CBSE Class 11 Chemistry Chapter 4 Notes: Chemical Bonding and Molecular Structure

Chapter Overview

Chemical Bonding and Molecular Structure is the fourth chapter of CBSE Class 11 Chemistry and one of the most important for both board exams and competitive exams. It explains why atoms combine, how bonds form, and how the arrangement of bonds determines molecular shape and properties.

This chapter covers the Kossel-Lewis approach to bonding, ionic and covalent bonds, VSEPR theory for predicting molecular shapes, hybridization for explaining observed geometries, molecular orbital theory for understanding bond order and magnetic properties, and hydrogen bonding. Master this chapter and you will understand the structure behind every reaction in chemistry.

1. Kossel-Lewis Approach to Chemical Bonding

1.1 Octet Rule

Atoms tend to gain, lose, or share electrons to achieve a noble gas configuration (8 electrons in the valence shell, except H and He which need 2). This is called the octet rule.

Lewis symbols: Valence electrons of an atom are represented by dots around the element symbol.

1.2 Formal Charge

Formal charge helps determine the most stable Lewis structure:

Rules for choosing the best Lewis structure: 1. The structure with the smallest formal charges (closest to zero) is preferred. 2. Negative formal charge should reside on the more electronegative atom. 3. Like charges on adjacent atoms should be avoided.

2. Ionic Bond (Electrovalent Bond)

An ionic bond is formed by the complete transfer of electrons from a metal atom (low IE) to a non-metal atom (high electron affinity), resulting in electrostatic attraction between cations and anions.

Example — Formation of NaCl:

Conditions favourable for ionic bond formation: 1. Low ionization enthalpy of the metal atom. 2. High electron gain enthalpy (more negative) of the non-metal atom. 3. High lattice energy of the resulting ionic compound.

Properties of ionic compounds:

• High melting and boiling points (strong electrostatic forces).
• Conduct electricity in molten state and in aqueous solution.
• Soluble in polar solvents (water), insoluble in non-polar solvents.
• Hard and brittle.

3. Covalent Bond

A covalent bond is formed by the sharing of electron pairs between two atoms (usually non-metals).

Types of covalent bonds:

• Single bond: One shared pair (e.g., H-H, Cl-Cl)
• Double bond: Two shared pairs (e.g., O=O, C=O)
• Triple bond: Three shared pairs (e.g., N≡N, C≡C)

Coordinate (Dative) bond: A covalent bond in which both electrons of the shared pair come from one atom.

Fajan's Rules (Polarization of Ionic Bonds):

4. Bond Parameters

4.1 Bond Length

The distance between the nuclei of two bonded atoms. Bond length depends on:

• Size of atoms (larger atoms = longer bonds)
• Bond order (triple < double < single)

4.2 Bond Angle

The angle between two bonds originating from the same atom. Determines molecular shape.

4.3 Bond Enthalpy (Bond Dissociation Energy)

The energy required to break one mole of bonds of a particular type in gaseous molecules.

• Larger bond enthalpy means a stronger bond.
• For polyatomic molecules, average bond enthalpy is used.

4.4 Bond Order

The number of chemical bonds between a pair of atoms.

Bond order is directly proportional to bond strength and inversely proportional to bond length.

4.5 Resonance

When a single Lewis structure cannot represent a molecule adequately, multiple structures (resonance structures or canonical forms) are drawn. The actual structure is a resonance hybrid of all contributing structures.

Example — Ozone (O₃):

The actual O-O bond length in ozone is intermediate between a single and a double bond.

Resonance energy: The difference between the actual energy of the molecule and the energy of the most stable contributing structure. The resonance hybrid is always more stable than any individual contributing structure.

5. VSEPR Theory

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular shapes based on the principle that electron pairs (bonding and lone pairs) around a central atom repel each other and arrange themselves to maximize separation.

Order of repulsion: Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair

Steps to determine molecular shape: 1. Count the number of valence electrons on the central atom. 2. Add one electron for each bonded atom (each bond). 3. Add or subtract electrons for ionic charges. 4. Divide total by 2 to get electron pairs. 5. Determine the electron-pair geometry from the number of pairs. 6. Subtract lone pairs to get the molecular shape.

VSEPR Geometry Table

6. Hybridization

Hybridization is the concept of mixing atomic orbitals of the same atom to form new orbitals (hybrid orbitals) of equal energy and identical shape, which can overlap more effectively with orbitals of other atoms.

6.1 Types of Hybridization

6.2 Hybridization with Lone Pairs

Lone pairs occupy more space than bonding pairs, compressing bond angles. Each additional lone pair reduces the bond angle by approximately 2-2.5° from the ideal geometry.

6.3 Hybridization in Carbon Compounds

Key rule: Count the number of sigma bonds + lone pairs on the central atom to determine hybridization.

7. Molecular Orbital Theory (MOT)

MOT treats electrons in molecules as belonging to the entire molecule, not individual atoms. Atomic orbitals combine to form molecular orbitals.

