Chemical Bonding and Molecular Structure

Physical Chemistry Weightage: 3-4 Questions (12-16 Marks) JEE Unit 3

“Welcome back! These notes assume you have already read the NCERT chapter and know the basic definitions of ionic and covalent bonds, hybridisation and molecular orbitals, so they skip the textbook story and keep what JEE Main actually asks. Chemical Bonding is the heaviest scoring chapter in the paper: it regularly gives three to four questions, and almost all of them are pattern questions on shapes, hybridisation, bond order, magnetism, dipole moment and a handful of famous exceptions. Learn the shape table, the molecular orbital filling routine and the exception list below and this chapter becomes one of the quickest 12 to 16 marks in the paper.”

— SCORECHEM ACADEMIC TEAM

1. Ionic Bond and Fajans' Rules

Favourable for an ionic bond: low ionisation enthalpy of the metal, very negative electron gain enthalpy of the non-metal, and high lattice enthalpy (rises with charge, falls with ionic size). Lattice enthalpy, not electronegativity difference alone, decides stability of the solid.
Fajans' rules say when an ionic bond gains covalent character: small, highly charged cation; large, highly charged anion; and a cation with an 18-electron (pseudo-noble-gas) shell.

Fig. 1: Fajans' rules. More covalent means lower melting point, lower solubility, deeper colour.

Consequences to quote: covalent character up means lower melting point (BeCl₂ < MgCl₂ < ... < BaCl₂ but CaF₂ > CaCl₂ > CaBr₂ > CaI₂), lower solubility in water (Ag₂S < Ag₂O; Fe(OH)₃ < Fe(OH)₂), and deeper colour (AgCl white, AgBr pale yellow, AgI yellow; PbCl₂ white, PbI₂ yellow; SnCl₂ white, SnI₂ black).
Higher oxidation state, more covalent: SnCl₄ > SnCl₂, PbCl₄ > PbCl₂. This is why UF₆ is more covalent than UF₄ (U⁶⁺ is smaller and more polarising than U⁴⁺). It looks backwards to many students because fluorides are "ionic".

⚠️ JEE Trap: UF₄ vs UF₆. Fajans' rule compares oxidation states of the SAME metal: the higher state (UF₆, +6) is more covalent. Also remember the anomaly in halogen bond dissociation enthalpy: Cl₂ > Br₂ > F₂ > I₂, because lone-pair repulsion on the tiny F atoms weakens F–F.

Isomorphism needs the same formula type AND isostructural ions of similar size: MgSO₄.7H₂O and ZnSO₄.7H₂O, alums, BaSO₄ and KMnO₄ (both tetrahedral anions). NaNO₃ and NaClO₃ are NOT isomorphous: NO₃⁻ is planar (sp²) but ClO₃⁻ is pyramidal (sp³).

2. Lewis Structures, Formal Charge, Resonance

Formal charge FC = (valence e⁻) − (non-bonding e⁻) − ½(bonding e⁻). Structures with the smallest FCs, and negative FC on the more electronegative atom, are preferred.
HNO₃ (O atoms: OH, =O, O⁻): FC of N = 5 − 0 − 4 = +1; OH oxygen 0; =O oxygen 0; terminal O⁻ = 6 − 6 − 1 = −1. O₃: central O = 6 − 2 − 3 = +1; single-bonded terminal O = −1; double-bonded terminal O = 0.
Octet-rule exceptions (JEE always asks the count):

Type	Examples
Incomplete octet	BeH₂ (4e⁻), BF₃, BCl₃, AlCl₃ (6e⁻), LiCl
Odd electron	NO (11), NO₂ (17), ClO₂ (19)
Expanded octet	PF₅, PCl₅, SF₆, SF₄, ClF₃, XeF₂, IF₇, H₂SO₄
Obeys octet (do not count)	CO₂, CH₄, SiF₄, C₂H₆, CHCl₃, CBr₄, NH₃, SCl₂

Resonance and bond order: BO = (total bonds between the two atoms over all structures) / (number of structures). CO₃²⁻, NO₃⁻: 1.33; O₃, NO₂⁻, benzene, RCOO⁻: 1.5. Resonance energy is the extra stability of the hybrid over the most stable single structure. Equal bond lengths in the hybrid (all C–O in carbonate identical) are the proof.
Coordinate bond: both electrons come from one atom (NH₄⁺, H₃O⁺, O₃, HNO₃, [Cu(NH₃)₄]²⁺).

