Introduction
Why do atoms combine? Atoms bond together to reach a lower energy, more stable state. The driving force is the tendency to achieve a stable noble gas configuration. Chemical bonding is one of the most important chapters in NEET Chemistry, contributing 3 to 5 questions most years, making it the highest-weightage Class 11 chemistry chapter.
The most tested areas are hybridisation (identifying sp, sp², sp³, etc. for given molecules), VSEPR shapes, bond angles, MO theory bond order and magnetism, and hydrogen bonding effects.
Ionic Bonding
An ionic bond forms when one atom transfers electrons to another, creating oppositely charged ions (cation and anion) that attract each other electrostatically. This typically happens between a metal (low IE, gives electrons) and a non-metal (high electron affinity, accepts electrons).
Example: Na (2,8,1) loses 1 electron to form Na⁺ (2,8). Cl (2,8,7) gains 1 electron to form Cl⁻ (2,8,8). Both achieve the octet. NaCl forms with a strong electrostatic attraction.
Lattice energy is the energy released when one mole of an ionic compound is formed from its gaseous ions. Higher lattice energy means a more stable, harder ionic compound with a higher melting point.
Covalent Bonding and Lewis Structures
A covalent bond forms when two atoms share electron pairs. Lewis dot structures show the valence electrons of atoms and the bonds between them.
Drawing Lewis Structures
- Count the total valence electrons of all atoms in the molecule.
- Arrange atoms: the least electronegative atom is usually the central atom (except H, which is always terminal).
- Connect atoms with single bonds (2 electrons each). Subtract those electrons from the total.
- Complete octets of outer (terminal) atoms first with lone pairs.
- Place remaining electrons on the central atom.
- If the central atom has fewer than 8 electrons, form double or triple bonds using lone pairs from terminal atoms.
Common Lewis Structures to Know
| Molecule | Valence electrons | Bonds | Lone pairs on central atom |
|---|---|---|---|
| H₂O | 8 | 2 O-H single bonds | 2 lone pairs on O |
| NH₃ | 8 | 3 N-H single bonds | 1 lone pair on N |
| CO₂ | 16 | 2 C=O double bonds | 0 lone pairs on C |
| N₂ | 10 | 1 N≡N triple bond | 1 lone pair on each N |
| PCl₅ | 40 | 5 P-Cl single bonds | 0 lone pairs on P |
| SF₆ | 48 | 6 S-F single bonds | 0 lone pairs on S |
Formal Charge
Formal charge helps identify the most stable Lewis structure. It is the hypothetical charge on an atom in a molecule assuming electrons are shared equally.
The best Lewis structure is the one where formal charges are minimised and any negative formal charge is on the more electronegative atom.
Example: In CO₂ with the structure O=C=O, formal charges: C = 4 − 0 − ½(8) = 0. O = 6 − 4 − ½(4) = 0. All zero, so this is the best structure.
Exceptions to the Octet Rule
1. Electron-deficient molecules (incomplete octet)
Some molecules have a central atom with fewer than 8 electrons in their Lewis structure.
- BF₃: B has only 6 electrons. B can accept electron pairs (Lewis acid).
- AlCl₃: Al has only 6 electrons. Forms dimers (Al₂Cl₆) to overcome this.
- BeH₂: Be has only 4 electrons.
2. Odd-electron molecules (radical species)
Molecules with an odd number of electrons cannot satisfy the octet for all atoms.
- NO (11 electrons): nitrogen has 7 electrons around it; it is a radical.
- NO₂ (17 electrons): odd-electron molecule.
- ClO₂: another example.
3. Expanded octet (more than 8 electrons)
Elements from Period 3 and beyond can accommodate more than 8 electrons around the central atom by using d orbitals.
- PCl₅: P has 10 electrons (5 bonds × 2). Hybridisation sp³d.
- SF₆: S has 12 electrons (6 bonds × 2). Hybridisation sp³d².
- ClF₃: Cl has 10 electrons (3 bonds + 2 lone pairs × 2).
- XeF₄: Xe has 12 electrons. Hybridisation sp³d².
Note: Period 2 elements (C, N, O, F) cannot expand their octet because they have no available d orbitals in their valence shell.
