Chemical Bonding and Molecular Structure

Kössel-Lewis Approach to Chemical Bonding

Lewis Symbols & Octet Rule

• Core definition: Atoms combine by transferring or sharing valence electrons to attain a stable outer octet (ns²np⁶), resembling noble gases. Lewis symbols represent valence electrons as dots around the atomic symbol, indicating group valence.
• Formal Charge (F.C.): Helps select the lowest energy structure (smallest F.C. is most stable). F.C. = (Valence e^-) - (Non-bonding e^-) - ½(Shared e^-).
• Exceptions/Crucial detail: The octet rule is not universal.
  • Incomplete Octet: Central atom has <8 electrons (e.g., LiCl, BeH₂, BCl₃, AlCl₃, BF₃). Act as Lewis acids.
  • Odd-electron molecules: Octet not satisfied for all atoms (e.g., NO, NO₂).
  • Expanded Octet: Central atom has >8 electrons, utilizing d-orbitals (e.g., PF₅, SF₆, H₂SO₄).

PYQ Problem 1:
Amongst the following, the total number of species NOT having eight electrons around central atom in its outer most shell, is: NH₃, AlCl₃, BeCl₂, CCl₄, PCl₅ NEET 2023
(1) 1 (2) 2 (3) 3 (4) 4

Solution:
Step 1 - Analyze Octet: NH₃ (8 e^-), CCl₄ (8 e^-).
Step 2 - Identify Exceptions: AlCl₃ (6 e^-, incomplete), BeCl₂ (4 e^-, incomplete), PCl₅ (10 e^-, expanded).
Conclusion: 3 species do not obey the octet rule. Option (3) is correct.

PYQ Problem 2:
Which of the following is electron-deficient? NEET 2013
(1) (CH₃)₂ (2) (SiH₃)₂ (3) (BH₃)₂ (4) PH₃

Solution:
Step 1 - Define Electron Deficient: Molecules with central atoms having less than 8 valence electrons.
Step 2 - Evaluate Options: C, Si, and P compounds here have complete octets. Boron in BH₃ has 6 e^-.
Conclusion: (BH₃)₂ is electron-deficient. Option (3) is correct.

PYQ Problem 3:
Which one of the following formulae does not correctly represent the bonding capacities of the two atoms involved? AIPMT 1990
(1) H-P-H (with H above and below P)
(2) F-F
(3) O ← N=O (with O-H on N)
(4) H-C=C=O-H

Solution:
Step 1 - Valency Check: Carbon has a maximum valency of 4.
Step 2 - Analyze Structures: In Option (4), the central carbon is forming 5 bonds (=C= and -O-), which violates its bonding capacity.
Conclusion: Option (4) is incorrect.

Ionic or Electrovalent Bond

Formation & Lattice Enthalpy

• Core definition: Formed by the electrostatic attraction between positive and negative ions.
• Reaction condition: Formation is favored by low Ionization Enthalpy of the metal and high negative Electron Gain Enthalpy of the non-metal.
• Lattice Enthalpy: The energy required to completely separate one mole of a solid ionic compound into gaseous ions. It stabilizes the crystal lattice. Lattice energy ∝ Charge / Size.

PYQ Problem 4:
Among the following, which compound will show the highest lattice energy? AIPMT 1993
(1) KF (2) NaF (3) CsF (4) RbF

Solution:
Step 1 - Lattice Energy Dependency: Lattice energy increases as the size of ions decreases (for identical charges).
Step 2 - Compare Cations: Order of cation size: Na⁺ < K⁺ < Rb⁺ < Cs⁺.
Conclusion: NaF has the smallest cation, hence the highest lattice energy. Option (2) is correct.

PYQ Problem 5:
The sequence of ionic mobility in aqueous solution is: AIPMT 2008
(1) Na⁺ > K⁺ > Rb⁺ > Cs⁺
(2) K⁺ > Na⁺ > Rb⁺ > Cs⁺
(3) Cs⁺ > Rb⁺ > K⁺ > Na⁺
(4) Rb⁺ > K⁺ > Cs⁺ > Na⁺

Solution:
Step 1 - Hydration and Mobility: Smaller cations undergo greater hydration, resulting in a larger hydrated ionic radius, which lowers mobility.
Step 2 - Trend: Size order: Na⁺ < K⁺ < Rb⁺ < Cs⁺. Hydrated size order: Na⁺_aq > K⁺_aq > Rb⁺_aq > Cs⁺_aq.
Conclusion: Mobility order is Cs⁺ > Rb⁺ > K⁺ > Na⁺. Option (3) is correct.

