Chapter 4: Chemical Bonding and Molecular Structure

Chemical Bonding Introduction

Key Concepts

• Chemical bond: Attractive force holding atoms or ions together.
• Atoms interact due to:
• Net decrease in energy from attractive forces exceeding repulsive forces.
• Tendency to achieve noble gas electron configuration.

Solved Examples

• Ex 1: Why do atoms form bonds?
  Atoms form bonds to lower their energy and achieve stable electron configurations.

Practice Set

• Q1: What is a chemical bond?
  A) Repulsive force between atoms
  B) Attractive force holding atoms together
  C) Force that breaks molecules
  D) Force only in ionic compounds
• Q2: Which factor drives atoms to bond?
  A) Increase in energy
  B) Achieving noble gas configuration
  C) Random electron movement
  D) Nuclear reactions
• Q3: What happens when attractive forces are greater than repulsive forces?
  A) Atoms repel
  B) Atoms bond
  C) Atoms ionize
  D) Atoms remain isolated

Answer Key

• Q1: B - Chemical bond is an attractive force holding atoms together.
• Q2: B - Atoms bond to achieve noble gas electron configuration.
• Q3: B - When attraction exceeds repulsion, atoms form bonds.

Memory Aids

• Bonding = Energy decrease + Noble gas stability.
• Atoms bond to fill outer shells.

Kössel—Lewis Approach

Key Concepts

• Atoms combine by:
• Transfer of valence electrons (electrovalency) forming ionic bonds.
• Sharing of valence electrons forming covalent or coordinate bonds.
• Octet rule: Atoms tend to complete 8 electrons in valence shell (duplet for H).
• Coordinate (dative) bond: Both shared electrons come from one atom.

Solved Examples

• Ex 1: Identify bond type in NaCl.
  Na transfers electron to Cl → ionic bond.
• Ex 2: Identify bond in NH₃ + H⁺ → NH₄⁺.
  Coordinate bond formed by NH₃ donating lone pair to H⁺.

Practice Set

• Q1: What type of bond forms by electron transfer?
  A) Covalent
  B) Ionic
  C) Coordinate
  D) Metallic
• Q2: When shared electrons come equally from both atoms, the bond is:
  A) Ionic
  B) Coordinate
  C) Covalent
  D) Metallic
• Q3: Which bond involves one atom donating both electrons?
  A) Ionic
  B) Covalent
  C) Coordinate
  D) Hydrogen

Answer Key

• Q1: B - Electron transfer forms ionic bonds.
• Q2: C - Equal sharing forms covalent bonds.
• Q3: C - Coordinate bonds have both electrons from one atom.

Memory Aids

• Ionic = Transfer, Covalent = Share equally, Coordinate = Share from one atom.
• Octet rule guides bonding types.

Covalent Bond

Key Concepts

• Formed by mutual sharing of electron pairs between atoms.
• Each atom contributes at least one electron to shared pair.
• Atoms achieve 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: Identify bond type in H₂.
  Two H atoms share one pair → single covalent bond.
• Ex 2: Bond type in O₂.
  Two O atoms share two pairs → double bond.

Practice Set

• Q1: A single covalent bond involves how many shared electrons?
  A) 1
  B) 2
  C) 4
  D) 6
• Q2: Double bond means sharing of:
  A) 1 pair
  B) 2 pairs
  C) 3 pairs
  D) 4 pairs
• Q3: Triple bond involves sharing of:
  A) 2 pairs
  B) 3 pairs
  C) 1 pair
  D) 4 pairs

Answer Key

• Q1: B - Single bond shares 2 electrons (1 pair).
• Q2: B - Double bond shares 2 pairs (4 electrons).
• Q3: B - Triple bond shares 3 pairs (6 electrons).

Memory Aids

• Single = 1 pair, Double = 2 pairs, Triple = 3 pairs shared electrons.
• Shared pairs = bond strength increases.

Lewis Structures

Key Concepts

• Valence electrons shown as dots around atoms.
• Shared electron pairs between atoms represent bonds.
• Lewis dot structures depict bonding and lone pairs.
• Lewis representations use lines for bonds (1 line = 2 electrons).
• Charges shown for ions with brackets and signs.

Solved Examples

• Ex 1: Draw Lewis structure of H₂O.
  O shares electrons with two H atoms; lone pairs on O shown as dots.
• Ex 2: Lewis structure of NH₄⁺.
  N shares electrons with four H atoms; positive charge shown.

Practice Set

• Q1: In Lewis structures, dots represent:
  A) Protons
  B) Valence electrons
  C) Neutrons
  D) Atomic nuclei
• Q2: A single line in Lewis representation means:
  A) 1 electron
  B) 2 electrons
  C) 3 electrons
  D) 4 electrons
• Q3: How is charge shown in Lewis structures for ions?
  A) No indication
  B) Brackets and charge sign
  C) Color coding
  D) Extra dots

Answer Key

• Q1: B - Dots represent valence electrons.
• Q2: B - Single line represents 2 shared electrons.
• Q3: B - Charges shown with brackets and signs.

Memory Aids

• Dots = valence electrons; lines = bonds.
• Charge = brackets + sign.

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.
• Example: Ozone (O₃) formal charges: central O +1, one end O 0, other end O -1.

