Why is water ($\text{H}_2\text{O}$) a bent molecule with a bond angle of $104.5^\circ$ while carbon dioxide ($\text{CO}_2$) is perfectly linear at $180^\circ$? Why is liquid oxygen attracted to a magnetic field despite having an even number of valence electrons? Why are axial bonds longer than equatorial bonds in $\text{PCl}_5$? Chemical Bonding and Molecular Structure forms Chapter 4 of the CBSE Class 11 Chemistry syllabus and is the single highest-weightage chapter in JEE Main, JEE Advanced, and NEET.
This comprehensive guide covers the Kossel-Lewis octet rule, formal charge calculations, VSEPR theory molecular geometries, Valence Bond Theory ($\sigma$ vs $\pi$ bonds), hybridization ($sp, sp^2, sp^3, sp^3d, sp^3d^2$), Molecular Orbital Theory (MOT) LCAO diagrams, bond order formulas, $\text{O}_2$ paramagnetism proof, Hydrogen bonding, and five step-by-step solved entrance exam questions.
- 1. Kossel-Lewis Octet Rule & Formal Charge Calculation
- 2. VSEPR Theory & Master Molecular Geometry Table
- 3. Valence Bond Theory (VBT) & Hybridization Types
- 4. Molecular Orbital Theory (MOT) & Paramagnetism of O2
- 5. Intermolecular vs Intramolecular Hydrogen Bonding
- 6. Solved Entrance Exam Questions (JEE / NEET)
- 7. Frequently Asked Questions (FAQ)
1. Kossel-Lewis Octet Rule & Formal Charge Calculation
Lewis proposed that atoms combine to achieve a stable 8-electron octet in their outer valence shell ($\text{ns}^2 \text{np}^6$), resembling noble gas electronic structures.
\text{FC} = V - L - \frac{1}{2} S
Where:
• $V =$ Total number of valence electrons in free atom.
• $L =$ Total number of non-bonding (lone pair) electrons.
• $S =$ Total number of shared (bonding) electrons.
Example (Ozone $\text{O}_3$):
• Central Oxygen atom (1): $V = 6, L = 2, S = 6 \implies \text{FC} = 6 - 2 - \frac{1}{2}(6) = \mathbf{+1}$.
• Double-bonded Oxygen atom (2): $V = 6, L = 4, S = 4 \implies \text{FC} = 6 - 4 - \frac{1}{2}(4) = \mathbf{0}$.
• Single-bonded Oxygen atom (3): $V = 6, L = 6, S = 2 \implies \text{FC} = 6 - 6 - \frac{1}{2}(2) = \mathbf{-1}$.
2. VSEPR Theory & Master Molecular Geometry Table
Valence Shell Electron Pair Repulsion (VSEPR) Theory predicts the 3D shapes of molecules based on electron pair repulsions around the central atom:
Effect of Lone Pairs: Lone pairs occupy larger spatial domain on central atom, distorting standard bond angles (e.g. $\text{CH}_4$ tetrahedral $109.5^\circ \rightarrow \text{NH}_3$ pyramidal $107^\circ \rightarrow \text{H}_2\text{O}$ bent $104.5^\circ$).
