Chemical Bonding & Molecular Structure
Detailed Class 11 Notes for CBSE | ISC | WBCHSE – Principles, Types, Shapes, Theories, MOT, Practice MCQs
1. Introduction
Chemical Bond: A force of attraction that holds two or more atoms together in a molecule or compound, making them behave as a single unit.
Understanding bond formation explains why atoms combine, molecular shapes, and physical properties of substances.
2. Types of Chemical Bonds
(a) Ionic (Electrovalent) Bond
- Formed by complete transfer of electrons from a metal to a non-metal.
- Results in ions held by strong electrostatic forces.
Example: NaCl (Na → Na+ + e–; Cl + e– → Cl–)
Ionic compounds are usually crystalline solids with high melting points and conduct electricity in molten/solution state.
(b) Covalent Bond
- Formed by mutual sharing of electrons between non-metal atoms to complete their octet.
- Single, double, or triple bonds: H2, O2, N2
Cl2: Cl· + ·Cl → Cl:Cl (single covalent bond)
Covalent compounds are generally soft, low melting, and poor conductors.
(c) Coordinate (Dative) Bond
- Both bonding electrons are supplied by one atom (donor).
- Example: NH4+ formation, SO2, O3
3. Octet Rule and Its Limitations
- Atoms attain stability by acquiring 8 electrons in their valence shell (duplet for H, He).
- Failures: Odd electron species (NO), expanded octet (PF5), incomplete octet (BF3)
[Insert Lewis structures: BF3, PF5, NO Here]
4. Lewis Structures and Resonance
- Lewis structure: Shows valence electrons as dots/crosses. Shared pair = line; lone pair = dots.
- Formal Charge: The charge assigned to an atom in a given Lewis structure
Formal charge = (Valence e–) – (Lone e–) – ½ (Bonding e–) - Resonance: When two or more valid structures are possible; real molecule is a hybrid.
[CO32- resonance structures]
5. Theories of Chemical Bonding
5.1 Valence Bond Theory (VBT)
- Bonds form when atomic orbitals overlap; electron pairs localized between atoms.
- Types: Sigma (σ) – head-on overlap; Pi (π) – sidewise overlap.
[Sigma and Pi bond formation diagrams]
5.2 Hybridization
- Mixing of atomic orbitals to form new hybrid orbitals of equal energy.
sp, sp2, sp3 etc. for linear, trigonal planar, tetrahedral shapes.
Examples: CH4 – sp3, BF3 – sp2, BeCl2 – sp
5.3 Molecular Orbital Theory (MOT)
MOT was developed by Hund and Mulliken and explains bonding through the formation of molecular orbitals extending over the entire molecule, not just between two atoms.
- Atomic orbitals combine to form Molecular Orbitals (MOs): The number of MOs formed equals the number of atomic orbitals combined.
- Bonding vs. Antibonding:
- Bonding MOs (lower energy): Constructive overlap, electron density between nuclei increases stability (denoted σ, π).
- Antibonding MOs (higher energy): Destructive overlap, electron density between nuclei decreases (denoted σ*, π*).
- Filling of MOs:
- MOs are filled in order of increasing energy (Aufbau principle), with each orbital holding 2 electrons with opposite spins.
- Bonding orbitals are filled before antibonding orbitals.
- Types of Molecular Orbitals:
- σ (sigma): Along internuclear axis (head-on overlap of s-s, s-p, or p-pz).
- π (pi): Side-to-side overlap of p orbitals (p-px, p-py).
Bond Order (BO) = ½ × [Number of electrons in bonding MOs − Number in antibonding MOs]
BO = ( Nbonding − Nantibonding )/2
BO = ( Nbonding − Nantibonding )/2
Energy Level Diagram for Homonuclear Diatomics
[Insert Molecular Orbital Energy Level Diagram for B2 to N2 and O2 to F2 here]
- Order of orbital energies:
For Z ≤ 7 (Li2, Be2, B2, C2, N2):
σ1s < σ*1s < σ2s < σ*2s < π2px = π2py < σ2pz < π*2px = π*2py < σ*2pz -
For Z ≥ 8 (O2, F2, Ne2):
σ1s < σ*1s < σ2s < σ*2s < σ2pz < π2px = π2py < π*2px = π*2py < σ*2pz
Applications of MOT
- Explains bond order, magnetism, and stability of molecules.
