ChemistryCHEM 02
Chemical Bonding
Hybridisation, VSEPR theory, molecular orbital theory, hydrogen bonding, dipole moment
Concept Core
Essential theory — everything NCERT tests on Chemical Bonding
VSEPR THEORY
Shape depends on electron pairs (bonding + lone pairs) around central atom. Lone pairs occupy more space → greater repulsion → compressed bond angles.
2 BP: linear (180°). 3 BP: trigonal planar (120°). 4 BP: tetrahedral (109.5°). With 1 LP: pyramidal (107°). 2 LP: bent (104.5°).
NH₃: pyramidal. H₂O: bent. BF₃: planar. CH₄: tetrahedral. PCl₅: trigonal bipyramidal. SF₆: octahedral.
HYBRIDISATION
sp: 2 hybrid orbitals, linear (180°). sp²: 3 orbitals, planar (120°). sp³: 4 orbitals, tetrahedral (109.5°). sp³d: 5, trigonal bipyramidal. sp³d²: 6, octahedral.
BeCl₂ = sp. BF₃ = sp². CH₄ = sp³. PCl₅ = sp³d. SF₆ = sp³d². XeF₄ = sp³d².
MOLECULAR ORBITAL THEORY
LCAO: Atomic orbitals combine → molecular orbitals. Bonding MO (σ, π) = lower energy. Antibonding (σ*, π*) = higher energy.
Bond order = ½(Nb − Na). BO=0: doesn't exist. BO>0: stable. Higher BO = shorter, stronger bond.
O₂: BO=2, paramagnetic (2 unpaired e in π*). N₂: BO=3 (strongest N≡N). He₂: BO=0 (doesn't exist).
HYDROGEN BONDING
F–H···F, O–H···O, N–H···N. Requires high EN + small size + lone pair.
Intermolecular H-bonding: higher BP, viscosity (HF, H₂O, NH₃ higher BP than expected). Intramolecular: o-nitrophenol (lower BP than p-nitrophenol).
FAJAN'S RULES & DIPOLE MOMENT
Fajan's rules: covalent character increases with smaller cation, larger anion, high charge. LiF more ionic; LiI more covalent.
Dipole moment: μ = q×d. BF₃ = 0 (symmetric). CCl₄ = 0. H₂O ≠ 0 (bent). NH₃ ≠ 0. CO₂ = 0 (linear).
Fact & Formula Vault
High-yield facts, numbers, and formulas
Hybridisation Table
sp: BeCl₂, CO₂, linear
sp²: BF₃, NO₃⁻, trigonal planar
sp³: CH₄, NH₃, H₂O
sp³d: PCl₅ (trigonal bipyramidal)
sp³d²: SF₆, XeF₄ (octahedral)
Bond Order
N₂: BO=3 (triple bond)
O₂: BO=2 (double, paramagnetic)
F₂: BO=1
NO: BO=2.5 (paramagnetic)
He₂: BO=0 (doesn't exist)
Key Dipoles
CO₂: μ=0 (linear symmetric)
BF₃: μ=0 (trigonal symmetric)
H₂O: μ≠0 (bent)
NH₃: μ≠0 (pyramidal)
CCl₄: μ=0 (tetrahedral symmetric)
Worked Examples
NEET-style questions solved step-by-step
EASYShape of H₂O molecule and bond angle:▾
Shape of H₂O molecule and bond angle:
H₂O has sp³ hybridisation at O. 2 bonding pairs + 2 lone pairs. VSEPR → bent/angular shape. Bond angle ≈ 104.5° (less than tetrahedral 109.5° due to 2 lone pairs).
MEDIUMO₂ is paramagnetic because it has:▾
O₂ is paramagnetic because it has:
2 unpaired electrons in degenerate π*₂p orbitals (Hund's rule in MOs). MOT correctly predicts O₂ paramagnetism; VBT fails to explain this.
HARDBond order of NO molecule using MOT:▾
Bond order of NO molecule using MOT:
MO configuration of NO: σ1s²σ*1s²σ2s²σ*2s²σ2p²π2p⁴π*2p¹. N_b=8, N_a=3. BO = (8−3)/2 = 2.5. Paramagnetic (1 unpaired e in π*).
Mistake DNA
Common NEET traps for this chapter
⚠ BF₃ is electron deficient
BF₃ has only 3 bond pairs, no lone pair on B — it is an electron deficient molecule. It is a Lewis acid.
✓ Fix: BF₃: sp², trigonal planar, μ=0. Accepts electron pairs.
⚠ H-bonding and boiling point
HF has LOWER bp than H₂O despite stronger H-bonds because H₂O forms more H-bonds per molecule (4 per molecule vs ~2 for HF).
✓ Fix: H₂O > HF for bp because more H-bonds per molecule
⚠ Dipole moment of CCl₄
CCl₄ has polar C-Cl bonds but symmetric tetrahedral shape → net dipole = 0. Lone pairs and bonding pairs are symmetric.
✓ Fix: Symmetric molecule: μ = 0 even if individual bonds are polar.
Chapter Intelligence
Exam data and last-minute strategy
NEET Frequency
3–4 Q/year. VSEPR shapes, hybridisation, bond order (O₂ paramagnetic), H-bonding effects on bp, dipole moment of key molecules.
High-Yield
sp³ = tetrahedral. sp = linear. sp² = planar. O₂ paramagnetic (BO=2). N₂: BO=3. H₂O bent (104.5°). CO₂ and BF₃: μ=0.
Strategy
Make a shape table: molecule, hybridisation, geometry, bond angle. Know O₂ and N₂ bond orders. These are the most frequently appearing question types.
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