7.1 Bonding and Antibonding Orbitals

7.2 Rules for Combining Atomic Orbitals

1. Linear combination: Only atomic orbitals of comparable energy combine effectively. 2. Symmetry matching: Only orbitals with the same symmetry about the internuclear axis combine. 3. Maximum overlap: Greater overlap means stronger bonding.

7.3 Order of Filling (Second Period Homonuclear Diatomic Molecules)

For Li₂, Be₂, B₂, C₂, N₂:

For O₂, F₂ (and Ne₂):

The difference is in the order of σ2pz and π2px/π2py, which swaps depending on the extent of s-p mixing.

7.4 Bond Order and Magnetic Nature

Key points about O₂:

• O₂ has two unpaired electrons in π* orbitals, making it paramagnetic.
• This explains why liquid oxygen is attracted to a magnet — a fact that VBT cannot explain but MOT can.

Stability and bond order:

• Bond order > 0: molecule is stable
• Bond order = 0: molecule does not exist (e.g., He₂)
• Higher bond order = shorter bond length and higher bond dissociation energy

8. Hydrogen Bonding

A hydrogen bond is a special type of dipole-dipole interaction where a hydrogen atom bonded to a highly electronegative atom (F, O, or N) is attracted to another electronegative atom.

Conditions for hydrogen bonding: 1. Hydrogen must be bonded to a highly electronegative atom (F, O, or N). 2. The electronegative atom must have a lone pair.

8.1 Types of Hydrogen Bonding

8.2 Consequences of Hydrogen Bonding

1. High boiling point of water: H₂O (100°C) vs H₂S (-60°C) — despite O and S being in the same group, water has a much higher boiling point due to extensive H-bonding. 2. Ice floats on water: In ice, H₂O molecules form an open hexagonal structure with empty spaces, making ice less dense than liquid water. 3. Low density of ice vs water: Hydrogen bonds in ice create a cage-like structure with voids. 4. Solubility of alcohols in water: Alcohols can form H-bonds with water, making them soluble. 5. Anomalous properties of HF, NH₃, H₂O: Higher melting/boiling points than expected from molecular mass trends.

9. Important Definitions

10. Common Mistakes to Avoid

1. Confusing hybridization with electron-pair geometry — Hybridization determines orbital arrangement; VSEPR determines molecular shape (which accounts for lone pairs). NH₃ has sp³ hybridization (like CH₄) but trigonal pyramidal shape (not tetrahedral). 2. Forgetting lone pairs in VSEPR — Always count both bonding and lone pairs. H₂O has 2 bonding + 2 lone pairs = 4 electron pairs = sp³ hybridization, but bent shape. 3. Writing wrong MO filling order — The filling order differs for O₂/F₂ vs earlier diatomic molecules (Li₂ to N₂). The σ2pz and π2px/π2py swap positions. 4. Confusing paramagnetic and diamagnetic — Paramagnetic = has unpaired electrons (attracted to magnetic field). Diamagnetic = all electrons paired (weakly repelled). O₂ is paramagnetic (2 unpaired in π*). 5. Applying the octet rule to all molecules — Molecules like PCl₅ (10 electrons around P) and SF₆ (12 electrons around S) exceed the octet. The octet rule applies mainly to second-period elements. 6. Confusing intermolecular and intramolecular hydrogen bonding — Intermolecular occurs between molecules (e.g., water, HF); intramolecular occurs within one molecule (e.g., ortho-nitrophenol). Intermolecular raises boiling point; intramolecular may lower it. 7. Counting bond order incorrectly in MOT — Bond order = (Nb - Na)/2, where Nb is bonding electrons and Na is antibonding electrons. Higher bond order = stronger bond = shorter bond length.

11. Exam Tips

• For Lewis structure questions: Draw the structure, count valence electrons, then calculate formal charge to find the most stable structure.
• For VSEPR questions: Count electron pairs (bonding + lone), determine electron geometry, then remove lone pairs for molecular shape.
• For hybridization questions: Count sigma bonds + lone pairs on the central atom. 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d².
• For MOT questions: Write the electronic configuration, count bonding and antibonding electrons, then calculate bond order. Check for unpaired electrons to determine paramagnetism.
• For hydrogen bonding questions: Look for H bonded to F, O, or N. Intermolecular H-bonding increases boiling point; intramolecular H-bonding does not.
• Common 1-mark questions: Define hybridization, state Fajan's rules, name the shape of given molecules, define bond order.
• Common 3-mark questions: Explain VSEPR theory with examples, draw MO diagrams for O₂/N₂, differentiate ionic and covalent bonds.
• Common 5-mark questions: Explain hybridization types with examples, explain MOT with bond order calculations, discuss hydrogen bonding and its consequences.

