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Chemical Bonding & Molecular Structure

References

Compiled from: NCERT Chemistry Part-I, Class XI (Chemical Bonding and Molecular Structure) • J.D. Lee (Concise Inorganic) • O.P. Tandon • V.K. Jaiswal • MS Chouhan practice sets • Previous JEE/NEET papers. Target exams: Boards • JEE Main • JEE Advanced • NEET.

Visuals & Diagrams: All key NCERT diagrams (Lewis symbols, Born–Haber cycle, Resonance hybrids, Dipole vector additions, VSEPR 3D geometries, Potential energy curve, Orbital overlap types, σ/π bonds, Hybridisation boxes & flowchart, LCAO MO wavefunctions, and MO energy levels for $N_2$ & $O_2$) are rendered directly as clean, scalable inline vector diagrams with dark/light mode support.

1. Kössel–Lewis Approach & the Octet Rule

Matter is made up of one or different types of elements. Except noble gases, atoms under normal conditions do not exist independently. A group of atoms held together as a stable species is called a molecule. The attractive force holding various constituents (atoms, ions, etc.) together in different chemical species is defined as a chemical bond. Every system tends to attain stability by lowering its potential energy.

1.1 The Kössel–Lewis Foundations

G.N. Lewis's Cubical Atom & Lewis Symbols (1916)
W. Kössel's Postulates on Electrovalent Bonding (1916)

1.2 The Octet Rule & Types of Covalent Bonds

The Octet Rule (Kössel & Lewis, 1916): Atoms combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing valence electrons in order to achieve an octet in their valence shell.

The Covalent Bond (Langmuir, 1919 Refinement): Irving Langmuir refined Lewis's model by introducing the concept of a covalent bond formed by the sharing of electron pairs between combining atoms. Each atom contributes at least one electron to the shared pair, and both atoms attain the nearest noble-gas electronic configuration.

(I) Single Covalent Bond Formation: $\mathrm{Cl_2}$, $\mathrm{H_2O}$, and $\mathrm{CCl_4}$

When two combining atoms share one electron pair, they are joined by a single covalent bond. In $\mathrm{Cl_2}$, each chlorine atom ($[\mathrm{Ne}]3s^2 3p^5$) contributes one electron to the shared pair, completing an octet ($8\mathrm{e^-}$) for both atoms. In $\mathrm{H_2O}$, hydrogen attains a stable duplet of $2\mathrm{e^-}$ while oxygen attains an octet of $8\mathrm{e^-}$. In $\mathrm{CCl_4}$, the central carbon shares four electron pairs with four chlorine atoms so that all five atoms achieve stable octets.

Lewis Dot Structures and Covalent Bonds in Cl2, H2O, and CCl4

Figure 1.1: Covalent single bond formation in $\mathrm{Cl_2}$ and attainment of duplet/octet in $\mathrm{H_2O}$ and $\mathrm{CCl_4}$.

(II) Double Covalent Bond: $\mathrm{CO_2}$ and $\mathrm{C_2H_4}$ (Ethene)

If two combining atoms share two pairs of electrons, the covalent bond between them is called a double bond. In carbon dioxide ($\mathrm{CO_2}$), the carbon atom shares two electron pairs with each of the two oxygen atoms ($:\!\ddot{\mathrm{O}} = \mathrm{C} = \ddot{\mathrm{O}}\!:$). In ethene ($\mathrm{C_2H_4}$), the two carbon atoms share two pairs of electrons ($\mathrm{C=C}$ double bond) and each carbon shares one pair with two hydrogen atoms ($\mathrm{C-H}$ single bonds).

Lewis Structures of Double Bonds in CO2 and C2H4

Figure 1.2: Representation of double covalent bonds in carbon dioxide ($\mathrm{CO_2}$) and ethene ($\mathrm{C_2H_4}$).

