JEE/Chemistry/Chemical Bonding

Inorganic Chemistry · Must Do · Chapter Notes & Practice

Chemical Bonding — JEE Main & Advanced Notes

Explain structure and properties through ionic/covalent bonding, VSEPR, hybridisation, resonance, dipole moment and molecular orbital theory.

VSEPRhybridisationMOTresonance
Learning resources and corrections: check assumptions, work through the examples, and report any unclear step.

1. Begin with electrons and a Lewis structure

Count the valence electrons of all atoms, add one per negative charge and subtract one per positive charge. Connect the atoms, distribute electrons, then check formal charges and the total electron count. Hydrogen needs a duet; an octet is a useful main-group guide, with electron-deficient and odd-electron exceptions.

Formal charge = valence electrons of the free atom − nonbonding electrons − half the bonding electrons. The sum of formal charges must equal the species' charge. Formal charge is bookkeeping within a Lewis structure, not the same as oxidation state or measured partial charge.

A single bond contains one σ bond; a double contains one σ and one π; a triple contains one σ and two π. A multiple bond still counts as one region around a central atom in a VSEPR shape calculation.

2. Predict shape without a false angle shortcut

Count bonding regions and lone pairs around the central atom. The electron-domain arrangement includes both; molecular shape describes the positions of atoms. There is no fixed two-degree decrease per lone pair. Bond angles depend on the atoms, bonds and electron distribution.

  • AX₂: linear; ideal angle 180°.
  • AX₃: trigonal planar; ideal angle 120°. AX₂E is bent.
  • AX₄: tetrahedral; ideal angle about 109.5°. AX₃E is trigonal pyramidal; AX₂E₂ is bent.
  • Five domains: trigonal bipyramidal arrangement. AX₄E is seesaw; AX₃E₂ is T-shaped; AX₂E₃ is linear.
  • Six domains: octahedral arrangement. AX₅E is square pyramidal; AX₄E₂ is square planar.

In the five-domain model, lone pairs prefer equatorial positions to reduce 90° interactions. With two lone pairs in an octahedral arrangement, opposite positions give the square-planar case. X counts bonded atoms and E counts central lone pairs.

3. Resonance, hybridisation and polarity

Resonance contributors keep the same atom connectivity and differ in electron placement. The molecule does not jump back and forth between drawings. Equivalent contributors can explain equal bond lengths, such as the three C–O bonds in carbonate.

For simple main-group localised bonding, two, three and four electron domains commonly correspond to sp, sp² and sp³ descriptions. These are bonding models. The familiar sp³d and sp³d² labels for hypervalent examples are exam bookkeeping conventions; they do not establish substantial d-orbital participation in the real bond.

To assess molecular polarity, combine geometry with the directions and magnitudes of bond dipoles. Equal surrounding atoms in a symmetric geometry can give cancellation. A lone pair is a prompt to inspect shape, not a universal proof of polarity: XeF₄ has lone pairs but its square-planar bond dipoles cancel.

4. Molecular orbitals and bond order

Use an energy ordering appropriate to the species, then apply Pauli and Hund. In the standard second-period homonuclear diagrams, π2p lies below σ2p for B₂, C₂ and N₂; σ2p lies below π2p for O₂ and F₂. Include antibonding orbitals, marked with an asterisk, when filling.

Bond order = (bonding electrons − antibonding electrons)/2. Unpaired electrons produce paramagnetism. For O₂, the two highest electrons occupy separate π* orbitals; adding or removing electrons changes both bond order and the pairing.

Compare bond lengths using bond order within a related series, not as an absolute law across unrelated elements. A basic MO diagram giving zero bond order predicts no ordinary net covalent bond in that model; it is not proof that weakly bound species can never exist.

5. Ionic character and hydrogen bonding

Ionic and covalent character form a continuum. In a simple electrostatic comparison of similar lattices, larger charge products and smaller interionic separation increase the magnitude of attraction. Do not compare lattice energies using charge alone when structure and size differ.

Fajans' reasoning asks whether a cation strongly distorts an anion's electron cloud: small, highly charged cations and large, easily polarised anions favour greater covalent character.

Common school-level hydrogen-bond donors contain H bonded to N, O or F; an acceptor needs an available electron-rich site. Hydrogen bonding can occur within a molecule or between molecules. Intermolecular association can raise boiling points, but neither melting-point order nor an exact boiling point follows from counting hydrogen bonds alone.

6. Six worked bonding problems

1. Audit the carbonate electron count

Find the valence-electron total in CO₃²⁻ and formal charges in a contributor with one C=O and two C–O bonds.