Resonance
When a single Lewis structure cannot adequately represent a molecule because the electrons are delocalized over multiple bonds, we draw multiple resonance structures. The actual molecule is a hybrid (superposition) of all resonance structures.
Key examples:
- Ozone (O₃): Two resonance structures (double bond on left or right oxygen). The actual bond length (128 pm) is intermediate between single (148 pm) and double (121 pm).
- Carbonate ion (CO₃²⁻): Three equivalent resonance structures. Bond order = 4/3.
- Benzene (C₆H₆): Two Kekulé structures. All C-C bonds are equivalent (bond length = 139 pm, between single and double). All bond angles = 120°.
- NO₃⁻ (nitrate): Three resonance structures. Bond order = 4/3.
- SO₃: Three resonance structures. Bond order = 4/3.
Resonance stabilises molecules. The actual energy is lower than any single resonance structure. This extra stability is called resonance energy (or delocalization energy). Benzene's resonance energy is about 150 kJ/mol.
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VSEPR Theory and Molecular Geometry
Select the number of bonding pairs and lone pairs on the central atom to predict the molecular geometry.
Bonding pairs: 4
Lone pairs: 0
Molecular geometry
Tetrahedral
Hybridization
sp³
Bond angle
109.5°
Description
Four bonding pairs pointing to the corners of a tetrahedron.
Steric number: 4 (4 bonding + 0 lone)
Examples: CH₄, NH₄⁺, CCl₄
Valence Shell Electron Pair Repulsion (VSEPR) theory predicts the 3D shape of a molecule based on repulsions between electron pairs (both bonding pairs and lone pairs) around the central atom.
Key rules:
- Electron pairs (bonding or lone) arrange to minimise repulsion.
- Lone pair (LP) repulsion > bonding pair (BP) repulsion. So lone pairs compress bond angles.
- LP-LP > LP-BP > BP-BP in repulsion strength.
- Double bonds are treated as a single region of high electron density for VSEPR purposes.
VSEPR Shapes Table
| Total e⁻ pairs | Bonding pairs | Lone pairs | Shape | Bond angle | Example |
|---|---|---|---|---|---|
| 2 | 2 | 0 | Linear | 180° | BeCl₂, CO₂ |
| 3 | 3 | 0 | Trigonal planar | 120° | BF₃, SO₃ |
| 3 | 2 | 1 | Bent (V-shaped) | < 120° | SO₂, SnCl₂ |
| 4 | 4 | 0 | Tetrahedral | 109.5° | CH₄, SiCl₄ |
| 4 | 3 | 1 | Trigonal pyramidal | 107° | NH₃, PCl₃ |
| 4 | 2 | 2 | Bent (V-shaped) | 104.5° | H₂O, H₂S |
| 5 | 5 | 0 | Trigonal bipyramidal | 90°, 120° | PCl₅ |
| 5 | 4 | 1 | See-saw | < 90°, <120° | SF₄ |
| 5 | 3 | 2 | T-shaped | < 90° | ClF₃ |
| 5 | 2 | 3 | Linear | 180° | XeF₂, I₃⁻ |
| 6 | 6 | 0 | Octahedral | 90° | SF₆ |
| 6 | 5 | 1 | Square pyramidal | ≈ 90° | IF₅, BrF₅ |
| 6 | 4 | 2 | Square planar | 90° | XeF₄, [PtCl₄]²⁻ |
Bond Angle Compression by Lone Pairs
Lone pairs take up more space than bonding pairs, compressing adjacent bond angles.
- CH₄: 109.5° (no lone pairs on C)
- NH₃: 107° (1 lone pair on N compresses from 109.5°)
- H₂O: 104.5° (2 lone pairs on O compress further)
Valence Bond Theory
The Valence Bond (VB) theory explains bonding as the overlap of atomic orbitals.
- Sigma (σ) bond: Formed by head-on (axial) overlap of orbitals. Electron density is concentrated along the internuclear axis. Allows free rotation around the bond.
- Pi (π) bond: Formed by lateral (sideways) overlap of p orbitals above and below the internuclear axis. Restricts rotation. Found only in double and triple bonds (in addition to sigma bonds).
