Chemical Bonding and Molecular Structure
Chemical Bonding and Molecular Structure for JEE Main & Advanced
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:
- Smaller cation (high charge density, high polarizing power)
- Larger anion (more polarizable)
- Higher charge on either ion
- 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
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.
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.
Ionic bond is formed between:
Lattice energy increases when:
Most covalent character (by Fajan's rules):
Born-Haber cycle is based on:
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):
- Count total valence electrons
- Determine central atom (least electronegative, except H)
- Connect with single bonds
- Distribute remaining electrons as lone pairs (outer atoms first, then central)
- 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
Draw Lewis structure of SO₂ and find formal charges.
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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.
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.
Coordinate bond is shown by:
Number of valence electrons in PCl₅:
The molecule with incomplete octet on central atom:
Resonance hybrid is:
The formal charge on N in NH₄⁺:
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.
| SN | Geometry | Examples (no lone pair) | Bond angle |
|---|---|---|---|
| $2$ | Linear | BeCl₂, CO₂ | $180°$ |
| $3$ | Trigonal planar | BF₃, BCl₃, NO₃⁻ | $120°$ |
| $4$ | Tetrahedral | CH₄, NH₄⁺ | $109.5°$ |
| $5$ | Trigonal bipyramidal | PCl₅ | $90°, 120°$ |
| $6$ | Octahedral | SF₆ | $90°$ |
| $7$ | Pentagonal bipyramidal | IF₇ | $72°, 90°$ |
Shapes (with lone pairs):
| SN | LP | Shape | Example | Bond angle |
|---|---|---|---|---|
| 4 | 1 | Pyramidal | NH₃ | $107°$ |
| 4 | 2 | Bent (V-shape) | H₂O | $104.5°$ |
| 5 | 1 | See-saw | SF₄ | $\sim 90°, 102°$ |
| 5 | 2 | T-shape | ClF₃, BrF₃ | $\sim 87°$ |
| 5 | 3 | Linear | XeF₂, I₃⁻ | $180°$ |
| 6 | 1 | Square pyramidal | BrF₅ | $\sim 90°$ |
| 6 | 2 | Square planar | XeF₄ | $90°$ |
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.
| Hybridization | Geometry | Bond angle | Example |
|---|---|---|---|
| sp | Linear | $180°$ | BeCl₂, CO₂, C₂H₂ |
| sp² | Trigonal planar | $120°$ | BF₃, ethene |
| sp³ | Tetrahedral | $109.5°$ | CH₄, NH₃, H₂O |
| sp³d | Trigonal bipyramidal | $90°, 120°$ | PCl₅ |
| sp³d² | Octahedral | $90°$ | SF₆ |
| sp³d³ | Pentagonal bipyramidal | $72°, 90°$ | IF₇ |
| dsp² | Square 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.
Determine hybridization, shape and bond angle of NH₃, H₂O, BF₃, SF₄.
Show solution
| Molecule | LP + BP | Hybridization | Geometry | Shape | Bond Angle |
|---|---|---|---|---|---|
| NH₃ | 1 + 3 = 4 | sp³ | Tetrahedral | Pyramidal | $107°$ |
| H₂O | 2 + 2 = 4 | sp³ | Tetrahedral | Bent | $104.5°$ |
| BF₃ | 0 + 3 = 3 | sp² | Trigonal planar | Trigonal planar | $120°$ |
| SF₄ | 1 + 4 = 5 | sp³d | TBP | See-saw | $\sim 90°, 102°$ |
Why is bond angle in H₂O ($104.5°$) less than in NH₃ ($107°$)?
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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.
Hybridization of C in CH₄:
Geometry of XeF₄:
Bond angle in H₂O:
The molecule with $sp^3d^2$ hybridization:
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.
| Molecule | Dipole Moment | Reason |
|---|---|---|
| H₂O | $1.85$ D | Bent, not symmetric |
| NH₃ | $1.46$ D | Pyramidal, not symmetric |
| BF₃ | $0$ | Trigonal planar, symmetric |
| CO₂ | $0$ | Linear, dipoles cancel |
| CH₄ | $0$ | Tetrahedral, symmetric |
| NF₃ | $0.23$ D | Pyramidal 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.
Calculate BO and magnetic nature of N₂.
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$N_2$ 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.
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.
Bond order of O₂:
O₂ is:
Dipole moment of CO₂:
Hydrogen bonds are formed between H and:
NH₃ has greater dipole moment than NF₃ because:
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