Solved Examples

• Ex 1: Calculate formal charge on central O in O₃:
  6 - 2 - 1/2(6) = +1
• Ex 2: Formal charge on end O with 4 lone electrons and 4 bonding electrons:
  6 - 4 - 1/2(4) = 0

Practice Set

• Q1: Formal charge formula is:
  A) Valence - Lone pairs - Bonding electrons
  B) Valence - Lone pairs - 1/2 Bonding electrons
  C) Valence + Lone pairs + Bonding electrons
  D) Valence - Bonding electrons
• Q2: Which structure is most stable?
  A) Highest formal charges
  B) Lowest formal charges
  C) Random charges
  D) No charges
• Q3: Formal charge on atom with 5 valence electrons, 2 lone electrons, and 6 bonding electrons?
  A) 0
  B) +1
  C) -1
  D) +2

Answer Key

• Q1: B - Formal charge = Valence - Lone pairs - 1/2 Bonding electrons.
• Q2: B - Lowest formal charges indicate stability.
• Q3: A - 5 - 2 - 1/2(6) = 5 - 2 - 3 = 0.

Memory Aids

• Formal charge = Valence - Lone pairs - half bonding electrons.
• Choose structure with minimal formal charges.

Limitations of Octet Rule

Key Concepts

• Does not apply to atoms with less than 4 valence electrons (e.g., Be, B).
• Fails for molecules with expanded octet (e.g., PF₅).
• Does not explain molecules with odd electrons (e.g., NO, ClO₂).
• Cannot explain paramagnetism of O₂ (two unpaired electrons).
• Noble gases can form compounds (e.g., XeF₂), violating inertness assumption.
• Does not predict molecular shape, stability, or energy.
• Electron-deficient molecules (e.g., BeCl₂, BF₃) behave as Lewis acids.

Solved Examples

• Ex 1: Why PF₅ violates octet rule?
  P has 10 valence electrons, exceeding octet.
• Ex 2: Why O₂ is paramagnetic?
  Has two unpaired electrons, not explained by octet rule.

Practice Set

• Q1: Octet rule fails for:
  A) Molecules with even electrons
  B) Molecules with odd electrons
  C) Ionic compounds
  D) Metals
• Q2: Which molecule violates octet rule?
  A) PF₃
  B) PF₅
  C) NH₃
  D) CH₄
• Q3: Electron-deficient molecules act as:
  A) Lewis bases
  B) Lewis acids
  C) Reducing agents
  D) Oxidizing agents

Answer Key

• Q1: B - Octet rule fails for molecules with odd electrons.
• Q2: B - PF₅ has expanded octet.
• Q3: B - Electron-deficient molecules are Lewis acids.

Memory Aids

• Octet rule exceptions: Odd electrons, expanded octet, incomplete octet.
• Electron-deficient = 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 1e⁻ to form Na⁺; Cl (3s²3p⁵) gains 1e⁻ to form Cl⁻. Ionic bond forms by electrostatic attraction.

Practice Set

• Q1: Ionic bonds form due to
  • A) Sharing of electrons
  • B) Electrostatic attraction between ions
  • C) Metallic bonding
  • D) Hydrogen bonding
• Q2: Which pair is most likely to form an ionic bond?
  • A) Na and Cl
  • B) O and O
  • C) H and H
  • D) C and H
• Q3: Stability of ionic compounds depends mainly on
  • A) Octet rule fulfillment
  • B) Lattice enthalpy
  • C) Ionization energy only
  • D) Electron gain enthalpy only

Answer Key

• Q1: B — Ionic bonds arise from electrostatic attraction between oppositely charged ions.
• Q2: A — Na (low I.E.) and Cl (high negative electron gain enthalpy) form ionic bonds.
• Q3: B — Lattice enthalpy determines ionic compound stability.

Memory Aids

• Ionic bond = Transfer of electrons + Electrostatic attraction.
• Low I.E. metal + High electron affinity non-metal = Ionic bond.
• Stability ∝ Lattice enthalpy, not just octet.

Lattice Enthalpy

Definition

• Lattice enthalpy is the energy required to separate 1 mole of ionic solid into gaseous ions.

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 / ionic size.

Solved Examples

• Ex 1: Compare lattice enthalpy of MgO and NaCl.
  Mg²⁺ and O²⁻ have higher charges than Na⁺ and Cl⁻, so MgO has higher lattice enthalpy.

Practice Set

• Q1: Lattice enthalpy is the energy required to
  • A) Form ionic bonds
  • B) Separate ionic solid into gaseous ions
  • C) Break covalent bonds
  • D) Ionize atoms
• Q2: Which factor increases lattice enthalpy?
  • A) Larger ionic size
  • B) Lower ionic charge
  • C) Higher ionic charge
  • D) Lower melting point
• Q3: Between CaCl₂ and KCl, which has higher lattice enthalpy and why?
  • A) KCl, due to smaller ions
  • B) CaCl₂, due to higher charge on Ca²⁺
  • C) Both equal
  • D) KCl, due to higher melting point

Answer Key

• Q1: B — Lattice enthalpy is energy to separate ionic solid into gaseous ions.
• Q2: C — Higher ionic charge increases lattice enthalpy.
• Q3: B — CaCl₂ has higher lattice enthalpy due to Ca²⁺ charge.

Memory Aids

• Lattice enthalpy ↑ with ↑ ionic charge.
• Lattice enthalpy ↑ with ↓ ionic size.
• MgO > NaCl in lattice enthalpy due to charge difference.

Bond Length

Definition and Measurement

• Bond length is the equilibrium distance between nuclei of two bonded atoms.
• Expressed in Angstrom (1 C5 = 10B0B1 m) or picometer (1 pm = 10B0B2 m).
• In covalent molecules, bond length = sum of covalent radii of bonded atoms.

Covalent and van der Waals Radii

• Covalent radius: Half the distance between nuclei of two similar atoms bonded covalently.
• Van der Waals radius: Half the distance between nuclei of two similar atoms in non-bonded contact.

