Chemical Bonding

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Chemical Bonding explains why atoms combine, what holds them together, and how the type of bond influences physical and chemical properties. This chapter is arguably the most important in inorganic chemistry — it explains molecular geometry, reactivity, bond strength, and intermolecular forces. Nearly every JEE chemistry question requires understanding bonding.

Key Bonding Concepts for Quick Recall:

• Octet Rule: Atoms tend to gain, lose, or share electrons to achieve 8 electrons in their valence shell. Exceptions: H (2e⁻), expanded octet (PCl₅, SF₆, XeF₄), incomplete octet (BeCl₂, BF₃, Li₂), odd-electron molecules (NO, NO₂, ClO₂).
• Ionic Bond: Transfer of electrons → cation + anion. Lattice energy ∝ (z⁺z⁻)/r₀. Born-Haber cycle: ΔH_lattice = ΔH_atomization + IE − EA − ΔH_sublimation (for metals).
• Covalent Bond: Sharing of electron pairs. Polar covalent when electronegativity difference ΔEN > 0 (Fajans’ rules: small cation + large anion → covalent character).
• Bond Parameters: Bond order = (Nb − Na)/2 where Nb = bonding, Na = antibonding. Bond energy ∝ bond order, ∝ 1/bond length.
• VSEPR: Repulsion between electron pairs → molecular geometry. AXₙEₘ formula. LP–LP > LP–BP > BP–BP repulsion. lone pairs compress bond angles.
• Formal Charge: FC = valence e⁻ − (non-bonding e⁻ + ½ bonding e⁻). Most stable Lewis structure has minimum negative FC on more electronegative atom.

⚡ Exam tip: When asked to compare ionic character of LiCl vs CsCl: smaller cation (Li⁺) has higher charge density, polarizes the anion more → greater covalent character → LiCl is more covalent than CsCl. So LiCl has lower melting point than CsCl.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Chemical Bonding — JEE Chemistry Study Guide

1. Ionic Bonding — Lattice Energy and Born-Haber Cycle

Ionic bonds form between atoms with large electronegativity difference (typically ΔEN > 1.7). Energy changes in Born-Haber cycle (for NaCl):

Step 1: Na(s) → Na(g); ΔH = +109 kJ/mol (sublimation) Step 2: Na(g) → Na⁺(g) + e⁻; ΔH = +496 kJ/mol (IE₁) Step 3: ½Cl₂(g) → Cl(g); ΔH = +122 kJ/mol (bond dissociation) Step 4: Cl(g) + e⁻ → Cl⁻(g); ΔH = −349 kJ/mol (EA) Step 5: Na⁺(g) + Cl⁻(g) → NaCl(s); ΔH = −786 kJ/mol (lattice energy)

ΔH_f = +109 + 496 + 122 − 349 − 786 = −408 kJ/mol

Kapustinskii Equation (for salts without known crystal structure): U = (−121000 · z⁺z⁻ · ν)/(r₊ + r₋)) × (1 − 0.345/r₊ + r₋) kJ/mol where ν = number of ions per formula unit, r in pm.

Fajans’ Rules — Predicting covalent character:

1. Small cation (high charge density) → more covalent
2. Large anion (distortable/polarizable) → more covalent
3. Cation with high charge → more covalent (Al³⁺ > Mg²⁺ > Na⁺)
4. Electronic configuration: cations with pseudo-noble gas config (e.g., Ag⁺, Zn²⁺) are more covalent than those with noble gas config

Example: CuCl is covalent but NaCl is ionic. Why? Cu⁺ is small (98 pm) with incomplete d-subshell; Na⁺ is larger (102 pm) with noble gas config.

2. Covalent Bonding — Lewis Theory and VSEPR

Lewis dot structures: represent bonding with shared pairs (line) and lone pairs (dots).

Writing Lewis structures — systematic approach:

1. Count total valence electrons
2. Identify central atom (least electronegative, except H is never central)
3. Connect with single bonds
4. Complete octets of terminal atoms
5. Place remaining electrons on central atom
6. If central atom has fewer than 8, form multiple bonds

Resonance: When two or more valid Lewis structures exist for the same molecule. Actual structure is a hybrid of all resonance forms. Example: Ozone O₃. Two structures with one double bond and one single bond. O–O bond length is intermediate (127 pm, between O–O single 148 pm and O=O double 121 pm).