- Paramagnetism of O2: O2 has two unpaired electrons in π*2p MOs, explaining its observed paramagnetism, which cannot be explained by VBT.
- Predicts existence/non-existence of molecules (e.g., He2 does not exist as BO=0).
Examples:
For O2:
Configuration (Z=8):
σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 π2py2 π*2px1 π*2py1
Bonding electrons = 8, Antibonding = 4
BO = (8 − 4)/2 = 2
For O2:
Configuration (Z=8):
σ1s2 σ*1s2 σ2s2 σ*2s2 σ2pz2 π2px2 π2py2 π*2px1 π*2py1
Bonding electrons = 8, Antibonding = 4
BO = (8 − 4)/2 = 2
Summary of Bond Order and Magnetism using MOT:
N2 (BO=3) – Diamagnetic
O2 (BO=2) – Paramagnetic
O2– (BO=1.5) – Paramagnetic (1 unpaired)
He2 (BO=0) – Does not exist
N2 (BO=3) – Diamagnetic
O2 (BO=2) – Paramagnetic
O2– (BO=1.5) – Paramagnetic (1 unpaired)
He2 (BO=0) – Does not exist
[Insert diagrams showing electron filling in MOs for N2, O2, He2]
6. Polarity and Dipole Moment
- Electronegativity: Tendency to attract shared electrons (Pauling scale).
- Covalent bonds with differing EN create dipole moments.
- μ (dipole moment) = q × r, q = charge, r = bond length (unit: Debye)
Water is highly polar due to bent shape; CO2 is nonpolar despite polar bonds (linear geometry cancels dipoles).
7. Molecular Shapes: VSEPR Theory
- Electron pairs (bonding and lone) arrange to minimize mutual repulsion.
- Lone pairs occupy more space than bond pairs.
- Common geometries:
- Tetrahedral (sp3): 4 BP, 0 LP – CH4
- Pyramidal: 3 BP, 1 LP – NH3
- Angular/Bent: 2 BP, 2 LP – H2O
- Trigonal Planar (sp2): 3 BP, 0 LP – BF3
- Linear (sp): 2 BP – BeCl2, CO2
[VSEPR shapes: CH4, NH3, H2O, BF3, BeCl2]
8. Formal Charge and Oxidation Number
- Formal charge: theoretical charge in a Lewis structure.
- Oxidation number: actual/ionic charge; often differs from formal charge.
9. Bond Parameters
- Bond length: Distance between nuclei; shorter for higher bond order.
- Bond angle: Angle between two bonds at the atom.
- Bond order: Higher = stronger, shorter bond. Order: single (1), double (2), triple (3).
- Bond energy: Energy to break one mole of bonds in gas phase.
Increasing bond order increases bond energy and decreases bond length.
10. Exceptions and Special Cases
- Expanded Octet in PF5, SF6
- Odd Electron Molecule: NO, NO2
- Incomplete Octet: BeCl2, BF3
- Back bonding: Seen in BF3, CO
Always analyze exceptions to predict molecular structures accurately in exams!