Solved Examples

Ex 1: Calculate bond length if covalent radii of atoms A and B are 70 pm and 80 pm respectively.

• Bond length = 70 pm + 80 pm = 150 pm

Practice Set

• Q1: Bond length is measured in?
  • A. Meter
  • B. Angstrom
  • C. Liter
  • D. Kelvin
• Q2: Van der Waals radius is half the distance between two atoms in?
  • A. Same molecule bonded
  • B. Different molecules non-bonded
  • C. Ionic bond
  • D. Metallic bond
• Q3: If covalent radius of atom X is 60 pm and atom Y is 90 pm, bond length XY is?
  • A. 150 pm
  • B. 30 pm
  • C. 90 pm
  • D. 60 pm

Answer Key

• Q1: B - Angstrom is standard unit for bond length.
• Q2: B - Van der Waals radius applies to non-bonded atoms in different molecules.
• Q3: A - Bond length = sum of covalent radii = 60 + 90 = 150 pm.

Memory Aids

• Bond length = sum of covalent radii.
• Van der Waals radius > covalent radius.

Bond Angle

Definition and Importance

• Average angle between bonded orbitals in molecules with three or more atoms.
• Indicates spatial distribution of orbitals around central atom.

Solved Examples

Ex 1: Water molecule has bond angle approximately 104.5B0.

• Shows bent shape due to lone pairs on oxygen.

Practice Set

• Q1: Bond angle is the angle between?
  • A. Two nuclei
  • B. Two bonded orbitals
  • C. Two electrons
  • D. Two molecules
• Q2: Which molecule has bond angle close to 109.5B0?
  • A. CO₂
  • B. CH₄
  • C. NH₃
  • D. H₂O
• Q3: Bond angle decreases due to?
  • A. Lone pairs on central atom
  • B. More bonded atoms
  • C. Higher atomic number
  • D. Ionic bonding

Answer Key

• Q1: B - Bond angle is between bonded orbitals.
• Q2: B - Methane (CH₄) has tetrahedral angle ~109.5B0.
• Q3: A - Lone pairs repel more, reducing bond angle.

Memory Aids

• Tetrahedral angle = 109.5B0, trigonal planar = 120B0, linear = 180B0.
• Lone pairs reduce bond angle.

Bond Enthalpy

Definition and Characteristics

• Energy required to break 1 mole of a specific bond in gaseous state (bond dissociation energy).
• Expressed in kJ mol^-1.
• Average bond enthalpy used for bonds in polyatomic molecules due to varying environments.
• Example: H₂O O-H bonds average = (502 + 427)/2 = 464.5 kJ mol^-1.
• CO has highest bond enthalpy (1070 kJ mol^-1) among diatomic molecules.

Solved Examples

Ex 1: Calculate average bond enthalpy of O-H bonds in water if first bond = 502 kJ/mol and second = 427 kJ/mol.

• Average = (502 + 427)/2 = 464.5 kJ/mol

Practice Set

• Q1: Bond enthalpy is energy to break bonds in?
  • A. Solid state
  • B. Liquid state
  • C. Gaseous state
  • D. Aqueous solution
• Q2: Which molecule has highest bond enthalpy?
  • A. N₂
  • B. CO
  • C. C₂
  • D. CN
• Q3: Average bond enthalpy of 4 C-H bonds in CH₄ is?
  • A. Same for all bonds
  • B. Different for each bond
  • C. Average value used
  • D. Zero

Answer Key

• Q1: C - Bond enthalpy is measured in gaseous state.
• Q2: B - CO has highest bond enthalpy (1070 kJ/mol).
• Q3: C - Average bond enthalpy is used for polyatomic molecules.

Memory Aids

• Higher bond order 1 bond enthalpy 1 bond length.
• Triple bonds > double bonds > single bonds in strength.

Bond Order

Definition and Correlation

• Number of bonds between two atoms in a molecule.
• Isoelectronic species have same bond order.
• Examples: F₂ and O₂^2- bond order = 1; CO and NO⁺ bond order = 3.
• Higher bond order means higher bond enthalpy and shorter bond length.

Solved Examples

Ex 1: Calculate bond order of O₂^2-.

• Bond order = 1 (single bond).

Practice Set

• Q1: Bond order is number of?
  • A. Atoms
  • B. Bonds between two atoms
  • C. Electrons
  • D. Molecules
• Q2: Bond order of CO is?
  • A. 1
  • B. 2
  • C. 3
  • D. 4
• Q3: Increase in bond order causes?
  • A. Increase bond length
  • B. Decrease bond enthalpy
  • C. Increase bond enthalpy
  • D. No change

Answer Key

• Q1: B - Bond order is number of bonds between two atoms.
• Q2: C - CO has bond order 3 (triple bond).
• Q3: C - Higher bond order increases bond enthalpy.

Memory Aids

• Bond order = (total bonds in resonance structures) / (number of structures).
• Higher bond order = stronger, shorter bond.

Resonance Structures

Concept and Features

• Molecules with multiple Lewis structures called resonance forms.
• Actual structure is resonance hybrid, intermediate of all forms.
• Resonance energy = difference between actual and most stable form.
• Contributing structures have same atomic positions and electron count.
• Hybrid is more stable than any single resonance form.
• More resonance forms of similar energy increase stability.

Solved Examples

Ex 1: Calculate bond order of carbonate ion (CO₃^2-).

• 3 resonance structures with 4 bonds total (one double, two single).
• Bond order = 4 bonds / 3 structures = 1.33.

