TalentJR

Chemical Bonding and Molecular Structure – Class 11 Chemistry Chapter 4 Notes & NCERT PDF

Class 11ChemistryChapter 4NCERT book: Chemistry Part I - Class 11

Notes and a simple summary of Chapter 4 of the NCERT Class 11 Chemistry book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

🎧 Listen to the content

Chapter 4: Chemical Bonding and Molecular Structure

Introduction

Why do atoms combine, and why only in certain ratios? Why is H₂O bent but CO₂ straight? This unit builds the ideas of chemical bonding in order: the octet rule and Lewis structures, ionic bonds, bond parameters, the VSEPR model for shapes, valence bond theory with hybridisation, molecular orbital theory, and hydrogen bonding.

Key Concepts

4.1 Kössel–Lewis Approach

  • Octet rule: atoms combine by transferring or sharing electrons to reach eight electrons in the valence shell (noble-gas configuration).
  • Covalent bond (Langmuir, Lewis): a shared electron pair. Single, double and triple bonds share one, two and three pairs.
  • Lewis structures, step by step: (1) add all valence electrons, adding one per negative charge and subtracting one per positive charge; (2) choose a skeleton with the least electronegative atom in the centre; (3) put single bonds, then complete octets, converting lone pairs into multiple bonds where needed.
  • Formal charge = (valence electrons in free atom) − (non-bonding electrons) − ½(bonding electrons). The structure with the smallest formal charges is the most stable. In O₃, the central O has +1, one end O has −1, and the other 0.
  • Limitations of the octet rule: incomplete octet (BeCl₂, BCl₃), odd-electron molecules (NO, NO₂), and expanded octet (PF₅, SF₆, H₂SO₄). It also cannot explain shapes or relative stability.

    • 4.2 Ionic (Electrovalent) Bond

    Formed by complete electron transfer between a metal of low ionization enthalpy and a non-metal of high (negative) electron gain enthalpy. Ionic compounds form 3-D crystal lattices. Lattice enthalpy is the energy needed to separate one mole of a solid ionic compound into gaseous ions (NaCl: 788 kJ mol⁻¹); a high lattice enthalpy makes ionic compounds stable.

    4.3 Bond Parameters

  • Bond length: equilibrium distance between nuclei (measured by X-ray/electron diffraction); covalent radius is half the distance between two identical bonded atoms.
  • Bond angle: angle between bonding orbitals around the central atom (H–O–H in water is 104.5°).
  • Bond enthalpy: energy to break one mole of bonds in the gas phase: H₂ 435.8, O₂ 498, N₂ 946 kJ mol⁻¹. Larger bond enthalpy means a stronger bond.
  • Bond order: number of bonds between two atoms (H₂ 1, O₂ 2, N₂ 3). Higher bond order means higher bond enthalpy and shorter bond length.
  • Resonance: when one Lewis structure is not enough, the true structure is a resonance hybrid of canonical forms. Examples: O₃, CO₃²⁻ (all three C–O bonds equal), CO₂ (C–O length 115 pm, between C=O 121 pm and C≡O 110 pm).
  • Polarity: a bond between atoms of different electronegativity is polar. Dipole moment μ = charge × distance, in debye (D). Symmetric molecules such as CO₂, BF₃, CH₄ and CCl₄ have μ = 0; H₂O (1.85 D) and NH₃ (1.47 D) are polar. NF₃ (0.23 D) is much less polar than NH₃ because in NF₃ the lone-pair moment opposes the bond moments, while in NH₃ they add up. Fajans' rules: small, highly charged cations and large anions increase covalent character.

    • 4.4 VSEPR Theory

    Shape depends on the number of electron pairs (bond pairs + lone pairs) around the central atom, which repel and move as far apart as possible. Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.

  • 2 pairs → linear (BeCl₂, 180°)
  • 3 pairs → trigonal planar (BF₃, 120°); with 1 lone pair → bent (SO₂)
  • 4 pairs → tetrahedral (CH₄, 109.5°); 1 lone pair → trigonal pyramidal (NH₃, 107°); 2 lone pairs → bent (H₂O, 104.5°)
  • 5 pairs → trigonal bipyramidal (PCl₅); with lone pairs → see-saw (SF₄), T-shaped (ClF₃)
  • 6 pairs → octahedral (SF₆); 1 lone pair → square pyramidal (BrF₅); 2 lone pairs → square planar (XeF₄)

    • 4.5 Valence Bond Theory

    A bond forms when half-filled atomic orbitals overlap; greater overlap gives a stronger bond. In H₂ formation the potential energy falls to a minimum at 74 pm. Sigma (σ) bonds form by head-on (axial) overlap; pi (π) bonds by sideways overlap of p orbitals. σ bonds are stronger than π bonds; a double bond is 1σ + 1π and a triple bond is 1σ + 2π.

