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Class 11 Chemistry Notes

Chemical Bonding and Molecular Structure Class 11 Notes

Complete, exam-ready notes on chemical bonding: why atoms bond, the octet rule and Lewis structures, ionic and covalent bonds, VSEPR molecular shapes, hybridisation and bond parameters — written for CBSE, JEE and NEET revision.

Class11SubjectChemistryCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is chemical bonding in one line?

Atoms bond by sharing or transferring electrons to reach a stable electronic configuration — the octet rule — forming ionic, covalent or metallic bonds.

Octet Rule and Lewis Structures

Octet rule

Atoms tend to bond so that each achieves eight electrons in its valence shell (two for hydrogen). Lewis structures show bonds as shared electron pairs and lone pairs as dots, using only valence electrons.

Exceptions matter

BeCl₂ (4 electrons), BCl₃ (6) and the expanded octets of PCl₅ and SF₆ break the octet rule. Exam questions use these to test whether you apply the rule mechanically.

Ionic (Electrovalent) Bond

Ionic bond

A bond formed by complete transfer of electrons from a metal to a non-metal, e.g. NaCl. It is strong, directional in crystal packing, and stable because the lattice energy released exceeds the ionisation + hydration cost.

  • Favoured by low ionisation enthalpy, high electron gain enthalpy and high lattice energy.
  • Fajan's rules: small cations and large, polarisable anions increase covalent character (e.g. LiCl is more covalent than NaCl).
  • Ionic compounds conduct electricity in molten or aqueous state, not as solids.

Covalent Bond and Polarity

Covalent bond

A pair of electrons shared between two atoms. Single, double and triple bonds share 2, 4 and 6 electrons respectively. When the atoms differ in electronegativity, the shared pair is pulled closer to the more electronegative atom — a polar bond.

μ=q×d,% ionic character=μobsμionic×100\mu = q \times d,\qquad \%\,\text{ionic character} = \frac{\mu_{\text{obs}}}{\mu_{\text{ionic}}}\times100
Dipole moment

VSEPR Theory — Molecular Shapes

  • Valence pairs arrange themselves to minimise repulsion; lone pairs repel more strongly than bond pairs, squeezing bond angles.
  • Linear (2 pairs, 180°) — BeCl₂ and CO₂.
  • Trigonal planar (3 pairs, 120°) — BF₃.
  • Tetrahedral (4 pairs, 109.5°) — CH₄; with one lone pair NH₃ becomes trigonal pyramidal (107°), with two lone pairs H₂O is bent (104.5°).

Angle order

Bond angle decreases as lone pairs grow: CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°). Lone-pair — lone-pair repulsion beats lone-pair — bond-pair, which beats bond-pair — bond-pair.

Hybridisation of Atomic Orbitals

Hybridisation

SN=bond pairs+lone pairs\text{SN} = \text{bond pairs} + \text{lone pairs}

Mixing atomic orbitals creates equivalent hybrid orbitals with specific geometry: sp (linear, 180°), sp² (trigonal planar, 120°), sp³ (tetrahedral, 109.5°), sp³d (trigonal bipyramidal), sp³d² (octahedral). The shape is set by the steric number SN.

  • CH₄, NH₃, H₂O — sp³ with SN = 4.
  • BCl₃ — sp² with SN = 3.
  • BeCl₂ and C₂H₂ (acetylene) — sp with SN = 2.
  • Greater s-character shortens and strengthens the bond: sp bonds are shorter than sp² bonds, which are shorter than sp³ bonds.

Bond Parameters, Resonance and Bond Order

Bond order

BO=Nb−Na2\text{BO} = \frac{N_b - N_a}{2}

In molecular orbital theory, bond order is half the number of bonding electrons minus anti-bonding electrons. O₂ has BO = 2 and is paramagnetic; N₂ has BO = 3 (the strongest common bond). Resonance spreads electron density over equivalent structures — e.g. carbonate ion — and the bond order is the average over contributing structures.

FC=V−L−B2\text{FC} = V - L - \frac{B}{2}
Formal charge (V = valence, L = lone pair e⁻, B = bonding e⁻)

Solved Examples

Example: Draw the Lewis structure of CO2\text{CO}_2 and state its molecular shape.

Solution: Carbon shares four electrons — a double bond with each oxygen. No lone pairs on carbon, so the shape is linear (sp hybridised) with bond angle 180°.

Example: Why is the H₂O bond angle 104.5° and not 109.5°?

Solution: Water is sp³ hybridised (SN = 4) with two lone pairs on oxygen. Lone-pair — lone-pair repulsion is stronger than bond-pair repulsion, so the two O–H bonds are pushed closer from 109.5° to 104.5°.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

Dipole moment

μ=q×d\mu = q \times d

Formal charge

FC=V−L−B2\text{FC} = V - L - \frac{B}{2}

Bond order (MOT)

BO=Nb−Na2\text{BO} = \frac{N_b - N_a}{2}

Percent ionic character

μobsμionic×100\frac{\mu_{\text{obs}}}{\mu_{\text{ionic}}}\times100

Steric number

SN=bond pairs+lone pairs\text{SN} = \text{bond pairs} + \text{lone pairs}

Valence electrons

valence e−=group number\text{valence e}^- = \text{group number}

Water bond angle

∠HOH=104.5∘\angle \text{HOH} = 104.5^{\circ}

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • Octet rule exceptions: BeCl₂, BCl₃, PCl₅, SF₆.
  • Lone pairs repel more than bond pairs: CH₄ 109.5° > NH₃ 107° > H₂O 104.5°.
  • sp linear, sp² trigonal planar, sp³ tetrahedral — match SN to shape.
  • Bond order of O₂ is 2 (paramagnetic), N₂ is 3.
  • Greater s-character gives shorter, stronger bonds.
  • Formal charges in a Lewis structure must add up to the molecule's overall charge.

FAQ

Common questions

What is the octet rule?

Atoms bond to attain eight electrons in their valence shell (two for hydrogen). It guides Lewis structures but has exceptions like BeCl₂, BCl₃ and the expanded octets of PCl₅ and SF₆.

What is the difference between ionic and covalent bonds?

Ionic bonds form by complete electron transfer (metal + non-metal, e.g. NaCl) and give lattice structures; covalent bonds form by electron sharing between non-metals and give discrete molecules or networks.

Why is the H₂O bond angle 104.5°?

Oxygen in water is sp³ hybridised (SN = 4), but two lone pairs repel the bond pairs more strongly, compressing the O–H bonds from the ideal 109.5° to 104.5°.

How do you calculate bond order?

In molecular orbital theory, bond order = (number of bonding electrons − number of anti-bonding electrons)/2. N₂ has 3, O₂ has 2, and higher bond order means shorter, stronger bonds.

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