Practice Set

• Q1: Resonance structures differ in?
  • A. Atomic positions
  • B. Electron arrangement
  • C. Number of atoms
  • D. Molecular formula
• Q2: Resonance hybrid is?
  • A. One resonance form
  • B. Average of all resonance forms
  • C. Unstable structure
  • D. Imaginary molecule
• Q3: Bond order in resonance molecules is calculated by?
  • A. Total bonds / number of atoms
  • B. Total bonds / number of resonance structures
  • C. Number of atoms / bonds
  • D. Number of resonance structures / bonds

Answer Key

• Q1: B - Resonance structures differ in electron arrangement only.
• Q2: B - Hybrid is average of all resonance forms.
• Q3: B - Bond order = total bonds / number of resonance structures.

Memory Aids

• Resonance structures have same atom positions, differ in electron placement.
• Hybrid structure is real, more stable.

Polarity of Bonds

Types of Covalent Bonds

• Non-polar covalent bond: Between identical atoms; electron pair shared equally.
• Polar covalent bond: Between different atoms; electron pair shifted towards more electronegative atom.
• Dipole moment (BC) = charge (q) D7 distance (d).
• Measured in Debye (D); 1 D = 3.33 D7 10^-30 C m.
• Dipole moment is a vector pointing from positive to negative end.

Dipole Moment and Molecular Geometry

• Non-polar diatomic molecules have zero dipole moment.
• Molecules with symmetrical geometry have zero net dipole moment.
• Irregular geometry or lone pairs cause non-zero dipole moment.
• Net dipole moment: BC = BC₁² + BC₂² + 2BC₁BC₂cosB8 (bond angle B8).

Solved Examples

Ex 1: Explain why NH₃ has higher dipole moment than NF₃.

• In NH₃, dipoles point towards N and lone pair reinforces dipole.
• In NF₃, dipoles point towards F but lone pair opposes dipole, reducing net moment.

Practice Set

• Q1: Dipole moment is zero in?
  • A. HCl
  • B. O₂
  • C. NH₃
  • D. HF
• Q2: Dipole moment unit is?
  • A. Coulomb
  • B. Debye
  • C. Newton
  • D. Joule
• Q3: Net dipole moment depends on?
  • A. Bond moments and bond angle
  • B. Atomic masses
  • C. Number of atoms
  • D. Molecular weight

Answer Key

• Q1: B - O₂ is non-polar with zero dipole moment.
• Q2: B - Dipole moment measured in Debye.
• Q3: A - Depends on bond moments and bond angle.

Memory Aids

• Dipole moment vector points from positive to negative.
• Lone pairs affect net dipole moment.
• Symmetry cancels dipole moments.

Fajans' Rule

Polarization and Covalent Character

• Polarization: distortion of anion electron cloud by cation.
• Smaller cation size and higher charge increase polarizing power.
• Cations with ns² np⁶ nd¹⁰ configuration polarize more than ns² np⁶.
• Larger anions are more polarizable.
• Greater polarization increases covalent character in ionic bonds.

Solved Examples

Ex 1: Which cation has greater polarizing power: Ag⁺ or K⁺?

• Ag⁺ has ns² np⁶ nd¹⁰ configuration and smaller size, so greater polarizing power.

Practice Set

• Q1: Polarizing power increases with?
  • A. Larger cation size
  • B. Smaller cation size
  • C. Lower cation charge
  • D. Larger anion size
• Q2: Which anion is more polarizable?
  • A. F^-
  • B. I^-
  • C. Cl^-
  • D. Br^-
• Q3: According to Fajans' rule, covalent character increases with?
  • A. Decreased polarization
  • B. Increased polarization
  • C. Larger cation size
  • D. Smaller anion size

Answer Key

• Q1: B - Smaller cation size increases polarizing power.
• Q2: B - Larger anions like I^- are more polarizable.
• Q3: B - Increased polarization increases covalent character.

Memory Aids

• Polarizing power 1 with smaller size and higher charge of cation.
• Larger anions are more easily polarized.

Valence Shell Electron Pair Repulsion Theory

Fundamentals

• Developed by Nyholm and Gillespie in 1957.
• Molecules adopt shapes that keep electron pairs as far apart as possible.
• Shape depends on number of valence shell electron pairs (bonded and lone pairs) around central atom.
• Electron pair repulsion order: lone pair–lone pair (lp–lp) > lone pair–bond pair (lp–bp) > bond pair–bond pair (bp–bp).
• Electron pairs arrange to minimize repulsion and maximize distance.
• Applicable to all resonance structures of a molecule.

Shapes and Bond Angles Examples

• SF₄ (Sulphur Tetrafluoride): Trigonal bipyramidal shape; axial F–S–F bond angle ~173° (near linear).
• SF₆ (Sulphur Hexafluoride): Octahedral shape; all F–S–F bond angles 90°.
• NH₃ (Ammonia): Tetrahedral electron geometry; molecular shape trigonal pyramidal; H–N–H bond angle 107.8° due to one lone pair.
• H₂O (Water): Tetrahedral electron geometry; bent molecular shape; H–O–H bond angle 104.5° due to two lone pairs.

Solved Examples

• Ex 1: Predict shape and bond angle of NH₃.
• Step 1: Count electron pairs around N: 3 bond pairs + 1 lone pair = 4 pairs.
• Step 2: Electron geometry is tetrahedral.
• Step 3: Molecular shape is trigonal pyramidal due to one lone pair.
• Step 4: Bond angle slightly less than 109.5°, observed 107.8°.

