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APEX CLASS ESTD 2018

IITJEE CHEMISTRY: VSEPR Geometries, Hybridization & Molecular Orbital Theory Visualizer

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Ashish Kumar Bahekar

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APEX CLASS • IITJEE Chemical Bonding Lab

VSEPR Geometries, Hybridization & Molecular Orbital Theory (MOT)

Master Steric Number Shapes, Bent’s Rule (sp³d), s-p Mixing (≤14e⁻ vs >14e⁻), and 14-Electron Bond Order!

🔬 Steric Number SN = ½ [V + M – C + A]. Select Hybridization Family:
1. CH₄ (4 BP + 0 LP) C Tetrahedral (109°28′) Dipole μ = 0 D (Symmetric) 2. NH₃ (3 BP + 1 LP) •• N Trigonal Pyramidal (107°) μ(NH₃) > μ(NF₃) [JEE Favorite!] 3. H₂O (2 BP + 2 LP) •• •• O Bent / Angular (104.5°) Strong LP–LP Repulsion! 🎯 VSEPR Repulsion Order: LP–LP > LP–BP > BP–BP! 1. PCl₅ (5 BP + 0 LP) Trigonal Bipyramidal daxial > deq ! 2. SF₄ (4 BP + 1 LP) •• See-Saw Shape 1 LP at Equatorial (120°) 3. ClF₃ (3 BP + 2 LP) Bent T-Shape (87.5°) 2 LPs at Equatorial 4. XeF₂ / I₃⁻ (2 BP + 3 LP) LINEAR (180°!) All 3 LPs at Equatorial! 🏆 Bent’s Rule (sp³d): Lone pairs & double bonds ALWAYS go to Equatorial; Electronegative atoms go to Axial! 1. SF₆ (6 BP + 0 LP) Regular Octahedral (90°) All 6 S-F bonds are EQUAL length! 2. BrF₅ / XeOF₄ (5 BP + 1 LP) Square Pyramidal (<90°) Polar Molecule (μ ≠ 0) 3. XeF₄ / ICl₄⁻ (4 BP + 2 LP) Square Planar (90°) 2 LPs Trans (180°) ⇒ μ = 0 D! ⚡ sp³d³ Bonus: IF₇ is Pentagonal Bipyramidal, XeF₆ is Distorted Octahedral! 1. Drago’s Rule (Pure p-Orbitals, ~90°!) • Hydrides of Group 15 & 16 (Period ≥ 3): PH₃, AsH₃, SbH₃ and H₂S, H₂Se, H₂Te • NO Hybridization takes place! Bonding uses almost pure orthogonal p-orbitals ⇒ Angle ≈ 90°–93°! • Lone pair sits in stereochemically inert pure s-orbital (Why Basicity: NH₃ >> PH₃ > AsH₃ > SbH₃!) 2. Odd-Electron Radicals • CF₃• vs CH₃• Radical: CH₃• is sp² (Planar, 120°), whereas CF₃• is sp³ (Pyramidal) due to high EN of F! • NO₂ (134°, sp² Bent) & ClO₂ (118°, sp²/sp³ Bent) • Solid State Ionic Forms: PCl₅(s) = [PCl₄]⁺(sp³) + [PCl₆]⁻(sp³d²) PBr₅(s) = [PBr₄]⁺(sp³) + Br⁻ 🎯 Bond Angle Master: NO₂⁺ (180°, sp Linear) > NO₂ (134°, Odd e⁻) > NO₂⁻ (115°, sp² with full Lone Pair)!
Steric Number Formula
SN = ½ [V + M – C + A]
M = Monovalent Atoms (H, X)
% s-Character & EN
sp(50%) > sp²(33%) > sp³(25%)
More %s = Shorter, Stronger Bond!
Zero Dipole (μ = 0) List
CO₂, BF₃, CCl₄, PCl₅, SF₆
Also XeF₂ (Linear) & XeF₄ (Sq Planar)
Hydride Angle Order
NH₃(107°) > PH₃(93.6°) > AsH₃
H₂O(104.5°) > H₂S(92°) > H₂Se
🧬 Bond Order = ½ (Nbonding – Nantibonding). Select Diagram:
