1. Chapter at a Glance
Atoms join together to reach a more stable, lower-energy arrangement. This chapter explains how they do it: ionic bonds by electron transfer, covalent bonds by sharing, and the theories that predict the shapes and properties of molecules — Lewis structures, VSEPR theory, valence bond theory with hybridisation, and molecular orbital theory. It also covers bond parameters, dipole moment, resonance and hydrogen bonding. This is one of the most important chapters of NEET Chemistry, giving two or three questions every year.
- Shape and hybridisation of molecules and ions (VSEPR).
- Bond order, magnetic nature and stability from MO theory (O₂, N₂ and their ions).
- Dipole moment and polarity; which molecules are non-polar.
- Formal charge, resonance and bond angles.
- Fajans' rules, lattice enthalpy and hydrogen bonding.
2. Octet Rule, Ionic Bond and Lewis Structures
- Kossel–Lewis approach: atoms combine to achieve a noble-gas configuration (an octet, or a duplet for H).
- Ionic bond: formed by complete transfer of electrons from a metal (low ionisation enthalpy) to a non-metal (high electron gain enthalpy). Its stability comes mainly from the lattice enthalpy released when ions pack into a crystal.
- Covalent bond: formed by sharing electron pairs. Single, double and triple bonds share one, two and three pairs.
- Formal charge = (valence electrons) − (non-bonding electrons) − ½(bonding electrons). The best Lewis structure has the smallest formal charges.
- Limitations of the octet rule: incomplete octet (BeCl₂, BF₃), odd-electron molecules (NO, NO₂), and expanded octet (PCl₅, SF₆, H₂SO₄).
An ionic bond gains covalent character when the cation is small and highly charged, the anion is large, and the cation has a pseudo-noble-gas (18-electron) configuration. So LiI is more covalent than LiF, and AlCl₃ is more covalent than NaCl.
3. Bond Parameters and Resonance
- Bond length decreases as bond order increases: C–C (154 pm) > C=C (134 pm) > C≡C (120 pm).
- Bond enthalpy increases with bond order.
- Bond angle depends on hybridisation and on lone-pair repulsion.
- Resonance: when one Lewis structure cannot describe a molecule, the real structure is a hybrid of several canonical forms. In O₃ and CO₃²⁻ all the equivalent bonds have the same length. Resonance lowers the energy of the molecule.
4. VSEPR Theory: Shapes of Molecules
Electron pairs around the central atom repel each other and arrange themselves as far apart as possible. The order of repulsion is lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.
| Bond pairs + lone pairs | Shape | Example | NEET Note |
|---|---|---|---|
| 2 + 0 | Linear | BeCl₂, CO₂ | 180° |
| 3 + 0 | Trigonal planar | BF₃ | 120° |
| 2 + 1 | Bent (V-shaped) | SO₂ | < 120° |
| 4 + 0 | Tetrahedral | CH₄, NH₄⁺ | 109.5° |
| 3 + 1 | Trigonal pyramidal | NH₃ | 107° |
| 2 + 2 | Bent | H₂O | 104.5° |
| 5 + 0 | Trigonal bipyramidal | PCl₅ | Axial bonds longer than equatorial |
| 4 + 1 | See-saw | SF₄ | Lone pair equatorial |
| 3 + 2 | T-shaped | ClF₃ | — |
| 2 + 3 | Linear | XeF₂, I₃⁻ | Three lone pairs equatorial |
| 6 + 0 | Octahedral | SF₆ | 90° |
| 5 + 1 | Square pyramidal | BrF₅ | — |
| 4 + 2 | Square planar | XeF₄ | Lone pairs opposite |
5. Valence Bond Theory and Hybridisation
- A covalent bond forms by overlap of half-filled orbitals. End-on (axial) overlap gives a σ bond; side-on overlap of p orbitals gives a weaker π bond.
- Hybridisation: orbitals of similar energy on the central atom mix to form an equal number of identical hybrid orbitals.
- Quick rule: number of hybrid orbitals = σ bonds + lone pairs on the central atom (the steric number).
| Steric number | Hybridisation | Geometry | Example |
|---|---|---|---|
| 2 | sp | Linear | BeCl₂, C₂H₂ |
| 3 | sp² | Trigonal planar | BF₃, C₂H₄ |
| 4 | sp³ | Tetrahedral | CH₄, NH₃, H₂O |
| 5 | sp³d | Trigonal bipyramidal | PCl₅ |
| 6 | sp³d² | Octahedral | SF₆ |
6. Molecular Orbital Theory
- Atomic orbitals combine (LCAO) to form bonding (lower energy) and antibonding (higher energy, marked *) molecular orbitals.