(III) Triple Covalent Bond: $\mathrm{N_2}$ and $\mathrm{C_2H_2}$ (Ethyne)

When combining atoms share three electron pairs, a triple bond is formed. In the nitrogen molecule ($\mathrm{N_2}$), each nitrogen atom ($2s^2 2p^3$) contributes three electrons, forming three shared pairs ($:\!\mathrm{N} \equiv \mathrm{N}\!:$) and completing an $8\mathrm{e^-}$ octet on both atoms. In ethyne ($\mathrm{C_2H_2}$), a triple bond connects the two carbon atoms ($\mathrm{C \equiv C}$), and single bonds connect each carbon to a hydrogen atom ($\mathrm{H-C \equiv C-H}$).

Lewis Structures of Triple Bonds in N2 and C2H2

Figure 1.3: Triple bond sharing in nitrogen ($\mathrm{N_2}$) and ethyne ($\mathrm{C_2H_2}$) molecules.

1.3 Step-by-Step Method for Drawing Lewis Structures

The 4-Step Master Algorithm
  1. Count total valence electrons ($n_{\text{total}}$): Add the valence electrons of all constituent atoms. For anions, add $1$ electron for each unit of negative charge. For cations, subtract $1$ electron for each unit of positive charge. $$n_{\text{total}} = \sum (\text{Valence } e^-) + (\text{Negative Charge}) - (\text{Positive Charge})$$
  2. Skeletal framework selection: Write the skeletal structure by placing the least electronegative atom in the central position (e.g., in $\mathrm{NF_3}$ and $\mathrm{CO_3^{2-}}$, $\mathrm{N}$ and $\mathrm{C}$ occupy central positions). Note: Hydrogen ($\mathrm{H}$) and Fluorine ($\mathrm{F}$) always occupy terminal positions.
  3. Allocate single bonds & complete terminal octets: Place one shared pair of electrons ($\mathrm{single\ bond}$) between each adjacent pair of bonded atoms. Use the remaining electrons to satisfy octets ($8\mathrm{e^-}$, or $2\mathrm{e^-}$ for $\mathrm{H}$) on terminal atoms first. Remaining electron pairs are placed as lone pairs on the central atom.
  4. Form multiple bonds for electron-deficient centres: If the central atom does not have an octet, convert one or more lone pairs from surrounding terminal atoms into double or triple bonds. Verify the final structure using Formal Charge calculations.

Table 1.1: The Lewis Representation of Important Molecules & Polyatomic Ions (NCERT Table 4.1)

The standard Lewis representations for neutral molecules and polyatomic ions showing single, double, coordinate, and coordinate-covalent linkages:

NCERT Table 4.1 Lewis Representation of Molecules and Ions

Table 4.1: Lewis dot representations of $\mathrm{H_2}$, $\mathrm{O_2}$, $\mathrm{O_3}$, $\mathrm{NF_3}$, $\mathrm{CO_3^{2-}}$, and $\mathrm{HNO_3}$.

Classic NCERT Worked Problems (Step-by-Step Solutions)

NCERT Problem 4.1 Write the Lewis dot structure of the Carbon Monoxide ($\mathrm{CO}$) molecule.

Step 1: Count total valence electrons: $\mathrm{C} (2s^2 2p^2) \implies 4\mathrm{e^-}$; $\mathrm{O} (2s^2 2p^4) \implies 6\mathrm{e^-}$. Total $= 4 + 6 = \mathbf{10\text{ valence electrons}}$.

Step 2: Skeletal structure: $\mathrm{C} \quad \mathrm{O}$.

Step 3: Draw a single shared pair ($\mathrm{C : O}$) and complete the octet on oxygen with 3 lone pairs ($\mathrm{:\!C} - \overset{\bullet\bullet}{\underset{\bullet\bullet}{\mathrm{O}}}\mathrm{:}$). This leaves $2\mathrm{e^-}$ as a lone pair on carbon. Oxygen has 8 electrons, but carbon has only 4 electrons.

