NeetCrackers Chemistry Notes
Complete Notes

Chemical Bonding and Molecular Structure

Every topic explained simply, with practice questions and answers after each one.

📑 Topics Covered

  1. Kössel-Lewis Approach & the Octet Rule
  2. Ionic (Electrovalent) Bond
  3. Covalent Bond & Lewis Structures
  4. Bond Parameters
  5. Resonance
  6. Polarity of Bonds & Dipole Moment
  7. VSEPR Theory & Molecular Shapes
  8. Valence Bond Theory & Orbital Overlap
  9. Hybridisation
  10. Molecular Orbital Theory
  11. Hydrogen Bonding

Kössel-Lewis Approach & the Octet Rule

G.N. Lewis proposed that atoms combine to achieve a stable electronic configuration, usually resembling that of the nearest noble gas — with 8 electrons in the outermost shell (the octet rule). Hydrogen and helium are exceptions, being stable with just 2 electrons (the duplet rule).

Lewis Symbols

A Lewis symbol represents an element's chemical symbol surrounded by dots equal to the number of valence electrons. Bonds form when atoms share or transfer these valence electrons to complete their octets.

Limitations of the Octet Rule

Some stable molecules don't follow it: $\text{BF}_3$ has only 6 electrons around boron (incomplete octet); $\text{SF}_6$ has 12 electrons around sulphur (expanded octet); and odd-electron species like $\text{NO}$ can never satisfy it exactly.

Q. Why is BF₃ considered an exception to the octet rule?
Ans: Boron in BF₃ has only 6 electrons around it (3 bond pairs, no lone pair) instead of the usual 8 — an incomplete octet — yet the molecule is stable.

Ionic (Electrovalent) Bond

Definition

An ionic bond forms when one atom completely transfers one or more electrons to another, creating oppositely charged ions that are held together by strong electrostatic attraction. Typically forms between a metal (low ionisation enthalpy) and a non-metal (high electron gain enthalpy).

Factors Favouring Ionic Bond Formation

Lattice Enthalpy Trend $$\text{Lattice enthalpy} \propto \dfrac{(\text{charge on cation})(\text{charge on anion})}{\text{distance between ions}}$$

Higher ionic charges and smaller ionic sizes give stronger, more stable ionic lattices.

Q. Between NaCl and MgO, which has a higher lattice enthalpy, and why?
Ans: MgO has a higher lattice enthalpy — Mg²⁺ and O²⁻ carry higher charges (2+ and 2−) than Na⁺ and Cl⁻ (1+ and 1−), and lattice enthalpy increases with the product of ionic charges.

Covalent Bond & Lewis Structures

Definition

A covalent bond forms when two atoms share one or more pairs of electrons, typically between two non-metals of similar electronegativity. A single shared pair gives a single bond, two pairs a double bond, and three pairs a triple bond.

Steps to Draw a Lewis Structure

  1. Count the total valence electrons of all atoms in the molecule/ion (add extra electrons for negative charge, remove for positive charge).
  2. Identify the central atom (usually the least electronegative, excluding hydrogen) and arrange other atoms around it.
  3. Connect atoms with single bonds first, then distribute remaining electrons as lone pairs to complete octets, starting with outer atoms.
  4. If the central atom still lacks an octet, convert lone pairs on outer atoms into double/triple bonds.

Formal Charge

Formal Charge Formula $$\text{FC} = (\text{Valence electrons}) - (\text{Non-bonding electrons}) - \tfrac{1}{2}(\text{Bonding electrons})$$

Used to identify the most stable Lewis structure among several possibilities — the structure with formal charges closest to zero, and negative charges on the more electronegative atoms, is usually preferred.

Q. Calculate the formal charge on the oxygen atom in the ozone (O₃) structure where the central O has a double bond to one O and a single bond to the other.
Ans: For the central O (2 bonding pairs = 4 bonds if counted as double+single, i.e., 3 bonds total, 6 bonding electrons, 1 lone pair = 2 non-bonding electrons): FC = 6 − 2 − (6/2) = 6 − 2 − 3 = +1.

Bond Parameters

ParameterMeaningTrend
Bond lengthEquilibrium distance between the nuclei of two bonded atomsDecreases as bond order increases (triple < double < single)
Bond angleAngle between two adjacent bonds at a common atomDepends on hybridisation & lone pair repulsion
Bond enthalpyEnergy needed to break one mole of a particular bond in the gaseous stateIncreases as bond order increases
Bond orderNumber of bonds (single/double/triple) between two atomsHigher bond order = shorter, stronger bond
Q. Arrange the carbon-carbon bonds in ethane (C–C), ethene (C=C), and ethyne (C≡C) in order of increasing bond length.
Ans: Ethyne (C≡C) < Ethene (C=C) < Ethane (C–C) — bond length decreases as bond order (and hence bond strength) increases.