12. Quick Revision Notes

• Octet rule: atoms tend to have 8 valence electrons (2 for H/He)
• Ionic bond: transfer of electrons; metal + non-metal; electrostatic attraction
• Covalent bond: sharing of electrons; non-metal + non-metal
• Coordinate bond: both electrons from one atom (e.g., NH₄⁺)
• Fajan's rules: small cation, large anion, high charge → more covalent character
• VSEPR: electron pairs repel → arrange to minimize repulsion; LP-LP > LP-BP > BP-BP
• Hybridization: sp (2, 180°), sp² (3, 120°), sp³ (4, 109.5°), sp³d (5, 90°/120°), sp³d² (6, 90°)
• MOT: bonding (σ, π) stabilized; antibonding (σ, π) destabilized
• Bond order = (Nb - Na)/2; > 0 = stable; higher = stronger/shorter
• O₂ is paramagnetic (2 unpaired electrons in π*) — key MOT prediction
• N₂ has bond order 3 (triple bond), O₂ has bond order 2 (double bond)
• Hydrogen bonding: H must be bonded to F, O, or N
• Intermolecular H-bonding: raises boiling point (H₂O > H₂S)
• Intramolecular H-bonding: within same molecule (o-nitrophenol)
• Ice floats because H-bonding creates open hexagonal structure with voids
• Formal charge = valence e⁻ - non-bonding e⁻ - ½(bonding e⁻)

13. Practice Questions

1. Draw the Lewis structure of CO₃²⁻ and calculate the formal charge on each atom. 2. Predict the shape and bond angle of (a) SF₄ (b) ClF₃ (c) XeF₄. 3. What is the hybridization of the central atom in (a) PCl₅ (b) SF₆ (c) H₂O (d) C₂H₂? 4. Write the molecular orbital configuration of N₂ and calculate its bond order. 5. Why does O₂ exhibit paramagnetism? Explain using molecular orbital theory. 6. Explain why H₂O has a higher boiling point than H₂S despite O and S being in the same group. 7. Arrange in order of increasing bond length: O₂, O₂⁺, O₂⁻. 8. Differentiate between sigma and pi bonds with examples. 9. Why is BeCl₂ linear but SF₂ is bent? 10. Explain the formation of a coordinate bond in NH₄⁺ with a diagram.

Quick Answers to Practice Questions

1. CO₃²⁻: Three resonance structures with one C=O double bond and two C-O single bonds. Formal charge on C = 4 - 0 - 4 = 0; on O(double bond) = 6 - 4 - 2 = 0; on O(single bonds) = 6 - 6 - 1 = -1 (each). Net charge = 0 + 0 + (-1) + (-1) = -2. 2. (a) SF₄: see-saw (4 BP + 1 LP from 5 electron pairs) (b) ClF₃: T-shaped (3 BP + 2 LP) (c) XeF₄: square planar (4 BP + 2 LP) 3. (a) PCl₅: sp³d (b) SF₆: sp³d² (c) H₂O: sp³ (d) C₂H₂: sp (each C) 4. N₂: KK(σ2s)²(σ2s)²(π2px)²(π2py)²(σ2pz)²; Nb=10, Na=4; Bond order = (10-4)/2 = 3 5. O₂ has two unpaired electrons in the π2px and π*2py antibonding orbitals. Unpaired electrons make it paramagnetic. VBT cannot explain this; only MOT predicts it correctly. 6. H₂O has strong intermolecular hydrogen bonding (O-H···O) which requires more energy to break. H₂S cannot form hydrogen bonds effectively because S is less electronegative than O. 7. O₂⁺ (bond order 2.5) < O₂ (bond order 2) < O₂⁻ (bond order 1.5) — higher bond order means shorter bond length, so O₂⁺ has the shortest bond. 8. Sigma bond: head-on overlap along internuclear axis; stronger; formed first. Pi bond: sideways overlap perpendicular to internuclear axis; weaker; formed after sigma. Single bond = 1 sigma; double = 1 sigma + 1 pi; triple = 1 sigma + 2 pi. 9. Be in BeCl₂ has 2 bonding pairs, 0 lone pairs → sp → linear. S in SF₂ has 2 bonding pairs + 2 lone pairs → sp³ → bent. 10. NH₃ has a lone pair on N. H⁺ has no electrons. N donates its lone pair to form a coordinate (dative) bond with H⁺, giving NH₄⁺ where all four N-H bonds are identical due to resonance.

14. Final Revision Checklist

Before the exam, make sure you can:

• Draw Lewis structures and calculate formal charges
• Explain ionic and covalent bond formation with examples
• Apply VSEPR theory to predict molecular shapes and bond angles
• Determine hybridization (sp, sp², sp³, sp³d, sp³d²) from molecular formula
• Write MOT configurations for diatomic molecules (especially O₂ and N₂)
• Calculate bond order and determine paramagnetic/diamagnetic nature
• Explain hydrogen bonding and its effects on physical properties
• Differentiate sigma and pi bonds
• Draw resonance structures and explain resonance stabilization
• Apply Fajan's rules to predict covalent character in ionic compounds