Practice Set

• Q1: What is the shape of SF₆ molecule?
  A. Tetrahedral
  B. Octahedral
  C. Trigonal bipyramidal
  D. Square planar
• Q2: Which repulsion is strongest according to VSEPR theory?
  A. Bond pair–bond pair
  B. Lone pair–bond pair
  C. Lone pair–lone pair
  D. Bond pair–free radical
• Q3: What is the approximate H–O–H bond angle in water?
  A. 109.5°
  B. 104.5°
  C. 120°
  D. 90°

Answer Key

• Q1: B. Octahedral — SF₆ has 6 bonded atoms arranged octahedrally with 90° bond angles.
• Q2: C. Lone pair–lone pair — Lone pairs repel more strongly than bond pairs.
• Q3: B. 104.5° — Water has two lone pairs reducing bond angle from 109.5° to 104.5°.

Memory Aids

• Repulsion strength mnemonic: "lp–lp > lp–bp > bp–bp".
• Shapes by electron pairs: 2-linear, 3-trigonal planar, 4-tetrahedral, 5-trigonal bipyramidal, 6-octahedral.
• Lone pairs reduce bond angles from ideal geometry.

Geometry Based on VSEPR Theory

Electron Pair Count and Molecular Shapes

• Molecules are classified by total electron pairs (bonded + lone) around central atom.
• General formula: AX_nE_m, where A = central atom, X = bonded atoms, E = lone pairs.
• Shapes vary with number of electron pairs and lone pairs.
• Examples:
• AX₂: Linear (e.g., BeCl₂)
• AX₃: Trigonal planar (e.g., BF₃)
• AX₂E: Bent (e.g., SO₂)
• AX₄: Tetrahedral (e.g., CH₄)
• AX₃E: Trigonal pyramidal (e.g., NH₃)
• AX₂E₂: Bent (e.g., H₂O)

Solved Examples

• Ex 1: Determine shape of molecule with formula AX₃E.
• Step 1: Total electron pairs = 4 (3 bonded + 1 lone).
• Step 2: Electron geometry is tetrahedral.
• Step 3: Molecular shape is trigonal pyramidal.

Practice Set

• Q1: What is the shape of AX₂E₂ molecule?
  A. Linear
  B. Bent
  C. Trigonal planar
  D. Tetrahedral
• Q2: Which formula corresponds to trigonal bipyramidal shape?
  A. AX₄E
  B. AX₅
  C. AX₃E₂
  D. AX₆
• Q3: For AX₃E₂, what is the molecular shape?
  A. T-shaped
  B. Trigonal planar
  C. Linear
  D. Square planar

Answer Key

• Q1: B. Bent — Two bonded atoms and two lone pairs cause bent shape.
• Q2: B. AX₅ — Five bonded atoms form trigonal bipyramidal geometry.
• Q3: A. T-shaped — Three bonded atoms and two lone pairs give T-shaped molecule.

Memory Aids

• AX_nE_m notation helps predict shape quickly.
• Electron pairs = bonded atoms + lone pairs.
• More lone pairs distort ideal geometry, reducing bond angles.

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 stability.
• Minimum energy at equilibrium (intermolecular) distance; greater energy decrease means stronger bond.

Solved Examples

• Ex 1: Explain why bond formation releases energy.
  - Overlapping valence orbitals lower system energy.
  - System stabilizes at minimum potential energy.

Practice Set

• Q1: Which electrons participate in bond formation?
  A. Core electrons
  B. Valence electrons
  C. Both core and valence
  D. None
• Q2: At equilibrium bond distance, potential energy is:
  A. Maximum
  B. Zero
  C. Minimum
  D. Undefined
• Q3: Which scientist contributed to Valence Bond Theory?
  A. Heitler and London
  B. Dalton
  C. Rutherford
  D. Bohr

Answer Key

• Q1: B - Only valence electrons form bonds.
• Q2: C - Energy is minimum at equilibrium distance.
• Q3: A - Heitler and London introduced the theory.

Memory Aids

• Valence electrons = Bond makers; core electrons stay put.
• Energy minimum = Stable bond distance.
• Heitler & London = Valence Bond pioneers.

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; examples: s–s, s–p, p–p.
• Pi (π) bond: Lateral overlap perpendicular to internuclear axis; only p orbitals.

Strength and Properties

• σ bonds have greater overlap extent; stronger than π bonds.
• Multiple bonds: one σ bond + one or more π bonds.
• Bond energy order: Single < Double < Triple.
• Bond length order: Triple < Double < Single.
• Reactivity order: Single < Double < Triple.
• Maximum covalent bonds between two atoms: three.
• π bond never forms alone; always with σ bond.
• Head-on overlap stronger than lateral overlap.
• Free rotation possible about σ bond; restricted about π bond.

Solved Examples

• Ex 1: Identify bond types in ethene (C₂H₄).
  - C–C bond: one σ + one π bond.
  - C–H bonds: σ bonds.

Practice Set

• Q1: Which bond involves head-on overlap?
  A. π bond
  B. σ bond
  C. Both
  D. None
• Q2: Which bond type allows free rotation?
  A. π bond
  B. σ bond
  C. Both
  D. None
• Q3: Arrange bonds by increasing bond length:
  A. Single < Double < Triple
  B. Triple < Double < Single
  C. Double < Single < Triple
  D. Triple < Single < Double

Answer Key

• Q1: B - σ bond is head-on overlap.
• Q2: B - σ bond allows free rotation.
• Q3: B - Triple bonds are shortest, single longest.

Memory Aids

• σ = Straight (head-on), π = Parallel (sideways).
• Multiple bonds = 1 σ + π(s).
• Rotation free only around σ bonds.
• Bond length: Triple < Double < Single.