Resonance

Definition

Resonance occurs when a single Lewis structure cannot accurately represent a molecule's actual structure, so it's described as a hybrid of two or more valid structures (resonance structures/canonical forms) that differ only in the position of electrons, not atoms. The real molecule is a blend — a resonance hybrid — more stable than any single contributing structure.

Classic example: ozone (O₃), where the two O–O bonds are experimentally found to be identical in length (intermediate between a single and double bond) — explained by two resonance structures with the double bond on alternating sides.

Quick Tip

Resonance structures are not different molecules interconverting rapidly — they don't actually exist separately; only the resonance hybrid is real.

Q. Why do both C–O bonds in the carbonate ion (CO₃²⁻) have the same, intermediate bond length rather than one being a clear single bond and the others double bonds?
Ans: Because CO₃²⁻ is a resonance hybrid of three equivalent structures, each with the double bond on a different oxygen — the actual structure is an average of all three, giving all three C–O bonds equal, intermediate length and strength.

Polarity of Bonds & Dipole Moment

When two atoms of different electronegativity form a covalent bond, the shared electron pair is pulled closer to the more electronegative atom, creating partial charges ($\delta^+$ and $\delta^-$) — this is a polar covalent bond.

Dipole Moment $$\mu = q \times d$$

where $q$ = magnitude of partial charge, $d$ = distance between the charges. Measured in Debye (D); $1\ \text{D} = 3.33 \times 10^{-30}\ \text{C m}$. Dipole moment is a vector quantity — pointing from the positive to the negative end.

Net Dipole Moment of a Molecule

The overall dipole moment is the vector sum of all individual bond dipoles. Even if individual bonds are polar, a molecule can be non-polar overall if the bond dipoles cancel due to symmetry — e.g., $\text{CO}_2$ (linear, dipoles cancel) is non-polar despite having polar C=O bonds, while $\text{H}_2\text{O}$ (bent) is polar.

Q. Both CCl₄ and CHCl₃ contain polar C–Cl bonds. Why is CCl₄ non-polar while CHCl₃ is polar?
Ans: CCl₄ is perfectly tetrahedral and symmetric, so the four C–Cl bond dipoles cancel out exactly, giving zero net dipole moment. In CHCl₃, one Cl is replaced by H, breaking the symmetry, so the bond dipoles don't cancel and the molecule has a net dipole moment.

VSEPR Theory & Molecular Shapes

Core Idea

Valence Shell Electron Pair Repulsion (VSEPR) theory predicts molecular shape based on the idea that electron pairs (bonding and lone) around a central atom arrange themselves as far apart as possible to minimise repulsion. Repulsion strength follows: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.

Bond pairs + Lone pairsShapeBond angleExample
2 + 0Linear180°BeCl₂
3 + 0Trigonal planar120°BF₃
2 + 1Bent<120°SO₂
4 + 0Tetrahedral109.5°CH₄
3 + 1Trigonal pyramidal<109.5°NH₃
2 + 2Bent (V-shaped)<109.5°H₂O
5 + 0Trigonal bipyramidal90° & 120°PCl₅
6 + 0Octahedral90°SF₆
Q. NH₃ and H₂O both have a nearly tetrahedral arrangement of electron pairs, but their bond angles (107° and 104.5°) are both less than the ideal 109.5°, and H₂O's is smaller than NH₃'s. Why?
Ans: NH₃ has 1 lone pair and H₂O has 2 lone pairs. Since lone pair–bond pair repulsion is stronger than bond pair–bond pair repulsion, each lone pair squeezes the bond angle down from the ideal 109.5°. With two lone pairs, H₂O experiences more of this compression than NH₃ with just one, so its bond angle is pushed down further.

Valence Bond Theory & Orbital Overlap

Valence Bond Theory (VBT) explains covalent bond formation as the overlap of atomic orbitals from two atoms, each contributing one unpaired electron. Greater orbital overlap gives a stronger bond.

Sigma (σ) and Pi (π) Bonds

Types of Overlap

A sigma bond forms by head-on (axial) overlap of orbitals along the internuclear axis — allows free rotation and is generally stronger. A pi bond forms by sideways (lateral) overlap of parallel p-orbitals above and below the internuclear axis — restricts rotation and is generally weaker than a sigma bond. A single bond is always one σ bond; a double bond is one σ + one π; a triple bond is one σ + two π.

Q. Why can't the two carbon atoms in ethene (C₂H₄) rotate freely around the C=C bond, unlike the C–C bond in ethane?
Ans: The C=C double bond consists of one σ bond and one π bond. The π bond forms from sideways overlap of p-orbitals, which would break if the atoms rotated relative to each other — so rotation around a double bond is restricted, unlike a single σ bond in ethane, which allows free rotation.