Step 4: Shift two lone pairs from oxygen into the interatomic region to form a triple bond: $$\mathbf{:\!C \equiv O\!: \quad \text{or} \quad :\!C \leftarrow O\!:}$$ Both $\mathrm{C}$ and $\mathrm{O}$ now satisfy the octet rule ($8\mathrm{e^-}$ each).

NCERT Problem 4.2 Write the Lewis structure of the Nitrite Ion ($\mathrm{NO_2^-}$).

Step 1: Count total valence electrons: $\mathrm{N} (2s^2 2p^3) \implies 5\mathrm{e^-}$; $2 \times \mathrm{O} (2s^2 2p^4) \implies 12\mathrm{e^-}$; negative charge ($-1$) $\implies 1\mathrm{e^-}$. Total $= 5 + 12 + 1 = \mathbf{18\text{ valence electrons}}$.

Step 2: Skeletal structure: $\mathrm{O} - \mathrm{N} - \mathrm{O}$.

Step 3: Distribute single bonds ($\mathrm{O : N : O}$) using $4\mathrm{e^-}$, and complete the octets on terminal oxygen atoms using $12\mathrm{e^-}$. The remaining $2\mathrm{e^-}$ form a lone pair on nitrogen ($\mathrm{N}$). Nitrogen now has only 6 electrons ($3\text{ pairs}$).

Step 4: Shift one lone pair from an oxygen atom to make a $\mathrm{N=O}$ double bond: $$\mathbf{\left[ \overset{\bullet\bullet}{\mathrm{O}} = \overset{\bullet\bullet}{\mathrm{N}} - \overset{\bullet\bullet}{\underset{\bullet\bullet}{\mathrm{O}}}\mathrm{:} \right]^- \longleftrightarrow \left[ \mathrm{:}\overset{\bullet\bullet}{\underset{\bullet\bullet}{\mathrm{O}}} - \overset{\bullet\bullet}{\mathrm{N}} = \overset{\bullet\bullet}{\mathrm{O}} \right]^-}$$ All atoms now possess complete octets.

Visual Solution: Lewis Structures of $\mathrm{CO}$ and $\mathrm{NO_2^-}$ (NCERT Problems 4.1 & 4.2)

Lewis Structures of CO and Nitrite Ion NO2-

Figure 1.4: Complete shared-electron diagrams for carbon monoxide ($\mathrm{CO}$) and nitrite ion ($\mathrm{NO_2^-}$).

1.4 Formal Charge — The Stability Decider

In polyatomic molecules/ions, the overall net charge is possessed by the ion as a whole. However, it is feasible to assign a formal charge to each individual atom in the Lewis structure. The formal charge of an atom in a molecule/ion is defined as the difference between the number of valence electrons of that atom in an isolated (free) state and the number of electrons assigned to that atom in the Lewis structure.

Formal Charge Formula (NCERT Definition) $$\text{Formal Charge (F.C.)} = \left[\begin{array}{c} \text{Total number of valence} \\ \text{electrons in free atom } (V) \end{array}\right] - \left[\begin{array}{c} \text{Total number of non-bonding} \\ \text{lone pair electrons } (L) \end{array}\right] - \frac{1}{2}\left[\begin{array}{c} \text{Total number of bonding} \\ \text{shared electrons } (S) \end{array}\right]$$ $$FC = V - L - \frac{1}{2}S$$
⭐ WORKED NCERT BENCHMARK: FORMAL CHARGES IN OZONE ($\mathrm{O_3}$)

Consider the Lewis structure of Ozone ($\mathrm{O_3}$): $\quad \overset{\bullet\bullet}{\mathrm{O}}_1 = \overset{\bullet}{\mathrm{O}}_2 - \overset{\bullet\bullet}{\underset{\bullet\bullet}{\mathrm{O}}}_3\mathrm{:}$

Significance of Formal Charge

1.5 Limitations of the Octet Rule (Direct Exam Question)

Although the octet rule is successful in explaining the bonding of organic and main group compounds, it has three major types of exceptions:

