JEE Main & Advanced

Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure for JEE Main & Advanced

1
Module 1

Ionic and Covalent Bonding

Ionic Bond, Lattice Energy, Born-Haber Cycle, Fajan's RulesTopic 1

Chemical Bond: Force of attraction between atoms holding them together. Octet rule (Lewis): Atoms tend to attain 8 electrons (or 2 for H, He) in outer shell.

Ionic (Electrovalent) Bond: Formed by complete transfer of electrons from metal (low IE) to non-metal (high EN). Examples: NaCl, MgO, CaF₂.

Conditions favoring ionic bond:

  • Low IE of metal (s-block, especially heavy alkali metals)
  • High EN of non-metal (halogens, oxygen)
  • High lattice energy

Lattice Energy ($U$): Energy released when 1 mole of ionic crystal is formed from gaseous ions. Magnitude depends on: $$U \propto \frac{q_1 q_2}{r_+ + r_-}$$

Higher for: (i) higher charges, (ii) smaller ionic radii.

Born-Haber Cycle: Thermodynamic cycle to calculate lattice energy using Hess's law. For NaCl:

$$\Delta_f H^\ominus = \Delta_{\text{sub}} H_{Na} + \frac{1}{2}\Delta_{\text{diss}} H_{Cl_2} + IE_{Na} + EA_{Cl} + U$$

Solving for $U$ gives lattice energy.

Fajan's Rules: Predict ionic vs covalent character. Covalent character increases when:

  1. Smaller cation (high charge density, high polarizing power)
  2. Larger anion (more polarizable)
  3. Higher charge on either ion
  4. Cation with $d$-electrons (e.g., $Cu^+$ more polarizing than $Na^+$ of similar size)

Polarizing power $\propto$ charge/radius² of cation. Polarizability of anion $\propto$ size and charge of anion.

Examples:

  • LiCl is more covalent than NaCl (smaller Li⁺)
  • AgCl is more covalent than NaCl (Ag⁺ has pseudo-noble configuration; $d$-electrons)
  • AlCl₃ is more covalent than NaCl (higher charge on Al³⁺)

Properties of Ionic Compounds:

  • High melting and boiling points
  • Hard and brittle
  • Soluble in polar solvents (water), insoluble in nonpolar
  • Conduct electricity in molten state or aqueous solution
Worked Examples
1

Among NaCl, NaBr, NaI, which has the most covalent character?

Show solution

Same cation (Na⁺), but anion size: I⁻ > Br⁻ > Cl⁻. Larger anion is more polarizable. By Fajan's rules, NaI has the most covalent character.

Final Answer: NaI.

2

Using Born-Haber cycle data, find lattice energy of NaCl. ($\Delta_f H = -411$, $\Delta_{\text{sub}}$Na = $108$, $\frac{1}{2}\Delta_{\text{diss}}$Cl₂ = $122$, IE(Na) = $496$, EA(Cl) = $-349$ kJ/mol)

Show solution

$$-411 = 108 + 122 + 496 + (-349) + U$$ $$-411 = 377 + U \implies U = -788 \text{ kJ/mol}$$

Final Answer: $U = -788$ kJ/mol.

✎ Self-Check — 5 questions0 / 5
Q1.

Ionic bond is formed between:

Q2.

Lattice energy increases when:

Q3.

Most covalent character (by Fajan's rules):

Q4.

Born-Haber cycle is based on:

Q5.

Highest lattice energy:

Covalent Bond, Lewis Structures, Resonance, Formal ChargeTopic 2

Covalent Bond: Formed by sharing of electrons between atoms of similar EN. Lewis (1916) — atoms share electron pairs to attain noble gas configuration.

Types of Covalent Bonds:

  • Single bond: 1 shared pair (e.g., H-H)
  • Double bond: 2 shared pairs (e.g., O=O)
  • Triple bond: 3 shared pairs (e.g., N≡N)

Coordinate (Dative) Bond: Both electrons of the shared pair come from one atom. e.g., NH₄⁺, H₃O⁺, NH₃→BF₃.

Lewis Structure (Steps):

  1. Count total valence electrons
  2. Determine central atom (least electronegative, except H)
  3. Connect with single bonds
  4. Distribute remaining electrons as lone pairs (outer atoms first, then central)
  5. Form multiple bonds if octet incomplete on central atom

Limitations of Octet Rule:

  • Incomplete octet: BeCl₂ (4 e⁻ on Be), BF₃ (6 e⁻ on B), LiCl
  • Expanded octet: PF₅, SF₆, IF₇ (central atom can use $d$-orbitals)
  • Odd-electron molecules: NO, NO₂, ClO₂
  • Predicts equal bond properties for some species, but reality differs

Formal Charge: $$\text{FC} = (\text{valence electrons}) - (\text{lone pair electrons}) - \frac{1}{2}(\text{bonding electrons})$$

Sum of formal charges = total charge on species. The most stable Lewis structure has formal charges closest to zero, and negative formal charge on the more electronegative atom.

Resonance: When a molecule cannot be represented by a single Lewis structure; actual structure is a hybrid of contributing forms. Real molecule has lower energy than any single structure (resonance stabilization).