Hybridisation

Definition and Features

• Hybridisation: Mixing of atomic orbitals of similar energy in the valence shell to form new equivalent hybrid orbitals.
• Number of hybrid orbitals = Number of atomic orbitals mixed.
• Hybrid orbitals have equal energy and shape.
• 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 = Number of monovalent atoms attached
• C = Charge on cation
• A = Charge on anion

Important Conditions

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

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

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 SO₄^2- ion is?
  • A. sp³
  • B. sp³d
  • C. sp³d²
  • D. sp²

Answer Key

• Q1: C; Carbon forms 4 sigma bonds, so sp³ hybridisation.
• Q2: C; sp³d involves 5 orbitals hybridised.
• Q3: C; SO₄^2- has 6 electron pairs, so sp³d² hybridisation.

Memory Aids

• Hybrid orbitals count = number of orbitals mixed.
• sp = 2 orbitals, linear; sp² = 3 orbitals, trigonal planar; sp³ = 4 orbitals, tetrahedral.
• Use formula H = 1/2 (V + M – C + A) for quick hybridisation.

Hybridisation in Molecules with Bond Pairs Only

Types and Shapes

• sp: 2 hybrid orbitals, linear shape, 180° bond angle (e.g., CO₂).
• sp²: 3 hybrid orbitals, trigonal planar, 120° bond angle (e.g., BF₃).
• sp³: 4 hybrid orbitals, tetrahedral, 109.5° bond angle (e.g., CH₄).
• sp³d: 5 hybrid orbitals, trigonal bipyramidal, 90° and 120° bond angles (e.g., PCl₅).
• sp³d²: 6 hybrid orbitals, octahedral, 90° bond angle (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 is?
  • A. 90°
  • B. 109.5°
  • C. 120°
  • D. 180°
• Q2: Which molecule has trigonal bipyramidal shape?
  • A. SF₆
  • B. PCl₅
  • C. BF₃
  • D. NH₃
• Q3: Number of hybrid orbitals in octahedral geometry?
  • A. 4
  • B. 5
  • C. 6
  • D. 3

Answer Key

• Q1: B; sp³ hybridisation gives tetrahedral shape with 109.5° bond angle.
• Q2: B; PCl₅ has trigonal bipyramidal shape (sp³d).
• Q3: C; Octahedral geometry involves 6 hybrid orbitals (sp³d²).

Memory Aids

• Remember shapes by hybridisation type: sp-linear, sp²-trigonal planar, sp³-tetrahedral.
• Trigonal bipyramidal = 5 orbitals, octahedral = 6 orbitals.

Hybridisation in Molecules with Lone Pairs and Bond Pairs

Effect of Lone Pairs on Shape

• Count bond pairs (X) and lone pairs (E) around central atom (AX_mE_n).
• Predict hybridisation from total electron pairs.
• Lone pairs occupy more space, reducing bond angles.
• Actual molecular shape differs from ideal geometry due to lone pairs.
• Examples: NH₃ (trigonal pyramidal), H₂O (bent).

Solved Examples

Ex 1: Determine shape and hybridisation of NH₃.

• Bond pairs = 3, Lone pairs = 1
• Total pairs = 4 → sp³ hybridisation
• Shape: Trigonal pyramidal
• Bond angle: ~107° (less than 109.5° due to lone pair)

Practice Set

• Q1: Shape of H₂O molecule is?
  • A. Linear
  • B. Bent
  • C. Trigonal planar
  • D. T-shaped
• Q2: Hybridisation of central atom in SF₄ is?
  • A. sp³
  • B. sp³d
  • C. sp³d²
  • D. sp²
• Q3: Which molecule has T-shaped geometry?
  • A. ClF₃
  • B. PCl₅
  • C. NH₃
  • D. BF₃

Answer Key

• Q1: B; H₂O has 2 bond pairs and 2 lone pairs, bent shape.
• Q2: B; SF₄ has 5 electron pairs, sp³d hybridisation.
• Q3: A; ClF₃ has 3 bond pairs and 2 lone pairs, T-shaped.

Memory Aids

• Lone pairs reduce bond angles and alter shape.
• AX₂E₂ = bent; AX₃E = trigonal pyramidal; AX₃E₂ = T-shaped.

d-Orbital Involvement in Hybridisation

Orbital Types in Hybridisation

• d_x²-y² orbital involved in dsp² hybridisation.
• d_z² orbital involved in sp³d hybridisation.
• d_x²-y² and d_z² orbitals involved in sp³d² (d²spp³) hybridisation.
• d_xy, d_yz, d_xz orbitals involved in sp³d³ hybridisation.

Solved Examples

Ex 1: Identify d orbitals in sp³d hybridisation of PCl₅.

• Involves d_z² orbital along with s and p orbitals.

Practice Set

• Q1: Which d orbital is involved in dsp² hybridisation?
  • A. d_xy
  • B. d_x²-y²
  • C. d_z²
  • D. d_xz
• Q2: sp³d² hybridisation involves which d orbitals?
  • A. d_xy, d_yz, d_xz
  • B. d_x²-y², d_z²
  • C. d_z² only
  • D. None
• Q3: sp³d³ hybridisation includes which d orbitals?
  • A. d_xy, d_yz, d_xz
  • B. d_x²-y², d_z²
  • C. d_xy only
  • D. None

Answer Key

• Q1: B; d_x²-y² orbital is involved in dsp² hybridisation.
• Q2: B; sp³d² involves d_x²-y² and d_z² orbitals.
• Q3: A; sp³d³ includes d_xy, d_yz, d_xz orbitals.

Memory Aids

• d_x²-y² in dsp², d_z² in sp³d.
• sp³d² uses d_x²-y² and d_z².
• sp³d³ uses d_xy, d_yz, d_xz.

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 equals number of combining AOs.
• Two AOs combine to form one bonding MO and one antibonding MO.
• Bonding MOs have lower energy and greater stability than antibonding MOs.
• MOs represent electron probability distribution around nuclei.
• MOs fill according to Aufbau principle, Pauli’s exclusion, and Hund’s rule.

