JEE Main Chemistry · Chapter 3

Chemical Bonding and Molecular Structure

Ionic and covalent bonds, VSEPR shapes, hybridisation, molecular orbital theory, dipole moment and hydrogen bonding.

NCERT Class 11JEE Main: 1–2 Qs per shift 8 min read

1. Chapter at a Glance

Atoms combine to form molecules because bonding lowers their energy. This chapter explains why atoms bond (octet rule, lattice energy), how they bond (ionic, covalent, coordinate, hydrogen bonds) and what shape the resulting molecules take (VSEPR theory, hybridisation), and uses molecular orbital theory to explain bond order and magnetism. It is a favourite JEE chapter — 1 to 2 questions every shift — because it links structure with properties.

What JEE asks most
  • Shape and hybridisation of molecules and ions (XeF₄, ClF₃, SF₄, PCl₅, I₃⁻ …).
  • Bond order, bond length and magnetic nature from MO theory (O₂, N₂, their ions).
  • Dipole moment: which molecules are polar or non-polar.
  • Comparison of bond angles; number of σ and π bonds; hydrogen bonding effects.

2. Ionic and Covalent Bonds

  • Octet rule (Kössel–Lewis): atoms tend to gain, lose or share electrons to get eight electrons in the valence shell. Exceptions: incomplete octet (BeCl₂, BF₃), expanded octet (PCl₅, SF₆), odd-electron molecules (NO, NO₂).
  • Ionic bond: transfer of electrons from a metal to a non-metal. Favoured by low ionisation enthalpy, high electron gain enthalpy and high lattice enthalpy.
  • Covalent bond: sharing of electron pairs. A coordinate bond is a covalent bond in which both electrons come from one atom (e.g. in NH₄⁺).
  • Fajans' rules: covalent character in an ionic compound increases with a small, highly charged cation, a large anion, and a cation with a pseudo-noble-gas (18-electron) configuration.
  • Formal charge = (valence electrons) − (non-bonding electrons) − ½(bonding electrons). The best Lewis structure has the lowest formal charges.

3. VSEPR Theory and Shapes

Electron pairs around the central atom repel each other and arrange themselves as far apart as possible. Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Lone pairs therefore squeeze bond angles.

Electron pairs (BP + LP)HybridisationExampleShapeBond angle
2 + 0spBeCl₂, CO₂Linear180°
3 + 0sp²BF₃Trigonal planar120°
4 + 0sp³CH₄Tetrahedral109.5°
3 + 1sp³NH₃Pyramidal≈107°
2 + 2sp³H₂OBent (V-shape)≈104.5°
5 + 0sp³dPCl₅Trigonal bipyramidal90°, 120°
4 + 1sp³dSF₄See-saw—
3 + 2sp³dClF₃T-shaped—
2 + 3sp³dXeF₂, I₃⁻Linear180°
6 + 0sp³d²SF₆Octahedral90°
5 + 1sp³d²BrF₅Square pyramidal—
4 + 2sp³d²XeF₄Square planar90°
Finding hybridisation quickly

Steric number = ½ [V + M − C + A], where V = valence electrons of the central atom, M = number of monovalent atoms attached, C = positive charge, A = negative charge. Steric number 2, 3, 4, 5, 6 → sp, sp², sp³, sp³d, sp³d².

4. Molecular Orbital Theory

Atomic orbitals combine to form bonding (lower energy) and antibonding (higher energy, marked *) molecular orbitals.

