1. Introduction & Octet Rule

Atoms combine to attain stability by completing their octet (8 electrons in valence shell), similar to noble gases.

Exceptions to Octet Rule:

Incomplete Octet: Central atom has < 8 e^- (e.g., LiCl, BeH₂, BCl₃).
Expanded Octet (Hypervalent): Central atom has > 8 e^- (e.g., PCl₅, SF₆). Possible due to d-orbitals.
Odd Electron Species: NO, NO₂.

Types of Bonds

Ionic Bond	Covalent Bond
Complete transfer of electrons.	Sharing of electrons.
Metal (Low IE) + Non-metal (High EA).	Non-metal + Non-metal.
Favored by High Lattice Energy.	Can be Polar or Non-polar.

Formal Charge (F.C.):

F.C. = V - L - ½S

(V=Valence e^-, L=Lone pair e^-, S=Shared e^-)

2. Bond Parameters

Bond Order: Number of bonds between two atoms. B.O. ∝ Stability ∝ Bond Energy ∝ 1/Bond Length.
Resonance: When a single Lewis structure cannot explain all properties. The actual structure is a Resonance Hybrid. (e.g., in Ozone, both O-O bond lengths are identical).
Dipole Moment (μ): Measure of polarity. μ = q × d. Unit: Debye (D).
Resultant μ = √(μ₁² + μ₂² + 2μ₁μ₂ cosθ)

3. VSEPR Theory

Valence Shell Electron Pair Repulsion Theory: Shapes depend on the number of valence electron pairs around the central atom.

Repulsion Order:

Lone Pair-Lone Pair > LP-BP > Bond Pair-Bond Pair

[Image of VSEPR molecular geometries chart]

Pairs (BP+LP)	Geometry	Shape (if LP exists)	Example
2 (2+0)	Linear	Linear	BeCl₂
3 (2+1)	Trigonal Planar	Bent / V-shape	SO₂
4 (3+1)	Tetrahedral	Pyramidal	NH₃
4 (2+2)	Tetrahedral	Bent / V-shape	H₂O
5 (4+1)	Trigonal Bipyramidal	See-Saw	SF₄
5 (3+2)	Trigonal Bipyramidal	T-Shape	ClF₃

4. VBT & Hybridization

Sigma (σ) vs Pi (π) Bond:

σ-bond: Head-on overlap (s-s, s-p, p-p). Stronger. Free rotation possible.
π-bond: Sideways overlap (p-p). Weaker. No rotation. Always present with σ.

Hybridization

Intermixing of atomic orbitals to form new equivalent hybrid orbitals.

How to find (Steric Number):

Z = ½ [V + M - C + A]

V = Valence e^-, M = Monovalent atoms, C/A = Charge

sp (Z=2): Linear (180°). e.g., BeCl₂, C₂H₂.
sp² (Z=3): Trigonal Planar (120°). e.g., BCl₃, C₂H₄.
sp³ (Z=4): Tetrahedral (109.5°). e.g., CH₄.
sp³d (Z=5): Trigonal Bipyramidal. e.g., PCl₅.
sp³d² (Z=6): Octahedral. e.g., SF₆.
sp³d³ (Z=7): Pentagonal Bipyramidal. e.g., IF₇.

5. Hydrogen Bonding

Electrostatic attraction between H atom covalently bonded to highly electronegative atom (F, O, N) and another electronegative atom.

Inter-molecular	Between different molecules. Effect: Increases BP, MP, viscosity. Ex: H₂O, HF.
Intra-molecular	Within same molecule. Effect: Decreases BP (prevents association). Ex: o-Nitrophenol.

[Image of Hydrogen bonding in water and ice]

Numericals & HOTS

Q1. Predicting Hybridization

Determine the hybridization and geometry of the central atom in XeF₄ and I₃^-.

Solution:
1. XeF₄: H = 1/2 [8 + 4] = 6 → sp³d².
Geometry: Octahedral; Shape: Square Planar (due to 2 lone pairs).

