⚗️ CBSE · Class 11 · Chemistry · Chapter 4

Chemical Bonding and
Molecular Structure

Complete chapter resources for CBSE Class 11 Chemistry — topic breakdown, key concepts, Lewis structures, VSEPR theory, hybridisation, molecular orbital theory, sample questions, and instant AI question paper generation.

5Topics
8–10Board marks
8Sample questions
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Key Concepts — Chapter 4
  • Hybridisation: H = ½(V + M − C + A)
  • Formal charge: FC = V − N − B/2
  • Bond order (MOT): BO = (Nb − Na) / 2
  • VSEPR notation: AXnEm (X=bonds, E=lone pairs)
  • Dipole moment: μ = q × d (Debye, D)
  • H-bond condition: Small highly electronegative atom (N, O, F) + H
Chapter Overview

What this chapter covers

Chapter 4 of NCERT Class 11 Chemistry explains why and how atoms combine to form molecules. It opens with the concept of chemical bond — atoms attain stable electronic configurations (octet rule) by sharing, transferring, or delocalising electrons. The chapter covers three fundamental bond types: ionic bonds (formed by complete electron transfer), covalent bonds (electron sharing), and coordinate or dative bonds (one atom donates both electrons). Lewis dot structures are the primary tool for representing bonding and lone-pair distribution, and formal charges help identify the most stable resonance form.

Two major theories extend Lewis structures to three-dimensional shapes. VSEPR (Valence Shell Electron Pair Repulsion) theory predicts molecular geometry by minimising repulsion among all electron pairs — lone pairs repel more strongly than bonding pairs, explaining the distorted angles in NH₃ (107°) and H₂O (104.5°). Valence Bond Theory (VBT) introduces the concept of orbital hybridisation (sp, sp², sp³, sp³d, sp³d²) and distinguishes between sigma (σ) bonds formed by head-on orbital overlap and pi (π) bonds formed by lateral overlap.

Molecular Orbital Theory (MOT) goes further by combining atomic orbitals to form bonding and antibonding molecular orbitals. It correctly predicts the paramagnetism of O₂ and the instability of He₂ — results that VBT cannot explain. Bond order = (Nb − Na) / 2, where Nb and Na are electrons in bonding and antibonding MOs respectively. The chapter concludes with hydrogen bonding (intermolecular and intramolecular) and its role in determining boiling points, solubility, and biological structures.

Topics

What's inside Chapter 4

As per NCERT Class 11 Chemistry Part 1 (CBSE syllabus)

Topic 1
Ionic and Covalent Bonds — Lewis Structures
Formation of ionic bonds (electronegativity difference > 1.7), covalent bonds, and coordinate bonds. Drawing Lewis dot structures, applying the octet rule, formal charges, and resonance structures.
Topic 2
VSEPR Theory and Molecular Geometry
Predicting 3D shapes using electron-pair repulsion: linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral. Effect of lone pairs on bond angles. Shape vs. electron geometry distinction.
Topic 3
Valence Bond Theory and Hybridisation
Orbital overlap concept, sigma and pi bonds, and hybridisation types — sp (linear, 180°), sp² (trigonal planar, 120°), sp³ (tetrahedral, 109.5°), sp³d, sp³d². Linking hybridisation to geometry.
Topic 4
Molecular Orbital Theory (MOT)
Formation of bonding and antibonding MOs by LCAO. MO energy diagrams for homodiatomic molecules (H₂, He₂, N₂, O₂, F₂). Bond order calculation; magnetic properties (dia vs. paramagnetic).
Topic 5
Polarity, Dipole Moment & Hydrogen Bonding
Electronegativity difference and bond polarity; dipole moment (μ = q × d) and molecular polarity. Intermolecular and intramolecular hydrogen bonding; effect on physical properties like boiling point and solubility.
Connections

How this chapter fits in

Useful for setting question difficulty and cross-chapter papers.

Builds on
Ch 2 · Structure of Atom
Quantum numbers, orbital shapes, electron configurations
Ch 3 · Classification of Elements
Electronegativity trends, atomic radii, ionisation energy
Chapter 4 Chemical
Bonding
Leads to
Ch 9 · Hydrogen
Hydrogen bonding explains water's anomalous properties
Class 12 · Coordination Compounds
Coordinate bonding, crystal field theory, hybridisation of complex ions
Board Blueprint

Marks & question-type breakdown

Typical pattern based on CBSE Class 11 Chemistry board papers from the last five years.

Question type Marks Typical count What's usually tested
MCQ / Assertion–Reason 1 2–3 Bond order, hybridisation identification, magnetic character of a molecule
Very Short Answer 2 1–2 Lewis structure, formal charge, VSEPR shape with bond angle
Short Answer 3 1 Hybridisation + shape + bond angle for a given molecule; dipole moment comparison
Long Answer / MOT 5 1 MO energy level diagram, bond order, magnetic property, stability comparison
Total (approximate) 8–10 5–7 Weightage varies across paper sets and years
Sample Questions

8 sample questions — generated by MarksZen AI

Aligned to CBSE Class 11 Chemistry Chapter 4. Covers all question types across Easy, Medium, and Hard difficulty.