Examples of Resonance:

  • O₃ (ozone): two resonance structures, equal bond lengths
  • CO₃²⁻: three equivalent structures; all C–O bonds equal
  • Benzene: two Kekulé structures (alternating double bonds); actual bonds are intermediate
  • NO₂⁻, NO₃⁻, $\text{SO}_3$

Conditions for Resonance:

  • Atoms in same position
  • Same number of paired and unpaired electrons
  • Differ only in placement of $\pi$ electrons or lone pairs
Worked Examples
1

Draw Lewis structure of SO₂ and find formal charges.

Show solution

Total valence electrons: $6 + 6 \times 2 = 18$. Central atom S. Structure: O=S=O or O–S=O with appropriate lone pairs. Most accepted: S with one S=O double bond, one S–O single bond, one lone pair. Resonance gives equal S–O bonds.

For structure O=S=O (linear-style, but bent in 3D with lone pair): FC on S = $6 - 2 - 6 = -2$ (no, let me recompute with correct lewis)

In resonance hybrid structure O–S(+1)=O with each O having FC, sum is zero. Geometry bent due to lone pair on S.

Final Answer: SO₂ has bent shape; two equivalent S–O bonds via resonance.

2

Determine which contributes most: structures of $\text{CO}_2$.

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Lewis: O=C=O is the major (all atoms with formal charge $0$, complete octets). Other resonance forms: O≡C–O⁻ and ⁻O–C≡O have higher formal charges on O (less stable).

Final Answer: O=C=O is the major contributor.

✎ Self-Check — 5 questions0 / 5
Q1.

Coordinate bond is shown by:

Q2.

Number of valence electrons in PCl₅:

Q3.

The molecule with incomplete octet on central atom:

Q4.

Resonance hybrid is:

Q5.

The formal charge on N in NH₄⁺:

2
Module 2

VSEPR, Hybridization, MOT and Intermolecular Forces

VSEPR, Hybridization, Bond ParametersTopic 1

VSEPR Theory (Valence Shell Electron Pair Repulsion):

  • Electron pairs around central atom orient to minimize repulsion
  • Order of repulsion: Lone pair – lone pair (lp-lp) > lp-bp > bp-bp
  • Lone pairs occupy more space than bond pairs → distort bond angles

Steric Number (SN) = Lone pairs + bonded atoms (sigma bonds only) = Total number of electron domains.

SNGeometryExamples (no lone pair)Bond angle
$2$LinearBeCl₂, CO₂$180°$
$3$Trigonal planarBF₃, BCl₃, NO₃⁻$120°$
$4$TetrahedralCH₄, NH₄⁺$109.5°$
$5$Trigonal bipyramidalPCl₅$90°, 120°$
$6$OctahedralSF₆$90°$
$7$Pentagonal bipyramidalIF₇$72°, 90°$

Shapes (with lone pairs):

SNLPShapeExampleBond angle
41PyramidalNH₃$107°$
42Bent (V-shape)H₂O$104.5°$
51See-sawSF₄$\sim 90°, 102°$
52T-shapeClF₃, BrF₃$\sim 87°$
53LinearXeF₂, I₃⁻$180°$
61Square pyramidalBrF₅$\sim 90°$
62Square planarXeF₄$90°$

Bond Angle Variation: As EN of central atom decreases or lone pairs increase, bond angle decreases. e.g., NH₃ ($107°$) > PH₃ ($94°$) > AsH₃ ($91°$).

Hybridization: Mixing of atomic orbitals to form new equivalent hybrid orbitals.

HybridizationGeometryBond angleExample
spLinear$180°$BeCl₂, CO₂, C₂H₂
sp²Trigonal planar$120°$BF₃, ethene
sp³Tetrahedral$109.5°$CH₄, NH₃, H₂O
sp³dTrigonal bipyramidal$90°, 120°$PCl₅
sp³d²Octahedral$90°$SF₆
sp³d³Pentagonal bipyramidal$72°, 90°$IF₇
dsp²Square planar$90°$[Ni(CN)₄]²⁻, XeF₄

Steric Number → Hybridization: SN 2 → sp; 3 → sp²; 4 → sp³; 5 → sp³d; 6 → sp³d².

Bond Length: Distance between nuclei in bonded atoms. Order: single > double > triple. Bond Energy: Energy required to break bond. Order: triple > double > single.

Worked Examples
1

Determine hybridization, shape and bond angle of NH₃, H₂O, BF₃, SF₄.

Show solution
MoleculeLP + BPHybridizationGeometryShapeBond Angle
NH₃1 + 3 = 4sp³TetrahedralPyramidal$107°$
H₂O2 + 2 = 4sp³TetrahedralBent$104.5°$
BF₃0 + 3 = 3sp²Trigonal planarTrigonal planar$120°$
SF₄1 + 4 = 5sp³dTBPSee-saw$\sim 90°, 102°$
2

Why is bond angle in H₂O ($104.5°$) less than in NH₃ ($107°$)?

Show solution

H₂O has 2 lone pairs; NH₃ has 1. More lone pairs cause greater compression of bond pairs (lp-lp > lp-bp repulsions), reducing bond angle further in H₂O.

