Chemical bonding for JEE Main

Chemical bonding is one of the highest-return chapters in JEE Main chemistry. Its questions are short (predict a shape, compare bond orders, pick the molecule with zero dipole moment) and the same reasoning returns in p-block, coordination compounds and organic chemistry. A few rules, applied carefully, answer most of them.

The official 2026 syllabus lists the Kossel–Lewis approach, ionic bonding with the factors affecting it and the calculation of lattice enthalpy, electronegativity, Fajans' rules, dipole moment, VSEPR theory and shapes of simple molecules, valence bond theory, hybridisation involving s, p and d orbitals, resonance, molecular orbital theory for homonuclear diatomic molecules with bond order, bond length and bond energy, and an elementary idea of metallic and hydrogen bonding.

Ionic bonding and lattice enthalpy

An ionic bond forms readily when the metal has a low ionisation enthalpy, the non-metal has a strongly negative electron gain enthalpy, and the lattice enthalpy is large. Lattice enthalpy is the energy needed to separate one mole of a solid ionic compound into gaseous ions. It rises with higher ionic charges and smaller ions: MgO has a far larger lattice enthalpy than NaCl.

Lattice enthalpy cannot be measured directly, so it is calculated from a Born–Haber cycle. For NaCl, the formation enthalpy equals the sum of the steps:

ΔfH = ΔsubH(Na) + IE(Na) + ½ΔbondH(Cl₂) + ΔegH(Cl) − ΔlatticeH

Covalent character and Fajans' rules

No bond is purely ionic. A cation distorts (polarises) the electron cloud of the anion, and the bond gains covalent character when:

• the cation is small and highly charged (Al³⁺ more than Na⁺)
• the anion is large (I⁻ more than F⁻)
• the cation has a pseudo-noble-gas (18-electron) configuration, as Cu⁺ and Ag⁺ do

So AlCl₃ is more covalent than MgCl₂, which is more covalent than NaCl, and LiI is more covalent than LiF.

Lewis structures and formal charge

Formal charge = valence electrons − non-bonding electrons − ½(bonding electrons). The best Lewis structure keeps formal charges small and puts negative charge on the more electronegative atom. In O₃, the central O has a formal charge of +1 and one end O has −1.

VSEPR and hybridisation

Count the steric number (atoms bonded to the central atom plus its lone pairs). A quick formula for the number of electron pairs: ½(V + M − c + a), where V is the central atom's valence electrons, M the number of monovalent atoms attached, c the positive charge and a the negative charge.

Rules that decide the details:

• Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.
• In a trigonal bipyramid, lone pairs go to the equatorial positions.
• Bond angles fall as lone pairs increase: CH₄ 109.5° > NH₃ about 107° > H₂O about 104.5°.

Resonance and bond order

When one Lewis structure cannot describe a molecule, the real structure is a hybrid. All the bonds that resonate are identical. Average bond order = (total bonds from the central atom to the resonating atoms)/(number of those atoms): CO₃²⁻ and NO₃⁻ have 4/3, O₃ has 1.5, and each C–C bond in benzene has 1.5.

Dipole moment

μ = q × d, measured in debye (1 D = 3.336 × 10⁻³⁰ C m). Bond dipoles add as vectors, so symmetrical molecules have zero net moment: CO₂, BF₃, CH₄, CCl₄, PCl₅, SF₆, XeF₂ and XeF₄.

• NH₃ has a much larger dipole moment than NF₃. In NH₃ the lone pair's moment adds to the N–H bond moments; in NF₃ it opposes the N–F bond moments.
• Percentage ionic character = (observed μ/μ for a full unit charge at the same distance) × 100.

Molecular orbital theory

Atomic orbitals combine into bonding and antibonding molecular orbitals. Bond order = ½(N_b − N_a). A higher bond order means a shorter, stronger bond; a bond order of zero means the molecule does not exist. Unpaired electrons make the species paramagnetic.

Filling order: for up to 14 electrons (B₂, C₂, N₂), the π2p orbitals lie below σ2p. For O₂ and F₂, σ2p lies below π2p. Either order gives the same bond orders.

Metallic and hydrogen bonding

In a metal, positive ions sit in a sea of delocalised valence electrons, which explains conductivity and malleability.