Bond Parameters

Covalent Character in Ionic Bonds (Fajans' Rules)

• Core concept: No bond is 100% ionic. Ionic bonds acquire covalent character depending on polarization.
• Fajans' Rules: Covalent character increases with:
  • Smaller cation size.
  • Larger anion size.
  • Greater charge on either ion.
  • Pseudo-noble gas configuration of cation ((n-1)d¹⁰ns⁰) is more polarizing than noble gas configuration.

PYQ Problem 6:
Which of the following compounds has the lowest melting point? AIPMT 2011
(1) CaF₂ (2) CaCl₂ (3) CaBr₂ (4) CaI₂

Solution:
Step 1 - Fajans' Rule Application: More covalent character = lower melting point.
Step 2 - Anion Size: Covalent character increases with larger anion size: I^- > Br^- > Cl^- > F^-.
Conclusion: CaI₂ is most covalent and has the lowest melting point. Option (4) is correct.

Dipole Moment & Polarity

• Core formula: Dipole Moment (μ) = Q × r, expressed in Debye (D). It is a vector quantity.
• Crucial detail: Symmetrical molecules (e.g., BeF₂, BF₃, CCl₄, CO₂) have a net dipole moment of zero because individual bond dipoles cancel out.
• Exception/Comparison: NH₃ (μ = 1.47 D) has a higher dipole moment than NF₃ (μ = 0.23 D). In NH₃, the orbital dipole of the lone pair and the N-H bond dipoles are in the same direction. In NF₃, the N-F bond dipoles oppose the lone pair dipole.

PYQ Problem 7:
Which of the following molecules has "NON ZERO" dipole moment value? NEET 2024
(1) CCl₄ (2) HI (3) CO₂ (4) BF₃

Solution:
Step 1 - Identify Symmetry: CCl₄, CO₂, and BF₃ are highly symmetrical and their bond dipoles cancel out (μ = 0).
Step 2 - Identify Polar Molecule: HI is a heteronuclear diatomic molecule with electronegativity difference, hence polar.
Conclusion: HI has a non-zero dipole moment. Option (2) is correct.

PYQ Problem 8:
The electronegativity difference between N and F is greater than that between N and H yet the dipole moment of NH₃ (1.5 D) is larger than that of NF₃ (0.2 D). This is because: AIPMT 2006
(1) In NH₃ as well as in NF₃ the atomic dipole and bond dipole are in the same direction
(2) In NH₃, the atomic dipole and bond dipole are in the same direction whereas in NF₃ these are in opposite directions
(3) In NH₃ as well as NF₃ the atomic dipole and bond dipole are in opposite directions
(4) In NH₃, the atomic dipole and bond dipole are in the opposite directions whereas, in NF₃ these are in the same directions

Solution:
Step 1 - Direction of Dipoles: In NH₃, bond moments point towards N (same direction as lone pair). In NF₃, bond moments point towards F (opposite to lone pair).
Conclusion: Option (2) is correct.

Bond Order, Length, and Resonance

• Core concept: Bond Order is the number of bonds between two atoms. As bond order increases, bond enthalpy increases and bond length decreases.
• Resonance: A single Lewis structure cannot explain all experimental parameters (e.g., in O₃, both O-O bonds are identical at 128 pm, intermediate between single and double bonds). Resonance averages bond characteristics and stabilizes the molecule.

PYQ Problem 9:
The correct order of C-O bond length among: CO, CO₃^2-, CO₂ is: AIPMT 2007
(1) CO < CO₃^2- < CO₂
(2) CO < CO₂ < CO₃^2-
(3) CO₃^2- < CO₂ < CO
(4) CO₂ < CO₃^2- < CO

Solution:
Step 1 - Determine Bond Order: CO has a triple bond (B.O. = 3). CO₂ has double bonds (B.O. = 2). CO₃^2- has resonance (2 single, 1 double), yielding B.O. = 4/3 = 1.33.
Step 2 - Apply Trend: Bond length ∝ 1 / Bond Order.
Conclusion: Length order: CO < CO₂ < CO₃^2-. Option (2) is correct.

Valence Shell Electron Pair Repulsion (VSEPR) Theory

Postulates & Geometry

• Core concept: Molecular shape is determined by repulsions between valence shell electron pairs. Electron pairs position themselves to minimize repulsion.
• Repulsion Order: Lone pair - Lone pair (lp-lp) > Lone pair - Bond pair (lp-bp) > Bond pair - Bond pair (bp-bp). This causes bond angles to compress (e.g., CH₄ = 109.5°, NH₃ = 107°, H₂O = 104.5°).