- Bond order = ½(Nb − Na). Bond order 0 means the molecule does not exist (He₂). Higher bond order means a shorter, stronger bond.
- Molecules with unpaired electrons are paramagnetic; with all electrons paired, diamagnetic.
- For B₂, C₂ and N₂ (up to 14 electrons), π2p orbitals lie below σ2p. For O₂ and F₂, σ2p lies below π2p.
| Species | Total electrons | Bond order | Magnetic nature |
|---|---|---|---|
| H₂ | 2 | 1 | Diamagnetic |
| He₂ | 4 | 0 | Does not exist |
| N₂ | 14 | 3 | Diamagnetic |
| O₂ | 16 | 2 | Paramagnetic (2 unpaired) |
| O₂⁺ | 15 | 2.5 | Paramagnetic |
| O₂⁻ | 17 | 1.5 | Paramagnetic |
| O₂²⁻ | 18 | 1 | Diamagnetic |
Bond order of oxygen species: O₂⁺ (2.5) > O₂ (2) > O₂⁻ (1.5) > O₂²⁻ (1); bond length follows the reverse order.
7. Dipole Moment and Hydrogen Bonding
- Dipole moment μ = charge × distance (unit: debye, D). It is a vector: symmetric molecules such as CO₂, BF₃, CH₄ and CCl₄ have μ = 0 even though their bonds are polar.
- NH₃ (1.47 D) has a larger dipole moment than NF₃ (0.23 D): in NH₃, the lone pair moment adds to the bond moments; in NF₃ it opposes them.
- Hydrogen bond: attraction between H bonded to F, O or N and another highly electronegative atom. Intermolecular H-bonding raises boiling points (H₂O is a liquid, H₂S a gas) and causes association (HF). Intramolecular H-bonding (o-nitrophenol) lowers the boiling point and solubility.
8. Formula Master Table
| Quantity | Formula |
|---|---|
| Formal charge | V − N − ½B |
| Steric number / hybrid orbitals | σ bonds + lone pairs |
| Quick hybridisation count | ½[V + M − c + a] |
| Bond order (MOT) | ½(Nb − Na) |
| Dipole moment | μ = q × d (1 D = 3.336 × 10⁻³⁰ C m) |
| Percentage ionic character | (μobserved/μionic) × 100 |
| Spin-only magnetism | unpaired e⁻ ⇒ paramagnetic |
In the quick count, V = valence electrons of the central atom, M = monovalent atoms attached, c = cationic charge, a = anionic charge. For example, NH₄⁺: ½(5 + 4 − 1) = 4 ⇒ sp³.
9. Master Summary Table
| Theory / idea | Explains | NEET Note |
|---|---|---|
| Lewis / octet | Simple bonding | Fails for BF₃, PCl₅, NO |
| VSEPR | Shapes and angles | lp–lp repulsion is strongest |
| Valence bond | Overlap; σ and π bonds | σ stronger than π |
| Hybridisation | Geometry | Steric number rule |
| MO theory | Bond order, magnetism | O₂ is paramagnetic |
| Dipole moment | Polarity | Symmetric ⇒ μ = 0 |
| Hydrogen bonding | High boiling points | F, O, N only |
| Fajans' rules | Covalent character | Small, highly charged cation |
10. Mind Map
- Ionic
- Covalent
- Hydrogen bond
- Lewis, formal charge
- Resonance
- Bond parameters
- VSEPR
- Hybridisation
- Dipole moment
- Bond order
- Para/diamagnetism
- Stability
11. Tips and Tricks
- For a species with total electrons 14 or fewer, bond order is quickest from the N₂ pattern; for 15–20 electrons use the O₂ pattern.
- Electrons 16 (O₂) and 10 (B₂) give paramagnetic molecules — a favourite NEET question.
- More s-character in a hybrid ⇒ larger bond angle and higher electronegativity (sp > sp² > sp³).
- Number of σ bonds = number of single bonds + one for each double and triple bond; each extra bond in a multiple bond is π.
- Calling XeF₂ bent (it is linear, with three equatorial lone pairs).
- Calling SF₄ tetrahedral (it is see-saw).
- Saying CCl₄ is polar because its C–Cl bonds are polar.
- Forgetting that O₂ is paramagnetic even though its Lewis structure shows all electrons paired.
