Chemical Bonding

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Chemical Bonding — Key Facts for Makerere University (Uganda) Core concept: Chemical bonding explains how atoms combine to form molecules and compounds. The three main types are ionic, covalent, and metallic bonding High-yield points: Lewis structures, VSEPR theory, hybridization, bond polarity, and intermolecular forces ⚡ Exam tip: Drawing Lewis structures and predicting molecular geometry using VSEPR are perennial exam questions

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🟡 Standard — Regular Study (2d–2mo)

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Chemical Bonding — Makerere University (Uganda) Study Guide

1. Types of Chemical Bonds

Ionic (Electrovalent) Bonding

Ionic bonds form when one atom transfers electrons to another, producing oppositely charged ions held together by electrostatic attraction.

Characteristics of Ionic Compounds:

• High melting and boiling points (strong electrostatic forces)
• Conduct electricity when molten or dissolved in water (ions are mobile)
• Usually soluble in polar solvents like water
• Form crystalline lattices (e.g., NaCl, MgO, CaCl₂)

Example — Formation of NaCl: Na (1s² 2s² 2p⁶ 3s¹) → Na⁺ (1s² 2s² 2p⁶) + e⁻ Cl (1s² 2s² 2p⁶ 3s² 3p⁵) + e⁻ → Cl⁻ ([Ar]) Na⁺ and Cl⁻ attract each other → NaCl crystal lattice

Example — Formation of MgO: Mg (1s² 2s² 2p⁶ 3s²) → Mg²⁺ (1s² 2s² 2p⁶) + 2e⁻ O (1s² 2s² 2p⁴) + 2e⁻ → O²⁻ ([Ne])

Born-Haber Cycle: Energy changes in ionic bond formation:

1. Sublimation of solid metal → gaseous atoms
2. Dissociation of diatomic non-metal molecules → gaseous atoms
3. Ionization of metal atoms → gaseous ions
4. Electron affinity of non-metal atoms
5. Lattice energy (energy released when gaseous ions form solid lattice)

ΔH_f = ΔH_sub + ½ΔH_dis + IE + EA + U

Covalent Bonding

Covalent bonds form when atoms share electron pairs. The shared electrons count toward the octet of both atoms.

Single bond: 2 shared electrons (σ bond only) — e.g., H₂, Cl₂, CH₄ Double bond: 4 shared electrons (1 σ + 1 π bond) — e.g., O₂, CO₂, C₂H₄ Triple bond: 6 shared electrons (1 σ + 2 π bonds) — e.g., N₂, C₂H₂, HCN

Coordinate (Dative) Bond: A covalent bond where both electrons come from the same atom. Example: NH₄⁺ — three N–H bonds are regular covalent bonds; the fourth N–H bond is a coordinate bond where N donates its lone pair to H⁺.

Metallic Bonding

Metallic bonds form between metal atoms in a lattice of positive ions surrounded by a “sea” of delocalized electrons.

• Electrons are free to move throughout the structure
• Explains electrical conductivity, thermal conductivity, malleability, and ductility
• Alloys are mixtures of metals with modified properties

2. Lewis Structures (Electron Dot Structures)

Octet Rule

Atoms tend to gain, lose, or share electrons to achieve 8 electrons in their valence shell (except H = 2 electrons, B = 6 electrons).

Steps for Drawing Lewis Structures:

1. Count total valence electrons
2. Draw a skeletal structure (usually the least electronegative atom in center)
3. Place single bonds between atoms
4. Complete octets of surrounding atoms
5. Place remaining electrons on the central atom
6. If central atom lacks octet, form multiple bonds by sharing lone pairs
7. Calculate and verify formal charges

Formal Charge = Valence e⁻ − (Non-bonding e⁻ + ½ Bonding e⁻)

Example — CO₂: Total valence e⁻: C (4) + 2O (6×2) = 16 O=C=O (double bonds to each O, central C has no lone pairs) Formal charges: C=0, O=0 (resonance: O=C=O is the major contributor)

Example — CO₃²⁻ (carbonate ion): Total valence e⁻: C (4) + 3O (6×3) + 2 (charge) = 24 Resonance structures: one double-bonded O and two single-bonded O⁻, delocalized.

