Key Diagrams in JEE Main Chemistry You Must Know

JEE Main Chemistry papers consistently include questions that require you to recall or interpret a specific molecular structure, energy diagram, or phase diagram. These visual/spatial questions reward students who have built a mental image library. This guide catalogues the diagrams with the highest exam frequency, tells you the one or two critical features to note about each, and explains the question types they generate.

Molecular Structure Diagrams

XeF₂: linear (sp³d, 3 lone pairs, 2 bond pairs — VSEPR gives linear shape despite sp³d hybridisation). XeF₄: square planar (sp³d², 2 lone pairs equatorial and 2 axial positions from VSEPR — lone pairs go equatorial in bipyramidal arrangement, giving square planar). XeF₆: distorted octahedral (one lone pair). PCl₅: trigonal bipyramidal (equatorial bonds shorter/stronger than axial). SF₄: see-saw shape (1 lone pair in equatorial position). ClF₃: T-shaped (2 lone pairs equatorial). ICl₂⁻: linear (3 lone pairs on I). I₃⁻: linear (same as XeF₂). JEE uses these in "identify the correct structure/hybridisation/bond angle" MCQs.

Critical features to note: (1) in trigonal bipyramidal arrangements, lone pairs ALWAYS go to equatorial positions (less repulsion at 120° than at 90°); (2) bond angles decrease when lone pairs replace bonding pairs (109.5° → 107° in NH₃ → 104.5° in H₂O); (3) resonance averages out double bonds (SO₃ has 3 equal S-O bonds despite the Lewis structure showing one double and two single). Take a free chemical bonding mock. See our VSEPR and chemical bonding guide.

MO Diagrams and Bond Order

Molecular orbital diagrams for diatomic molecules: H₂ (BO = 1), He₂ (BO = 0, not stable), O₂ (BO = 2, paramagnetic — 2 unpaired electrons in degenerate π* orbitals), N₂ (BO = 3, triple bond). O₂ paramagnetic result (from MO theory) contradicts Lewis structure prediction (all paired) — this is the major MO theory exam point. F₂ (BO = 1), Ne₂ (BO = 0). For O₂ MO: σ1s², σ*1s², σ2s², σ*2s², σ2pz², π2px² = π2py², π*2px¹ = π*2py¹ — the two unpaired electrons in degenerate π* orbitals cause paramagnetism. Bond order = (bonding e⁻ − antibonding e⁻)/2 = (10−6)/2 = 2.

Reaction Coordinate Diagrams

Energy vs reaction coordinate (reaction profile): reactants at one level, transition state at the maximum (activation energy Ea = E_TS − E_reactants), products at another level. For exothermic reactions: products are lower than reactants (ΔH < 0). For endothermic: products higher (ΔH > 0). Ea (forward) − Ea (reverse) = ΔH. Catalyst lowers both Ea (forward) and Ea (reverse) by the same amount, without changing ΔH or the position of equilibrium. JEE question: given a catalysed and uncatalysed energy profile, find the ratio of rate constants or identify the effect on equilibrium constant K (answer: K is unchanged by catalyst).

Phase Diagrams for Water and CO₂

Water phase diagram: triple point at 0.006 atm, 0.01°C. Critical point at 218 atm, 374°C. The solid-liquid boundary has negative slope (ice melts under pressure) — anomalous compared to most substances. CO₂ phase diagram: triple point at 5.11 atm, −56.6°C; critical point at 73 atm, 31°C. At 1 atm, CO₂ sublimes (solid → gas directly at −78.5°C — dry ice). JEE asks about phase transitions at specific P and T: is CO₂ a liquid at 1 atm (no, it sublimes)? Can liquid water exist below 0°C (yes, if pressure is high enough — see ice skating explanation). For the electrochemistry diagrams see our electrochemistry guide and for organic reaction energy diagrams see our reaction mechanisms guide.