JEE Chemistry — Chemical Bonding Complete Chapter Guide

Chapter Overview & Weightage

Chemical Bonding is one of those chapters where JEE rewards deep understanding over rote memorisation. The concepts here — hybridization, MOT, VSEPR — keep appearing in both straightforward and twisted forms across Main and Advanced.

Weightage: 6–8% in JEE Main (typically 2–3 questions per paper). In JEE Advanced, bonding concepts appear embedded inside inorganic and coordination chemistry questions rather than standalone. Treat this chapter as a foundation investment — weak bonding concepts will hurt you in d-block, p-block, and organic chemistry too.

Year	JEE Main Questions	Topics Covered
2024	2	MOT (O₂ bond order), VSEPR (geometry of XeF₄)
2023	3	Hybridization, dipole moment, hydrogen bonding
2022	2	Fajan’s rules, lattice energy comparison
2021	2	MOT bond order, VSEPR geometry
2020	3	Born-Haber cycle, hybridization, dipole moment

The pattern is clear: MOT bond order + VSEPR geometry + hybridization identification together account for nearly 60% of Chemical Bonding questions in JEE Main.

Key Concepts You Must Know

Prioritised by exam frequency — start from the top.

Tier 1 (High frequency — must master)

Hybridization determination: sp, sp², sp³, sp³d, sp³d² for main-group compounds including exceptions (PCl₅, SF₆, XeF₄)
VSEPR geometry: electron pair geometry vs molecular geometry distinction; lone pair repulsion order (lp–lp > lp–bp > bp–bp)
MOT bond order formula: Bond Order = (Nb − Na) / 2; magnetic character from unpaired electrons
Hydrogen bonding: intermolecular vs intramolecular; effect on boiling point, solubility, viscosity

Tier 2 (Medium frequency — know well)

Fajan’s rules: predicting ionic vs covalent character; polarising power vs polarisability
Dipole moment: resultant in symmetric vs asymmetric molecules; zero dipole despite polar bonds (CO₂, BF₃, CCl₄)
Lattice energy trends: Born-Haber cycle as a Hess’s Law application
VSEPR for hypervalent molecules: PCl₅ (trigonal bipyramidal), IF₇ (pentagonal bipyramidal)

Tier 3 (Lower frequency but asked in Advanced)

MOT for diatomics: filling order changes between N₂ and O₂; why O₂ is paramagnetic
Resonance and formal charge: identifying most stable resonance structure
Bent rule: orbitals with more s-character prefer electronegative substituents

Important Formulas

\text{Bond Order} = \frac{N_b - N_a}{2}

where $N_b$ = electrons in bonding MOs, $N_a$ = electrons in antibonding MOs.

When to use: Any question asking about bond length, bond strength, or magnetic nature of a diatomic species (O₂, N₂, NO, CO, etc.). Higher bond order → shorter bond length → higher bond dissociation energy.

\text{Hybridization} = \frac{1}{2}\left[V + H - C + An\right]

where $V$ = valence electrons of central atom, $H$ = monovalent atoms attached, $C$ = charge if cationic, $An$ = charge if anionic.

When to use: Quick hybridization determination for any molecule without drawing the full structure. The result gives total electron pairs (bond pairs + lone pairs).

U \propto \frac{Z^+ \cdot Z^-}{r_0}

where $Z^+, Z^-$ are ionic charges and $r_0$ is interionic distance.

When to use: Comparing lattice energies across a series (MgO vs NaCl, or NaF vs NaI). Charge has a quadratic effect — doubling charge quadruples lattice energy, so MgO >> NaCl.

\mu = q \times d

Units: Debye (D). For a molecule, the net dipole is the vector sum of all bond dipoles.

When to use: Identifying whether a molecule has a net dipole (decides polarity, solubility in water). A molecule can have polar bonds and still be non-polar if bond dipoles cancel by symmetry.

Solved Previous Year Questions

PYQ 1 — Bond Order and Magnetic Nature (JEE Main 2024 Shift 1)

Question: The bond order and magnetic nature of O₂⁻ respectively are:

(A) 1.5, paramagnetic (B) 1.5, diamagnetic (C) 2.0, paramagnetic (D) 2.0, diamagnetic

Solution:

O₂ has 16 electrons. O₂⁻ (superoxide ion) has 17 electrons.

MOT filling for O₂ (16e):

\sigma_{1s}^2, \sigma_{1s}^{*2}, \sigma_{2s}^2, \sigma_{2s}^{*2}, \sigma_{2p}^2, \pi_{2p}^4, \pi_{2p}^{*2}

For O₂⁻, add 1 more electron to $\pi_{2p}^*$ :

\pi_{2p}^{*3} \text{ (one orbital gets the extra electron)}

$N_b = 2 + 4 + 2 = 8$ (counting only valence shell), $N_a = 2 + 3 = 5$

\text{Bond Order} = \frac{8-5}{2} = 1.5

The $\pi_{2p}^*$ level has 3 electrons — one orbital has a single unpaired electron. So O₂⁻ is paramagnetic.

Answer: (A)

Many students forget to count both the 1s and 2s electrons in $N_b$ and $N_a$ , then get fractional numbers that don’t make sense. The cleaner approach: work only with valence electrons (n=2 shell) and the well-known O₂ MOT diagram, then just add/remove electrons from the top.