Total = 4 + 3 × 6 + 2 = 24 electrons. Carbon: 4 − 0 − 8/2 = 0. Double-bonded oxygen: 6 − 4 − 4/2 = 0. Each single-bonded oxygen: 6 − 6 − 2/2 = −1. Total = −2. Three equivalent contributors give mean C–O bond order 4/3 in the resonance model.

2. Separate electron arrangement from shape

What are the electron-domain arrangement and molecular shape of ClF₃?

There are 28 valence electrons. Three Cl–F bonds and the terminal F octets leave four electrons on Cl: two lone pairs. Five domains give a trigonal-bipyramidal arrangement. Two equatorial lone pairs leave a T-shaped molecule.

3. Check polarity despite lone pairs

Does XeF₄ have a permanent molecular dipole in its ideal equilibrium geometry?

Its 36 valence electrons leave two lone pairs on Xe after four bonds and the F octets. Six domains give an octahedral arrangement; opposite lone pairs leave square-planar XeF₄. The four equal Xe–F bond dipoles cancel, giving zero net permanent dipole.

4. Compare oxygen ions

Find bond orders and unpaired-electron counts for O₂⁺, O₂ and O₂⁻ using the standard MO filling.

The bonding valence-electron count is 8. Antibonding counts are 3, 4 and 5 respectively. Bond orders are 2.5, 2 and 1.5. The π* occupancies give 1, 2 and 1 unpaired electrons. All three are paramagnetic in this model; O₂⁺ is predicted to have the shortest bond of this set.

5. Count σ and π bonds

How many σ and π bonds are in H–C≡C–H?

The C≡C bond contributes one σ and two π bonds. The two C–H bonds each contribute one σ. Total: 3 σ and 2 π. Each carbon has two bonding directions and an sp description.

6. Diagnose an invalid angle calculation

A learner subtracts 2° for each of water's two lone pairs from 109.5° and reports 105.5°. Is that a justified prediction?

No. VSEPR gives a bent shape, not a fixed subtraction rule. Water's gas-phase angle is about 104.5°. Use a supplied or established angle when a numerical value is needed; molecular identity matters.

7. Self-check and revision route

Why does a double bond count as one VSEPR region?

The bonded atom lies in one direction from the central atom. The σ and π components do not represent two separate bonded-atom directions.

Are formal charge and oxidation state interchangeable?

No. Formal charge divides bonding electrons equally; oxidation-state bookkeeping assigns them using electronegativity rules. Use the definition asked for.

Does a larger bond order always mean a stronger bond between any two species?

No. Bond order is most useful within comparable species. Atom size, orbital overlap and other electronic effects prevent a universal ranking of unrelated bonds.

Revision sequence: electron count → Lewis structure → domains and shape → polarity → MO filling where needed. Revisit atomic structure and orbital filling, then browse the Chemistry chapter directory.

Further reading: molecular shape and polarity and molecular orbital theory. These notes and worked problems were independently written for LearnAtMyPlace.

Verified question bank

Chemical Bonding questions with answers

Open any of these 38 quality-checked JEE questions to review all four options, the correct answer, and the worked explanation.

  1. Question 1 · EasyLewis Structure
  2. Question 2 · EasyOctet Rule
  3. Question 3 · EasyBond Order
  4. Question 4 · EasyVSEPR
  5. Question 5 · EasyVSEPR
  6. Question 6 · MediumHybridisation
  7. Question 7 · MediumHybridisation
  8. Question 8 · MediumHybridisation
  9. Question 9 · MediumVSEPR shape
  10. Question 10 · MediumDipole Moment
  11. Question 11 · MediumElectronegativity
  12. Question 12 · HardFormal Charge
  13. Question 13 · HardResonance
  14. Question 14 · HardVSEPR lone pairs
  15. Question 15 · MediumMOT
  16. Question 16 · MediumMOT Magnetism
  17. Question 17 · HardMOT Bond Order
  18. Question 18 · EasyIonic Bond
  19. Question 19 · MediumBond Length
  20. Question 20 · MediumPolarity
  21. Question 21 · MediumHydrogen Bond
  22. Question 22 · MediumCoordinate Bond
  23. Question 23 · HardExpanded Octet
  24. Question 24 · HardBent molecule
  25. Question 25 · MediumElectron deficiency
  26. Question 26 · HardVSEPR
  27. Question 27 · HardMolecular Orbital
  28. Question 28 · HardComparison
  29. Question 29 · HardExceptions
  30. Question 30 · HardDipole moment
  31. Question 31 · HardLattice energy
  32. Question 32 · MediumHydrogen bonding effect
  33. Question 33 · HardResonance
  34. Question 34 · EasyBasic
  35. Question 35 · MediumHybridisation
  36. Question 36 · MediumVSEPR
  37. Question 37 · HardBond energy
  38. Question 38 · HardMOT stability
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