3. VSEPR Shapes and Bond Angles

Repulsion order: lp–lp > lp–bp > bp–bp. Multiple bonds count as one region but take more space. Learn the table by (SN / lp) code rather than by names.

Fig. 2: SN = sigma bonds + lone pairs on the central atom. Shape follows the lone pairs.

Fig. 3: In trigonal bipyramidal (sp³d) electron geometry, lone pairs fill equatorial slots.

sp³d rules: lone pairs equatorial; the more electronegative atom axial (Bent's rule); double bonds equatorial. Consequences: SF₄ (see-saw), ClF₃ (T), XeF₂ (linear), XeO₂F₂ (see-saw with the two O equatorial and the two F axial; F–Xe–F is about 180°).
Unequal bond lengths: axial bonds are longer than equatorial in PF₅, SF₄ and ClF₃. All bonds are equal in BF₄⁻, SiF₄, XeF₄, SF₆.
Bond angle orders that JEE loves:
- NO₂⁺ (180°) > NO₂ (134°) > NO₂⁻ (115°): fewer electrons in the odd/lone position, more angle.
- NH₃ (107°) > PH₃ (93.6°) and H₂O (104.5°) > H₂S (92°): on going down the group the bond pair moves away from the central atom and hybridisation is replaced by nearly pure p orbitals.
- Cl₂O (111°) > H₂O (104.5°) > OF₂ (103°): with electronegative F the bond pair is drawn away from O, so bp–bp repulsion falls and lp–lp closes the angle.
- BF₃ (120°) > PF₃ (97.8°) > ClF₃ (87.5°, below 90° because two lone pairs squeeze the axial F).
Solid-state exceptions: PCl₅(s) = [PCl₄]⁺[PCl₆]⁻; PBr₅(s) = [PBr₄]⁺Br⁻; XeF₆(s) = [XeF₅]⁺F⁻; I₂Cl₆ shows [ICl₂]⁺ and [ICl₄]⁻.

⚠️ JEE Trap: Counting bonds instead of regions. ClF₃ has 3 bonds but 5 electron pairs (T-shaped, sp³d, polar). BrF₅, XeOF₄ and IF₅ are square pyramidal; XeF₄ and ICl₄⁻ are square planar. Always write SN = (valence e⁻ of the central atom ± charge − e⁻ used by monovalent atoms) / 2 for a quick start.

4. Hybridisation and Bond Parameters

SN	Hybridisation	% s	Geometry
2	sp	50	Linear
3	sp²	33.3	Trigonal planar
4	sp³	25	Tetrahedral
4	dsp²	0 (uses inner d)	Square planar
5	sp³d	20	Trigonal bipyramidal
6	sp³d²	16.7	Octahedral
7	sp³d³	14.3	Pentagonal bipyramidal

Trends from s-character: more s means shorter bond, larger bond angle, higher electronegativity of carbon and stronger acid character of C–H. C–H lengths: sp (106 pm) < sp² (108 pm) < sp³ (109 pm).
Bond length order: triple < double < single for the same atoms. For different single bonds compare atomic radii. Example order: C–H (109 pm) < C≡N (116 pm) < C=O (123 pm) < C–O (143 pm).
Species to memorise: planar sp²: BF₃, NO₃⁻, CO₃²⁻, NO₂⁻, NO₂, carbocations (CH₃⁺). sp³ with different shapes: NH₃, H₂O, ClO₃⁻, ClO₂⁻. Tetrahedral around S in S₂O₃²⁻, S₂O₇²⁻, SO₄²⁻; only SO₃²⁻ is pyramidal in that family.
sigma and pi counting: each single bond is one σ; a double bond is 1σ + 1π; a triple bond is 1σ + 2π. CH₂=CH–C≡N has 6 σ and 3 π bonds.
Silicates and P oxides: P₄O₆ has 6 P–O–P links; P₄O₁₀ adds 4 terminal P=O. SiO₂ is a 3D network of sp³ silicon with only Si–O single bonds.

5. Molecular Orbital Theory

Fig. 4: MO energy order (1s levels omitted). Only the σ(2pz) and π(2p) positions swap.