| Molecule Type | Bond Pairs (bp) | Lone Pairs (lp) | Electron Geometry | Molecular Shape | Hybridization & Examples |
|---|---|---|---|---|---|
| $AX_2$ | 2 | 0 | Linear | Linear ($180^\circ$) | $sp$ ($\text{BeCl}_2, \text{CO}_2, \text{C}_2\text{H}_2$) |
| $AX_3$ | 3 | 0 | Trigonal Planar | Trigonal Planar ($120^\circ$) | $sp^2$ ($\text{BF}_3, \text{AlCl}_3, \text{C}_2\text{H}_4$) |
| $AX_2E$ | 2 | 1 | Trigonal Planar | Bent / V-shaped ($<120^\circ$) | $sp^2$ ($\text{SO}_2, \text{O}_3, \text{PbCl}_2$) |
| $AX_4$ | 4 | 0 | Tetrahedral | Tetrahedral ($109.5^\circ$) | $sp^3$ ($\text{CH}_4, \text{CCl}_4, \text{NH}_4^+$) |
| $AX_3E$ | 3 | 1 | Tetrahedral | Trigonal Pyramidal ($107^\circ$) | $sp^3$ ($\text{NH}_3, \text{PCl}_3, \text{H}_3\text{O}^+$) |
| $AX_2E_2$ | 2 | 2 | Tetrahedral | Bent / Angular ($104.5^\circ$) | $sp^3$ ($\text{H}_2\text{O}, \text{H}_2\text{S}, \text{OF}_2$) |
| $AX_5$ | 5 | 0 | Trigonal Bipyramidal | Trigonal Bipyramidal | $sp^3d$ ($\text{PCl}_5, \text{PF}_5, \text{AsF}_5$) |
| $AX_4E$ | 4 | 1 | Trigonal Bipyramidal | Seesaw | $sp^3d$ ($\text{SF}_4, \text{SeF}_4$) |
| $AX_3E_2$ | 3 | 2 | Trigonal Bipyramidal | T-Shaped | $sp^3d$ ($\text{ClF}_3, \text{BrF}_3$) |
| $AX_2E_3$ | 2 | 3 | Trigonal Bipyramidal | Linear ($180^\circ$) | $sp^3d$ ($\text{XeF}_2, \text{I}_3^-$) |
| $AX_6$ | 6 | 0 | Octahedral | Octahedral ($90^\circ$) | $sp^3d^2$ ($\text{SF}_6, \text{SeF}_6, \text{PF}_6^-$) |
| $AX_5E$ | 5 | 1 | Octahedral | Square Pyramidal | $sp^3d^2$ ($\text{BrF}_5, \text{IF}_5$) |
| $AX_4E_2$ | 4 | 2 | Octahedral | Square Planar ($90^\circ$) | $sp^3d^2$ ($\text{XeF}_4, \text{ICl}_4^-$) |
• 3 Equatorial $\text{P-Cl}$ bonds lie in one plane at $120^\circ$ angles ($2.02\text{ \Acirc}$).
• 2 Axial $\text{P-Cl}$ bonds lie perpendicular above and below plane at $90^\circ$ angles ($2.40\text{ \Acirc}$).
Reason: Axial bonds experience 3 strong $90^\circ$ repulsions from equatorial bond pairs, making axial bonds longer and weaker than equatorial bonds!
• Pi ($\pi$) Bond: Formed by sideways (lateral/parallel) overlap of $p$-orbitals perpendicular to internuclear axis. Weaker bond, restricts rotation.
3. Molecular Orbital Theory (MOT) & Paramagnetism of O2
Molecular Orbital Theory (MOT) (Hund & Mulliken) states that atomic orbitals of comparable energy and proper symmetry combine linearly (LCAO) to form molecular orbitals ($\text{BMO}$ and $\text{ABMO}$) belonging to the entire molecule.
$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$.
2. Energy Order for $Z > 7$ ($\text{O}_2, \text{F}_2, \text{Ne}_2$):
$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \mathbf{\sigma 2p_z} < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$.
3. Bond Order Formula:
\text{Bond Order (BO)} = \frac{1}{2} (N_b - N_a)
Where $N_b =$ number of bonding electrons, $N_a =$ number of antibonding electrons.
Relations: $\text{Bond Order} \propto \text{Bond Energy} \propto \frac{1}{\text{Bond Length}}$.
Why Oxygen ($\text{O}_2$) is Paramagnetic
Valence Bond Theory failed to explain why liquid oxygen is attracted to a magnet. MOT provides the definitive proof:
• Bond Order: $\text{BO} = \frac{1}{2} (10 - 6) = \mathbf{2}$ (Double bond $\text{O}=\text{O}$).
• Magnetic Property: Has 2 unpaired electrons in degenerate antibonding $\pi^* 2p_x$ and $\pi^* 2p_y$ orbitals ⇒ PARAMAGNETIC!
4. Intermolecular vs Intramolecular Hydrogen Bonding
Hydrogen Bond is the attractive electrostatic force binding hydrogen atom of one molecule to an electronegative atom ($\text{F}, \text{O}, \text{N}$) of the same or another molecule.
| Feature | Intermolecular Hydrogen Bonding | Intramolecular Hydrogen Bonding |
|---|---|---|
| Definition | H-bond formed between two different molecules of the same or different substances. | H-bond formed within the same single molecule between hydrogen and electronegative atom. |
| Examples | $\text{H}_2\text{O}$, $\text{HF}$, $\text{NH}_3$, Ethanol ($\text{C}_2\text{H}_5\text{OH}$), $p$-nitrophenol. | $o$-nitrophenol, Salicylaldehyde, $o$-hydroxybenzoic acid. |
| Boiling Point Effect | Increases Boiling Point significantly due to molecular association. | Lowers Boiling Point (steam volatile) as no inter-particle association occurs. |
| Water Solubility | High solubility in water. | Lower solubility in water. |
5. Solved Entrance Exam Questions (JEE / NEET)
In $\text{O}_3$ Lewis structure: $\text{O}(1) = \text{O}(2) - \text{O}(3)^-$.