PYQ 2 — VSEPR Geometry (JEE Main 2023 Shift 2)

Question: The correct geometry and hybridization for XeF₄ are:

(A) Tetrahedral, sp³ (B) Square planar, sp³d² (C) See-saw, sp³d (D) Trigonal bipyramidal, sp³d

Solution:

Xe in XeF₄:

Valence electrons on Xe = 8
4 F atoms form 4 bond pairs
Remaining electrons on Xe = 8 − 4×1 = 4 → 2 lone pairs

Total electron pairs = 4 (bp) + 2 (lp) = 6 → octahedral electron geometry → hybridization is sp³d²

Now place the 2 lone pairs for minimum repulsion: they go axial positions (not adjacent to each other). The 4 F atoms are in the equatorial plane → molecular geometry is square planar.

Answer: (B)

A key observation: when lone pairs are in an octahedral arrangement, they always occupy positions that are 180° apart (axial-axial or trans positions) to minimise lp–lp repulsion. This rule consistently gives the right answer for XeF₄, XeF₂ (linear), and ICl₄⁻.

PYQ 3 — Fajan’s Rules (JEE Main 2022 Session 1)

Question: Arrange the following in increasing order of covalent character: AlF₃, AlCl₃, AlBr₃, AlI₃

Solution:

All four have Al³⁺ as the cation — same polarising power. The difference is the anion’s polarisability.

Polarisability of anion increases down the group: F⁻ < Cl⁻ < Br⁻ < I⁻

Larger, more diffuse electron cloud = easier to distort = more covalent character.

\text{Covalent character: AlF}_3 < \text{AlCl}_3 < \text{AlBr}_3 < \text{AlI}_3

This is why AlI₃ is a molecular compound (exists as Al₂I₆ dimer) while AlF₃ is a typical ionic salt with a high melting point.

Fajan’s rules questions often ask you to compare across the same cation with different anions (as above) OR the same anion with different cations. For cation comparison: higher charge and smaller size → more polarising → more covalent. Li⁺ compounds are more covalent than Na⁺ compounds of the same anion.

Difficulty Distribution

For JEE Main Chemical Bonding questions, the typical split is:

Difficulty	% of Questions	What It Looks Like
Easy	40%	Direct hybridization, VSEPR of common molecules (NH₃, H₂O, BF₃)
Medium	45%	MOT bond order of ions, dipole moment comparison, Fajan’s rules ordering
Hard	15%	Bent rule application, resonance + formal charge in exotic molecules, Born-Haber cycle calculation

The good news: Chemical Bonding in JEE Main rarely goes beyond Medium difficulty. The Hard questions appear more in Advanced, and even there, they’re usually hybrids with coordination chemistry.

Expert Strategy

Week 1 of preparation: Get VSEPR and hybridization locked in completely. These two together give you easy marks with minimal effort. Practice 20 molecules until you can determine geometry and hybridization in under 30 seconds each.

Week 2: Do MOT for the 10 most important species: H₂, He₂, Li₂, B₂, C₂, N₂, O₂, F₂, NO, CO. Know the filling order change at N₂ (where $\pi_{2p}$ comes before $\sigma_{2p}$ ) and memorise that O₂ is paramagnetic — this is a classic trap.

The 10-molecule MOT shortcut: You don’t need to redo MOT from scratch for every question. Memorise the bond orders for neutral diatomics (H₂=1, He₂=0, Li₂=1, B₂=1, C₂=2, N₂=3, O₂=2, F₂=1, Ne₂=0). For ions, just add/subtract electrons from the highest occupied MO and adjust bond order by ±0.5.

Born-Haber cycle: Don’t memorise it as a formula. Draw the energy ladder from scratch each time — elements at bottom, ionic compound at top. The cycle must close (sum = 0). This approach means you can answer any variation without blanking on formula order.

Hydrogen bonding questions in JEE are mostly about consequences: boiling point anomalies (HF, H₂O, NH₃), ortho vs para nitrophenol separation, viscosity of glycerol. Make a quick table: intramolecular H-bonding lowers BP, intermolecular raises it.

Common Traps

Trap 1 — Dipole moment of CO₂ vs SO₂: Students often call both non-polar because they “look similar.” CO₂ is linear → bond dipoles cancel → μ = 0. SO₂ is bent (lone pair on S) → bond dipoles don’t cancel → μ ≠ 0. The lone pair on the central atom changes everything.

Trap 2 — Hybridization of BeCl₂ vs H₂O: Both have 4 electron pairs… wait, no. BeCl₂ has only 2 bond pairs and no lone pairs on Be → sp, linear. Students incorrectly assign sp³. Always count lone pairs on the central atom after bonding is done.

Trap 3 — “More ionic character” vs “more covalent character”: Questions sometimes ask for the most ionic compound in a series. This is the opposite of Fajan’s rules application. LiF is the most ionic alkali metal halide (small cation, small anion = minimum polarisation). Students who only memorise “LiI is more covalent” sometimes flip this in the heat of the exam.

Trap 4 — Formal charge vs actual charge: In resonance structures, the “best” structure minimises formal charges and places negative formal charge on the more electronegative atom. This is not the same as the actual partial charge from electronegativity. JEE Advanced questions sometimes test whether you can distinguish these two concepts in the same molecule.

One last pattern worth noting: JEE Main 2019–2024 has asked about the geometry of XeF₂, XeF₄, XeOF₄ at least once each. These three xenon fluorides are a guaranteed return investment — 3 molecules, 3 distinct geometries (linear, square planar, square pyramidal), and they appear consistently in the exam.