Example — NO₃⁻ (nitrate ion): Total valence e⁻: N (5) + 3O (6×3) + 1 (charge) = 24 Resonance structures with one N=O double bond and two N–O⁻ single bonds, with π electrons delocalized over all three N–O bonds.

3. VSEPR Theory (Valence Shell Electron Pair Repulsion)

VSEPR predicts molecular geometry based on the fact that electron pairs (bonding and non-bonding) repel each other and arrange to maximize separation.

Steric Number (SN) = Number of Bonding Pairs + Number of Lone Pairs

SN	Electron Pair Geometry	Molecular Geometry	Bond Angles
2	Linear	Linear	180°
3	Trigonal planar	Trigonal planar	120°
3	Trigonal planar	Bent	~117°
4	Tetrahedral	Tetrahedral	109.5°
4	Tetrahedral	Trigonal pyramidal	~107°
4	Tetrahedral	Bent	~104.5°
5	Trigonal bipyramidal	Trigonal bipyramidal	90°, 120°
5	Trigonal bipyramidal	Seesaw	~90°, ~120°
5	Trigonal bipyramidal	T-shaped	90°
5	Trigonal bipyramidal	Linear	180°
6	Octahedral	Octahedral	90°
6	Octahedral	Square pyramidal	~90°
6	Octahedral	Square planar	90°

Lone Pair Repulsion Order

Lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair This is why H₂O (2 lone pairs) has a smaller bond angle (104.5°) than NH₃ (1 lone pair, 107°).

Examples:

• CH₄ (SN=4, 0 lone pairs): Tetrahedral, 109.5°
• NH₃ (SN=4, 1 lone pair): Trigonal pyramidal, ~107°
• H₂O (SN=4, 2 lone pairs): Bent, ~104.5°
• CO₂ (SN=2, 0 lone pairs): Linear, 180°
• BF₃ (SN=3, 0 lone pairs): Trigonal planar, 120°
• PCl₅ (SN=5, 0 lone pairs): Trigonal bipyramidal
• SF₆ (SN=6, 0 lone pairs): Octahedral

4. Hybridization

Hybridization is the mixing of atomic orbitals to form new hybrid orbitals for bonding.

Hybridization	Orbitals Mixed	Geometry	Examples
sp³	one s + three p	Tetrahedral	CH₄, NH₃, H₂O
sp²	one s + two p	Trigonal planar	BF₃, C₂H₄, CO₃²⁻
sp	one s + one p	Linear	BeCl₂, C₂H₂, CO₂
sp³d	one s + three p + one d	Trigonal bipyramidal	PCl₅
sp³d²	one s + three p + two d	Octahedral	SF₆

Example — CH₄ (methane): Carbon (ground state): 1s² 2s² 2p² (2 unpaired → only 2 bonds) Carbon (excited state): 1s² 2s¹ 2p³ (4 unpaired → 4 bonds) Hybridization: 2s + 2px + 2py + 2pz → four sp³ orbitals Each sp³ orbital overlaps with H 1s orbital → tetrahedral CH₄

Example — C₂H₄ (ethene): Each C is sp² hybridized (3 regions of electron density around each C) The unhybridized p orbital on each C overlaps sideways → π bond Result: planar molecule with 120° H-C-H and H-C=C angles

Example — C₂H₂ (ethyne): Each C is sp hybridized (2 regions of electron density) Two unhybridized p orbitals on each C form two π bonds Result: linear molecule, 180° H-C≡C-H

5. Polarity of Bonds and Molecules

Bond Polarity

Bond polarity arises from electronegativity differences.

• Pure covalent: EN difference ≈ 0 (e.g., Cl–Cl: EN diff = 0)
• Polar covalent: EN diff 0.1–1.7 (e.g., H–Cl: EN diff = 0.9)
• Ionic: EN diff > 1.7 (e.g., Na–Cl: EN diff = 2.1)

Dipole Moment

A molecule has a dipole moment if it has an uneven charge distribution.

