Chapter 4: Chemical Bonding and Molecular Structure

Chemical Bonding Introduction

Key Concepts

• Chemical bond: Attractive force holding atoms/ions together.
• Atoms interact due to energy decrease from net attractive forces & desire for noble gas configuration.

Solved Examples

• Ex 1: Why do atoms form bonds? To lower energy and achieve stable electron configuration.

Practice Set

• Q1: What is a chemical bond?
  A. Repulsive force
  B. Attractive force
  C. Neutral force
  D. None
  
• Q2: Atoms combine to achieve:
  A. Maximum energy
  B. Noble gas configuration
  C. Random state
  D. Unstable state
  
• Q3: Which factor drives atoms to bond?
  A. Increase in energy
  B. Decrease in energy
  C. No change
  D. None
  

Answer Key

• Q1: B - Chemical bond is an attractive force holding atoms together.
• Q2: B - Atoms combine to achieve noble gas configuration.
• Q3: B - Bonding decreases energy making the system stable.

Memory Aids

• Atoms bond to lower energy and mimic noble gases.
• Think "bond = attraction + stability".

Kössel—Lewis Approach

Key Concepts

• Atoms combine by transferring or sharing valence electrons to complete octet/duplet.
• Octet rule: Atoms tend to have 8 electrons in outer shell (2 for H).
• Two ways to achieve stable configuration:
  • Electrovalency: Complete electron transfer → ionic bond.
  • Sharing electrons → covalent or coordinate (dative) bond.
• Covalent bond: Equal sharing of electrons.
• Coordinate bond: One atom donates both electrons.

Solved Examples

• Ex 1: Identify bond type in NaCl. Answer: Ionic (electrovalent) bond by electron transfer.
• Ex 2: Identify bond in H₂. Answer: Covalent bond by equal sharing.

Practice Set

• Q1: What bond forms by electron transfer?
  A. Covalent
  B. Ionic
  C. Coordinate
  D. Metallic
  
• Q2: When electrons are shared equally, bond is:
  A. Ionic
  B. Covalent
  C. Coordinate
  D. Hydrogen
  
• Q3: Coordinate bond involves:
  A. Both atoms donating electrons
  B. One atom donating both electrons
  C. Electron transfer
  D. No electron sharing
  

Answer Key

• Q1: B - Ionic bond forms by electron transfer.
• Q2: B - Covalent bond involves equal sharing.
• Q3: B - Coordinate bond has one atom donating both electrons.

Memory Aids

• Octet rule: Complete 8 electrons (except H: 2).
• Ionic = transfer, Covalent = share equally, Coordinate = share unequally.

Covalent Bond

Key Concepts

• Formed by mutual sharing of electron pairs between atoms.
• Each atom contributes at least one electron to shared pair.
• Atoms attain noble gas configuration by sharing electrons.
• Types of covalent bonds:
  • Single bond: 1 shared pair
  • Double bond: 2 shared pairs
  • Triple bond: 3 shared pairs

Solved Examples

• Ex 1: Number of bonds in O₂? Answer: Double bond (2 pairs shared).
• Ex 2: Number of bonds in N₂? Answer: Triple bond (3 pairs shared).

Practice Set

• Q1: Covalent bond involves:
  A. Electron transfer
  B. Electron sharing
  C. No electron movement
  D. Ionic interaction
  
• Q2: A triple bond shares how many electron pairs?
  A. 1
  B. 2
  C. 3
  D. 4
  
• Q3: Which molecule has a double bond?
  A. H₂
  B. O₂
  C. N₂
  D. He
  

Answer Key

• Q1: B - Covalent bond is electron sharing.
• Q2: C - Triple bond shares 3 pairs.
• Q3: B - O₂ has a double bond.

Memory Aids

• Single = 1 pair, Double = 2 pairs, Triple = 3 pairs shared.
• Remember: Covalent = share electrons.

Lewis Structures

Key Concepts

• Valence electrons shown as dots around atoms.
• Shared pairs shown as dots between atoms or lines.
• Lewis structures depict bonding and lone pairs.
• Octet rule visualized by electron pairs around atoms.
• For ions, brackets and charges indicate electron gain/loss.

Solved Examples

• Ex 1: Draw Lewis structure of H₂O showing lone pairs and bonds.

Practice Set

• Q1: Lewis dots represent:
  A. Protons
  B. Valence electrons
  C. Neutrons
  D. Nucleus
  
• Q2: In Lewis structure, a line represents:
  A. Lone pair
  B. Bonding pair
  C. Proton
  D. Neutron
  
• Q3: What do brackets with charge indicate?
  A. Neutral molecule
  B. Ion with gained/lost electrons
  C. Isotope
  D. None
  

Answer Key

• Q1: B - Dots show valence electrons.
• Q2: B - Line represents bonding pair.
• Q3: B - Brackets with charge show ions.

Memory Aids

• Dots = valence electrons, Lines = bonds.
• Brackets + charge = ion.

Formal Charge

Key Concepts

• Formal charge = Valence electrons - Non-bonding electrons - 1/2 Bonding electrons.
• Used to find most stable Lewis structure.
• Lowest energy structure has smallest formal charges.