Formation of Molecular Orbitals (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: Determine bonding and antibonding MOs from two atomic orbitals \(\psi_A\) and \(\psi_B\).

• Bonding MO: \(\sigma = \psi_A + \psi_B\) (lower energy)
• Antibonding MO: \(\sigma^* = \psi_A - \psi_B\) (higher energy)

Practice Set

• Q1: What is the number of molecular orbitals formed when two atomic orbitals combine?
  A. One
  B. Two
  C. Three
  D. Four
• Q2: Which molecular orbital has lower energy?
  A. Bonding MO
  B. Antibonding MO
  C. Non-bonding MO
  D. Excited MO
• Q3: The antibonding molecular orbital formed by subtraction of wave functions is denoted as:
  A. σ
  B. π
  C. σ*
  D. π

Answer Key

• Q1: B. Two; each pair of atomic orbitals forms one bonding and one antibonding MO.
• Q2: A. Bonding MO; it has lower energy and greater stability.
• Q3: C. σ*; antibonding MOs are denoted with an asterisk.

Memory Aids

• "Add for bond" – addition of wave functions forms bonding MO.
• "Subtract to repel" – subtraction forms antibonding MO.
• Number of MOs = number of AOs combined.

Molecular Orbital Energy Order

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

• σ1s < σ*1s < σ2s < σ*2s < σ2p_z < π2p_x = π2p_y < π*2p_x = π*2p_y < σ*2p_z

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

• σ1s < σ*1s < σ2s < σ*2s < π2p_x = π2p_y < σ2p_z < π*2p_x = π*2p_y < σ*2p_z

Gerade (g) and Ungerade (u) Orbitals

• Gerade (g): Symmetrical about center; all σ bonding MOs and π* antibonding MOs.
• Ungerade (u): Unsymmetrical; all σ* antibonding MOs and π bonding MOs.

Solved Examples

Ex 1: Identify the symmetry of σ bonding and σ* antibonding MOs.

• σ bonding MO: gerade (g)
• σ* antibonding MO: ungerade (u)

Practice Set

• Q1: Which MO has higher energy in O₂?
  A. σ2p_z
  B. π2p_x
  C. π*2p_x
  D. σ*2p_z
• Q2: In lighter diatomic molecules, which MO comes before σ2p_z?
  A. π2p_x
  B. σ*2p_z
  C. π*2p_x
  D. σ2s
• Q3: Which symbol represents an ungerade orbital?
  A. g
  B. u
  C. *
  D. π

Answer Key

• Q1: D. σ*2p_z; antibonding orbitals have higher energy.
• Q2: A. π2p_x; π orbitals come before σ2p_z in lighter molecules.
• Q3: B. u; ungerade orbitals are denoted by 'u'.

Memory Aids

• "g" for symmetrical (gerade), "u" for unsymmetrical (ungerade).
• In lighter molecules, π orbitals fill before σ2p_z.
• Antibonding orbitals marked with asterisk (*) and often ungerade.

Bond Order and Magnetic Properties

Bond Order Calculation

• Bond order (B.O.) = ½ (N_b - N_a)
• N_b = Number of bonding electrons
• N_a = Number of antibonding electrons
• B.O. > 0: molecule is stable and exists.
• B.O. = 0 or < 0: molecule is unstable or does not exist.
• Bond order ∝ Bond energy ∝ Stability ∝ 1 / Bond length
• Higher bond order means higher bond dissociation energy and shorter bond length.

Magnetic Properties

• Diamagnetic: All electrons paired.
• Paramagnetic: Presence of unpaired electrons.

Solved Examples

Ex 1: Calculate bond order for O₂ with 10 bonding and 6 antibonding electrons.

• B.O. = ½ (10 - 6) = 2 (stable molecule)

Practice Set

• Q1: What is the bond order if N_b = 8 and N_a = 4?
  A. 1
  B. 2
  C. 3
  D. 4
• Q2: A molecule with bond order zero 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. 2; B.O. = ½ (8 - 4) = 2.
• Q2: B. Unstable; bond order zero means no stable bond.
• Q3: B. Bond dissociation energy; increases with bond order.

Memory Aids

• Bond order formula: half the difference of bonding and antibonding electrons.
• Bond order > 0 means stable molecule.
• Diamagnetic = all paired; paramagnetic = unpaired electrons.

Bonding in Homonuclear Diatomic Molecules

Molecular Orbital Configurations

• H₂: σ1s², Bond order = ½(2 – 0) = 1, diamagnetic
• H₂⁺: σ1s¹, Bond order = ½(1 – 0) = 0.5, paramagnetic
• H₂⁻: σ1s²σ*1s¹, Bond order = ½(2 – 1) = 0.5, paramagnetic
• He₂: σ1s²σ*1s², Bond order = 0, molecule does not exist
• Li₂: σ1s²σ*1s²σ2s², Bond order = 1, diamagnetic
• C₂: σ1s²σ*1s²σ2s²σ*2s²π2p_x²=π2p_y², Bond order = 2, diamagnetic
• O₂: σ1s²σ*1s²σ2s²σ*2s²σ2p_z²π2p_x²=π2p_y²π*2p_x¹=π*2p_y¹, Bond order = 2, paramagnetic
• O₂⁺: Same as O₂ but one less electron in π* orbitals, Bond order = 2.5, paramagnetic
• O₂⁻: One extra electron in π* orbitals, Bond order = 1.5, paramagnetic

Solved Examples

Ex 1: Calculate bond order of O₂ molecule.