    4.6 Hybridisation

    Atomic orbitals of similar energy mix to give equivalent hybrid orbitals.

  • sp: linear, 180° (BeCl₂, C₂H₂)
  • sp²: trigonal planar, 120° (BCl₃, C₂H₄)
  • sp³: tetrahedral, 109.5° (CH₄; NH₃ and H₂O with lone pairs)
  • sp³d: trigonal bipyramidal (PCl₅, axial bonds longer than equatorial)
  • sp³d²: octahedral (SF₆)

    • 4.7 Molecular Orbital Theory

    Atomic orbitals combine by LCAO to form bonding (lower energy) and antibonding (higher energy) molecular orbitals. Combining orbitals must have similar energy, the same symmetry about the molecular axis, and maximum overlap.

  • Filling order up to N₂: σ1s, σ1s, σ2s, σ2s, (π2pₓ = π2p_y), σ2p_z, (π2pₓ = π2p_y), σ*2p_z. For O₂ and F₂, σ2p_z lies below the π2p orbitals.
  • Bond order = ½(N_b − N_a). A positive value means the molecule is stable; zero means it does not exist (He₂, Ne₂).
  • Bond orders: H₂ 1, He₂ 0, Li₂ 1, C₂ 2, N₂ 3, O₂ 2, F₂ 1.
  • Magnetism: O₂ has two unpaired electrons in π* orbitals, so it is paramagnetic — something Lewis theory could not explain.

    • 4.9 Hydrogen Bonding

    An H atom bonded to a highly electronegative atom (F, O, N) is attracted to a lone pair on another electronegative atom (shown by a dotted line). Intermolecular H-bonds (HF, water) raise boiling points; intramolecular H-bonds occur within one molecule (o-nitrophenol).

    Worked Examples

  • Lewis structure of CO. Valence electrons = 4 + 6 = 10. A single bond with octets leaves too few electrons, so form a triple bond: :C≡O: (one lone pair on each atom).
  • Lewis structure of NO₂⁻. Electrons = 5 + 2(6) + 1 = 18. Skeleton O–N–O; one N=O double bond, one N–O single bond, a lone pair on N; the ion is a resonance hybrid of two equivalent forms.
  • Formal charges in O₃. Central O: 6 − 2 − ½(6) = +1; double-bonded end O: 6 − 4 − ½(4) = 0; single-bonded end O: 6 − 6 − ½(2) = −1.
  • Carbonate ion (CO₃²⁻). 24 electrons; three equivalent canonical forms; all C–O bonds are identical in reality, so it is a resonance hybrid.
  • Bond order of O₂, O₂⁺ and O₂⁻. O₂: ½(10 − 6) = 2; O₂⁺: ½(10 − 5) = 2.5; O₂⁻: ½(10 − 7) = 1.5. Stability: O₂⁺ > O₂ > O₂⁻.
  • Shape of NH₃. 4 electron pairs (3 bond pairs + 1 lone pair), sp³ → trigonal pyramidal; the lone pair squeezes the angle from 109.5° to 107°.

    • Formula Sheet

  • Formal charge = V − L − ½B
  • Bond order (MO) = ½(N_b − N_a)
  • Dipole moment μ = q × d (1 D = 3.33564 × 10⁻³⁰ C m)
  • Repulsion order: lp–lp > lp–bp > bp–bp
  • Bond angles: BeCl₂ 180°, BF₃ 120°, CH₄ 109.5°, NH₃ 107°, H₂O 104.5°

    • Exam Focus

  • Explain why BeH₂ has zero dipole moment but H₂O does not. BeH₂ is linear, so the bond dipoles cancel; H₂O is bent, so they add to 1.85 D.
  • Using MO theory, show that O₂ is paramagnetic. Write the configuration σ1s²σ1s²σ2s²σ2s²σ2p_z²π2pₓ²π2p_y²π2pₓ¹π2p_y¹; there are two unpaired electrons and the bond order is 2.
  • Distinguish σ and π bonds. Axial vs sideways overlap; free rotation vs restricted rotation; σ is stronger; a π bond forms only in addition to a σ bond.