Polar molecules: H₂O, NH₃, HCl, CO Non-polar molecules: CO₂ (linear, dipoles cancel), CH₄ (tetrahedral, dipoles cancel), N₂ (identical atoms)

⚡ Key rule: A molecule with polar bonds is polar if the dipoles do not cancel out due to symmetry.

6. Intermolecular Forces

Van der Waals Forces (Weakest)

• London dispersion forces: Instantaneous dipole-induced dipole attractions; present in ALL molecules; strength increases with molecular size/mass.
• Debye forces: Permanent dipole-induced dipole; polar molecules attract non-polar molecules.

Dipole-Dipole Interactions

Between polar molecules; stronger than London forces. Example: HCl molecules attract each other via δ+–δ− interactions.

Hydrogen Bonding (Strongest IMF for small molecules)

Occurs when H is bonded to N, O, or F (highly electronegative atoms). Hydrogen bond: F–H···F, O–H···O, N–H···N

Examples of H-bonding effects:

• Water (H₂O) has unexpectedly high boiling point (100°C) compared to H₂S (−60°C)
• Ammonia (NH₃) boiling point: −33°C
• HF dimer: (HF)₂ exists as a associated liquid

Effect of IMF on Boiling Points

Higher IMF = higher boiling point. Order of increasing boiling points for hydrides of Group 14: CH₄ (−162°C) < SiH₄ (−112°C) < GeH₄ (−88°C) < SnH₄ (−52°C) [London dispersion increases with molar mass]. But for Group 16: H₂O (100°C) >> H₂S (−60°C) [H-bonding in H₂O dominates]

7. Exam-Style Questions & Tips

Common exam question patterns at Makerere:

1. “Draw the Lewis structure of [species] and state the formal charges”
2. “Predict the shape and bond angle of [molecule] using VSEPR theory”
3. “Explain the hybridization of carbon in [species]”
4. “Compare and contrast ionic, covalent, and metallic bonding”
5. “Explain why H₂O has a higher boiling point than H₂S”
6. “Draw the resonance structures of [species] (e.g., benzene, nitrate)”
7. “Predict whether [molecule] is polar or non-polar”

⚡ Exam tips:

• When drawing Lewis structures, always count total electrons first
• Formal charge helps determine the correct resonance structure — the structure with the smallest formal charges closest to zero is preferred
• For VSEPR: count ALL electron pairs (bonding AND lone pairs)
• Lone pairs repel more strongly than bond pairs — this affects bond angles
• In multiple-choice questions, check both the geometry AND the bond angle

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Advanced Bonding Concepts

Molecular Orbital (MO) Theory

MO theory describes bonding in terms of molecular orbitals formed by linear combinations of atomic orbitals (LCAO).

Bonding orbital (σ, π): Lower in energy, electrons in bonding orbitals stabilize the molecule. Antibonding orbital (σ, π)**: Higher in energy, electrons in antibonding orbitals destabilize the molecule.

Bond Order = ½(Bonding e⁻ − Antibonding e⁻)

Example — H₂: 2 atomic orbitals → 2 molecular orbitals (σ and σ*) H₂ has 2 electrons, both in σ (bonding) Bond order = ½(2 − 0) = 1 → single bond ✓

Example — He₂: Bond order = ½(2 − 2) = 0 → does not exist ✓

Example — O₂: O (Z=8): 1s² 2s² 2p⁴ MO configuration: σ1s² σ1s² σ2s² σ2s² σ2pz² π2px² π2py² π2px¹ π2py¹ Bond order = ½(10 − 6) = 2 → double bond O₂ is paramagnetic (has unpaired electrons in π* orbitals) — this explains its reactivity.

Example — N₂: MO configuration: σ1s² σ1s² σ2s² σ2s² σ2pz² π2px² π2py² Bond order = ½(10 − 4) = 3 → triple bond ✓ N₂ has no unpaired electrons (diamagnetic) — consistent with its inertness.

