Chemical Bonding & Molecular Structure

01

Kössel–Lewis Approach to Bonding

Atoms bond to attain the stable noble gas electronic configuration.

The Octet Rule

Atoms combine by losing, gaining, or sharing electrons so that each atom acquires eight electrons in its outermost shell to achieve the stable configuration of the nearest noble gas. Hydrogen and helium are exceptions, as they only need 2 electrons (the duplet rule).

NEET recall Exceptions to the octet rule include the incomplete octet (BeCl₂, BF₃), the expanded octet (PCl₅, SF₆), and odd-electron species (NO, NO₂).

Lewis Symbols & Structures

A Lewis (dot) symbol shows the valence electrons of an atom as dots around its symbol. Follow these rules for drawing a Lewis structure of a molecule or ion:

Count the total valence electrons of all atoms (add the charge magnitude for anions, and subtract it for cations).
Identify the central atom. This is usually the least electronegative, non-hydrogen atom.
Connect atoms with single bonds (2 electrons each) to form the skeleton.
Distribute the remaining electrons as lone pairs to satisfy the octets of the outer atoms first.
If the central atom lacks an octet, convert the lone pairs on the outer atoms into additional bonds (multiple bonds).

Formal Charge

Formal charge helps you choose the most reasonable Lewis structure among several possibilities:

$$ \text{F.C.} = [\text{Valence Electrons}] - [\text{Non-bonding Electrons}] - \frac{1}{2}[\text{Bonding Electrons}] $$

The structure with formal charges closest to zero, and a negative formal charge placed on the more electronegative atom, is the more stable (and preferred) representation.

02

Ionic (Electrovalent) Bond

Complete transfer of electrons from a metal to a non-metal, held together by electrostatic attraction.

Factors Favouring Ionic Bond Formation

Low ionization enthalpy of the metal atom (the cation forms easily).
High (negative) electron gain enthalpy of the non-metal atom (the anion forms easily).
High lattice enthalpy of the resulting crystal. The more exothermic this process is, the more stable the ionic compound becomes.

Lattice Enthalpy & Born–Landé Equation

Lattice enthalpy is the energy required to completely separate one mole of a solid ionic compound into its gaseous ions. It is estimated using this formula:

$$ U = -\dfrac{N_A M z^{+} z^{-} e^{2}}{4\pi \varepsilon_0 r_0}\left(1 - \dfrac{1}{n}\right) $$

where $N_A$ = Avogadro constant, $M$ = Madelung constant, $z^+, z^-$ = ionic charges, $r_0$ = inter-ionic distance, $n$ = Born exponent.

NEET recall Lattice enthalpy ∝ (charge product) and ∝ 1/(interionic distance). Higher charge combined with a smaller ionic radius leads to higher lattice enthalpy, which results in a higher melting point. This is why MgO (2+,2−) has a far higher melting point than NaCl (1+,1−).

Born–Haber Cycle

This is an indirect route to calculate lattice enthalpy via Hess's Law. You can sum the sublimation, ionization, dissociation, electron gain, and formation enthalpies to equal the lattice enthalpy step.

03

Covalent Bond & Lewis Structures

A shared pair of electrons holds two atoms together. Langmuir named this interaction the covalent bond.

Single Bond

One shared electron pair, e.g. H–H, Cl–Cl. The bond order is 1.

Multiple Bonds

A double bond has 2 shared pairs (O=O), and a triple bond has 3 shared pairs (N≡N). As the bond order increases, the bond strength increases and the bond length decreases.

Coordinate (Dative) Bond

Both shared electrons are donated by a single atom (the donor) to another (the acceptor), such as in $\text{NH}_4^+$ and $\text{H}_3\text{O}^+$. Once formed, a coordinate bond behaves identically to a normal covalent bond.

04

Bond Parameters

The measurable fingerprints of a bond: length, angle, enthalpy, and order.

Parameter	Definition	Trend / Note
Bond length	Equilibrium distance between nuclei of two bonded atoms	↓ as bond order ↑; ↑ down a group
Bond angle	Angle between orbitals containing bonding electron pairs around a central atom	Distorted by lone pairs (lp–lp > lp–bp > bp–bp repulsion)
Bond enthalpy	Energy needed to break one mole of bonds in gaseous state	↑ as bond order ↑ (triple > double > single)
Bond order	Number of bonds between two atoms	Higher order means a shorter, stronger bond

$$ \text{Bond order} = \dfrac{N_b - N_a}{2} $$

$N_b$ = electrons in bonding molecular orbitals, $N_a$ = electrons in antibonding molecular orbitals.

05

Resonance & Bond Polarity

When one Lewis structure isn't enough, and when a shared pair isn't shared equally.