Solved Examples

• Ex 1: Calculate formal charge on central O in O₃: 6 - 2 - 1/2(6) = +1.
• Ex 2: Calculate formal charge on end O in O₃: 6 - 4 - 1/2(4) = 0.

Practice Set

• Q1: Formal charge formula is:
  A. Valence - Lone pairs - 1/2 bonding
  B. Valence + Lone pairs + bonding
  C. Valence only
  D. None
  
• Q2: Formal charge helps to:
  A. Find highest energy structure
  B. Find lowest energy structure
  C. Ignore structure stability
  D. None
  
• Q3: Formal charge on O with 6 valence, 6 lone electrons, 2 bonding electrons is:
  A. -1
  B. 0
  C. +1
  D. +2
  

Answer Key

• Q1: A - Correct formula for formal charge.
• Q2: B - Helps find lowest energy structure.
• Q3: A - Calculation: 6 - 6 - 1/2(2) = -1.

Memory Aids

• Formal charge = valence - lone pairs - half bonding.
• Lowest formal charge = most stable structure.

Limitations of Octet Rule

Key Concepts

• Fails for atoms with less than 4 valence electrons (e.g., BeCl₂, BF₃).
• Fails for atoms with d-orbitals available (e.g., PF₅).
• Fails for molecules with odd electrons (e.g., NO, ClO₂).
• Does not explain paramagnetism of O₂.
• Noble gases can form compounds, contradicting inertness assumption.
• Does not explain molecular shape, stability, or energy.
• Electron deficient molecules act as Lewis acids.

Solved Examples

• Ex 1: PF₃ obeys octet, PF₅ does not due to d-orbitals.
• Ex 2: BeCl₂ has incomplete octet, acts as Lewis acid.

Practice Set

• Q1: Octet rule fails for:
  A. Atoms with <4 valence electrons
  B. Atoms with d-orbitals
  C. Molecules with odd electrons
  D. All above
  
• Q2: Molecules with incomplete octet are:
  A. Lewis bases
  B. Lewis acids
  C. Inert gases
  D. None
  
• Q3: Octet rule does not explain:
  A. Molecular shape
  B. Stability
  C. Paramagnetism
  D. All above
  

Answer Key

• Q1: D - All listed cases fail octet rule.
• Q2: B - Electron deficient molecules are Lewis acids.
• Q3: D - Octet rule fails to explain shape, stability, paramagnetism.

Memory Aids

• Octet rule exceptions: <4 valence, d-orbitals, odd electrons.
• Electron deficient molecules = Lewis acids.

Ionic or Electrovalent Bond

Definition and Formation

• An ionic bond is the electrostatic attraction between cations and anions formed by complete electron transfer to achieve noble gas configuration.
• Forms easily between elements with low ionization energy and high negative electron gain enthalpy.
• Stability depends on lattice enthalpy, not just octet completion in gaseous ions.

Solved Examples

• Ex 1: Formation of NaCl
  • Na (1s²2s²2p⁶3s¹) loses 1 electron → Na⁺
  • Cl (1s²2s²2p⁶3s²3p⁵) gains 1 electron → Cl⁻
  • Na⁺ and Cl⁻ attract to form NaCl ionic bond.

Practice Set

• Q1: Which pair is most likely to form an ionic bond?
  • A. Na and Cl
  • B. H and O
  • C. C and H
  • D. N and O
• Q2: Ionic bonds form due to
  • A. Sharing of electrons
  • B. Complete transfer of electrons
  • C. Metallic bonding
  • D. Hydrogen bonding
• Q3: Which factor favors ionic bond formation?
  • A. High ionization energy of metal
  • B. Low electron gain enthalpy of non-metal
  • C. Low ionization energy of metal and high negative electron gain enthalpy of non-metal
  • D. Similar electronegativities

Answer Key

• Q1: A - Na and Cl form ionic bond due to electron transfer.
• Q2: B - Ionic bonds form by complete electron transfer.
• Q3: C - Low ionization energy metal and high negative electron gain enthalpy non-metal favor ionic bond.

Memory Aids

• Low I.E. metal + high negative EGE non-metal = ionic bond.
• NaCl is classic ionic compound example.
• Stability depends on lattice enthalpy, not just octet.

Lattice Enthalpy

Definition

• Energy required to separate 1 mole of ionic solid into gaseous ions completely.

Factors Affecting Lattice Enthalpy

• Charge on ions: Higher charge → stronger attraction → higher lattice enthalpy.
  • Order: bi-bivalent > uni-bi or bi-uni > uni-univalent ions.
• Size of ions: Smaller ions → stronger attraction → higher lattice enthalpy.
  • Lattice enthalpy ∝ 1 / size of ions.

Solved Examples

• Ex 1: Compare lattice enthalpy of MgO and NaCl
  • Mg²⁺ and O²⁻ have higher charges than Na⁺ and Cl⁻.
  • MgO has higher lattice enthalpy due to higher ionic charges and smaller ion sizes.