Fill order: use the left ladder when the species has 14 electrons or fewer in total (B₂, C₂, N₂, N₂⁺, CN⁻ ...) and the right ladder for 15 or more (O₂, F₂, NO, O₂⁺ and their ions). Fill by Aufbau and Hund.
BO = ½(N_b − N_a), N_b counted without the core. BO > 0 means the species can exist. Bond strength ∝ BO ∝ 1/(bond length).
Isoelectronic shortcut: N₂, CO, CN⁻, NO⁺, C₂²⁻ all have 14 electrons and BO 3, diamagnetic.

Fig. 5: Bond order (BO) and magnetism from MO filling. Bond length rises as BO falls.

Ionise it, and the bond changes the other way: removing an electron from N₂ takes it from a bonding MO (N₂⁺: BO 2.5, weaker), but removing one from O₂ takes it from an antibonding MO (O₂⁺: BO 2.5, stronger). Bond lengths: O₂⁺ (112 pm) < O₂ (121) < O₂⁻ (128) < O₂²⁻ (149).
C₂ has two π bonds and no σ bond. B₂ has a single bond made of two half-π bonds and is paramagnetic (two unpaired e⁻ in degenerate π).
Half-integer species: H₂⁺, He₂⁺, H₂⁻ have BO 0.5 and are paramagnetic; He₂, Be₂, Ne₂ have BO 0 and do not exist.
Wave functions: bonding MO = ψ_A + ψ_B (constructive), antibonding σ* = ψ_A − ψ_B (destructive, node between nuclei).
Paramagnetic list: B₂, O₂, NO, NO₂, ClO₂, O₂⁺, O₂⁻, N₂⁺, N₂⁻, C₂⁻. Diamagnetic: C₂, N₂, F₂, O₂²⁻, CN⁻, CO, NO⁺.

⚠️ JEE Trap: Same bond order, different magnetism. A pair must match on BOTH: O₂⁺ (2.5, paramagnetic) and N₂⁻ (2.5, paramagnetic) qualify; O₂⁻ (1.5) and N₂⁺ (2.5) do not. Always write the electron count first: N₂⁻ has 15, O₂⁺ has 15 (both with one unpaired electron in the last orbital).

6. Dipole Moment

μ = q × d; 1 D = 3.336 × 10⁻³⁰ C m. Vector sum for two bond dipoles at angle θ: μ_R = √(P² + Q² + 2PQ cosθ). Percent ionic character = (μ_observed / μ_{100% ionic}) × 100, where μ_100% = e × d.
Zero dipole moment (symmetric): linear CO₂, CS₂, BeCl₂, BeF₂, XeF₂; trigonal planar BF₃, SO₃; tetrahedral CH₄, CCl₄, SiF₄; TBP PCl₅(g); octahedral SF₆; square planar XeF₄; and H₂, N₂.
Non-zero: bent H₂O, H₂S, SO₂, O₃; pyramidal NH₃, NF₃, XeO₃; SF₄, ClF₃, CHCl₃, HX.

Fig. 6: Why NH₃ has a much larger dipole moment than NF₃, though F is more electronegative than H.

Orders to remember: HF (1.91 D) > HCl (1.03) > HBr (0.79) > HI (0.38); CH₃Cl (1.86) > CH₂Cl₂ (1.60) > CHCl₃ (1.04) > CCl₄ (0). CH₃Cl > CH₃F because the longer C–Cl bond makes q × d larger. H₂O (1.85) > H₂S (0.97).
Aromatics and isomers: ortho > meta > para for disubstituted benzenes with the same groups (para = 0 for identical groups). cis usually > trans (trans-1,2-dichloroethene has zero moment); exceptions arise when donor and acceptor groups are present.
Worked example: HCl, d = 127 pm, μ_obs = 1.03 D. For 100% ionic, μ = 1.602 × 10⁻¹⁹ × 127 × 10⁻¹² = 2.03 × 10⁻²⁹ C m = 6.1 D. Ionic character = 1.03/6.1 = 17%.

