Must-Remember Formulas

Formal Charge

F.C. = V - (L + S/2)

Bond Order (MOT)

B.O. = (N_b - N_a) / 2

N_b = Bonding e-, N_a = Antibonding e-

Dipole Moment

μ = q × d (in Debye)

1 D = 3.335 × 10^-30 C·m

Hybridization Shortcut

H = 1/2 (V + M - C + A)

V=Valence, M=Monovalent, C=Cation, A=Anion

5-Step Exam Strategy

  1. 1
    MOT Questions: Always check if molecule is Z≤7 (N₂ type) or Z≥8 (O₂ type) to avoid filling orbitals in wrong sequence.
  2. 2
    VSEPR: Calculate lone pairs on central atom FIRST. Geometry depends on BP + LP, but shape only considers atoms.
  3. 3
    Fajan's Rule: Smaller cation + larger anion + higher charge = More covalent.
  4. 4
    Dipole Moment: Draw 3D structure to see if bond dipoles cancel out (e.g., BF₃ vs NH₃).
  5. 5
    H-Bonding: Ortho-isomers have intramolecular H-bonds (lower BP). Para/Meta have intermolecular (higher BP).

7-Day Chemical Bonding Mastery Roadmap

Day 1

Lewis & FC

  • • Lewis structures
  • • Octet exceptions
  • • Formal charge
Day 2

Ionic & Fajan

  • • Lattice enthalpy
  • • Born-Haber cycle
  • • Covalent character
Day 3

VSEPR

  • • Steric number
  • • Lone pair repulsion
  • • VSEPR chart
Day 4

Hybridization

  • • VBT postulates
  • • sp, sp2, sp3, sp3d
  • • Geometry & angles
Day 5

MOT

  • • Energy level diagrams
  • • Bond order
  • • Magnetic nature
Day 6

Bond Params

  • • Dipole moment
  • • H-bonding
  • • Resonance
Day 7

PYQs

  • • 80+ PYQs
  • • Revise MOT
  • • Formula sheet

Chemical Bonding and Molecular Structure — Complete JEE Study Guide

The Chemical Bonding and Molecular Structure is Chapter 4 of NCERT Class 11 Chemistry and forms the core of chemical physics concepts for JEE Main and Advanced. This chapter covers the nature of chemical bonds—ionic, covalent, and coordinate—along with advanced theories like VSEPR, Valence Bond Theory (VBT), and Molecular Orbital Theory (MOT).

The most important topics for JEE from this chapter are Molecular Orbital Theory (MOT), Hybridization, VSEPR shapes, Formal Charge, and Dipole Moment. MOT alone accounts for nearly 30% of all PYQs from this chapter, with questions specifically asking for bond order calculations and magnetic nature (paramagnetic vs diamagnetic) of species like O₂, N₂, NO, and CO. Hybridization and VSEPR theory account for another 25%, where students must identify the geometry of Xenon compounds (XeF₂, XeF₄, XeF₆) and interhalogen compounds (SF₄, ClF₃, IF₇).

Students should focus on understanding the difference between intermolecular and intramolecular hydrogen bonding, as it directly affects boiling points. Fajan's rule is another critical topic used to predict the covalent character of ionic bonds, which is highly tested in JEE Main. The shortcut formula for hybridization, H = 1/2(V + M - C + A), is a lifesaver for time-crunch situations.

For JEE Advanced, focus on the anomalous behaviors: why B₂ is paramagnetic, why NH₃ has a higher dipole moment than NF₃, and why o-nitrophenol is more volatile than p-nitrophenol. Drawing correct MO energy level diagrams for Z ≤ 7 vs Z ≥ 8 is essential, as confusing the order of π2p and σ2p orbitals is the most common mistake students make in the exam.