Practice Set

• Q1: Lattice enthalpy increases with
  • A. Decreasing ionic charge
  • B. Increasing ionic charge
  • C. Increasing ionic size
  • D. Decreasing ionic charge and size
• Q2: Which ionic compound has highest lattice enthalpy?
  • A. NaCl
  • B. KBr
  • C. MgO
  • D. CsI
• Q3: Lattice enthalpy is directly proportional to
  • A. Size of ions
  • B. Charge on ions
  • C. Number of ions
  • D. Atomic mass

Answer Key

• Q1: B - Higher ionic charge increases lattice enthalpy.
• Q2: C - MgO has highest lattice enthalpy due to 2+ and 2− ions.
• Q3: B - Lattice enthalpy ∝ charge on ions.

Memory Aids

• Charge ↑ → lattice enthalpy ↑
• Size ↑ → lattice enthalpy ↓
• MgO > NaCl in lattice enthalpy due to charge and size.

Valence Shell Electron Pair Repulsion Theory

Key Concepts

• Developed by Nyholm and Gillespie in 1957.
• Molecules adopt shapes that keep bonded atoms as far apart as possible around the central atom.
• Shape depends on the number of valence shell electron pairs (bonded and lone pairs) around the central atom.
• Electron pairs repel each other in the order: lone pair-lone pair (lp-lp) > lone pair-bond pair (lp-bp) > bond pair-bond pair (bp-bp).
• Electron pairs arrange themselves to minimize repulsion and maximize distance.
• Applicable to molecules with resonance structures.

Solved Examples

• Ex 1: Predict the shape of SF₄.
• SF₄ has 5 electron pairs (4 bonding, 1 lone pair) around S.
• Shape: Trigonal bipyramidal with one lone pair causing distortion.
• Bond angle between axial fluorines is 173° (almost linear).

Practice Set

• Q1: What is the bond angle in NH₃?
  • A. 109.5°
  • B. 107.8°
  • C. 104.5°
  • D. 90°
• Q2: Which molecule has an octahedral shape?
  • A. SF₄
  • B. SF₆
  • C. NH₃
  • D. H₂O
• Q3: Arrange the repulsion strength in decreasing order.
  • A. bp-bp > lp-bp > lp-lp
  • B. lp-lp > lp-bp > bp-bp
  • C. lp-bp > lp-lp > bp-bp
  • D. bp-bp > lp-lp > lp-bp

Answer Key

• Q1: B - NH₃ has a bond angle of 107.8° due to one lone pair reducing the ideal tetrahedral angle.
• Q2: B - SF₆ has an octahedral shape with 6 bonding pairs and 90° bond angles.
• Q3: B - Lone pair-lone pair repulsion is strongest, followed by lone pair-bond pair, then bond pair-bond pair.

Memory Aids

• Remember: "lp-lp > lp-bp > bp-bp" for repulsion strength.
• Trigonal bipyramidal shape has 5 electron pairs; lone pairs occupy equatorial positions.
• Octahedral shape has 6 bonding pairs with 90° angles.

Key Concepts

• Developed by Nyholm and Gillespie in 1957.
• Molecules adopt shapes that keep bonded atoms as far apart as possible around the central atom.
• Shape depends on the number of valence shell electron pairs (bonded and lone pairs) around the central atom.
• Electron pairs repel each other in the order: lone pair-lone pair (lp-lp) > lone pair-bond pair (lp-bp) > bond pair-bond pair (bp-bp).
• Electron pairs arrange themselves to minimize repulsion and maximize distance.
• Applicable to molecules with resonance structures.

Solved Examples

• Ex 1: Predict the shape of SF₄.
• SF₄ has 5 electron pairs (4 bonding, 1 lone pair) around S.
• Shape: Trigonal bipyramidal with one lone pair causing distortion.
• Bond angle between axial fluorines is 173° (almost linear).

Practice Set

• Q1: What is the bond angle in NH₃?
  • A. 109.5°
  • B. 107.8°
  • C. 104.5°
  • D. 90°
• Q2: Which molecule has an octahedral shape?
  • A. SF₄
  • B. SF₆
  • C. NH₃
  • D. H₂O
• Q3: Arrange the repulsion strength in decreasing order.
  • A. bp-bp > lp-bp > lp-lp
  • B. lp-lp > lp-bp > bp-bp
  • C. lp-bp > lp-lp > bp-bp
  • D. bp-bp > lp-lp > lp-bp

Answer Key

• Q1: B - NH₃ has a bond angle of 107.8° due to one lone pair reducing the ideal tetrahedral angle.
• Q2: B - SF₆ has an octahedral shape with 6 bonding pairs and 90° bond angles.
• Q3: B - Lone pair-lone pair repulsion is strongest, followed by lone pair-bond pair, then bond pair-bond pair.

Memory Aids

• Remember: "lp-lp > lp-bp > bp-bp" for repulsion strength.
• Trigonal bipyramidal shape has 5 electron pairs; lone pairs occupy equatorial positions.
• Octahedral shape has 6 bonding pairs with 90° angles.