Category Explanation & Electronic Feature NCERT Examples
1. Incomplete Octet of Central Atom Central atom has fewer than 8 valence electrons (electron-deficient compounds). Typically formed by elements with less than 4 valence electrons ($\mathrm{Li, Be, B}$). $\mathrm{LiCl}$ ($2\mathrm{e^-}$), $\mathrm{BeH_2}$ ($4\mathrm{e^-}$), $\mathrm{BeCl_2}$ ($4\mathrm{e^-}$), $\mathrm{BCl_3}$ ($6\mathrm{e^-}$), $\mathrm{BF_3}$ ($6\mathrm{e^-}$)
2. Odd-Electron Molecules Molecules with an odd total number of electrons where the octet rule cannot be satisfied for all atoms simultaneously. Nitric oxide ($\mathrm{NO}$, $11\text{ valence } \mathrm{e^-}$), Nitrogen dioxide ($\mathrm{NO_2}$, $17\text{ valence } \mathrm{e^-}$)
3. The Expanded Octet (Super Octet) Elements in Period 3 and beyond have empty $d$-orbitals ($3d, 4d$) available in addition to $s$ and $p$ orbitals, and can accommodate 10, 12, or 14 electrons around the central atom. $\mathrm{PF_5}$ ($10\mathrm{e^-}$), $\mathrm{SF_6}$ ($12\mathrm{e^-}$), $\mathrm{H_2SO_4}$ ($12\mathrm{e^-}$), $\mathrm{IF_7}$ ($14\mathrm{e^-}$), $\mathrm{PCl_5}$ ($10\mathrm{e^-}$)
4. Noble Gas Compounds The octet rule is based on the chemical inertness of noble gases, yet noble gases (especially Xenon and Krypton) combine with oxygen and fluorine to form stable compounds. $\mathrm{XeF_2, \ XeF_4, \ XeF_6, \ XeOF_2, \ KrF_2}$
5. Shape & Energy Silence The octet theory does not account for the 3D shape/geometry of molecules, nor does it explain the difference in bond energies and relative stabilities of molecules. — (Resolved later by VSEPR, VBT, and MOT)
✍ IN-TEXT PRACTICE 1.1 — NEET / Boards (Formal Charge Mastery)

Q. Assign formal charges to all atoms in the carbonate ion ($\mathrm{CO_3^{2-}}$) with one $\mathrm{C=O}$ double bond and two $\mathrm{C-O^-}$ single bonds.

Carbon atom ($\mathrm{C}$): $V = 4$, $L = 0$, $S = 8 \implies FC = 4 - 0 - \tfrac{1}{2}(8) = \mathbf{0}$.
Double-bonded Oxygen ($\mathrm{O_1}$): $V = 6$, $L = 4$, $S = 4 \implies FC = 6 - 4 - \tfrac{1}{2}(4) = \mathbf{0}$.
Each single-bonded Oxygen ($\mathrm{O_2, O_3}$): $V = 6$, $L = 6$, $S = 2 \implies FC = 6 - 6 - \tfrac{1}{2}(2) = \mathbf{-1}$.
Sum of Formal Charges: $0 + 0 + (-1) + (-1) = \mathbf{-2}$ (matches the net $-2$ charge on $\mathrm{CO_3^{2-}}$).

2. Ionic (Electrovalent) Bond & Lattice Enthalpy

An ionic bond is the electrostatic attraction holding oppositely charged ions together. The formation of an ionic compound involves three decisive thermodynamic factors:

Three Favourable Factors for Ionic Bond Formation
  1. Low Ionisation Enthalpy ($\Delta_i H$) of the metal: Less energy required to form cations ($\mathrm{M_{(g)} \to M^+_{(g)} + e^-}$).
  2. High Negative Electron Gain Enthalpy ($\Delta_{eg} H$) of the non-metal: Greater energy released when non-metal atoms accept electrons ($\mathrm{X_{(g)} + e^- \to X^-_{(g)}}$).
  3. High Lattice Enthalpy ($U$): Large amount of energy released when gaseous ions condense to form 1 mole of crystalline ionic solid ($\mathrm{M^+_{(g)} + X^-_{(g)} \to MX_{(s)}}$). High lattice energy easily overcomes endothermic ionisation enthalpy.
Lattice Enthalpy & Born-Haber Cycle for NaCl $$\Delta H_f^\circ(\mathrm{NaCl}) = \Delta_{\text{sub}}H^\circ(\mathrm{Na}) + \frac{1}{2}\Delta_{\text{diss}}H^\circ(\mathrm{Cl_2}) + \Delta_i H^\circ(\mathrm{Na}) + \Delta_{eg}H^\circ(\mathrm{Cl}) + U_{\text{lattice}}(\mathrm{NaCl})$$

For $\mathrm{NaCl}$: $\Delta H_f^\circ = -411.2\text{ kJ/mol}$, $\Delta_{\text{sub}}H = +108.4$, $\frac{1}{2}\Delta_{\text{diss}}H = +121$, $\Delta_i H = +495.6$, $\Delta_{eg}H = -348.6\text{ kJ/mol} \implies \mathbf{U_{\text{lattice}} = -788\text{ kJ/mol}}$.

Born–Haber Cycle for Formation of Solid Sodium Chloride ($\mathrm{NaCl_{(s)}}$)

$\mathrm{Na_{(s)} + \frac{1}{2}Cl_{2(g)}}$ Direct: $\Delta_f H^\circ = -411.2\text{ kJ/mol}$ (Enthalpy of Formation) $\mathrm{NaCl_{(s)} \text{ (Crystal)}}$ $\Delta_{\text{sub}}H = +108.4$ $\mathrm{Na_{(g)} + \frac{1}{2}Cl_{2(g)}}$ $\Delta_i H = +495.6$ $\mathrm{Na^+_{(g)} + e^- + \frac{1}{2}Cl_{2(g)}}$ $\frac{1}{2}\Delta_{\text{diss}}H = +121.0$ $\mathrm{Na^+_{(g)} + Cl_{(g)} + e^-}$ $\Delta_{eg}H = -348.6$ $\mathrm{Na^+_{(g)} + Cl^-_{(g)}}$ $U_{\text{lattice}} = -788\text{ kJ/mol}$ Hess's Law Verification: $\Delta_f H^\circ = \Delta_{\text{sub}}H + \Delta_i H + \frac{1}{2}\Delta_{\text{diss}}H + \Delta_{eg}H + U$ $-411.2 = 108.4 + 495.6 + 121 - 348.6 - 788$ ✓

3. Bond Parameters — Length, Angle, Enthalpy, Order, Resonance, Polarity

Parameter Meaning Key trends
Bond length Equilibrium distance between nuclei of bonded atoms (pm); sum of covalent radii Single > double > triple (C–C 154, C=C 134, C≡C 120 pm)
Bond angle Angle between orbitals/bonds at the central atom Decided by hybridisation + lone pairs (Section 5)
Bond enthalpy Energy to break 1 mol of a particular bond in the gas phase More bond order → stronger; in polyatomics use MEAN bond enthalpy (e.g., average of the two O–H breakings in water)
Bond order Number of shared pairs between two atoms ↑BO ⟹ ↑enthalpy, ↓length. Isoelectronic species share BO: N₂, CO, NO⁺ all have BO 3

3.1 Resonance & Resonance Hybrids

When a single Lewis structure cannot accurately depict all observed properties of a molecule or ion, the actual structure is represented by a resonance hybrid of several canonical forms (Lewis structures) that differ only in electron positions.

NCERT Problem 4.3 Explain the structure of the carbonate ion ($\mathrm{CO_3^{2-}}$) in terms of resonance.