2. I₃^-: H = 1/2 [7 + 2 + 1] = 5 → sp³d.
Geometry: Trigonal Bipyramidal; Shape: Linear (due to 3 lone pairs in equatorial positions).

Q2. Formal Charge in Ozone

Calculate the formal charge on the central oxygen atom in the Ozone (O₃) molecule.

Solution:
For central Oxygen:
Valence electrons (V) = 6
Lone pair electrons (L) = 2 (one pair)
Shared electrons (S) = 6 (one double bond, one single bond)
F.C. = 6 - 2 - 1/2(6) = 6 - 2 - 3 = +1.

Q3. Polarity Comparison

Why does NF₃ have a much lower dipole moment (0.24 D) than NH₃ (1.47 D) despite Fluorine being more electronegative than Hydrogen?

[Image comparing dipole moment vectors in NH3 and NF3]

Solution:
In NH₃, the N-H bond dipoles are directed towards Nitrogen, which is in the same direction as the lone pair dipole. They reinforce each other.

In NF₃, the N-F bond dipoles are directed towards Fluorine (away from Nitrogen), opposing the lone pair dipole. They partially cancel each other out.

Q4. MOT Application

Arrange the following in increasing order of bond length: O₂, O₂⁺, O₂^-, O₂^2-.

Solution:
Bond Order (B.O.) calculation via MOT:
O₂⁺: 2.5 | O₂: 2.0 | O₂^-: 1.5 | O₂^2-: 1.0

Concept: Bond Length ∝ 1 / Bond Order.
Higher the bond order, shorter the bond.
Order: O₂⁺ < O₂ < O₂^- < O₂^2-

Q5. Melting Point Paradox

Why is SnCl₄ a liquid at room temperature while SnCl₂ is a high-melting solid?

Solution:
According to Fajan's Rule, higher the charge on the cation, the greater its polarizing power and covalent character.

In SnCl₄, Sn is in +4 state (more covalent → weak intermolecular forces → liquid).
In SnCl₂, Sn is in +2 state (more ionic → strong lattice forces → solid).

Q6. Effect of Electronegativity

Compare the bond angles of NH₃ (107°) and PH₃ (93.6°). Why is there such a large difference?

Solution:
1. Electronegativity: N is more EN than P. Bond pairs in NH₃ are closer to N, causing higher BP-BP repulsion and opening the angle.

2. Drago's Rule: In PH₃, the central atom is large and uses almost pure p-orbitals for bonding instead of sp³ hybrid orbitals. Pure p-orbitals are at 90° to each other, hence the angle is close to 90°.

Q7. Bond Length in CO₃^2-

In the carbonate ion, there are two C-O single bonds and one C=O double bond. Yet experimental data shows all three bond lengths are equal (128 pm). Explain.

Solution:
This is due to Resonance. The double bond is not fixed on one Oxygen atom but is delocalized over all three Oxygen atoms. The actual structure is a resonance hybrid where each C-O bond has a partial double bond character (Bond Order = 1.33), making them identical.

Q8. Calculation of Bonds

Count the number of σ and π bonds in Tetracyanoethylene: (CN)₂C=C(CN)₂.

Solution:
Structure: Four -C≡N groups attached to C=C.
σ bonds: 5 (C-C) + 4 (C-N) = 9 σ.
π bonds: 1 (C=C) + 4 × 2 (C≡N) = 9 π.

Q9. Physical States

H₂O is a liquid while H₂S is a gas at room temperature, even though S has a higher molecular mass. Explain.

Solution:
Oxygen is highly electronegative and small in size, which allows H₂O molecules to form strong Intermolecular Hydrogen Bonds. This results in the association of molecules into a liquid state.

Sulphur has low electronegativity and cannot form H-bonds. H₂S molecules are held by weak Van der Waals forces, hence it exists as a gas.

Q10. Hydration vs Lattice Energy

Why is AgF soluble in water while AgI is insoluble?

Solution:
Solubility requires: Hydration Enthalpy > Lattice Enthalpy.