Type	Bond Pairs	Lone Pairs	Shape	Example
AB2E	2	1	Bent	SO2, O3
AB3E	3	1	Trigonal Pyramidal	NH3
AB2E2	2	2	Bent / V-shape	H2O
AB4E	4	1	See-saw	SF4
AB3E2	3	2	T-shape	ClF3
AB5E	5	1	Square Pyramidal	BrF5
AB4E2	4	2	Square Planar	XeF4

PYQ Problem 10:
Match List-I with List-II: NEET 2022
List-I (molecules) -> List-II (shape)
A. NH₃ -> I. square pyramidal
B. ClF₃ -> II. trigonal bipyramidal
C. PCl₅ -> III. trigonal pyramidal
D. BrF₅ -> IV. T shape
(1) A-III, B-IV, C-I, D-II
(2) A-II, B-III, C-IV, D-I
(3) A-III, B-IV, C-II, D-I
(4) A-I, B-IV, C-II, D-III

Solution:
Step 1 - Determine VSEPR Type: NH₃ (3bp, 1lp) = Trigonal Pyramidal. ClF₃ (3bp, 2lp) = T-shape. PCl₅ (5bp, 0lp) = Trigonal Bipyramidal. BrF₅ (5bp, 1lp) = Square Pyramidal.
Conclusion: A-III, B-IV, C-II, D-I. Option (3) is correct.

PYQ Problem 11:
Identify the incorrect statement about PCl₅: NEET 2019
(1) Three equatorial P-Cl bonds make an angle of 120° with each other
(2) Two axial P-Cl bonds make an angle of 180° with each other
(3) Axial P-Cl bonds are longer than equatorial P-Cl bonds
(4) PCl₅ molecule is non-reactive

Solution:
Step 1 - Analyze Geometry: PCl₅ is trigonal bipyramidal. Axial bonds suffer more repulsion from equatorial pairs, making them longer and weaker.
Step 2 - Reactivity: Because axial bonds are longer and weaker, PCl₅ is highly reactive.
Conclusion: Option (4) is incorrect.

Valence Bond Theory & Hybridisation

Orbital Overlap & Types of Bonds

• Core concept: Bonds form via partial interpenetration of half-filled atomic orbitals. Extent of overlap determines bond strength.
• σ vs π bonds:
  • Sigma (σ): Head-on/axial overlap (s-s, s-p, p-p). Stronger due to larger overlap extent.
  • Pi (π): Sideways/parallel overlap of unhybridized p-orbitals. Weaker. A double bond = 1 σ + 1 π; Triple bond = 1 σ + 2 π.

PYQ Problem 12:
Which of the following contains equal number of sigma and pi bonds? NEET 2015
(1) (CN)₂ (2) CH₂(CN)₂ (3) HCO₃^- (4) XeO₄

Solution:
Step 1 - Draw Structure of (CN)₂: N ≡ C - C ≡ N.
Step 2 - Count Bonds: Total σ bonds = 3. Total π bonds = 4. (Not equal).
Step 3 - Structure of XeO₄: Central Xe with 4 double bonds to Oxygen (O=Xe=O etc.). Total σ bonds = 4. Total π bonds = 4.
Conclusion: XeO₄ has 4 σ and 4 π bonds. Option (4) is correct.

Hybridisation

• Core definition: Hybridisation is the intermixing of slightly different energy atomic orbitals to form equivalent hybrid orbitals used in bond formation. Number of hybrid orbitals = number of atomic orbitals mixed.
• Geometries:
  • sp: Linear (BeCl₂, C₂H₂).
  • sp²: Trigonal planar (BCl₃, C₂H₄).
  • sp³: Tetrahedral (CH₄, C₂H₆).
  • sp³d: Trigonal bipyramidal (PCl₅).
  • sp³d²: Octahedral (SF₆).

PYQ Problem 13:
Which of the following organic compounds has same hybridisation as its combustion product (CO₂)? NEET 2014
(1) Ethyne (2) Ethene (3) Ethanol (4) Ethane

Solution:
Step 1 - Hybridisation of Combustion Product: CO₂ is O=C=O, which is sp hybridized.
Step 2 - Evaluate Options: Ethyne (C₂H₂) has C ≡ C triple bonds, which are sp hybridized. Ethene is sp², Ethane/Ethanol are sp³.
Conclusion: Ethyne is the correct answer. Option (1) is correct.