Bond order = ½ (Nb − Na)
  • Higher bond order ⇒ stronger and shorter bond. Bond order 0 ⇒ molecule does not exist (He₂).
  • Unpaired electrons in MOs ⇒ paramagnetic; all paired ⇒ diamagnetic. O₂ is paramagnetic — something the Lewis structure cannot explain.
  • Filling order for up to 14 electrons (B₂, C₂, N₂): σ1s, σ*1s, σ2s, σ*2s, (π2px = π2py), σ2pz, … For O₂ and F₂, σ2pz comes before the π2p pair.
O₂ (2p region) σ*2p π*2p ↑↑ π2p ↑↓↑↓ σ2p↑↓ Bond order = ½(8 − 4) = 2; two unpaired e⁻ → paramagnetic
O₂ has two unpaired electrons in the π* orbitals, which makes it paramagnetic.
SpeciesBond orderMagnetic nature
N₂3Diamagnetic
O₂2Paramagnetic (2 unpaired)
O₂⁺2.5Paramagnetic
O₂⁻ (superoxide)1.5Paramagnetic
O₂²⁻ (peroxide)1Diamagnetic
C₂2Diamagnetic (two π bonds)

5. Dipole Moment, Hydrogen Bonding and Resonance

  • Dipole moment μ = q × d (unit: debye). A molecule with polar bonds can still have μ = 0 if the bond dipoles cancel by symmetry: CO₂, BF₃, CH₄, CCl₄, PCl₅, SF₆, XeF₄.
  • NH₃ has a larger dipole moment than NF₃, because in NF₃ the lone-pair dipole opposes the bond dipoles.
  • Hydrogen bonding occurs when H is attached to F, O or N. Intermolecular H-bonding raises boiling points (H₂O is a liquid, H₂S a gas); intramolecular H-bonding (o-nitrophenol) lowers them.
  • Resonance: when one Lewis structure cannot describe a molecule, the real structure is a hybrid (O₃, CO₃²⁻, benzene). Resonance makes all equivalent bonds equal in length.
  • σ and π bonds: every single bond is σ; a double bond is 1σ + 1π; a triple bond is 1σ + 2π.

6. Formula Master Table

QuantityFormula / rule
Formal chargeV − L − ½B
Steric number½(V + M − C + A)
Bond order (MOT)½(Nb − Na)
Dipole momentμ = q × d
Percentage ionic character(μobserved/μionic) × 100
Bond order vs lengthbond order ↑ ⇒ bond length ↓, bond energy ↑
Repulsion orderlp–lp > lp–bp > bp–bp

7. Mind Map

Chemical Bonding
Bond types
  • Ionic, covalent, coordinate
  • Fajans' rules
  • Formal charge
Shapes
  • VSEPR
  • Hybridisation sp → sp³d²
  • Lone pairs squeeze angles
MO theory
  • Bond order
  • Para/diamagnetism
  • O₂ paramagnetic
Properties
  • Dipole moment
  • Hydrogen bonding
  • Resonance

8. Tips and Tricks

Shortcuts
  • Count electrons for MO questions: total electrons of the molecule/ion; for 10–14 electrons use the N₂-type order, for 15–20 use the O₂-type order.
  • Same number of electrons ⇒ same bond order (isoelectronic species): CO, N₂, CN⁻, NO⁺ all have bond order 3.
  • In trigonal bipyramidal shapes, lone pairs and more electronegative atoms go to equatorial positions.
  • For bond angles of hydrides down a group (NH₃ > PH₃ > AsH₃), the angle decreases as the central atom's electronegativity decreases.
Mistakes students make
  • Calling XeF₂ bent because of its lone pairs — its three lone pairs are equatorial, so the molecule is linear.
  • Assuming every molecule with polar bonds is polar.
  • Forgetting the charge of an ion while counting electrons for hybridisation or MO theory.

9. Important One-liners

  • A σ bond is stronger than a π bond formed between the same atoms.
  • π bonds are formed by sideways overlap and restrict rotation about the bond.
  • Bond length order: C–C > C=C > C≡C.
  • The ice structure is open because of hydrogen bonding, so ice is less dense than water.
  • He₂ does not exist because its bond order is zero.
  • B₂ is paramagnetic, while C₂ is diamagnetic.
  • Lattice enthalpy increases with higher ionic charges and smaller ionic sizes.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Medium

The shape of XeF₄ is:

  1. Tetrahedral
  2. Square planar
  3. See-saw
  4. Square pyramidal
Show solution

Xe has 8 valence electrons; 4 are used in bonds with F, leaving 2 lone pairs. 4 bond pairs + 2 lone pairs = 6 pairs → sp³d² (octahedral arrangement). The lone pairs sit opposite each other, so the shape is square planar.