The single Lewis structure based on the presence of two single bonds and one double bond between carbon and oxygen is inadequate because it represents unequal bonds. According to experimental findings, all three $\mathrm{C-O}$ bonds in $\mathrm{CO_3^{2-}}$ are equivalent. Therefore, the carbonate ion is best described as a resonance hybrid of three canonical forms $\mathrm{(I, II, III)}$:

Resonance in Carbonate Ion ($\mathrm{CO_3^{2-}}$): Canonical Forms & Resonance Hybrid

C :O: :O:⁻ :O:⁻ (I) C :O:⁻ :O: :O:⁻ (II) C :O:⁻ :O:⁻ :O: (III) ≡ C O⁻⅔ O⁻⅔ O⁻⅔ Resonance Hybrid All C–O bonds = 128 pm (B.O. = 1.33)
NCERT Problem 4.4 Explain the structure of the Carbon Dioxide ($\mathrm{CO_2}$) molecule in terms of resonance.

The experimentally determined carbon-to-oxygen bond length in $\mathrm{CO_2}$ is $\mathbf{115\text{ pm}}$. The lengths of a normal $\mathrm{C=O}$ double bond and $\mathrm{C\equiv O}$ triple bond are $121\text{ pm}$ and $110\text{ pm}$ respectively. Because $115\text{ pm}$ lies intermediate between $\mathrm{C=O}$ and $\mathrm{C\equiv O}$, $\mathrm{CO_2}$ is described as a hybrid of three resonance forms:

Resonance in Carbon Dioxide ($\mathrm{CO_2}$): Canonical Structures & Hybrid

:O⁺ C :O:⁻ (I) :O: C :O: (II) [Major] :O:⁻ C :O⁺ (III)
NCERT Misconceptions & Facts About Resonance

3.2 Bond Polarity & Dipole Moment

Dipole Moment Formula $$\text{Dipole Moment } (\mu) = \text{Charge } (Q) \times \text{Distance of separation } (r)$$

Unit: Debye (D), where $1\text{ D} = 3.33564 \times 10^{-30}\text{ C m}$. It is a vector quantity with magnitude and direction (crossed arrow $\mapsto$ pointing from positive to negative centre).

⭐ DIPOLE MOMENT BENCHMARKS & VECTOR SUMMATION

Vector Addition of Dipole Moments: $\mathrm{NH_3}$ (Reinforcing) vs $\mathrm{NF_3}$ (Opposing)

μ (lone pair) N H H H Resultant μ = 1.47 D $\mathrm{NH_3}$ (Dipoles Reinforce) μ (lone pair) N F F F Resultant μ = 0.24 D $\mathrm{NF_3}$ (Dipoles Oppose)

3.3 Partial Covalent Character of Ionic Bonds — Fajans' Rules

Covalent character in an ionic bond increases with polarisation of the anion by the cation. Covalency is favoured by:

✍ IN-TEXT PRACTICE 3.1 — JEE Main (Percent ionic character)

Q. HCl has bond length 127 pm and observed dipole moment 1.03 D. Find its percent ionic character. (e = 4.8 × 10⁻¹⁰ esu; 1 D = 10⁻¹⁸ esu·cm, or work in SI)

Fully ionic μ: $\mu_{ionic} = e\times d = 4.8\times10^{-10}\times1.27\times10^{-8} = 6.1$ D
% ionic: $\dfrac{1.03}{6.1}\times100 \approx \mathbf{17\%}$ — HCl is dominantly covalent.
✍ IN-TEXT PRACTICE 3.2 — NEET (Bond order via resonance)

Q. Find the average carbon–oxygen bond order in CO₃²⁻ and compare the C–O bond length in CO₃²⁻, CO₂ and CO.

BO(CO₃²⁻): 4 bonds ÷ 3 positions = 4/3 ≈ 1.33; BO(CO₂) = 2; BO(CO) = 3.
Length (inverse of BO): CO < CO₂ < CO₃²⁻.