Geometry Based on VSEPR Theory

Key Concepts

• Molecular shape depends on total electron pairs around the central atom (bonded + lone pairs).
• Electron pairs arrange to minimize repulsion, defining molecular geometry.
• Common geometries include linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral.
• Formulas like AX₂, AX₃E indicate bonded atoms (X) and lone pairs (E) around central atom (A).
• Examples: BeCl₂ (linear), CH₄ (tetrahedral), NH₃ (trigonal pyramidal), H₂O (bent).

Solved Examples

• Ex 1: Determine the shape of NH₃ using AX₃E notation.
• NH₃ has 3 bonded atoms and 1 lone pair (AX₃E).
• Shape: Trigonal pyramidal with bond angle 107.8°.

Practice Set

• Q1: What is the shape of a molecule with formula AX₂E₂?
  • A. Linear
  • B. Bent
  • C. Trigonal planar
  • D. Tetrahedral
• Q2: Which molecule is tetrahedral?
  • A. CO₂
  • B. CH₄
  • C. NH₃
  • D. SO₂
• Q3: For AX₅E, what is the molecular shape?
  • A. Trigonal bipyramidal
  • B. Square pyramidal
  • C. See-saw
  • D. Octahedral

Answer Key

• Q1: B - AX₂E₂ corresponds to bent shape (e.g., H₂O).
• Q2: B - CH₄ is tetrahedral with 4 bonded atoms and no lone pairs.
• Q3: C - AX₅E has see-saw shape due to one lone pair in trigonal bipyramidal arrangement.

Memory Aids

• AX_nE_m notation helps predict shape: A = central atom, X = bonded atoms, E = lone pairs.
• More lone pairs reduce bond angles from ideal geometry.
• Common shapes: linear (2 pairs), trigonal planar (3 pairs), tetrahedral (4 pairs), trigonal bipyramidal (5 pairs), octahedral (6 pairs).

Valence Bond Theory

Main Postulates

• Bond forms by interaction of valence electrons; core electrons do not participate.
• Only valence electrons lose their identity during bond formation.
• Bond formation releases energy, providing bond stability.
• Molecule has minimum energy at equilibrium (intermolecular) distance; greater energy decrease means stronger bond.

Energy vs Internuclear Distance

• Graph shows potential energy changes as atomic nuclei distance varies.
• Close nuclei cause steep energy rise due to repulsion.
• Minimum energy at optimal distance \( r_0 \) indicates stable molecule formation.
• Energy increases if nuclei separate beyond \( r_0 \), atoms stop interacting.
• Minimum point balances attraction and repulsion forces.

Solved Examples

• Ex 1: Explain why bond formation releases energy.
  - Valence electrons overlap, lowering system energy.
  - Energy release stabilizes the molecule.

Practice Set

• Q1: Which electrons participate in bond formation?
  A. Core electrons
  B. Valence electrons
  C. Both
  D. None
• Q2: What happens to energy at equilibrium bond distance?
  A. Maximum
  B. Minimum
  C. Zero
  D. Infinite
• Q3: Why is bond formation stable?
  A. Energy absorption
  B. Energy release
  C. No energy change
  D. Electron loss

Answer Key

• Q1: B - Only valence electrons form bonds.
• Q2: B - Energy is minimum at equilibrium distance.
• Q3: B - Bond formation releases energy, stabilizing molecule.

Memory Aids

• Valence electrons = Bond makers, core electrons = spectators.
• Energy minimum = stable bond distance.

Types of Overlapping and Nature of Covalent Bond

Orbital Overlapping

• Atomic orbitals overlap to form molecular orbitals when atoms approach.
• Overlap can be positive, negative, or zero based on orbital phase and orientation.
• Sigma (σ) bond: Head-on overlap along internuclear axis (s-s, s-p, p-p).
• Pi (π) bond: Lateral overlap perpendicular to internuclear axis, only p-orbitals.

Strength of Sigma and Pi Bonds

• Sigma bonds have greater overlap extent; stronger than pi bonds.
• Multiple bonds include one sigma and one or more pi bonds.
• Bond energy order: Single < Double < Triple.
• Bond length order: Triple < Double < Single.
• Reactivity order: Single < Double < Triple.

Additional Points

• Maximum covalent bonds between two atoms = 3.
• Pi bond never forms alone; always with sigma bond.
• Head-on overlap stronger than lateral overlap.
• Free rotation possible about sigma bond, not pi bond.

Solved Examples

• Ex 1: Identify bond types in ethene (C₂H₄).
  - C-C double bond = 1 sigma + 1 pi bond.
  - C-H bonds = sigma bonds.

Practice Set

• Q1: Sigma bond forms by:
  A. Lateral overlap
  B. Head-on overlap
  C. No overlap
  D. Random overlap
• Q2: Which bond is stronger?
  A. Pi bond
  B. Sigma bond
  C. Both equal
  D. None
• Q3: Which is true about pi bonds?
  A. Form alone
  B. Form with sigma bond
  C. Allow free rotation
  D. Stronger than sigma

Answer Key

• Q1: B - Sigma bonds form by head-on overlap.
• Q2: B - Sigma bonds are stronger due to greater overlap.
• Q3: B - Pi bonds always accompany sigma bonds.