A hydrogen bond forms when H bonded to F, O or N is attracted to a lone pair on another such atom. Intermolecular hydrogen bonding raises boiling points (H₂O, HF and NH₃ boil far higher than expected from their molar masses). Intramolecular hydrogen bonding does the opposite: o-nitrophenol is more volatile than p-nitrophenol.

Worked problems

Problem 1 (numerical answer): Born–Haber cycle. For NaCl, ΔsubH(Na) = 108, IE(Na) = 496, ½ΔbondH(Cl₂) = 121, ΔegH(Cl) = −349 and ΔfH(NaCl) = −411 kJ/mol. Find the lattice enthalpy in kJ/mol.

−411 = 108 + 496 + 121 − 349 − ΔlatticeH.
ΔlatticeH = 108 + 496 + 121 − 349 + 411 = 787 kJ/mol.

Problem 2: predicting shapes. Find the shapes of XeF₄, I₃⁻ and ClF₃.

XeF₄: ½(8 + 4) = 6 pairs, 4 bonded: square planar.
I₃⁻: ½(7 + 2 + 1) = 5 pairs, 2 bonded, lone pairs equatorial: linear.
ClF₃: ½(7 + 3) = 5 pairs, 3 bonded, two lone pairs equatorial: T-shaped.

Problem 3: MO comparison. Arrange O₂, O₂⁺, O₂⁻ and O₂²⁻ in order of bond length, and name the diamagnetic one.

Bond orders: O₂⁺ 2.5, O₂ 2, O₂⁻ 1.5, O₂²⁻ 1. Bond length rises as bond order falls:
O₂⁺ < O₂ < O₂⁻ < O₂²⁻. Only O₂²⁻ is diamagnetic.

Problem 4: percentage ionic character. HCl has a dipole moment of 1.03 D and a bond length of 127 pm. Find its percentage ionic character.

For a full charge: μ = 1.6 × 10⁻¹⁹ × 127 × 10⁻¹² = 2.03 × 10⁻²⁹ C m, which is (2.03 × 10⁻²⁹)/(3.336 × 10⁻³⁰) ≈ 6.1 D.
Ionic character = 1.03/6.1 × 100 ≈ 17%.

Traps to watch

• Counting lone pairs wrongly on the central atom. Always find the steric number first.
• Assuming all bonds are equal in a trigonal bipyramid. Axial and equatorial bonds differ in PCl₅.
• Using the O₂ filling order for B₂, C₂ or N₂. The bond order stays the same, but the orbital holding the last electron changes, and for B₂ and C₂ so does the predicted magnetism (B₂ would wrongly come out diamagnetic).
• "Polar bonds mean a polar molecule." Symmetry can cancel them, as in CO₂ and BF₃.

Practice set

1. What is the shape of ClF₃?
2. Find the bond order of O₂⁻.
3. Give the hybridisation of S in SF₄ and of P in PCl₅.
4. Which is paramagnetic: N₂, B₂, C₂ or F₂?
5. Find the formal charge on N in NH₄⁺.
6. Arrange NaCl, MgCl₂ and AlCl₃ in order of increasing covalent character.
7. Which boils at a higher temperature, o-nitrophenol or p-nitrophenol, and why?
8. What is the average C–O bond order in CO₃²⁻?

Answers

1. T-shaped (5 pairs, 2 lone pairs equatorial).
2. O₂⁻ has 17 electrons: 10 bonding and 7 antibonding, so ½(10 − 7) = 1.5.
3. sp³d for both: SF₄ has 4 bonds and 1 lone pair; PCl₅ has 5 bonds.
4. B₂, with two unpaired electrons in the degenerate π2p orbitals.
5. 5 − 0 − ½(8) = +1.
6. NaCl < MgCl₂ < AlCl₃, as the cation becomes smaller and more highly charged.
7. p-nitrophenol, which forms intermolecular hydrogen bonds. The ortho isomer bonds within itself.
8. Four bonds shared over three positions: 4/3 ≈ 1.33.

What to do next

• Fill the VSEPR table from memory with two examples in each cell.
• Write MO configurations for every species from H₂ to Ne₂, with bond order and magnetism.
• Revise periodic properties for the electronegativity and size arguments used here.
• Then move to chemical thermodynamics, which uses bond enthalpies.