Hybridisation

Definition

Hybridisation is the mixing of atomic orbitals of similar energy on the same atom to form new, equivalent hybrid orbitals, better suited for bond formation and matching the observed molecular geometry.

TypeOrbitals MixedGeometryExample
sp1 s + 1 pLinear (180°)BeCl₂, C₂H₂
sp²1 s + 2 pTrigonal planar (120°)BF₃, C₂H₄
sp³1 s + 3 pTetrahedral (109.5°)CH₄, NH₃, H₂O
sp³d1 s + 3 p + 1 dTrigonal bipyramidalPCl₅
sp³d²1 s + 3 p + 2 dOctahedralSF₆
Solved Example

Determine the hybridisation of the central atom in NH₃.

Nitrogen has 3 bond pairs (to 3 H atoms) + 1 lone pair = 4 electron domains → sp³ hybridisation, with a tetrahedral electron arrangement (though the molecular shape itself is trigonal pyramidal because one position is a lone pair).

Q. What is the hybridisation of carbon in CO₂, and why?
Ans: sp hybridisation. Carbon forms 2 sigma bonds (to each oxygen) with no lone pairs, giving 2 electron domains — sp hybridisation gives the linear geometry (180°) that CO₂ actually has.

Molecular Orbital Theory

Core Idea

Unlike VBT, Molecular Orbital Theory (MOT) treats electrons as belonging to the molecule as a whole, not to individual atoms. Atomic orbitals combine (via Linear Combination of Atomic Orbitals, LCAO) to form new molecular orbitals spread over the entire molecule.

Bonding & Antibonding Molecular Orbitals

Combining two atomic orbitals gives two molecular orbitals: a lower-energy bonding molecular orbital (BMO), formed by constructive overlap (in-phase addition), and a higher-energy antibonding molecular orbital (ABMO), formed by destructive overlap (out-of-phase subtraction, with a node between nuclei).

Conditions for Combination of Atomic Orbitals

Bond Order Formula $$\text{Bond order} = \tfrac{1}{2}(N_b - N_a)$$

where $N_b$ = electrons in bonding MOs, $N_a$ = electrons in antibonding MOs. Higher bond order means a shorter, stronger, more stable bond; a bond order of zero means the molecule cannot exist.

Magnetic Properties

If all electrons in the molecular orbitals are paired, the species is diamagnetic (weakly repelled by a magnetic field). If one or more molecular orbitals have unpaired electrons, the species is paramagnetic (attracted by a magnetic field) — this is how MOT correctly predicts that O₂ is paramagnetic, something VBT cannot explain.

Solved Example

Calculate the bond order of the nitrogen molecule, N₂ (14 electrons total).

Molecular orbital filling for N₂: $\sigma1s^2\, \sigma^*1s^2\, \sigma2s^2\, \sigma^*2s^2\, \pi2p_x^2\, \pi2p_y^2\, \sigma2p_z^2$ → bonding electrons = 10, antibonding electrons = 4.

Bond order $= \tfrac{1}{2}(10-4) = 3$ — consistent with the known N≡N triple bond.

Q. Using molecular orbital theory, explain why the O₂ molecule is paramagnetic.
Ans: Filling molecular orbitals for O₂'s 16 electrons places its last two electrons singly (with parallel spins) in the two degenerate π*2p antibonding orbitals, following Hund's rule. These unpaired electrons make O₂ paramagnetic — a result Valence Bond Theory fails to predict, since it shows all electrons in O₂ as paired.

Hydrogen Bonding

Definition & Cause

A hydrogen bond is a special, relatively weak attraction between a hydrogen atom (already covalently bonded to a highly electronegative atom — F, O, or N) and a lone pair on another highly electronegative atom, either on a nearby molecule or within the same molecule. Because F, O, and N are small and highly electronegative, they pull electron density strongly away from hydrogen, leaving it with a significant partial positive charge that then attracts a nearby lone pair.

Types of Hydrogen Bonding

Q. Water (H₂O) has a much higher boiling point than H₂S, even though both have a similar bent shape and S is directly below O in the periodic table. Explain why.
Ans: Oxygen is far more electronegative than sulphur and much smaller in size, so O–H bonds are highly polar and allow strong intermolecular hydrogen bonding between water molecules. Sulphur is larger and less electronegative, so H₂S molecules experience only weak van der Waals forces, not hydrogen bonding — giving water a much higher boiling point despite its lower molecular mass.

⚡ Formula Cheat-Sheet

Formal chargeV − N − ½B
Dipole momentμ = q × d
1 Debye3.33×10⁻³⁰ C·m
Bond order (MOT)½(Nᵦ − Nₐ)
sp hybridisationLinear, 180°
sp² hybridisationTrigonal planar, 120°
sp³ hybridisationTetrahedral, 109.5°
Repulsion order (VSEPR)lp-lp > lp-bp > bp-bp