Bond Length and Bond Order

Bond length decreases as bond order increases:

• C–C (single): 154 pm
• C=C (double): 134 pm
• C≡C (triple): 120 pm

Dipole Moment Values

μ = δ × d (charge × distance) Unit: Debye (D); 1 D = 3.336 × 10⁻³⁰ C·m

Molecule	μ (D)	Type
HCl	1.03	Polar
H₂O	1.85	Polar
NH₃	1.47	Polar
CO₂	0	Non-polar
CH₄	0	Non-polar
BF₃	0	Non-polar

Fajans’ Rule (Ionic Character)

For compounds with high ionic character, look for:

1. Small cation with high positive charge
2. Large anion with high negative charge
3. Cation with electronic configuration like a noble gas (not pseudo-noble gas)

Example: Compare MgO vs BaO — Mg²⁺ is smaller and has higher charge density, so MgO has more covalent character than expected for an ionic compound.

Valence Bond Theory vs Molecular Orbital Theory

Valence Bond Theory:

• Bonds form when atomic orbitals overlap
• Explains hybridization, bond angles, σ and π bonds
• Does not fully explain delocalization or paramagnetism of O₂

Molecular Orbital Theory:

• Electrons occupy molecular orbitals spanning the entire molecule
• Explains delocalization, resonance, bond order, magnetism
• More mathematically complex

Bonding in Complex Ions

Transition metal complexes involve coordinate bonding between metal ions and ligands.

• Metal ion: Central atom (usually d-block)
• Ligands: Electron pair donors (Lewis bases): NH₃, H₂O, Cl⁻, CN⁻, CO
• Coordination number: Number of ligand bonds to metal (2, 4, or 6 most common)

Examples:

• [Cu(NH₃)₄]²⁺: Cu²⁺ with 4 NH₃ ligands, square planar or tetrahedral
• [Fe(CN)₆]³⁻: Fe³⁺ with 6 CN⁻ ligands, octahedral

Crystal Field Theory

In octahedral complexes, d-orbitals split into two sets:

• t₂g (lower energy): dxy, dxz, dyz
• eg (higher energy): dz², dx²−y²

Crystal field splitting energy (Δ₀) determines:

• Color of complexes (visible light absorption)
• High spin vs low spin configurations
• Magnetic properties

Resonance (Delocalization)

Resonance occurs when no single Lewis structure adequately describes a molecule. Rules for resonance:

1. All resonance structures must have same total number of electrons
2. Only electrons (not nuclei) move between resonance forms
3. Each resonance structure must have valid Lewis structure (octet or expanded octet)
4. Resonance forms differ only in electron distribution, not in atomic positions

⚠️ Common misconception: Resonance does NOT mean the molecule rapidly switches between structures. The real structure is a hybrid (average) of all resonance forms.

Practice Problems

Q1: Draw the Lewis structures for: (a) SO₂ (b) H₂SO₄ (c) NCO⁻ (d) XeF₄ Q2: Predict the molecular geometry of: (a) BrF₅ (b) ClO₃⁻ (c) ICl₄⁻ (d) IF₃ Q3: Explain the hybridization of sulfur in SF₆ and explain why sulfur can expand its octet. Q4: Draw the MO diagram for N₂ and compare bond dissociation energy with that of N₂⁺. Q5: Explain why benzene (C₆H₆) is more stable than expected from its Kekulé structure.

Common Mistakes to Avoid

1. Forgetting lone pairs when determining molecular geometry: The shape you name is based on ATOM positions, not electron pairs. If there are 2 lone pairs and 2 bond pairs, the shape is bent (or V-shaped), not linear.
2. Confusing bond order with number of bonds: Bond order of 2 can be one double bond OR two equivalent single bonds in resonance.
3. Thinking expanded octets are impossible: Elements in Period 3+ (P, S, Cl, Xe) can have up to 12 electrons around them using d orbitals.
4. Mixing up VSEPR and hybridization: VSEPR predicts shape; hybridization explains how bonding occurs.
5. Forgetting that lone pairs occupy more space: They compress bond angles — H₂O (104.5°) < NH₃ (107°) < CH₄ (109.5°).

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