1. Central Oxygen (1): $V=6, L=2, S=6 \implies \text{FC} = 6 - 2 - \frac{1}{2}(6) = \mathbf{+1}$.
2. Double-bonded Oxygen (2): $V=6, L=4, S=4 \implies \text{FC} = 6 - 4 - \frac{1}{2}(4) = \mathbf{0}$.
3. Single-bonded Oxygen (3): $V=6, L=6, S=2 \implies \text{FC} = 6 - 6 - \frac{1}{2}(2) = \mathbf{-1}$.
4 single bonds with Fluorine atoms ($bp = 4$).
Remaining 4 electrons form 2 lone pairs ($lp = 2$).
Total electron pairs = $4 + 2 = 6 \implies$ $sp^3d^2$ Hybridization.
Electron geometry is Octahedral; 2 lone pairs occupy axial positions to minimize repulsions.
Molecular shape = Square Planar ($90^\circ$ bond angles).
• 3 equatorial $\text{P-Cl}$ bonds lie in equatorial plane at $120^\circ$ angles.
• 2 axial $\text{P-Cl}$ bonds lie along axis at $90^\circ$ angles to equatorial plane.
Axial bond pairs experience 3 strong repulsions at $90^\circ$ from equatorial bond pairs, whereas equatorial pairs experience only 2 repulsions at $90^\circ$. To minimize repulsions, axial bonds elongate to $2.40\text{ \Acirc}$ compared to $2.02\text{ \Acirc}$ for equatorial bonds!
1. $\text{O}_2^{2-}$ (18 e¯): $N_b=10, N_a=8 \implies \text{BO} = \frac{1}{2}(10-8) = \mathbf{1.0}$.
2. $\text{O}_2^-$ (17 e¯): $N_b=10, N_a=7 \implies \text{BO} = \frac{1}{2}(10-7) = \mathbf{1.5}$.
3. $\text{O}_2$ (16 e¯): $N_b=10, N_a=6 \implies \text{BO} = \frac{1}{2}(10-6) = \mathbf{2.0}$.
4. $\text{O}_2^+$ (15 e¯): $N_b=10, N_a=5 \implies \text{BO} = \frac{1}{2}(10-5) = \mathbf{2.5}$.
Since $\text{Stability} \propto \text{Bond Order}$: $\mathbf{\text{O}_2^{2-} (1.0) < \text{O}_2^- (1.5) < \text{O}_2 (2.0) < \text{O}_2^+ (2.5)}$.
• $p$-nitrophenol: Forms Intermolecular H-bonding linking adjacent molecules together into large polymeric clusters, requiring higher thermal energy to boil!
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7. Frequently Asked Questions (FAQ)
The VSEPR repulsive force order is: $\text{Lone Pair - Lone Pair (lp-lp)} > \text{Lone Pair - Bond Pair (lp-bp)} > \text{Bond Pair - Bond Pair (bp-bp)}$.
According to Molecular Orbital Theory, $\text{O}_2$ (16 electrons) has 2 unpaired electrons in degenerate antibonding molecular orbitals ($\pi^* 2p_x^1 = \pi^* 2p_y^1$), rendering it paramagnetic.
$\text{XeF}_4$ has $sp^3d^2$ hybridization with an Octahedral electron geometry and a Square Planar molecular shape (4 bonding pairs and 2 axial lone pairs at $90^\circ$).
Formal Charge is calculated using the formula: $\text{FC} = V - L - \frac{1}{2} S$, where $V$ is valence electrons, $L$ is lone pair electrons, and $S$ is shared bonding electrons.
Intermolecular H-bonding occurs between separate molecules (e.g. $\text{H}_2\text{O}, \text{C}_2\text{H}_5\text{OH}$), elevating boiling point. Intramolecular H-bonding occurs within a single molecule (e.g. $o$-nitrophenol), lowering boiling point.
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