Molecular Orbital Theory (MOT)

LCAO & Bonding

• Core concept: Atomic orbitals combine to form Molecular Orbitals (LCAO).
• Bonding MO (σ, π): Formed by constructive interference. Lower energy, stable, e^- density between nuclei.
• Antibonding MO (σ*, π*): Formed by destructive interference. Higher energy, unstable, has a nodal plane.
• Energy Level Sequence:
  • For O₂, F₂: σ1s < σ*1s < σ2s < σ*2s < σ2p_z < π2p_x = π2p_y < π*2p_x = π*2p_y < σ*2p_z.
  • For B₂, C₂, N₂ (due to 2s-2p mixing): σ1s < σ*1s < σ2s < σ*2s < π2p_x = π2p_y < σ2p_z < π*2p_x = π*2p_y < σ*2p_z.
• Bond Order Calculation: B.O. = ½(N_b - N_a).
• Magnetic Character: Unpaired electrons = Paramagnetic. All paired = Diamagnetic. O₂ is paramagnetic (2 unpaired e^- in π*).

PYQ Problem 14:
The correct order of energies of molecular orbitals of N₂ molecule is: NEET 2023
(1) σ 1s < σ* 1s < σ 2s < σ* 2s < (π 2p_x = π 2p_y) < (π* 2p_x = π* 2p_y) < σ 2p_z < σ* 2p_z
(2) σ 1s < σ* 1s < σ 2s < σ* 2s < (π 2p_x = π 2p_y) < σ 2p_z < (π* 2p_x = π* 2p_y) < σ* 2p_z
(3) σ 1s < σ* 1s < σ 2s < σ* 2s < σ 2p_z < (π 2p_x = π 2p_y) < (π* 2p_x = π* 2p_y) < σ* 2p_z
(4) σ 1s < σ* 1s < σ 2s < σ* 2s < σ 2p_z < π* 2p_x < (π 2p_x = π 2p_y) < (π* 2p_x = π* 2p_y)

Solution:
Step 1 - Check 2s-2p Mixing: N₂ exhibits 2s-2p mixing, so the energy of σ2p_z is higher than π2p_x and π2p_y.
Conclusion: The correct sequence is Option (2).

PYQ Problem 15:
Decreasing order of stability of O₂, O₂^-, O₂⁺ and O₂^2- is: NEET 2015
(1) O₂^- > O₂^2- > O₂⁺ > O₂
(2) O₂⁺ > O₂ > O₂^- > O₂^2-
(3) O₂^2- > O₂^- > O₂ > O₂⁺
(4) O₂ > O₂⁺ > O₂^2- > O₂^-

Solution:
Step 1 - Calculate Bond Orders:
O₂ (16 e^-): B.O. = (10 - 6)/2 = 2.0.
O₂⁺ (15 e^-): e^- removed from π*. B.O. = (10 - 5)/2 = 2.5.
O₂^- (17 e^-): e^- added to π*. B.O. = (10 - 7)/2 = 1.5.
O₂^2- (18 e^-): e^- added to π*. B.O. = (10 - 8)/2 = 1.0.
Step 2 - Relate to Stability: Stability ∝ Bond Order.
Conclusion: O₂⁺ (2.5) > O₂ (2.0) > O₂^- (1.5) > O₂^2- (1.0). Option (2) is correct.

Hydrogen Bonding

Conditions & Types

• Core concept: An attractive electrostatic force between a hydrogen atom covalently bonded to a highly electronegative atom (F, O, N) and another highly electronegative atom. Weaker than a covalent bond.
• Intermolecular H-Bonding: Occurs between different molecules (e.g., HF, H₂O, alcohols). Raises boiling points.
• Intramolecular H-Bonding: Occurs within the same molecule (e.g., o-nitrophenol, where H is trapped between two O atoms).

PYQ Problem 16:
Which one of the following compounds shows the presence of intramolecular hydrogen bond? NEET 2016
(1) Cellulose
(2) Concentrated acetic acid
(3) H₂O₂
(4) HCN

Solution:
Step 1 - Define Intramolecular H-Bond: Occurs within the same molecule.
Step 2 - Analyze Options: Cellulose is a polymer that contains numerous intra-chain (and inter-chain) hydrogen bonds stabilizing its structure. (Note: Though o-nitrophenol is the classic NCERT example, cellulose is structurally reliant on intramolecular H-bonds for its rigidity). Acetic acid forms intermolecular dimers. H₂O₂ and HCN form intermolecular H-bonds.
Conclusion: Cellulose forms intramolecular H-bonds. Option (1) is correct.

Select an Option