Every single bond is one sigma bond. A double bond = 1 sigma + 1 pi. A triple bond = 1 sigma + 2 pi.
Bond strength: σ > π (because end-on overlap is more effective than lateral).
Hybridisation
Select two elements to calculate their electronegativity difference and classify the bond.
Element A
EN = 2.20
—
Element B
EN = 3.98
Bond type
Ionic
The electronegativity difference is large. One atom effectively transfers its electron(s) to the other, forming ions. The bond is predominantly electrostatic.
|EN(H) − EN(F)| = |2.20 − 3.98| = 1.78
F is more electronegative (EN = 3.98). It pulls bonding electrons toward itself.
H +
F −
Arrow points toward the more electronegative atom
Classification thresholds
EN diff < 0.5: Nonpolar covalent
0.5 ≤ EN diff ≤ 1.7: Polar covalent
EN diff > 1.7: Ionic
Select a compound to see hybridization, sigma/pi bond count, lone pairs, and geometry of the central atom.
Central atom: Be in BeCl₂
Hybridization
sp
Geometry
Linear
Bond angle
180°
Sigma (σ) bonds
2
Pi (π) bonds
0
Lone pairs (central)
0
Key point
Be uses one s and one p orbital. Electron-deficient (only 4 electrons around Be).
Quick reference: steric number rule
Steric number = bonding pairs + lone pairs on central atom.
SN=2 → sp (linear) | SN=3 → sp² (planar) | SN=4 → sp³ (tetrahedral)
SN=5 → sp³d | SN=6 → sp³d²
All single bonds are sigma (σ). Double bond = 1σ + 1π. Triple bond = 1σ + 2π.
Hybridisation is the mixing of atomic orbitals to form a new set of equivalent orbitals called hybrid orbitals. Hybrid orbitals form sigma bonds and hold lone pairs. They do NOT form pi bonds.
sp Hybridisation
One s + one p orbital mix to form 2 sp hybrid orbitals. The 2 sp orbitals point in opposite directions (180°), giving linear geometry.
| Property | Value |
|---|---|
| Number of hybrid orbitals | 2 |
| Geometry | Linear |
| Bond angle | 180° |
| Unhybridised p orbitals | 2 (used for pi bonds) |
| Examples | BeCl₂, C₂H₂ (acetylene), CO₂, HCN, C in allenes |
sp² Hybridisation
One s + two p orbitals mix to form 3 sp² hybrid orbitals. They lie in a plane at 120° to each other.
| Property | Value |
|---|---|
| Number of hybrid orbitals | 3 |
| Geometry | Trigonal planar |
| Bond angle | 120° |
| Unhybridised p orbitals | 1 (for pi bond) |
| Examples | BF₃, C₂H₄ (ethylene), SO₃, CO₃²⁻, benzene (C₆H₆), graphite carbon |
sp³ Hybridisation
One s + three p orbitals mix to form 4 sp³ hybrid orbitals pointing to the corners of a tetrahedron at 109.5° (when all are bonding pairs).
| Property | Value |
|---|---|
| Number of hybrid orbitals | 4 |
| Geometry | Tetrahedral (all BP), trigonal pyramidal (1 LP), bent (2 LP) |
| Bond angle | 109.5° (CH₄), 107° (NH₃), 104.5° (H₂O) |
| Unhybridised p orbitals | 0 |
| Examples | CH₄, NH₃, H₂O, CCl₄, SiCl₄, PCl₃, H₂S, diamond carbon |
sp³d Hybridisation
One s + three p + one d orbital mix to form 5 sp³d orbitals arranged in a trigonal bipyramidal geometry.
- Examples: PCl₅, SF₄ (1 LP → see-saw), ClF₃ (2 LP → T-shape), XeF₂ (3 LP → linear)
- Note: in PCl₅, the 3 equatorial bonds (120°) and 2 axial bonds (90°) are not identical.
sp³d² Hybridisation
One s + three p + two d orbitals mix to form 6 sp³d² orbitals in an octahedral arrangement.