Answer: (B)

Q2Medium

Which of the following species is diamagnetic?

  1. O₂
  2. O₂⁺
  3. O₂⁻
  4. O₂²⁻
Show solution

O₂ (16 e⁻) has two unpaired electrons in π*. O₂⁺ has one unpaired electron, O₂⁻ has one, and O₂²⁻ (18 e⁻) has both π* orbitals filled — all paired, so it is diamagnetic.

Answer: (D)

Q3Easy

Which of the following molecules has a non-zero dipole moment?

  1. CO₂
  2. BF₃
  3. NH₃
  4. CCl₄
Show solution

CO₂ (linear), BF₃ (trigonal planar) and CCl₄ (tetrahedral) are symmetric, so their bond dipoles cancel. NH₃ is pyramidal with a lone pair, so its dipoles add up.

Answer: (C)

Q4Numerical

How many lone pairs are present on the central Xe atom in XeF₂?

Show solution

Xe has 8 valence electrons; 2 are used for the two Xe–F bonds, leaving 6 electrons = 3 lone pairs (steric number 5, sp³d, linear shape).

Answer: 3

Q5Medium

The correct order of bond angles is:

  1. H₂O > NH₃ > CH₄
  2. CH₄ > NH₃ > H₂O
  3. NH₃ > CH₄ > H₂O
  4. CH₄ = NH₃ = H₂O
Show solution

All three are sp³, but lone pairs compress the angle: CH₄ (0 lp) 109.5°, NH₃ (1 lp) ≈ 107°, H₂O (2 lp) ≈ 104.5°.

Answer: (B)

Q6Numerical

Find the total number of π bonds in acrylonitrile, CH₂=CH–C≡N.

Show solution

The C=C double bond has 1 π bond and the C≡N triple bond has 2 π bonds. Total = 3.

Answer: 3

Q7Hard

Which of the following pairs has the same shape and the same hybridisation of the central atom?

  1. XeF₂ and I₃⁻
  2. SF₄ and XeF₄
  3. BF₃ and NH₃
  4. ClF₃ and PCl₃
Show solution

XeF₂: 2 bp + 3 lp → sp³d, linear. I₃⁻: central I has 2 bp + 3 lp → sp³d, linear. They match.

SF₄ is sp³d see-saw while XeF₄ is sp³d² square planar; BF₃ is sp² planar while NH₃ is sp³ pyramidal; ClF₃ is sp³d T-shaped while PCl₃ is sp³ pyramidal.

Answer: (A)

Q8Hard

The correct order of bond order is:

  1. O₂²⁻ < O₂⁻ < O₂ < O₂⁺
  2. O₂⁺ < O₂ < O₂⁻ < O₂²⁻
  3. O₂ < O₂⁺ < O₂⁻ < O₂²⁻
  4. O₂⁻ < O₂²⁻ < O₂ < O₂⁺
Show solution

Bond orders: O₂²⁻ = 1, O₂⁻ = 1.5, O₂ = 2, O₂⁺ = 2.5. Adding electrons to π* orbitals lowers the bond order.

Answer: (A)

11. Quick Revision Summary

Remember before the exam
  • Steric number → hybridisation; lone pairs decide the final shape (see the VSEPR table).
  • Repulsion: lp–lp > lp–bp > bp–bp, so lone pairs reduce bond angles.
  • Bond order = ½(Nb − Na); higher bond order → shorter, stronger bond.
  • O₂ is paramagnetic; O₂²⁻ and N₂ are diamagnetic.
  • Symmetric molecules (CO₂, BF₃, CCl₄, XeF₄) have zero dipole moment.
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