Final Answer: Two lone pairs in H₂O cause more compression than one in NH₃.

✎ Self-Check — 5 questions0 / 5
Q1.

Hybridization of C in CH₄:

Q2.

Geometry of XeF₄:

Q3.

Bond angle in H₂O:

Q4.

The molecule with $sp^3d^2$ hybridization:

Q5.

Lone pair repulsion order:

Molecular Orbital Theory, Hydrogen Bonding, Dipole MomentTopic 2

Molecular Orbital Theory (MOT): Atomic orbitals (AOs) combine to form molecular orbitals (MOs). $N$ AOs → $N$ MOs ($N/2$ bonding + $N/2$ antibonding).

Bonding MO ($\sigma, \pi$): Lower energy than parent AOs (constructive interference). Antibonding MO ($\sigma^*, \pi^*$): Higher energy (destructive interference).

Energy Order:

For molecules up to N₂ (Li₂, Be₂, B₂, C₂, N₂): $$\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < \pi 2p_x = \pi 2p_y < \sigma 2p_z < \pi^*2p_x = \pi^*2p_y < \sigma^*2p_z$$

For O₂ and beyond (O₂, F₂, Ne₂): $$\sigma 1s < \sigma^*1s < \sigma 2s < \sigma^*2s < \sigma 2p_z < \pi 2p_x = \pi 2p_y < \pi^*2p_x = \pi^*2p_y < \sigma^*2p_z$$

Bond Order: $$\text{Bond order (BO)} = \frac{1}{2}(N_b - N_a)$$ where $N_b$ = bonding electrons, $N_a$ = antibonding electrons.

  • BO = $0$: molecule does not exist (e.g., He₂)
  • BO > $0$: molecule is stable
  • Higher BO → shorter bond, higher energy

Magnetic Behaviour:

  • All electrons paired → diamagnetic
  • Unpaired electrons → paramagnetic
  • MOT correctly predicts $O_2$ is paramagnetic (2 unpaired in $\pi^*$ orbitals)

Example: O₂ (16 electrons):

  • Config: $\sigma 1s^2\,\sigma^*1s^2\,\sigma 2s^2\,\sigma^*2s^2\,\sigma 2p_z^2\,\pi 2p_x^2 = \pi 2p_y^2\,\pi^*2p_x^1 = \pi^*2p_y^1$
  • BO = $(10-6)/2 = 2$
  • Paramagnetic (2 unpaired)

Hydrogen Bonding: Special dipole-dipole interaction when H is attached to highly EN atom (F, O, N). Donor: H-F, O-H, N-H. Acceptor: another F, O, N.

  • Intermolecular H-bonding: Between molecules. e.g., H₂O, NH₃, HF, alcohols, carboxylic acids.
  • Intramolecular H-bonding: Within molecule. e.g., o-nitrophenol, salicylaldehyde.

Effects: high boiling points, anomalous density of water (max at $4°$C), high heat of vaporization, ice less dense than water, life-sustaining properties of water.

Dipole Moment ($\mu$): Measure of polarity of bond/molecule. Unit: Debye (D); $1$ D = $3.336 \times 10^{-30}$ C·m. $$\mu = q \times d$$ Vector quantity; net dipole = vector sum of bond dipoles.

MoleculeDipole MomentReason
H₂O$1.85$ DBent, not symmetric
NH₃$1.46$ DPyramidal, not symmetric
BF₃$0$Trigonal planar, symmetric
CO₂$0$Linear, dipoles cancel
CH₄$0$Tetrahedral, symmetric
NF₃$0.23$ DPyramidal but lone pair dipole opposes N-F dipoles
CCl₄$0$Tetrahedral, symmetric

NF₃ < NH₃ in dipole moment despite F being more electronegative — because the lone pair on N points in the opposite direction in NF₃ but the same direction in NH₃ for the bond dipoles.

Worked Examples
1

Calculate BO and magnetic nature of N₂.

Show solution

$N_2$ has 14 e⁻. Configuration (using up-to-N₂ ordering): $\sigma 1s^2\,\sigma^*1s^2\,\sigma 2s^2\,\sigma^*2s^2\,\pi 2p_x^2 = \pi 2p_y^2\,\sigma 2p_z^2$. $N_b = 10$, $N_a = 4$. BO = $(10-4)/2 = 3$. All paired → diamagnetic.

Final Answer: BO = $3$; diamagnetic.

2

Why does water have higher boiling point than H₂S despite S being below O?

Show solution

H₂O molecules form strong intermolecular hydrogen bonds (O-H...O) due to high EN and small size of O. H₂S has weak van der Waals forces — S is larger, less EN, no H-bonding. Result: H₂O bp ($100°$C) >> H₂S bp ($-60°$C).

Final Answer: H₂O has H-bonding; H₂S does not.

✎ Self-Check — 5 questions0 / 5
Q1.

Bond order of O₂:

Q2.

O₂ is:

Q3.

Dipole moment of CO₂:

Q4.

Hydrogen bonds are formed between H and:

Q5.

NH₃ has greater dipole moment than NF₃ because:

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