≤ 14 e⁻ (B₂, C₂, N₂): s-p Mixing Active! E ↑ σ*₂pz π*₂px = π*₂py σ₂pz (Pushed UP!) ↑ ↑ π₂px = π₂py (B₂) Order: (π₂px = π₂py) < σ₂pz 🏆 3 Famous JEE Traps from s-p Mixing 1. Why B₂ (10 e⁻) is PARAMAGNETIC: Config ends in π₂px¹ π₂py¹ (2 unpaired, BO=1)! 2. Why C₂ (12 e⁻, BO=2) has ONLY π-Bonds: Config ends in π₂px² π₂py² (Both bonds are π-bonds!) 3. N₂ (14 e⁻, BO=3) → N₂⁺ (13 e⁻, BO=2.5): Electron leaves σ₂pz, so N₂⁺ has weaker σ-bond! 🎯 NCERT Highlight: In C₂ molecule (BO=2, Diamagnetic), the double bond consists of BOTH π-bonds because σ*₂s cancels σ₂s! > 14 e⁻ (O₂, F₂, Ne₂): NO s-p Mixing! σ*₂pz ↑ ↑ π*₂px¹ = π*₂py¹ (O₂!) ↑↓ ↑↓ π₂px = π₂py ↑↓ σ₂pz (Lowest!) Order: σ₂pz < (π₂px = π₂py) Why O₂ (16 e⁻) is Paramagnetic • Valence: σ₂pz² (π₂px² = π₂py²) (π*₂px¹ = π*₂py¹) • 2 Unpaired Electrons in degenerated π*! • Spin Magnetic Moment: μ = √8 = 2.83 BM • Bond Order of O₂: ½ (10 – 6) = 2.0 • Superoxide O₂⁻ (17 e⁻): 1 unpaired (Paramag) • Peroxide O₂²⁻ (18 e⁻): 0 unpaired (Diamag!) 🧲 Paramagnetic Even-Electron Species to Memorize: B₂ (10 e⁻) and O₂ / S₂ (16 e⁻) are Paramagnetic despite having even electrons! BO=3 BO=2 BO=1 BO=0 ★ 14 e⁻ Peak: BO = 3.0 (N₂, CO, CN⁻, NO⁺) 10e⁻(B₂): 1.0 12e⁻(C₂): 2.0 15e⁻(NO, O₂⁺): 2.5 16e⁻(O₂): 2.0 17e⁻(O₂⁻): 1.5 18e⁻(F₂, O₂²⁻): 1.0 8e⁻ (Be₂) 14 e⁻ 20e⁻ (Ne₂) 5-Second Bond Order Formula BO = 3.0 – 0.5 × | Total e⁻ – 14 | • Valid for ALL species from 10 e⁻ to 20 e⁻! • Every ±1 electron from 14 drops BO by 0.5 • BO = 0 means molecule does NOT exist: He₂ (4 e⁻), Be₂ (8 e⁻), Ne₂ (20 e⁻) → BO = 0! 🏆 Instant Magnetic Check: All ODD-electron species (11, 13, 15, 17 e⁻) + B₂ (10 e⁻) + O₂ (16 e⁻) are PARAMAGNETIC; all others are Diamagnetic! 1. O₂ Family Stability & Bond Length Series • O₂⁺ (15 e⁻, Dioxygenyl): BO = 2.5 (1 unpaired e⁻) • O₂ (16 e⁻, Dioxygen): BO = 2.0 (2 unpaired e⁻) • O₂⁻ (17 e⁻, Superoxide): BO = 1.5 (1 unpaired e⁻) • O₂²⁻(18 e⁻, Peroxide): BO = 1.0 (0 unpaired, Dia!) Bond Length Order: O₂⁺ < O₂ < O₂⁻ < O₂²⁻ (BL ∝ 1/BO)! 2. N₂ Series & The Famous CO⁺ (BO = 3.5!) Trap • N₂ Series: N₂ (BO = 3.0) > N₂⁺ (BO = 2.5) = N₂⁻ (BO = 2.5) (Note: N₂⁺ is MORE stable than N₂⁻ because N₂⁻ has e⁻ in π*!) • 🚨 JEE Advanced Exception: CO (14 e⁻) → CO⁺ (13 e⁻) In CO, highest HOMO is slightly antibonding (non-bonding σ*) ⇒ Removing 1 e⁻ INCREASES Bond Order to 3.5! Bond Length of CO⁺ (1.115 Å) < CO (1.128 Å)! ⚡ Equal Bond Order Tie-Breaker (N₂⁺ vs N₂⁻, both BO=2.5): The species with FEWER antibonding electrons (N₂⁺) is always MORE stable!
Bond Order Formula
BO = ½ (Nb – Na)
Higher BO = Higher BDE = Shorter BL
s-p Mixing Boundary
≤ 14 e⁻: π₂p < σ₂pz
> 14 e⁻ (O₂, F₂): σ₂pz < π₂p
Even-e⁻ Paramagnetic
B₂ (10 e⁻) & O₂ / S₂ (16 e⁻)
Each has 2 Unpaired Electrons (2.83 BM)
Polyatomic Bond Order
CO₃²⁻(1.33) | SO₄²⁻(1.5)
PO₄³⁻(1.25) | ClO₄⁻(1.75)
🔥 2 Bonus JEE Chemical Bonding Shortcuts:
1. Polyatomic Resonance Bond Order: BO = Total Bonds / Total Canonical Terminal Positions → NO₃⁻ / CO₃²⁻ = 4/3 = 1.33 | SO₄²⁻ = 6/4 = 1.5 | ClO₄⁻ = 7/4 = 1.75!
2. Fajans’ Rule (Covalent Character in Ionic Bonds): Small Cation + Large Anion + High Charge = Maximum Polarization = Higher Covalent Character & Lower Melting Point (e.g., Covalent character: LiCl > NaCl > KCl and AlCl₃ > MgCl₂ > NaCl)!
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