Resonance

Some molecules (like $\text{O}_3$, $\text{CO}_3^{2-}$, and benzene) cannot be represented by a single Lewis structure. The actual molecule is a weighted hybrid of two or more contributing structures (canonical forms) that differ only in electron placement, not atomic position. The resonance hybrid is always more stable than any single contributing structure.

Polarity of Covalent Bonds

Unequal sharing of the bonding pair, which happens because of a difference in electronegativity, creates a polar covalent bond with partial charges $\delta^+$ and $\delta^-$.

$$ \mu = q \times d $$

Dipole moment $\mu$ is measured in Debye (D); $1\ \text{D} = 3.33564 \times 10^{-30}\ \text{C m}$. It is a vector quantity, pointing from the positive to the negative end.

NEET recall The net dipole moment of a molecule depends on both the bond polarity and the molecular geometry. CO₂ (linear) has individual polar bonds, but the net μ = 0 due to symmetric cancellation. H₂O (bent) has μ ≈ 1.85 D.

$$ \%\ \text{ionic character} = \dfrac{\mu_{\text{observed}}}{\mu_{\text{fully ionic}}} \times 100 $$

06

VSEPR Theory

Valence Shell Electron Pair Repulsion dictates that electron pairs arrange to minimize mutual repulsion.

Postulates

The shape of a molecule depends on the number of valence shell electron pairs (bonding and lone) around the central atom.
Electron pairs orient themselves to minimize repulsion and maximize the distance between them.
A multiple bond is treated as a single effective electron pair (electron density) for shape calculation purposes.
The repulsion order is: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.

BP + LP	Shape	Bond angle	Example
2 + 0	Linear	180°	BeCl₂, CO₂
3 + 0	Trigonal planar	120°	BF₃, SO₃
2 + 1	Bent / angular	<120°	SO₂
4 + 0	Tetrahedral	109.5°	CH₄, SO₄²⁻
3 + 1	Trigonal pyramidal	~107°	NH₃
2 + 2	Bent / angular	~104.5°	H₂O
5 + 0	Trigonal bipyramidal	120° & 90°	PCl₅
4 + 1	See-saw	~173°, 120°, 90°	SF₄
3 + 2	T-shaped	~90°	ClF₃
2 + 3	Linear	180°	XeF₂, I₃⁻
6 + 0	Octahedral	90°	SF₆
5 + 1	Square pyramidal	~90°	BrF₅
4 + 2	Square planar	90°	XeF₄

NEET recall NH₃ (107°) vs H₂O (104.5°): both have tetrahedral electron geometries. However, water's two lone pairs compress the bond angle more than ammonia's single lone pair. This is a classic direct question in the exam.

07

Valence Bond Theory (VBT)

A covalent bond forms by the overlap of half-filled atomic orbitals of opposite spin.

Types of Overlap

Sigma (σ) bond

This is a head-on (axial) overlap along the internuclear axis using s–s, s–p, or p–p orbitals. Free rotation is possible around a σ bond.

Pi (π) bond

This is a sideways (lateral) overlap of parallel p-orbitals located above and below the internuclear axis. This restricts rotation and is weaker than a σ bond.

Strength of Overlap

Greater orbital overlap results in a stronger, shorter bond. The order of overlap strength is: s–s < s–p < p–p (σ) > p–p (π).

08

Hybridization

Mixing of atomic orbitals of similar energy on the same atom to form new, equivalent hybrid orbitals.

sp → 180° sp² → 120° sp³ → 109.5° sp³d → trigonal bipyramidal sp³d² → octahedral sp³d³ → pentagonal bipyramidal

Hybridization	Geometry	Examples
sp	Linear	BeCl₂, C₂H₂, CO₂
sp²	Trigonal planar	BCl₃, C₂H₄, SO₃
sp³	Tetrahedral	CH₄, NH₃, H₂O
sp³d	Trigonal bipyramidal	PCl₅
sp³d²	Octahedral	SF₆
sp³d³	Pentagonal bipyramidal	IF₇

Rules Governing Hybridization

Only orbitals of comparable energy from the same atom can mix.
The number of hybrid orbitals formed equals the number of atomic orbitals mixed.
Hybrid orbitals are equivalent in energy and shape, and they point towards the corners of a definite geometric figure.
They form only σ bonds or hold lone pairs, but they never form π bonds.

NEET recall Steric number = (bond pairs + lone pairs on the central atom). The patterns are: Steric number 2→sp, 3→sp², 4→sp³, 5→sp³d, 6→sp³d². This single shortcut solves most hybridization questions instantly.