F^- is very small, leading to extremely high hydration energy which overcomes the lattice energy of AgF.
I^- is large, and AgI has significant covalent character (Fajan's Rule), making the lattice very stable and hydration energy insufficient to break it.

Important Formulae & Trends

1. Formal Charge (F.C.)

F.C. = [V] - [L] - ½[S]

V = Valence electrons | L = Lone pair electrons | S = Shared electrons

2. Dipole Moment (μ)

μ = charge (q) × distance (d)

Resultant Dipole:

μ_res = √(μ₁² + μ₂² + 2μ₁μ₂cosθ)

3. Predicting Hybridization (H)

H = ½ [V + M - C + A]

V = Valence e^- on central atom
M = No. of Monovalent atoms (H, F, Cl, Br, I)
C = Cationic charge | A = Anionic charge

4. Fajan's Rule (Covalent Character)

Covalent Character ∝

1 / Size of Cation (Smaller cation = More covalent)
Size of Anion (Larger anion = More covalent)
Charge on Cation/Anion (Higher charge = More covalent)

20 Golden Facts (NEET)

1. Dipole Moment Paradox: NH₃ has a higher dipole moment than NF₃. In NH₃, orbital dipole and bond dipole are in same direction; in NF₃, they oppose each other.
2. Bond Angle Trends: Bond angle decreases as the number of Lone Pairs increases (VSEPR): CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°).
3. Resonance Hybrid: In Ozone (O₃) or Carbonate (CO₃^2-), all bond lengths are identical and intermediate between single and double bonds.
4. Hybridization of Solids: PCl₅ in gaseous state is sp³d (TBP), but in solid state it exists as [PCl₄]⁺ (sp³) and [PCl₆]^- (sp³d²).
5. Isoelectronic Molecules: Species with same no. of electrons often have same shape and bond order (e.g., CO and N₂ have B.O. = 3).
6. H-Bonding Strength: The order of H-bond strength is H—F > H—O > H—N, following the electronegativity of the atom.
7. Ortho- vs Para- Nitrophenol: o-Nitrophenol is steam volatile due to Intramolecular H-bonding, while p-Nitrophenol has Intermolecular H-bonding and higher boiling point.
8. SF₄ Shape: It has 4 BPs and 1 LP. Shape is See-Saw. The lone pair always occupies an equatorial position to minimize repulsion.
9. Back Bonding: BF₃ is a Lewis acid, but its acidity is less than BCl₃ due to pπ-pπ back bonding between B and F.
10. Symmetry & Dipole: Symmetrical molecules like BF₃, CCl₄, SF₆ have μ = 0, even if individual bonds are polar.
11. Bond length in PCl₅: Axial bonds are longer than equatorial bonds due to greater repulsion from equatorial bond pairs.
12. Hybridization in BeCl₂: In vapor phase it is sp (linear); in solid state, it polymerizes to form chloro-bridges with sp³ hybridization.
13. Lattice Enthalpy: Increases with higher ionic charges and smaller ionic radii (U ∝ q₁q₂/r).
14. Paramagnetism of O₂: Cannot be explained by VBT; Molecular Orbital Theory (MOT) explains it due to two unpaired electrons in antibonding orbitals.
15. Bond Order of CO⁺: Exceptional case; B.O. of CO is 3, but B.O. of CO⁺ is 3.5 (electron removed from antibonding orbital).
16. Solubility: Ionic compounds are soluble in polar solvents (high dielectric constant like H₂O) but insoluble in non-polar solvents.
17. Coordinate Bond: Also called Dative bond. Once formed, it is indistinguishable from a regular covalent bond.
18. Melting Point of NaCl: Very high because of strong non-directional electrostatic forces in the 3D crystal lattice.
19. Sigma vs Pi Strength: Sigma bond is always stronger than Pi bond because head-on overlap is more effective than lateral overlap.
20. Formal Charge Rule: The structure with the lowest formal charges on atoms is the most stable and preferred Lewis structure.

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