    • Key Terms

  • Octet rule (अष्टक नियम)
  • Covalent bond (सहसंयोजी बंध)
  • Ionic bond (आयनिक बंध)
  • Lattice enthalpy (जालक एन्थैल्पी)
  • Formal charge (औपचारिक आवेश)
  • Resonance (अनुनाद)
  • Bond order (आबंध कोटि)
  • Dipole moment (द्विध्रुव आघूर्ण)
  • Hybridisation (संकरण)
  • Sigma bond (सिग्मा बंध)
  • Pi bond (पाई बंध)
  • Molecular orbital (आण्विक कक्षक)
  • Paramagnetic (अनुचुंबकीय)
  • Hydrogen bond (हाइड्रोजन बंध)

    • Common Mistakes

  • Counting total electrons in a Lewis structure without adjusting for ionic charge (NO₂⁻ needs +1).
  • Saying NH₃ and H₂O are tetrahedral — the electron-pair geometry is tetrahedral, but the shapes are pyramidal and bent.
  • Assuming a polar bond always means a polar molecule (CO₂ and CCl₄ are non-polar).
  • Using the N₂ MO filling order for O₂ and F₂ (σ2p_z comes before π2p there).
  • Forgetting that every multiple bond contains exactly one σ bond.

    • 💡 Key Learning Points

    • ✓Write Lewis structures (including ions) and calculate formal charges to choose the most stable structure.
    • ✓Explain ionic bond formation and the role of lattice enthalpy.
    • ✓Use bond parameters — length, angle, enthalpy, order, resonance, dipole moment — to compare molecules.
    • ✓Predict molecular shapes with VSEPR theory and assign hybridisation (sp, sp², sp³, sp³d, sp³d²).
    • ✓Apply molecular orbital theory to find bond order and magnetic behaviour of homonuclear diatomics, and describe hydrogen bonding.

    👨‍🏫 Teaching Tips

    • →Use ball-and-stick or balloon models (two to six balloons tied together) to show VSEPR geometries physically.
    • →Practise Lewis structures with a fixed four-step routine on the board: count electrons → skeleton → single bonds → complete octets/add multiple bonds.
    • →Compare the dipole moments of NH₃ and NF₃ with vector arrows to show how lone pairs change polarity.
    • →Build MO energy-level diagrams for H₂ to O₂ as a progressive worksheet; let students discover that O₂ is paramagnetic.
    • →Demonstrate liquid oxygen (via video) being attracted to a magnet to connect MO theory with evidence.
    • →Relate hydrogen bonding to real life — ice floating on water, high boiling point of water compared with H₂S.
    • →Use a σ/π overlap animation to show why π bonds restrict rotation.

    📋 Assessment Questions

    1. Draw the Lewis structure of NO₂⁻ and assign formal charges.
    2. Predict the shape and bond angle of NH₃ and H₂O using VSEPR.
    3. Calculate the bond order of N₂, O₂ and O₂⁻ and compare their stability.
    4. Explain why CO₂ has zero dipole moment though its C=O bonds are polar.
    5. Differentiate between σ and π bonds with examples.

    Ask the AI tutor about Class 11 Chemistry Chapter 4

    Get question answers, important questions, a quick MCQ quiz or a simple explanation of “Chemical Bonding and Molecular Structure”. Free — sign in with Google.

    How the AI tutor works →

    👩‍🏫 Class 11 Chemistry classes near you — online or at home

    Prefer a real teacher? TalentJR teachers take live online classes anywhere in India and visit homes in selected cities. Tell us what you need and our team will call you with timings and fees.

    • Live online classesSmall groups or 1:1 with a teacher, from home, any city.
    • Teacher at homeA TalentJR teacher visits your home (selected cities).
    • At a centreClasses at a TalentJR partner centre near you, where available.

    Keep learning

    All chapters in Chemistry Part I - Class 11

    1. 1. Some Basic Concepts of Chemistry
    2. 2. Structure of Atom
    3. 3. Classification of Elements and Periodicity in Properties
    4. 4. Chemical Bonding and Molecular Structure
    5. 5. Thermodynamics
    6. 6. Equilibrium