Formal charge calculation: FC = Valence e⁻ − (Non-bonding e⁻ + ½ Bonding e⁻)

For CO₃²⁻: All three O atoms are equivalent in resonance hybrid. Each O has FC = −1. The double-bonded O has FC = 0.

VSEPR — Molecular Geometry:

Type	Formula	Bond pairs	Lone pairs	Shape
AB₂	BeCl₂	2	0	Linear
AB₃	BF₃	3	0	Trigonal planar
AB₄	CH₄	4	0	Tetrahedral
AB₃E	NH₃	3	1	Trigonal pyramidal
AB₂E₂	H₂O	2	2	Bent/V-shaped
AB₅	PCl₅	5	0	Trigonal bipyramidal
AB₄E	SF₄	4	1	See-saw
AB₃E₂	ClF₃	3	2	T-shaped
AB₂E₃	XeF₂	2	3	Linear
AB₆	SF₆	6	0	Octahedral
AB₅E	IF₅	5	1	Square pyramidal
AB₄E₂	XeF₄	4	2	Square planar

⚡ Exam tip: In VSEPR, lone pairs occupy more space than bond pairs. Bond angle in H₂O (104.5°) is less than tetrahedral (109.5°) because LP–LP repulsion compresses the H–O–H angle. Similarly, NH₃ has 107° (not 109.5°) because one lone pair compresses the H–N–H angle.

3. Valence Bond Theory (VBT)

According to VBT, a covalent bond forms when atomic orbitals on two atoms overlap. Types of overlap:

• Sigma (σ) bond: Head-on overlap along the internuclear axis. Maximum overlap → stronger bond. All single bonds are σ bonds.
• Pi (π) bond: Sideways overlap above and below the internuclear axis. Weaker than σ bond (less effective overlap). Double bond = 1 σ + 1 π; Triple bond = 1 σ + 2 π.

Hybridization — mixing of atomic orbitals to form new orbitals with specific geometry:

Hybridization	Geometry	Angle
sp³	Tetrahedral	109.5°
sp²	Trigonal planar	120°
sp	Linear	180°
sp³d	Trigonal bipyramidal	90°, 120°
sp³d²	Octahedral	90°

Examples:

• CH₄: C 2s² 2p² → sp³ hybridization, 4 equivalent σ bonds
• C₂H₄ (ethene): Each C is sp² hybridized, one unhybridized p orbital forms the π bond
• C₂H₂ (ethyne): Each C is sp hybridized, two unhybridized p orbitals form two π bonds
• PCl₅: P uses sp³d hybridization, trigonal bipyramidal. Axial bonds (longer, 219 pm) vs equatorial bonds (213 pm) — explained by greater s-character in equatorial orbitals (sp² vs sp³d).

⚡ Exam tip: In SF₄, the lone pair occupies an equatorial position (not axial) because equatorial positions have two 90° bond–bond repulsions while axial has three. Lone pair prefers less crowded equatorial spot to minimize repulsion.

4. Molecular Orbital Theory (MOT) — Beyond VBT

MOT treats electrons as delocalized over the entire molecule. Atomic orbitals combine to form molecular orbitals (MOs) — bonding (lower energy) and antibonding (higher energy).

Electron filling order (for diatomic molecules): σ1s < σ1s < σ2s < σ2s < π2pₓ = π2pᵧ < σ2pᵤ < π2pₓ = π2pᵧ < σ*2pᵤ

For O₂ and beyond: σ2pᵤ < π2pₓ = π2pᵧ < σ*2pᵤ (because σ2p is lower in energy than π2p for molecules with Z > 7).

Bond Order = ½(N_b − Nₐ)

• BO = 1 → single bond (e.g., H₂)
• BO = 2 → double bond (e.g., O₂²⁻, B₂)
• BO = 3 → triple bond (e.g., N₂)
• BO = 0 → no bond, unstable (e.g., He₂)

MOT for Heteronuclear Diatomics (like CN⁻, NO⁺): These behave like homonuclear diatomics — filling order follows the atomic orbital energies of the constituent atoms. CN⁻ has the same MO configuration as N₂ (bond order 3).