- Examples: SF₆, [PtCl₆]²⁻, IF₅ (1 LP → square pyramidal), XeF₄ (2 LP → square planar)
sp³d³ Hybridisation
- 7 orbitals → pentagonal bipyramidal geometry
- Example: IF₇
Quick Reference: Steric Number and Hybridisation
Steric number = number of bonding pairs + number of lone pairs on the central atom.
| Steric number | Hybridisation | Geometry |
|---|---|---|
| 2 | sp | Linear |
| 3 | sp² | Trigonal planar |
| 4 | sp³ | Tetrahedral |
| 5 | sp³d | Trigonal bipyramidal |
| 6 | sp³d² | Octahedral |
| 7 | sp³d³ | Pentagonal bipyramidal |
Bond Parameters
Bond Length
The equilibrium distance between the nuclei of two bonded atoms. Triple < double < single bond (for the same pair of atoms). Larger atoms form longer bonds.
- C-C: 154 pm. C=C: 134 pm. C≡C: 120 pm.
- Bond length decreases as bond order increases.
Bond Angle
The angle between two bonds at the central atom. Depends on hybridisation and number of lone pairs. Lone pairs reduce bond angles; electronegativity of substituents also affects angles slightly.
Bond Enthalpy
Energy required to break one mole of bonds in gaseous molecules. Higher bond order → higher bond enthalpy → stronger bond.
- C-C: 346 kJ/mol. C=C: 614 kJ/mol. C≡C: 839 kJ/mol.
- N-N: 163 kJ/mol. N=N: 418 kJ/mol. N≡N: 946 kJ/mol (very strong; N₂ is very stable).
- H-H: 436 kJ/mol. F-F: 155 kJ/mol (weak because of lone pair repulsion in F₂).
Bond Order
The number of bonds between two atoms. Single bond: order 1. Double bond: order 2. Triple bond: order 3. For resonance structures, fractional bond orders are possible.
Molecular Orbital (MO) Theory
MO theory treats electrons as belonging to the whole molecule, not individual atoms. Atomic orbitals combine to form molecular orbitals that are spread over the entire molecule.
Formation of Molecular Orbitals
When two AOs combine, they form two MOs:
- Bonding MO (σ or π): formed by constructive interference. Lower energy than the original AOs. Electrons in bonding MOs stabilise the molecule.
- Antibonding MO (σ* or π*): formed by destructive interference. Higher energy than the original AOs. Electrons in antibonding MOs destabilise the molecule.
MO Energy Level Diagram for Homonuclear Diatomic Molecules
For O₂ and F₂ (and heavier):
σ1s < σ*1s < σ2s < σ*2s < σ2p < π2p = π2p < π*2p = π*2p < σ*2p
For Li₂ to N₂ (lighter, where 2s-2p mixing occurs):
σ1s < σ*1s < σ2s < σ*2s < π2p = π2p < σ2p < π*2p = π*2p < σ*2p
Key difference: σ2p and π2p swap positions.
Bond Order in MO Theory
where Nb = electrons in bonding MOs, Na = electrons in antibonding MOs.
Bond order > 0: stable molecule. Bond order = 0: molecule does not exist. Higher bond order → shorter bond length → higher bond enthalpy.
Important MO Configurations and Properties
| Species | Total electrons | Bond order | Magnetic property | Stability |
|---|---|---|---|---|
| H₂ | 2 | 1 | Diamagnetic | Stable |
| He₂ | 4 | 0 | Diamagnetic | Does not exist |
| He₂⁺ | 3 | 0.5 | Paramagnetic | Unstable but exists momentarily |
| Li₂ | 6 | 1 | Diamagnetic | Stable |
| B₂ | 10 | 1 | Paramagnetic | Stable |
| C₂ | 12 | 2 | Diamagnetic | Stable |
| N₂ | 14 | 3 | Diamagnetic | Very stable |
| O₂ | 16 | 2 | Paramagnetic (2 unpaired e⁻) | Stable |
| F₂ | 18 | 1 | Diamagnetic | Stable |
| Ne₂ | 20 | 0 | Diamagnetic | Does not exist |
| NO | 15 | 2.5 | Paramagnetic | Stable |
| O₂⁺ | 15 | 2.5 | Paramagnetic | Bond length < O₂ |
| O₂⁻ | 17 | 1.5 | Paramagnetic | Bond length > O₂ |
| O₂²⁻ | 18 | 1 | Diamagnetic | Weakest O-O |
Paramagnetism means unpaired electrons are present (attracted to a magnetic field).Diamagnetism means all electrons are paired (weakly repelled).