4. VSEPR Theory — Predicting Shapes

Postulates (Sidgwick–Powell, refined by Nyholm–Gillespie)

4.1 The Master Shape Table (ABxEy)

Type bp lp Electron geometry Molecular shape Angle Examples
AB₂ 2 0 Linear Linear 180° BeCl₂, CO₂, HgCl₂
AB₃ 3 0 Trigonal planar Trigonal planar 120° BF₃, SO₃, NO₃⁻, CO₃²⁻
AB₂E 2 1 Trigonal planar Bent <120° (~119°) SO₂, O₃, SnCl₂, NO₂⁻
AB₄ 4 0 Tetrahedral Tetrahedral 109.5° CH₄, CCl₄, SO₄²⁻, NH₄⁺
AB₃E 3 1 Tetrahedral Trigonal pyramidal 107° NH₃, PCl₃, H₃O⁺
AB₂E₂ 2 2 Tetrahedral Bent (V-shape) 104.5° H₂O, OF₂, H₂S, SCl₂
AB₅ 5 0 Trigonal bipyramidal TBP 120° eq, 90° ax PCl₅, PF₅
AB₄E 4 1 TBP See-saw <120°, <90° SF₄
AB₃E₂ 3 2 TBP T-shape <90° ClF₃, BrF₃
AB₂E₃ 2 3 TBP Linear 180° XeF₂, I₃⁻
AB₆ 6 0 Octahedral Octahedral 90° SF₆, PF₆⁻
AB₅E 5 1 Octahedral Square pyramidal <90° BrF₅, IF₅, XeOF₄
AB₄E₂ 4 2 Octahedral Square planar 90° XeF₄, ICl₄⁻
⭐ TBP LONE-PAIR RULE + ANGLE LADDER

In trigonal bipyramidal arrangements, lone pairs ALWAYS occupy equatorial positions (only two 90° neighbours instead of three). Angle-squeeze ladder to memorise: CH₄ (109.5°) → NH₃ (107°) → H₂O (104.5°) — each extra lone pair bites ~2.5°. Down a group the angle falls further as the central atom grows: NH₃ > PH₃ > AsH₃; H₂O > H₂S.

VSEPR Molecular Shapes — All 11 Key Shapes with Bond Angles

SN 2–3 (No Lone Pairs) A B B Linear 180° BeCl₂, CO₂ A B B B Trigonal Planar 120° BF₃, SO₃ A lp B B Bent <120° SO₂, O₃, NO₂⁻ SN 4 A B B B B Tetrahedral 109.5° CH₄, CCl₄, NH₄⁺ A lp B B B Trigonal Pyramidal 107° NH₃, PCl₃ A lp lp B B Bent / V-shape 104.5° H₂O, H₂S SN 5 — Trigonal Bipyramidal Parent A B ax B ax B eq B B Trig. Bipyramidal PCl₅ — ax:90°, eq:120° A lp See-saw SF₄ (1 lp equatorial) SN 6 — Octahedral Parent A Octahedral 90° SF₆, PF₆⁻ A lp Square Pyramidal BrF₅, IF₅ lp–lp > lp–bp > bp–bp repulsion │ Lone pairs always occupy EQUATORIAL positions in TBP │ XeF₄ is Square Planar (2 lp trans)
✍ IN-TEXT PRACTICE 4.1 — NEET (Shape prediction)

Q. Predict the shapes of (a) XeF₄, (b) ClF₃, (c) I₃⁻ using VSEPR.

(a) XeF₄: Xe has 8 v.e.; 4 bonds + 2 lp → SN 6, AB₄E₂ = square planar.
(b) ClF₃: 7 v.e.; 3 bonds + 2 lp → SN 5, AB₃E₂ = T-shaped (lps equatorial).
(c) I₃⁻: central I: 7 + 1(charge) = 8 v.e.; 2 bonds + 3 lp → SN 5, AB₂E₃ = linear.