Key predictions of MOT:

• O₂ is paramagnetic (2 unpaired electrons in π*2p orbitals) — VBT cannot explain this
• B₂ is paramagnetic (2 unpaired electrons in π2p)
• Be₂ does not exist (BO = 0)
• N₂⁺ has BO = 2.5 (removes one bonding electron from N₂’s BO = 3)

⚡ Exam tip: For bond order questions, ALWAYS use the formula BO = ½(Σ bonding e⁻ − Σ antibonding e⁻). Don’t memorize. For O₂: σ1s² σ1s² σ2s² σ2s² σ2pᵤ² π2pₓ² π2pᵧ² π2pₓ¹ π2pᵧ¹. Bonding = 10, Antibonding = 6. BO = 2.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Chemical Bonding — Comprehensive JEE Advanced Notes

1. Advanced Covalent Bonding — Coordinate Bond and Formal Charge

A coordinate bond (dative covalent bond) is a covalent bond where both electrons come from the same atom. Examples: NH₄⁺, H₃O⁺, CO, NO₂⁺. Once formed, it’s indistinguishable from a regular covalent bond.

Formal charge in resonance — systematic: For SO₃ (non-planar D₃h symmetry in reality, but planar resonance): All three resonance structures have: S double bonded to two O atoms and single bonded to the third. The actual S–O bond order = 4/3 (partial double bond character = 1.33). S–O bond length = 141 pm (shorter than typical S–O single bond 155 pm).

Formal charge rules for stable Lewis structures (in order of priority):

1. Correct total charge
2. Minimum formal charges
3. Negative formal charge on more electronegative atom
4. No adjacent atoms with same sign FC

2. Hybridization — Going Beyond Simple sp/sp²/sp³

d-orbital participation: In PCl₅, P uses 3s, 3p orbitals, and two 3d orbitals for hybridization → sp³d. This requires the 3d orbitals to be close in energy to 3s and 3p — only possible for elements in and below the third period.

Why can’t second-period elements expand their octet? Second period (B, C, N, O, F) have no low-lying d orbitals (3d is too high in energy). So they strictly follow the octet rule. Third period and beyond (P, S, Cl, Se, Xe) have available 3d orbitals → expanded octet possible.

sp³d² hybridization in SF₆: S uses 3s, three 3p, and two 3d orbitals → forms 6 equivalent sp³d² hybrid orbitals directed toward corners of an octahedron. All S–F bonds are equivalent (axial/equatorial distinction disappears due to 90° symmetry of octahedron).

3. MOT — Complete MO Diagrams for Diatomics

H₂ (σ1s²): BO = 1, diamagnetic He₂ (σ1s² σ*1s²): BO = 0, does not exist Li₂ (σ1s² σ*1s² σ2s²): BO = 1, diamagnetic Be₂ (σ1s² σ1s² σ2s² σ2s²): BO = 0, not observed B₂ (σ1s² σ1s² σ2s² σ2s² π2pₓ¹ π2pᵧ¹): BO = 1, paramagnetic (2 unpaired) C₂ (σ1s² σ1s² σ2s² σ2s² σ2pᵤ² π2pₓ¹ π2pᵧ¹): BO = 2, diamagnetic N₂ (σ1s² σ1s² σ2s² σ2s² σ2pᵤ² π2pₓ² π2pᵧ²): BO = 3, diamagnetic. The HOMO is σ2pᵤ (in heteronuclear shift for Z ≤ 7, becomes bonding) — wait, correction: for N₂ (Z=7): σ2pᵤ is bonding and below π2p. The σ2pᵤ is actually the higher energy bonding orbital in this case. O₂ (σ1s² σ1s² σ2s² σ2s² σ2pᵤ² π2pₓ² π2pᵧ² π2pₓ¹ π2pᵧ¹): BO = 2, paramagnetic Ne₂: BO = 0

MO diagram for O₂ vs O₂⁻ vs O₂²⁻: O₂: BO = 2 (paramagnetic) O₂⁻ (superoxide): BO = 1.5 (paramagnetic, one extra electron in π* anti-bonding) O₂²⁻ (peroxide): BO = 1 (diamagnetic, fully paired π* electrons) O₂⁺ (dioxygenyl): BO = 2.5 (paramagnetic, one electron removed from π* anti-bonding)

Why is O₂ paramagnetic? MOT shows two unpaired electrons in degenerate π* orbitals. VBT predicts a double bond with all electrons paired — contradicts experiment.