Memory Aids

• Sigma = Straight (head-on), Pi = Parallel (sideways).
• Max 3 bonds between atoms.
• Rotation allowed only about sigma bonds.

Hybridisation

Definition and Features

• Atomic orbitals of similar energy mix to form new hybrid orbitals of equivalent energy.
• Number of hybrid orbitals = number of atomic orbitals mixed.
• Hybrid orbitals are equivalent in shape and energy.
• They orient in space to minimize electron pair repulsion.

Hybridisation Formula

H = 1/2 (V + M – C + A)

• H = number of hybrid orbitals (2 to 7) corresponding to sp, sp², sp³, sp³d, sp³d², sp³d³
• V = valence electrons of central atom
• M = monovalent atoms attached
• C = charge on cation
• A = charge on anion

Conditions for Hybridisation

• Only valence shell orbitals hybridise.
• Orbitals must have nearly equal energies.
• Electron promotion before hybridisation is not necessary.
• Both half-filled and filled orbitals can hybridise.

Solved Examples

Ex 1: Calculate hybridisation of central atom in BF₃.

• V = 3 (Boron valence electrons)
• M = 3 (three F atoms)
• C = 0, A = 0 (neutral molecule)
• H = 1/2 (3 + 3 - 0 + 0) = 3 → sp² hybridisation

Practice Set

• Q1: Hybridisation of carbon in CH₄ is?
  • A. sp
  • B. sp²
  • C. sp³
  • D. sp³d
• Q2: Number of hybrid orbitals formed in sp³d hybridisation?
  • A. 3
  • B. 4
  • C. 5
  • D. 6
• Q3: Hybridisation of central atom in SF₆ is?
  • A. sp³d
  • B. sp³d²
  • C. sp³d³
  • D. sp²

Answer Key

• Q1: C (sp³) - Carbon forms 4 bonds, so 4 hybrid orbitals.
• Q2: C (5) - sp³d involves 5 hybrid orbitals.
• Q3: B (sp³d²) - SF₆ has 6 bonds, requiring 6 orbitals.

Memory Aids

• sp: 2 orbitals, linear, 180°
• sp²: 3 orbitals, trigonal planar, 120°
• sp³: 4 orbitals, tetrahedral, 109.5°
• sp³d: 5 orbitals, trigonal bipyramidal, 90° & 120°
• sp³d²: 6 orbitals, octahedral, 90°

Hybridisation in Molecules with Bond Pairs Only

Types and Structures

• sp: 2 orbitals, linear shape, 180° bond angle (e.g., CO₂).
• sp²: 3 orbitals, trigonal planar, 120° (e.g., BF₃).
• sp³: 4 orbitals, tetrahedral, 109.5° (e.g., CH₄).
• sp³d: 5 orbitals, trigonal bipyramidal, 90° & 120° (e.g., PCl₅).
• sp³d²: 6 orbitals, octahedral, 90° (e.g., SF₆).

Solved Examples

Ex 1: Determine shape and bond angle of BeCl₂.

• Hybridisation: sp (2 orbitals)
• Shape: Linear
• Bond angle: 180°

Practice Set

• Q1: Bond angle in sp³ hybridised molecule?
  • A. 90°
  • B. 109.5°
  • C. 120°
  • D. 180°
• Q2: Shape of PCl₅ molecule?
  • A. Tetrahedral
  • B. Trigonal bipyramidal
  • C. Octahedral
  • D. Linear
• Q3: Hybridisation of CO₂?
  • A. sp
  • B. sp²
  • C. sp³
  • D. sp³d

Answer Key

• Q1: B (109.5°) - Tetrahedral bond angle for sp³.
• Q2: B (Trigonal bipyramidal) - PCl₅ has 5 bonds.
• Q3: A (sp) - CO₂ is linear with sp hybridisation.

Memory Aids

• Linear = 2 bonds, sp
• Tetrahedral = 4 bonds, sp³
• Trigonal bipyramidal = 5 bonds, sp³d

Hybridisation in Molecules with Lone and Bond Pairs

Structures and Effects

• Count bond pairs (X) and lone pairs (E) around central atom (AX_nE_m notation).
• Predict hybridisation from total electron pairs.
• Lone pairs affect bond angles and molecular shape.
• Shapes include bent, T-shaped, seesaw, square pyramidal, etc.
• Examples: NH₃ (sp³, trigonal pyramidal), H₂O (sp³, bent).

Solved Examples

Ex 1: Hybridisation of NH₃.

• 3 bond pairs + 1 lone pair = 4 electron pairs
• Hybridisation: sp³
• Shape: Trigonal pyramidal
• Bond angle: ~107°

Practice Set

• Q1: Shape of H₂O molecule?
  • A. Linear
  • B. Bent
  • C. Trigonal planar
  • D. T-shaped
• Q2: Hybridisation of central atom in SF₄?
  • A. sp³
  • B. sp³d
  • C. sp³d²
  • D. sp²
• Q3: Bond angle in NH₃?
  • A. 109.5°
  • B. 107°
  • C. 120°
  • D. 90°

Answer Key

• Q1: B (Bent) - Due to 2 bond pairs and 2 lone pairs in H₂O.
• Q2: B (sp³d) - SF₄ has 4 bonds + 1 lone pair.
• Q3: B (107°) - Slightly less than tetrahedral due to lone pair repulsion.