12. Important One-liners
- A σ bond is stronger than a π bond.
- The bond angle in NH₃ (107°) is smaller than in CH₄ (109.5°) because of lone-pair repulsion.
- BF₃ is trigonal planar and non-polar.
- He₂ does not exist because its bond order is zero.
- N₂ has the highest bond order (3) among the common diatomic molecules.
- Ice is less dense than water because of its open, hydrogen-bonded structure.
- Lattice enthalpy increases with higher ionic charges and smaller ions.
- In PCl₅, the axial bonds are longer than the equatorial bonds.
13. NEET-pattern Questions with Detailed Answers
These questions are original and based on concepts that NEET UG asks again and again.
The hybridisation of carbon in ethyne (C₂H₂) is:
- sp³
- sp²
- sp
- sp³d
Show solution
Each carbon has 2 σ bonds (one to H, one to C) and no lone pairs. A steric number of 2 means sp hybridisation, which gives a linear molecule. The two remaining p orbitals form two π bonds.
Answer: (C)
Which of the following molecules is paramagnetic?
- N₂
- O₂
- F₂
- H₂
Show solution
O₂ has two unpaired electrons in its π*2p orbitals, so it is paramagnetic. N₂, F₂ and H₂ have all their electrons paired, so they are diamagnetic.
Answer: (B)
The shape of XeF₄ is:
- Tetrahedral
- Square planar
- See-saw
- Square pyramidal
Show solution
Xe has 4 bond pairs and 2 lone pairs, a steric number of 6 (sp³d²). The two lone pairs sit opposite each other, leaving a square-planar arrangement of F atoms. See-saw is the shape of SF₄ (4 + 1), and square pyramidal is BrF₅ (5 + 1).
Answer: (B)
Which molecule has zero dipole moment?
- NH₃
- H₂O
- BF₃
- CHCl₃
Show solution
BF₃ is trigonal planar and symmetric, so its three B–F bond moments cancel. NH₃ and H₂O have lone pairs and non-symmetric shapes, and CHCl₃ has unequal bonds around carbon, so all three are polar.
Answer: (C)
The correct order of bond angles is:
- H₂O > NH₃ > CH₄
- CH₄ > NH₃ > H₂O
- NH₃ > CH₄ > H₂O
- CH₄ = NH₃ = H₂O
Show solution
All three are sp³, but lone pairs squeeze the bonds together. CH₄ has no lone pairs (109.5°), NH₃ has one (107°) and H₂O has two (104.5°).
Answer: (B)
The bond order of O₂⁻ (superoxide ion) is:
- 1
- 1.5
- 2
- 2.5
Show solution
O₂⁻ has 17 electrons: 10 bonding and 7 antibonding. Bond order = ½(10 − 7) = 1.5. The value 2 is for O₂, 2.5 is for O₂⁺, and 1 is for O₂²⁻.
Answer: (B)
NH₃ has a much larger dipole moment than NF₃, although fluorine is more electronegative than hydrogen. This is because:
- NF₃ is planar
- In NF₃ the lone-pair moment opposes the bond moments
- N–F bonds are non-polar
- NH₃ has hydrogen bonds within the molecule
Show solution
Both molecules are pyramidal. In NH₃, the N–H bond moments point towards N, the same direction as the lone-pair moment, so they add. In NF₃, the N–F moments point away from N, opposite to the lone pair, so they partly cancel. NF₃ is not planar, and N–F bonds are clearly polar.
Answer: (B)
Which pair has the same shape?
- XeF₂ and CO₂
- SF₄ and CCl₄
- ClF₃ and BF₃
- H₂O and BeCl₂
Show solution
XeF₂ (2 bond pairs + 3 lone pairs) and CO₂ are both linear. SF₄ is see-saw while CCl₄ is tetrahedral. ClF₃ is T-shaped while BF₃ is trigonal planar. H₂O is bent while BeCl₂ is linear.
Answer: (A)
14. Quick Revision Summary
- Ionic ↔ lattice enthalpy; Fajans: small, highly charged cation ⇒ covalent.
- VSEPR: lp–lp > lp–bp > bp–bp; learn the 13 shapes.
- Steric number 2, 3, 4, 5, 6 ⇒ sp, sp², sp³, sp³d, sp³d².
- Bond order ½(Nb − Na); O₂ paramagnetic; π2p below σ2p up to N₂.
- Symmetric molecules μ = 0; H-bonds need H on F, O or N.