4. Hydrogen Bonding — Nature and Consequences

Hydrogen bond forms when H is bonded to a highly electronegative atom (F, O, N) and interacts with a lone pair on another electronegative atom.

Energy: 10–40 kJ/mol (much weaker than covalent ~200–400 kJ/mol, but much stronger than van der Waals ~1–10 kJ/mol)

Types:

• Intermolecular: Water (H₂O), ammonia (NH₃), HF — explains high boiling points
• Intramolecular: o-nitrophenol (chelation) — lower boiling point than m/p isomers due to internal H-bonding

Hydrogen bonding in ice: Ice has an open tetrahedral structure (each water has 4 H-bonds). Density of ice < density of water → ice floats. On heating, H-bonds break → liquid water is more dense.

Why is water unique? Small size of F, O, N combined with high electronegativity creates very strong H-bonds. Fluorine is the most electronegative but HF has only linear chains. Water is most extensively hydrogen-bonded because oxygen can accept and donate 2 H-bonds simultaneously (tetrahedral geometry).

5. Van der Waals Forces — London Dispersion

All molecules (even nonpolar) experience London dispersion forces — instantaneous dipole-induced dipole interactions.

Dependence: Strength ∝ polarizability ∝ number of electrons ∝ molecular size Trend in halogens: F₂ < Cl₂ < Br₂ < I₂ (boiling points increase) In alkanes: CH₄ < C₂H₆ < C₃H₈ < … (molecular mass increases → more electrons → stronger London forces)

Dipole-dipole interactions (Keesom forces): Exist between polar molecules (HCl, SO₂, H₂S). Strength ∝ μ²/r³ where μ = dipole moment.

6. Dipole Moment — Vector Treatment

μ = δ × d (Debye = 3.33564 × 10⁻³⁰ C·m) For a molecule with symmetric charge distribution, μ = 0 (e.g., CO₂ linear O=C=O, BeCl₂, BF₃ trigonal planar, SF₆ octahedral).

For water (bent, 104.5°): μ = 1.85 D. The vector sum of two O–H dipoles doesn’t cancel because the angle is not 180°.

Independence of μ from electronegativity difference: CO₂ = 0 (linear, vectors cancel) though C and O have large electronegativity difference. NF₃ has μ = 0.24 D (small) while NH₃ has μ = 1.47 D (large). In NF₃, the N–F bond dipoles (F is more electronegative) point away from N and partially cancel. In NH₃, the N–H bond dipoles point toward N and reinforce each other.

7. Percentage Ionic Character

From dipole moment: % ionic character = (μ_obs / μ_calc for full charge transfer) × 100

For HCl: Observed μ = 1.08 D; if fully ionic, charge separation = 4.8 × 10⁻¹⁰ esu × 1.27 Å = 6.1 D % ionic character = (1.08/6.1) × 100 ≈ 17.7%

8. MOT for Polyatomic Molecules — Brief Introduction

In HOMO-LUMO theory (frontier molecular orbitals):

• HOMO: Highest Occupied MO — donates electron density
• LUMO: Lowest Unoccupied MO — accepts electron density
• In nucleophilic substitution, the nucleophile attacks the LUMO of the electrophile

Benzene MO diagram (advanced): Six p-orbitals combine to form 6 π MOs — 3 bonding (occupied), 3 antibonding (empty). All 6 carbon atoms contribute equally to each MO. The delocalized π system explains benzene’s exceptional stability.

9. MOT for Coordination Complexes — Crystal Field Theory Introduction

In octahedral complexes, d-orbitals split into:

• t₂g (dₓᵧ, dᵧᵤ, dᵤₓ): lower energy (−0.4 Δ₀)
• e_g (dᵤ², dᵤ²₋ᵧ²): higher energy (+0.6 Δ₀)

Strong field ligands (CN⁻, CO) cause large splitting → low spin complexes Weak field ligands (H₂O, F⁻) cause small splitting → high spin complexes

⚡ Exam tip: When asked to compare bond angles, check hybridization first (sp > sp² > sp³). But when comparing molecules with same hybridization (e.g., NH₃ vs NF₃ vs N(CH₃)₃), check lone pair repulsion and electronegativity. NF₃ has smaller bond angle (102°) than NH₃ (107°) because the more electronegative F atoms pull electron density away, reducing lone pair repulsion on the bond pairs.

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