Memory Aids

• Lone pairs reduce bond angles.
• AX₂E₂ = bent shape.
• AX₃E = trigonal pyramidal.
• d orbitals involved: d_x²-y² in dsp², d_z² in sp³d, d_x²-y² and d_z² in sp³d², d_xy, d_yz, d_xz in sp³d³.

Molecular Orbital Theory

Key Features

• Electrons in molecules occupy molecular orbitals (MOs).
• Atomic orbitals (AOs) of similar energy and symmetry combine to form MOs.
• MOs are polycentric, influenced by multiple nuclei.
• Number of MOs formed equals number of combining AOs.
• Two AOs combine to form one bonding MO and one antibonding MO.
• Bonding MO has lower energy and greater stability than antibonding MO.
• MOs represent electron probability distribution around nuclei.
• MOs fill according to Aufbau principle, Pauli’s exclusion, and Hund’s rule.

Formation of MOs (LCAO)

• MOs form by addition or subtraction of AO wave functions.
• Addition forms bonding MOs (σ or π) with lower energy.
• Subtraction forms antibonding MOs (σ* or π*) with higher energy.

Solved Examples

• Ex 1: Combine 1s orbitals of two H atoms: ψ1s + ψ1s forms bonding ψ MO; ψ1s - ψ1s forms antibonding ψ* MO.

Practice Set

• Q1: What is formed when two atomic orbitals combine?
  A. One bonding MO
  B. One antibonding MO
  C. One bonding and one antibonding MO
  D. No MO formed
• Q2: Which principle is NOT followed in filling molecular orbitals?
  A. Aufbau principle
  B. Hund's rule
  C. Pauli’s exclusion principle
  D. Le Chatelier's principle
• Q3: What type of MO is formed by subtraction of wave functions?
  A. Bonding MO
  B. Antibonding MO
  C. Non-bonding MO
  D. Hybrid MO

Answer Key

• Q1: C - Two MOs form: one bonding and one antibonding.
• Q2: D - Le Chatelier's principle is unrelated to MO filling.
• Q3: B - Subtraction forms antibonding MOs with higher energy.

Memory Aids

• MO formation: Addition = Bonding (lower energy), Subtraction = Antibonding (higher energy).
• Bonding MOs stabilize molecule; antibonding MOs destabilize.
• Follow Aufbau, Pauli, Hund for electron filling.

Key Features

• Electrons in molecules occupy molecular orbitals (MOs).
• Atomic orbitals (AOs) of similar energy and symmetry combine to form MOs.
• MOs are polycentric, influenced by multiple nuclei.
• Number of MOs formed equals number of combining AOs.
• Two AOs combine to form one bonding MO and one antibonding MO.
• Bonding MO has lower energy and greater stability than antibonding MO.
• MOs represent electron probability distribution around nuclei.
• MOs fill according to Aufbau principle, Pauli’s exclusion, and Hund’s rule.

Formation of MOs (LCAO)

• MOs form by addition or subtraction of AO wave functions.
• Addition forms bonding MOs (σ or π) with lower energy.
• Subtraction forms antibonding MOs (σ* or π*) with higher energy.

Solved Examples

• Ex 1: Combine 1s orbitals of two H atoms: 91s + 1s9 forms bonding c3 MO; 91s - 1s9 forms antibonding c3* MO.

Practice Set

• Q1: What is formed when two atomic orbitals combine?
  A. One bonding MO
  B. One antibonding MO
  C. One bonding and one antibonding MO
  D. No MO formed
• Q2: Which principle is NOT followed in filling molecular orbitals?
  A. Aufbau principle
  B. Hund's rule
  C. Pauli’s exclusion principle
  D. Le Chatelier's principle
• Q3: What type of MO is formed by subtraction of wave functions?
  A. Bonding MO
  B. Antibonding MO
  C. Non-bonding MO
  D. Hybrid MO

Answer Key

• Q1: C - Two MOs form: one bonding and one antibonding.
• Q2: D - Le Chatelier's principle is unrelated to MO filling.
• Q3: B - Subtraction forms antibonding MOs with higher energy.

Memory Aids

• MO formation: Addition = Bonding (lower energy), Subtraction = Antibonding (higher energy).
• Bonding MOs stabilize molecule; antibonding MOs destabilize.
• Follow Aufbau, Pauli, Hund for electron filling.

Molecular Orbital Energy Order

Energy Order for Homonuclear Diatomic Molecules (O₂, F₂, Ne₂)

• σ1s < σ*1s < σ2s < σ*2s < σ2p_z < σ2p_x = σ2p_y < σ*2p_x = σ*2p_y < σ*2p_z

Energy Order for Lighter Diatomic Molecules (B₂, C₂, N₂)

• σ1s < σ*1s < σ2s < σ*2s < σ2p_x = σ2p_y < σ2p_z < σ*2p_x = σ*2p_y < σ*2p_z

Solved Examples

• Ex 1: Arrange MOs for O₂: σ1s < σ*1s < σ2s < σ*2s < σ2p_z < σ2p_x = σ2p_y < σ*2p_x = σ*2p_y < σ*2p_z.