• Electrons in bonding orbitals = 10
• Electrons in antibonding orbitals = 6
• Bond order = ½(10 – 6) = 2

Ex 2: Determine magnetic property of H₂⁺.

• Electron configuration: σ1s¹
• Unpaired electron present → paramagnetic

Practice Set

Q1: What is the bond order of He₂ molecule?

• A. 1
• B. 0
• C. 2
• D. 0.5

Q2: Which molecule is paramagnetic?

• A. H₂
• B. Li₂
• C. O₂
• D. C₂

Q3: Calculate bond order of O₂⁻ ion.

• A. 1.5
• B. 2
• C. 2.5
• D. 1

Answer Key

Q1: B. 0 — Equal bonding and antibonding electrons cancel out, molecule unstable.

Q2: C. O₂ — Has unpaired electrons in π* orbitals, paramagnetic.

Q3: A. 1.5 — Bond order = ½(10 – 7) = 1.5, paramagnetic.

Memory Aids

• Bond order = ½ (bonding electrons – antibonding electrons)
• Paramagnetic if unpaired electrons present
• He₂ does not exist due to zero bond order
• O₂ and its ions differ by electron count in π* orbitals

Metallic Bonding

Elementary Idea

• Metallic bond: electrostatic attraction between positive metal ions (kernels) and delocalized mobile electrons.
• Holds metal atoms together in a lattice.
• Strength increases with more mobile electrons and higher effective nuclear charge.

Factors Favoring Metallic Bond Formation

• Low ionization enthalpy (easy electron loss)
• Free movement of valence electrons (delocalization)
• High positive charge on cation (stronger attraction)
• Small size of cation (closer packing)

Solved Examples

Ex 1: Why is sodium a good conductor of electricity?

• Has free valence electrons that move easily, enabling conduction.

Ex 2: Explain why Mg has stronger metallic bonding than Na.

• Mg has higher charge (+2) and more delocalized electrons than Na (+1), increasing bond strength.

Practice Set

Q1: Which factor does NOT favor metallic bonding?

• A. Low ionization enthalpy
• B. Large cation size
• C. Free movement of valence electrons
• D. High cation charge

Q2: Metallic bond strength increases with:

• A. Decrease in mobile electrons
• B. Increase in effective nuclear charge
• C. Increase in cation size
• D. Increase in ionization enthalpy

Q3: Which metal likely has the strongest metallic bond?

• A. Na (1 valence electron, +1 charge)
• B. Mg (2 valence electrons, +2 charge)
• C. K (1 valence electron, +1 charge, larger size)
• D. Ca (2 valence electrons, +2 charge, larger size than Mg)

Answer Key

Q1: B. Large cation size — Larger size weakens metallic bonding.

Q2: B. Increase in effective nuclear charge — Stronger attraction to electrons.

Q3: B. Mg — Higher charge and smaller size than Ca, stronger bond.

Memory Aids

• Metallic bond = metal ions + sea of electrons
• Stronger bond: more electrons, higher charge, smaller ions
• Low ionization enthalpy metals form metallic bonds easily

Hydrogen Bonding

Definition 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 molecules, e.g., HF, H₂O, ROH.
• Intramolecular hydrogen bonding: Occurs within the same molecule, e.g., o-nitrophenol; causes cyclization and reduces association.

HF molecules show hydrogen bonds as dashed lines between H^δ+ and F^δ-. This bond increases boiling point and reduces volatility.

Structural Examples

• o-Chlorophenol and o-Nitrophenol have -OH and substituents at ortho positions on benzene ring.
• Intramolecular H-bonding shown by dashed lines between H and electronegative atoms within molecule.

Effect on Physical Properties

• Double bonds count as one electron pair in geometry prediction.
• Intermolecular H-bonding increases solubility and boiling point; decreases volatility.
• Intramolecular H-bonding decreases solubility and boiling point; increases volatility.

Solved Examples

Ex 1: Explain why HF has a higher boiling point than HCl.

• HF forms strong intermolecular hydrogen bonds due to F's high electronegativity.
• HCl lacks hydrogen bonding; only weak van der Waals forces present.
• Strong H-bonds in HF require more energy to break, raising boiling point.

Ex 2: Why does o-nitrophenol show intramolecular hydrogen bonding but o-chlorophenol does not?

• In o-nitrophenol, -OH hydrogen bonds with adjacent NO₂ oxygen within the molecule.
• In o-chlorophenol, Cl is less electronegative and less capable of intramolecular H-bonding.
• Intramolecular H-bonding causes cyclization and reduces intermolecular association.

Practice Set

Q1: Which molecule exhibits intermolecular hydrogen bonding?

• A. CH₄
• B. H₂O
• C. CO₂
• D. CCl₄

Q2: Intramolecular hydrogen bonding is most likely in:

• A. m-Nitrophenol
• B. o-Nitrophenol
• C. p-Nitrophenol
• D. Phenol

Q3: Which property increases due to intermolecular hydrogen bonding?

• A. Volatility
• B. Boiling point
• C. Intramolecular cyclization
• D. Molecular weight

Answer Key

Q1: B. H₂O — Water molecules form strong intermolecular hydrogen bonds.

Q2: B. o-Nitrophenol — Ortho position allows intramolecular H-bonding.

Q3: B. Boiling point — Intermolecular H-bonding raises boiling point by increasing intermolecular forces.

Memory Aids

• H-bond forms only with F, O, N atoms.
• Intermolecular H-bonding ↑ solubility & boiling point, ↓ volatility.
• Intramolecular H-bonding causes cyclization, ↓ solubility & boiling point, ↑ volatility.
• Ortho-substitution favors intramolecular H-bonding.