09

Molecular Orbital Theory (MOT)

Atomic orbitals combine using LCAO to give molecular orbitals delocalized over the whole molecule.

Formation of Molecular Orbitals

Using the Linear Combination of Atomic Orbitals (LCAO) method, constructive overlap (addition) gives a lower-energy bonding MO (σ, π). Destructive overlap (subtraction) gives a higher-energy antibonding MO (σ^*, π^*), which features a node between the nuclei.

Atomic orbital
1s

+

Atomic orbital
1s

→

Bonding molecular orbital
σ1s

Atomic orbital
1s

-

Atomic orbital
1s

→

Antibonding molecular orbital
σ*1s

Atomic orbital
2p_z

+

Atomic orbital
2p_z

→

Bonding molecular orbital
σ2p_z

Atomic orbital
2p_z

-

Atomic orbital
2p_z

→

Antibonding molecular orbital
σ*2p_z

Atomic orbital
2p_x

+

Atomic orbital
2p_x

→

Bonding molecular orbital
π2p_x

Atomic orbital
2p_x

-

Atomic orbital
2p_x

→

Antibonding molecular orbital
π*2p_x

σ^*2p

π^*2p

π2p

σ2p

σ^*2s

σ2s

The order shown above applies to O₂, F₂, and Ne₂. For Li₂ through N₂, σ2p sits above π2p.

Energy Level Order

$$ \text{For } Z \le 7 \text{ (up to N}_2\text{):} $$ $$ \sigma_{2s} < \sigma^*_{2s} < (\pi_{2p_x}=\pi_{2p_y}) < \sigma_{2p_z} < (\pi^*_{2p_x}=\pi^*_{2p_y}) < \sigma^*_{2p_z} $$

$$ \text{For } Z > 7 \text{ (O}_2\text{, F}_2\text{, Ne}_2\text{):} $$ $$ \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < (\pi_{2p_x}=\pi_{2p_y}) < (\pi^*_{2p_x}=\pi^*_{2p_y}) < \sigma^*_{2p_z} $$

Bond Order & Key Examples

Species	Electron config (valence)	Bond order	Nature
H₂	σ1s²	1	Stable, diamagnetic
He₂	σ1s² σ^*1s²	0	Does not exist
N₂	σ2s² σ^*2s² π2p⁴ σ2p²	3	Very stable, diamagnetic
O₂	σ2s² σ^2s² σ2p² π2p⁴ π^2p²	2	Paramagnetic (2 unpaired e⁻ in π^*)

NEET recall The fact that O₂ is paramagnetic is one of MOT's biggest triumphs. The Lewis structure alone cannot explain this behavior, but MOT correctly predicts two unpaired electrons in the degenerate π* orbitals. This is a high-frequency exam question.

10

Hydrogen Bonding

A special dipole–dipole attraction created when hydrogen sits between two highly electronegative atoms.

Condition for Formation

Hydrogen must be covalently bonded to a small, highly electronegative atom (specifically F, O, or N). This leaves the H nucleus exposed enough to attract a lone pair on a neighbouring electronegative atom.

Intermolecular H-bond

This occurs between two different molecules, e.g. HF···HF, H₂O···H₂O. It raises the boiling point, viscosity, and surface tension significantly (leading to the anomalous behaviour of water, HF, and NH₃).

Intramolecular H-bond

This occurs within the same molecule and forms a ring, e.g. o-nitrophenol. It tends to lower the boiling point compared to the para-isomer, because the hydrogen isn't available for intermolecular association.

NEET recall Boiling point anomaly: H₂O > H₂Te > H₂Se > H₂S for group 16 hydrides. Even though H₂O has the lowest molar mass, its extensive hydrogen bonding pushes its boiling point far above the expected trend line.

✓

Quick Revision Sheet

Every formula and rule from this chapter in one glance, perfectly suited for the night before your exam.

Formal charge

$$ \text{F.C.} = V - N - \tfrac{B}{2} $$

Lattice enthalpy

$$ U \propto \dfrac{z^+ z^-}{r_0} $$

Bond order (general)

$$ \dfrac{N_b - N_a}{2} $$

Dipole moment

$$ \mu = q \times d\ (\text{in Debye}) $$

Repulsion order

lp–lp > lp–bp > bp–bp

Steric no. → hybridization

2→sp, 3→sp², 4→sp³, 5→sp³d, 6→sp³d²

O₂ magnetism

Paramagnetic → 2 unpaired e⁻ in π^* (MOT proof)

NH₃ vs H₂O angle

107° vs 104.5° → more lone pairs means a smaller angle

H-bond condition

H bonded to F, O, or N only

% ionic character

$$ \dfrac{\mu_{obs}}{\mu_{ionic}} \times 100 $$