7. Hydrogen Bonding and Forces

Feature	Intramolecular	Intermolecular
Where	inside one molecule; needs a strain-free 5- or 6-membered ring	between molecules
Effect on b.p. / volatility	lower b.p., steam volatile, less soluble in water	higher b.p., association, more soluble
Standard examples	o-nitrophenol, salicylaldehyde, chloral hydrate, Ni(DMG)₂, HSO₅⁻ (Caro's acid)	H₂O, ROH, HF, RCOOH dimers, p-nitrophenol

Strength: F–H···F is the strongest; energy 5 to 50 kJ/mol. HF forms zig-zag chains; KHF₂ contains linear [F–H–F]⁻; H₂O has the highest boiling point in its group because each molecule makes up to four H bonds.
Anomalies: ice is less dense than water (open cage); water has maximum density at 4 °C; Me₄N⁺OH⁻ is a much stronger base than Me₃NH⁺OH⁻ (H bond ties OH⁻ to N–H); acetylene dissolves in acetone through C–H···O=C bonding; ortho-hydroxybenzoic acid is more acidic than the para isomer because of the stabilised anion.
Van der Waals order: ion–dipole > dipole–dipole > ion–induced dipole > dipole–induced dipole > London. London forces grow with size and polarisability: He < Ne < Ar < Kr < Xe. Fluorocarbons boil unusually low because F holds its electrons tightly (tiny polarisability).

8. Special Bonding Cases

Electron-deficient 3c–2e bonds: B₂H₆ (two B–H–B banana bonds; four terminal H in one plane, the two bridge H above and below; no B–B bond) and Al₂(CH₃)₆. Al₂Cl₆ is not electron deficient: its bridges are ordinary 2c–2e coordinate bonds from Cl.
Back bonding (a lone pair donated into a vacant orbital of the neighbour) shortens the bond and cuts Lewis basicity/acidity: BF₃ < BCl₃ < BBr₃ < BI₃ in Lewis acidity, because efficient 2p–2p overlap in BF₃ satisfies boron best. Silicon compounds: N(SiH₃)₃ is planar and a weaker base than N(CH₃)₃; SiH₃NCO is linear while CH₃NCO is bent.
Ligands: N(CH₃)₃ and P(CH₃)₃ both act as ligands, but P(CH₃)₃ also accepts back-donation (π-acceptor). So bonding is not always the same with a transition metal.
Synergic bonding in metal carbonyls: CO → M σ donation plus M → CO π* back donation. The M–C bond strengthens and C–O weakens.
Azide: N₃⁻ is linear, symmetrical (N–N 116 pm each). PF₅ shows fast Berry pseudorotation, so NMR shows equivalent F.

9. How JEE Frames Questions

Statement I / Statement II: each has a hidden count or order (species with unequal bond lengths; bond dissociation order; covalent character trend). Verify each separately.
Count questions (numerical): number of species with zero dipole moment, one unpaired electron, exception to the octet rule, or pyramidal geometry. Write the list, tick each, then add.
Matching: shapes, hybridisation and Lewis structure of the species. Recall the (SN / lp) code, not the picture.
Bond angle and bond order orders: decide whether central atom, substituent electronegativity or lone-pair count is the deciding factor.

⚠️ JEE Trap: Odd-electron molecules and the octet rule. NO₂, NO and ClO₂ break the octet (odd electrons), BeH₂, BF₃, AlCl₃ are electron deficient and PCl₅, SF₆, H₂SO₄ are expanded. CO₂, CH₄, SiF₄, C₂H₆, CHCl₃ and CBr₄ obey the octet.

10. Quick Sheet and Checklist

Idea	Rule
Formal charge	valence − nonbonding − ½ bonding
Bond order (MO)	½(N_b − N_a)
Bond order (resonance)	total bonds / number of structures
Dipole	μ = q d; μ_R² = P² + Q² + 2PQ cosθ
Steric number	σ bonds + lone pairs
Lone pair order	lp–lp > lp–bp > bp–bp
Zero dipole	symmetric only

Before the exam, check you can:

Write the shape, hybridisation and dipole status of 20 species from the (SN / lp) table without hesitation.
Order O₂, O₂⁺, O₂⁻, O₂²⁻, N₂, N₂⁺, N₂⁻ by bond order and bond length and label each paramagnetic or diamagnetic.
Explain in one line why NH₃ > NF₃, CH₃Cl > CH₃F, and BF₃ < BCl₃ in Lewis acidity.
State which of NaNO₃/NaClO₃, UF₄/UF₆, and F₂/Cl₂ are the classic exceptions.