Q1 Easy 1 mark MCQ
The hybridisation of the central atom in BF₃ is: (a) sp (b) sp² (c) sp³ (d) sp³d
Q2 Easy 2 marks Short Answer
Draw the Lewis dot structure of CO₂. State the formal charge on each atom and identify the type of bonds present.
Q3 Medium 2 marks Short Answer
Using VSEPR theory, predict the shape and approximate bond angle of the following molecules: (i) NH₃ (ii) H₂O. Explain how lone pairs cause deviation from the ideal tetrahedral angle.
Q4 Medium 3 marks Short Answer
State the hybridisation, geometry, and bond angles for each of the following: (i) PCl₅ (ii) SF₆ (iii) XeF₂
Q5 Medium 3 marks Short Answer
Explain why the dipole moment of CO₂ is zero whereas that of SO₂ is non-zero, even though both contain polar bonds. Draw the structures to support your answer.
Q6 Hard 4 marks Word Problem
The boiling point of HF (19.5 °C) is much higher than expected for a compound of its molecular mass, while HCl boils at −85 °C. (i) Identify the intermolecular force responsible for HF's anomalously high boiling point. (ii) Explain why this force is stronger in HF than in HCl. (iii) Draw a diagram showing the hydrogen-bonded network in liquid HF.
Q7 Hard 5 marks Long Answer
Write the molecular orbital electronic configuration of O₂ and O₂⁻ (superoxide ion). (i) Calculate the bond order of each. (ii) Compare their bond lengths and bond energies. (iii) State whether each species is diamagnetic or paramagnetic. Justify your answer.
Q8 Hard 5 marks Long Answer
Explain the formation of N₂ molecule using Molecular Orbital Theory. (i) Write the MO electronic configuration and determine the bond order. (ii) Why is N₂ diamagnetic? (iii) Compare the stability of N₂ and N₂⁺. Which has the higher bond order? (iv) Account for the exceptionally high bond dissociation energy of N₂ (945 kJ/mol).
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Previous Year Questions

From CBSE board examinations

Actual questions from past Class 11 Chemistry board papers — Chemical Bonding chapter.

Board 20233 marks
Using VSEPR theory, explain the shapes of PCl₃ and PCl₅. Why does phosphorus form PCl₅ but nitrogen does not form NCl₅? (CBSE All India 2023)
Board 20222 marks
Write the molecular orbital configuration of O₂ molecule. Calculate its bond order and state whether it is diamagnetic or paramagnetic with reason. (CBSE Delhi 2022)
Board 20203 marks
Although the electronegativities of N and Cl are different, the dipole moment of NCl₃ is smaller than that of NH₃. Explain this in terms of the direction of bond moments and the lone pair. Also state the hybridisation and shape of both molecules. (CBSE 2020)
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FAQ

Questions teachers ask

How many marks does Chemical Bonding and Molecular Structure carry in the CBSE Class 11 Chemistry board exam? +
Chemical Bonding and Molecular Structure typically carries 8–10 marks in the CBSE Class 11 Chemistry board examination. Questions span MCQs (1 mark), short-answer questions (2–3 marks), and one long-answer question (5 marks). Topics like hybridisation, VSEPR theory, and molecular orbital theory are the most frequently tested.
What is the easiest way to determine the hybridisation of a central atom? +
Use the formula: Hybridisation = (V + M − C + A) / 2, where V = valence electrons of central atom, M = monovalent atoms bonded, C = positive charge on ion, A = negative charge on ion. For example, for SF6: V=6, M=6, charge=0 → (6+6)/2 = 6 → sp3d2. This shortcut works for all standard CBSE exam molecules.
How do you predict molecular shape using VSEPR theory in board exams? +
Count the total electron pairs (bonding + lone pairs) around the central atom. Lone pairs occupy equatorial positions and cause greater repulsion, distorting ideal geometry. For example, NH3 has 3 bonding pairs + 1 lone pair = tetrahedral electron geometry but trigonal pyramidal molecular shape with bond angle 107°. Board questions frequently ask for shape + bond angle together for 2–3 marks.
What is the difference between sigma and pi bonds and how do they relate to hybridisation? +
A sigma (σ) bond is formed by head-on (axial) overlap of orbitals and is the first bond in any bond order. A pi (π) bond is formed by lateral (sideways) overlap and always exists in addition to a sigma bond. In hybridisation: sp3 carbon has 4 sigma bonds (no pi), sp2 has 3 sigma + 1 pi, sp has 2 sigma + 2 pi. This directly links hybridisation to bond type — a very common 2-mark board question.
How do I generate a custom question paper for Chemical Bonding using MarksZen? +
Sign up for a free MarksZen account, choose CBSE Class 11 Chemistry, select Chapter 4 (Chemical Bonding and Molecular Structure), set your preferred question-type mix and total marks — the AI generates a complete board-aligned paper with answer key in under 2 minutes, ready for PDF export.