Practice Set

• Q1: Which MO has the lowest energy in O₂?
  A. σ*1s
  B. σ1s
  C. σ2p_z
  D. σ*2p_z
• Q2: In B₂, which MO comes just after σ*2s?
  A. σ2p_z
  B. σ2p_x = σ2p_y
  C. σ*2p_x
  D. σ*2p_z
• Q3: Which molecule follows the MO order with σ2p_x = σ2p_y before σ2p_z?
  A. O₂
  B. F₂
  C. N₂
  D. B₂

Answer Key

• Q1: B - σ1s is the lowest energy MO.
• Q2: B - σ2p_x = σ2p_y come after σ*2s in lighter diatomics.
• Q3: D - B₂ shows this MO energy order.

Memory Aids

• O₂, F₂, Ne₂: σ2p_z before σ2p_x, σ2p_y.
• B₂, C₂, N₂: σ2p_x, σ2p_y before σ2p_z.
• Remember by atomic number: lighter elements have different MO order.

Energy Order for Homonuclear Diatomic Molecules (O₂, F₂, Ne₂)

• c31s < c3*1s < c32s < c3*2s < c32p_z < c32p_x = c32p_y < c3*2p_x = c3*2p_y < c3*2p_z

Energy Order for Lighter Diatomic Molecules (B₂, C₂, N₂)

• c31s < c3*1s < c32s < c3*2s < c32p_x = c32p_y < c32p_z < c3*2p_x = c3*2p_y < c3*2p_z

Solved Examples

• Ex 1: Arrange MOs for O₂: c31s < c3*1s < c32s < c3*2s < c32p_z < c32p_x = c32p_y < c3*2p_x = c3*2p_y < c3*2p_z.

Practice Set

• Q1: Which MO has the lowest energy in O₂?
  A. c3*1s
  B. c31s
  C. c32p_z
  D. c3*2p_z
• Q2: In B₂, which MO comes just after c3*2s?
  A. c32p_z
  B. c32p_x = c32p_y
  C. c3*2p_x
  D. c3*2p_z
• Q3: Which molecule follows the MO order with c32p_x = c32p_y before c32p_z?
  A. O₂
  B. F₂
  C. N₂
  D. B₂

Answer Key

• Q1: B - c31s is the lowest energy MO.
• Q2: B - c32p_x = c32p_y come after c3*2s in lighter diatomics.
• Q3: D - B₂ shows this MO energy order.

Memory Aids

• O₂, F₂, Ne₂: c32p_z before c32p_x, c32p_y.
• B₂, C₂, N₂: c32p_x, c32p_y before c32p_z.
• Remember by atomic number: lighter elements have different MO order.

Gerade and Ungerade Molecular Orbitals

Symmetry and Notation

• Gerade (g): Symmetrical MOs about center; symbol ‘g’.
• Ungerade (u): Unsymmetrical MOs; symbol ‘u’.
• All σ bonding MOs are gerade.
• All σ antibonding MOs (σ*) are ungerade.
• All π antibonding MOs (π*) are gerade.
• All π bonding MOs are ungerade.

Solved Examples

• Ex 1: σ bonding MO in H₂ is gerade.
• Ex 2: σ* antibonding MO in H₂ is ungerade.

Practice Set

• Q1: What does gerade (g) indicate?
  A. Unsymmetrical MO
  B. Symmetrical MO
  C. Antibonding MO
  D. Non-bonding MO
• Q2: Which MO is ungerade?
  A. σ bonding
  B. σ* antibonding
  C. π* bonding
  D. π bonding
• Q3: π* MOs are generally?
  A. Gerade
  B. Ungerade
  C. Non-bonding
  D. Hybrid

Answer Key

• Q1: B - Gerade means symmetrical MO.
• Q2: B - σ* antibonding MOs are ungerade.
• Q3: A - π* antibonding MOs are gerade.

Memory Aids

• Gerade = g = symmetrical; Ungerade = u = unsymmetrical.
• σ bonding = gerade; σ* antibonding = ungerade.
• π bonding = ungerade; π* antibonding = gerade.

Bond Order and Stability

Bond Order Calculation

• Bond order (B.O.) = 1/2 (N_b - N_a)
• N_b = Number of bonding electrons.
• N_a = Number of antibonding electrons.
• If B.O. > 0, molecule is stable.
• If B.O. = 0 or < 0, molecule is unstable.
• Bond order ∝ Bond energy ∝ Stability ∝ 1 / Bond length.
• Higher bond order means higher bond dissociation energy and shorter bond length.
• Molecule is diamagnetic if all electrons are paired; paramagnetic if unpaired electrons present.

Solved Examples

• Ex 1: Calculate B.O. for O₂: N_b=10, N_a=6; B.O. = 1/2(10-6) = 2 (stable).