11. Practice MCQs (Answer & Explanation Below)
| MCQ | Options |
|---|---|
| 1. Which of the following forms an ionic bond? | a) H2 b) O2 c) NaCl d) Cl2 |
| 2. The number of lone pairs in NH3 is: | a) 0 b) 1 c) 2 d) 3 |
| 3. Shape of BF3 molecule is: | a) Tetrahedral b) Trigonal planar c) Bent d) Linear |
| 4. Which is a polar molecule? | a) CO2 b) CH4 c) H2 d) H2O |
| 5. The bond order of O2 molecule is: | a) 1 b) 2 c) 3 d) 4 |
| 6. Bond angle in water (H2O) is: | a) 90° b) 104.5° c) 120° d) 180° |
| 7. Which has a coordinate bond? | a) H2 b) NH4+ c) O2 d) H2O |
| 8. Which molecule has an expanded octet? | a) BH3 b) PF5 c) CO2 d) CH4 |
| 9. Valence electrons in BeCl2? | a) 4 b) 6 c) 8 d) 10 |
| 10. Resonance occurs in: | a) CH4 b) SO3 c) NH3 d) BeCl2 |
| 11. Which is not explained by VSEPR? | a) Lone pairs effect b) d-p π bonding c) Hybridization d) Bond length |
| 12. Which is paramagnetic? | a) N2 b) O2 c) CO2 d) HCl |
| 13. Molecule with highest bond order: | a) O2 b) N2 c) F2 d) H2 |
| 14. Which hybridization leads to linear shape? | a) sp b) sp2 c) sp3 d) d2sp3 |
| 15. Bond angle in CH4 is: | a) 90° b) 104.5° c) 120° d) 109.5° |
| 16. The term ‘Lewis acid’ refers to: | a) Proton donor b) Electron donor c) Electron pair acceptor d) None |
| 17. Example of molecule with bent shape: | a) CO2 b) H2O c) BeF2 d) BF3 |
| 18. Which does not obey the octet rule? | a) CH4 b) NH3 c) BeCl2 d) H2O |
| 19. The correct electron-dot structure for NO3–: | a) Single N−O bond only b) Double bonds only c) Resonance hybrid d) None |
| 20. Bond order for C≡N in HCN: | a) 1 b) 2 c) 3 d) 4 |
| 21. Presence of two unpaired electrons in O2 is explained by: | a) Lewis theory b) VBT c) MOT d) VSEPR |
| 22. Which has maximum number of lone pairs on central atom? | a) NH3 b) H2O c) BeCl2 d) PF5 |
| 23. CO2 is nonpolar because: | a) C has no electronegativity b) Shape is linear c) Bonds not polar d) None |
| 24. A triple bond consists of: | a) 3 σ bond b) 2 σ + 1 π c) 1 σ + 2 π d) 3 π |
| 25. The formal charge on O in O3 (central O): | a) +1 b) 0 c) -1 d) +2 |
| 26. The maximum covalency of nitrogen is: | a) 3 b) 4 c) 5 d) 6 |
| 27. The energy required to break a bond is: | a) Bond energy b) Ionization enthalpy c) Electron affinity d) Lattice energy |
| 28. Molecule with one lone pair and three bond pairs: | a) NH3 b) CH4 c) H2O d) BF3 |
| 29. Which is a planar molecule? | a) NH3 b) CH4 c) BeCl2 d) H2O |
| 30. Back bonding is seen in: | a) C2H4 b) BF3 c) H2O d) NH3 |
Answers & Explanations to MCQs
- c) NaCl
Because Na (metal) transfers electron to Cl (non-metal); forms strong ionic bond. - b) 1
NH3 has three bond pairs and one lone pair on N. - b) Trigonal planar
BF3 is sp2 hybridized (AX3), 120° angle. - d) H2O
Bent molecule with polar bonds, net dipole; CO2 is linear so overall nonpolar. - b) 2
O2 bond order from MOT is 2. - b) 104.5°
- b) NH4+
One bond (from N to H+) is coordinate. - b) PF5
Central P involves 10 valence electrons (expanded octet). - a) 4
Be(2e) + two Cl(1e each) = 4 valence electrons around Be. - b) SO3
SO3, like CO32-, exhibits resonance. - b) d-p π bondingVSEPR only deals with electron pair repulsions, not d-p π bonding.
- b) O2
MOT: O2 has unpaired electrons, so paramagnetic. - b) N2
N2 bond order is 3 (triple bond). - a) sp
sp hybridization yields 180° bond angle; linear shape (e.g., BeCl2, CO2). - d) 109.5°
- c) Electron pair acceptor
Lewis acid accepts an electron pair. - b) H2O
Bent shape; two lone pairs on O. - c) BeCl2
Be has only 4 electrons in its valence shell (violates octet). - c) Resonance hybrid
NO3– has 3 equivalent resonance structures. - c) 3
C≡N bond is a triple bond. - c) MOT
Molecular Orbital Theory explains the paramagnetism of O2. - b) H2O
2 bond pairs, 2 lone pairs on O. - b) Shape is linear
CO2 has polar bonds but is linear, dipoles cancel. - c) 1 σ + 2 π
Triple bond includes 1 sigma and 2 pi bonds (as in N2 or C≡N). - a) +1
For central O in O3; see formal charge calculation. - c) 5
Maximum covalency (e.g., in NH4+, N2O5 etc.) - a) Bond energy
- a) NH3
NH3: 3 bond pairs + 1 lone pair (on N). - c) BeCl2
BeCl2 is linear and planar. - b) BF3
B in BF3 gets electron density from F by back bonding.