Hydrogen Bonding
A hydrogen bond forms when a hydrogen atom covalently bonded to a highly electronegative atom (N, O, or F) experiences strong electrostatic attraction to another electronegative atom with a lone pair. It is not a covalent bond but is much stronger than van der Waals forces.
Types of Hydrogen Bonds
- Intermolecular hydrogen bonds: Between different molecules. Increases boiling point, melting point, viscosity, and surface tension. Examples: H₂O, HF, NH₃, alcohols, carboxylic acids.
- Intramolecular hydrogen bonds: Within the same molecule. Decreases boiling point (as intermolecular H-bonds are prevented). Examples: salicylaldehyde, o-nitrophenol, o-chlorophenol.
Effects of Hydrogen Bonding
- Water (H₂O) has an anomalously high boiling point (100°C) for its molecular mass, due to extensive H-bonding.
- HF has a higher boiling point than HCl despite lower molecular mass (H-bonding in HF).
- NH₃ has a higher boiling point than PH₃ (H-bonding in NH₃).
- Ice is less dense than liquid water because H-bonds in ice create an open cage structure with more empty space.
- Ortho-nitrophenol is more volatile (lower bp) than meta- and para-nitrophenol because it forms intramolecular H-bonds, reducing intermolecular associations.
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Worked NEET Problems
NEET-style problem · VSEPR and Shape
Question
Predict the shape and bond angle of XeF₄. Xenon has atomic number 54.
Solution
Step 1: Valence electrons on Xe = 8. Total from 4 F atoms contributing to bonds = 4 bonds = 4 × 2 = 8 bonding electrons. Total valence electrons = 8 (Xe) + 4 × 7 (F) = 36. Xe uses 8 in bonds; remainder on Xe = 36 − 8 − 4 × 6 = 36 − 8 − 24 = 4 electrons = 2 lone pairs.
Step 2: Total electron pairs on Xe = 4 (bonding) + 2 (lone) = 6. Hybridisation = sp³d².
Step 3: With 6 electron pairs in octahedral arrangement, 2 lone pairs prefer the equatorial positions (or opposite positions) to minimise repulsion. The 4 F atoms are in a square plane. Shape = square planar. Bond angle = 90°.
NEET-style problem · MO Theory: Bond Order
Question
Find the bond order and magnetic character of O₂ using MO theory.
Solution
O₂ has 16 electrons. MO filling (for O₂, using O/F order): σ1s² σ*1s² σ2s² σ*2s² σ2p² π2p² π2p² π*2p¹ π*2p¹ (2 electrons in antibonding π*).
Nb = 8 (σ1s², σ2s², σ2p², π2p², π2p²). Wait, let me recount including all core: σ1s² (2), σ*1s² (2), σ2s² (2), σ*2s² (2), σ2p² (2), π2p² π2p² (4), π*2p¹ π*2p¹ (2). Nb = 2+2+2+4 = 10. Na = 2+2+2 = 6. Bond order = (10−6)/2 = 2.
The two π* electrons go into separate π* orbitals with parallel spins (Hund's rule). Two unpaired electrons → O₂ is paramagnetic. This was one of MO theory's great successes: VB theory wrongly predicts O₂ to be diamagnetic.
NEET-style problem · Hybridisation
Question
Identify the hybridisation of the central atom in NH₃, BF₃, and PCl₅.
Solution
NH₃: N has 5 valence e⁻. Forms 3 bonds (3 bonding pairs) + 1 lone pair = 4 total electron pairs. Steric number = 4. Hybridisation = sp³. Shape = trigonal pyramidal.
BF₃: B has 3 valence e⁻. Forms 3 bonds + 0 lone pairs = 3 total electron pairs. Steric number = 3. Hybridisation = sp². Shape = trigonal planar.