Practice Set

• Q1: What is bond order if N_b = 8 and N_a = 4?
  A. 1
  B. 2
  C. 3
  D. 4
• Q2: If bond order is zero, molecule is:
  A. Stable
  B. Unstable
  C. Paramagnetic
  D. Diamagnetic
• Q3: Which property increases with bond order?
  A. Bond length
  B. Bond dissociation energy
  C. Number of antibonding electrons
  D. Paramagnetism

Answer Key

• Q1: B - B.O. = 1/2(8-4) = 2.
• Q2: B - B.O. = 0 means unstable molecule.
• Q3: B - Bond dissociation energy increases with bond order.

Memory Aids

• Bond order formula: 1/2 (bonding electrons - antibonding electrons).
• Higher bond order = stronger, shorter bond.
• Diamagnetic = all paired; paramagnetic = unpaired electrons.

Bond Order Calculation

• Bond order (B.O.) = 1/2 (N_b - N_a)
• N_b = Number of bonding electrons.
• N_a = Number of antibonding electrons.
• If B.O. > 0, molecule is stable.
• If B.O. = 0 or < 0, molecule is unstable.
• Bond order ∝ Bond energy ∝ Stability ∝ 1 / Bond length.
• Higher bond order means higher bond dissociation energy and shorter bond length.
• Molecule is diamagnetic if all electrons are paired; paramagnetic if unpaired electrons present.

Solved Examples

• Ex 1: Calculate B.O. for O₂: N_b=10, N_a=6; B.O. = 1/2(10-6) = 2 (stable).

Practice Set

• Q1: What is bond order if N_b = 8 and N_a = 4?
  A. 1
  B. 2
  C. 3
  D. 4
• Q2: If bond order is zero, molecule is:
  A. Stable
  B. Unstable
  C. Paramagnetic
  D. Diamagnetic
• Q3: Which property increases with bond order?
  A. Bond length
  B. Bond dissociation energy
  C. Number of antibonding electrons
  D. Paramagnetism

Answer Key

• Q1: B - B.O. = 1/2(8-4) = 2.
• Q2: B - B.O. = 0 means unstable molecule.
• Q3: B - Bond dissociation energy increases with bond order.

Memory Aids

• Bond order formula: 1/2 (bonding electrons - antibonding electrons).
• Higher bond order = stronger, shorter bond.
• Diamagnetic = all paired; paramagnetic = unpaired electrons.

Hydrogen Bonding

Concept and Types

• Hydrogen bond: Attractive force between hydrogen atom of one molecule and electronegative atom (F, O, N) of another molecule.
• Intermolecular hydrogen bonding: Occurs between two different molecules (e.g., HF, H₂O, ROH).
• 
  • H in HF has partial positive charge (H^\delta+), F has partial negative (F^\delta-).
  • Hydrogen bond shown as dashed line between H and F of different molecules.
  • Hydrogen bonding increases boiling point and reduces volatility.
• Intramolecular hydrogen bonding: Bond formed within the same molecule (e.g., o-nitrophenol), causing cyclization and preventing association.
• 
  • o-Chlorophenol and o-Nitrophenol have -OH and substituent groups at ortho positions.
  • Dashed lines show intramolecular hydrogen bonding in o-nitrophenol.
• Double bonds count as one electron pair in molecular geometry prediction.
• Intermolecular H-bonding ∝ Solubility ∝ Boiling point ∝ 1/Volatility.
• Intramolecular H-bonding ∝ 1/Solubility ∝ 1/Boiling point ∝ Volatility.

Solved Examples

• Ex 1: Explain hydrogen bonding in HF.
  • HF has polar covalent bond (H^\delta+-F^\delta-).
  • Positive H attracts negative F of another molecule forming H-bond.
  • H-bonds shown as dashed lines increase boiling point.
• Ex 2: Identify intramolecular H-bond in o-nitrophenol.
  • H of -OH forms bond with O of NO₂ group within same molecule.
  • This cyclization reduces intermolecular association.

Practice Set

• Q1: Which atom does hydrogen bond with?
  • A. Carbon
  • B. Fluorine
  • C. Hydrogen
  • D. Sodium
• Q2: Intramolecular hydrogen bonding occurs in:
  • A. HF
  • B. o-Nitrophenol
  • C. Water
  • D. Ethanol
• Q3: Which property increases due to intermolecular hydrogen bonding?
  • A. Volatility
  • B. Boiling point
  • C. Solubility decreases
  • D. Molecular weight

Answer Key

• Q1: B - Hydrogen bonds with electronegative atoms like fluorine, oxygen, or nitrogen.
• Q2: B - Intramolecular hydrogen bonding occurs within the same molecule, e.g., o-nitrophenol.
• Q3: B - Intermolecular hydrogen bonding increases boiling point by stronger molecular attraction.

Memory Aids

• H-bond forms only with F, O, N atoms.
• Intermolecular H-bonding ↑ solubility, boiling point; ↓ volatility.
• Intramolecular H-bonding ↓ solubility, boiling point; ↑ volatility.
• Double bonds count as one electron pair in shape prediction.