PCl₅: P has 5 valence e⁻. Forms 5 bonds + 0 lone pairs = 5 total electron pairs. Steric number = 5. Hybridisation = sp³d. Shape = trigonal bipyramidal.
Summary Cheat Sheet
| Concept | Key Formula / Fact |
|---|---|
| Formal charge | FC = V − N − B/2 (V=valence e⁻, N=lone pair e⁻, B=bonding e⁻) |
| VSEPR: lone pairs compress | Each lone pair reduces bond angle by ~2.5° |
| Hybridisation | Steric number 2=sp, 3=sp², 4=sp³, 5=sp³d, 6=sp³d² |
| Sigma vs Pi | All single bonds = σ. Double = 1σ+1π. Triple = 1σ+2π |
| Bond order (MO) | (Nb − Na)/2 |
| O₂ | Bond order = 2; paramagnetic (2 unpaired e⁻ in π*) |
| N₂ | Bond order = 3; diamagnetic |
| Hydrogen bond partners | N, O, F only (highly electronegative small atoms) |
| H-bond: intermolecular | Raises bp, mp (H₂O, HF, NH₃) |
| H-bond: intramolecular | Lowers bp; e.g. o-nitrophenol vs p-nitrophenol |
| Resonance: benzene | 6 π electrons; all C-C bonds equal (139 pm); sp² C |
| Exceptional shapes | XeF₄: square planar; ClF₃: T-shaped; XeF₂: linear |
Frequently asked questions
How do I quickly find the hybridisation of an atom in a molecule?
Calculate the steric number = (number of atoms bonded to the central atom) + (number of lone pairs on the central atom). Steric number 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d². For example, in H₂O, O has 2 bonds + 2 lone pairs = steric number 4 = sp³. In SO₂, S has 2 bonds + 1 lone pair = steric number 3 = sp².
Why is O₂ paramagnetic according to MO theory but appears diamagnetic in Lewis structures?
Lewis structures cannot capture this. In MO theory, the 16 electrons of O₂ fill up to the π*2p level. There are two degenerate π* orbitals, and by Hund's rule, each gets one electron with parallel spins. Two unpaired electrons means O₂ is paramagnetic. This was experimentally confirmed: liquid O₂ is attracted to magnets. It was one of the early triumphs of MO theory over VB theory.
What is the difference between bond order and bond length?
Bond order is the number of bonds between two atoms (can be fractional in resonance/MO theory). Bond length is the physical distance between the nuclei. Higher bond order means stronger, shorter bond. For example: C-C (order 1, length 154 pm), C=C (order 2, length 134 pm), C≡C (order 3, length 120 pm). In benzene, the C-C bond order is 1.5 and the bond length is 139 pm, between single and double.
What is the shape of NH₃ vs H₂O and why are the bond angles different?
Both N and O in NH₃ and H₂O are sp³ hybridised. NH₃ has 3 bonding pairs and 1 lone pair on N. Bond angle = 107° (lone pair pushes bonding pairs slightly closer). H₂O has 2 bonding pairs and 2 lone pairs on O. The 2 lone pairs push the bonding pairs further together, giving bond angle = 104.5°.
How many sigma and pi bonds are in a molecule of HCN?
H-C≡N: H-C is a single bond = 1 sigma. C≡N is a triple bond = 1 sigma + 2 pi. Total: 2 sigma bonds + 2 pi bonds. The C is sp hybridised (1 sigma bond to H, 1 sigma + 2 pi bonds to N, using all 2 sp orbitals for sigma bonds and 2 unhybridised p orbitals for the 2 pi bonds).
Why does ortho-nitrophenol have a lower boiling point than para-nitrophenol?
In ortho-nitrophenol, the OH group and the NO₂ group are close enough that an intramolecular hydrogen bond forms (within the same molecule), satisfying the H-bond donor and acceptor within one molecule. This prevents intermolecular hydrogen bonding between molecules. With fewer intermolecular H-bonds, less energy is needed to vaporise the compound, so its boiling point is lower. Para-nitrophenol cannot form intramolecular H-bonds (groups too far apart), so it relies on intermolecular H-bonding, raising its boiling point.
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