5. Valence Bond Theory — Overlap, σ and π Bonds

The VBT Picture (Heitler–London → Pauling)

Fig. 4.7 — Potential Energy Curve for the Formation of a Gaseous $\mathrm{H_2}$ Molecule

Potential Energy (kJ·mol⁻¹) → 0 Isolated atoms (H + H) Internuclear Distance (r in pm) → Unstable state (Parallel spins ↑↑ — Net Repulsion) Stable $\mathrm{H_2}$ Molecule (Opposite spins ↑↓ — Net Attraction) Bond Enthalpy = 435.8 kJ·mol⁻¹ Minimum PE at $r_0 = \mathbf{74\text{ pm}}$ 74 pm + + Approaching $\mathrm{H_2}$ Molecule ($r_0 = 74\text{ pm}$)

Fig. 4.9 — Types of Orbital Overlaps (Positive, Negative, Zero)

(i) Positive Overlap (Bonding) + + s–s (same phase) + + − s–p (in-phase) + − + − p–p axial (σ) (ii) Negative Overlap + − + s–p (out-of-phase → destabilises) (iii) Zero Overlap (No bond) + + − s–p (perpendicular → cancel: net = 0) + + − − p–p lateral (+/+) = π bond

5.1 σ vs π Bonds

Feature σ (sigma) bond π (pi) bond
Overlap Head-on (axial): s–s, s–p, p–p end-to-end Sideways (lateral): p–p parallel lobes
Electron cloud Symmetric about the bond axis Above and below the axis (nodal plane contains the axis)
Strength Stronger (larger overlap) Weaker; exists only alongside a σ bond
Rotation Free rotation possible Restricts rotation (cis/trans isomerism)
Count First bond of any pair 2nd and 3rd bonds (double = 1σ+1π; triple = 1σ+2π)
⭐ σ/π COUNTING DRILL

CO₂: 2σ + 2π. C₂H₄: 5σ + 1π. C₂H₂: 3σ + 2π. Benzene C₆H₆: 12σ + 3π. HCN: 2σ + 2π. Count σ = total bonds drawn as lines; π = extra strokes of double/triple bonds. Asked verbatim in NEET.

σ Bond vs π Bond — Orbital Overlap Comparison

σ (sigma) Bond — Head-on Overlap + − + − overlap region ←─ bond axis ─→ Electron density ON the axis s–s, s–p, or p–p end-to-end Free rotation possible │ Stronger bond FIRST bond formed in any pair π (pi) Bond — Lateral (Sideways) Overlap + − + − ← bond axis (nodal plane) → ↑ overlap above ↓ overlap below Electron density ABOVE + BELOW axis p–p parallel lobes (sideways) Restricts rotation │ Weaker bond 2nd/3rd bond only (alongside σ)

6. Hybridisation — The Geometry Engine

Concept

Hybridisation = intermixing of valence orbitals of nearly equal energy on the SAME atom to produce an equal number of identical hybrid orbitals with fixed directions. Hybrids overlap better than pure orbitals (stronger σ bonds) and their mutual repulsion fixes the geometry. Conditions: valence-shell orbitals of comparable energy; promotion of an electron is allowed but not required; even filled orbitals (future lone pairs) can occupy hybrids. Hybrid orbitals form σ bonds only — π bonds always use the leftover unhybridised p orbitals.

6.1 The Hybridisation Table

Hybridisation Orbitals mixed Geometry Angle Examples
sp 1s + 1p Linear 180° BeCl₂, C₂H₂, CO₂, HgCl₂
sp² 1s + 2p Trigonal planar 120° BCl₃, C₂H₄, SO₂*, graphite
sp³ 1s + 3p Tetrahedral 109.5° CH₄, NH₃*, H₂O*, diamond, NH₄⁺
sp³d 1s + 3p + 1d(z²) Trigonal bipyramidal 120°, 90° PCl₅ (2 axial bonds longer than 3 equatorial!)
sp³d² 1s + 3p + 2d Octahedral 90° SF₆
sp³d³ 1s + 3p + 3d Pentagonal bipyramidal 72°, 90° IF₇

*shape ≠ geometry when lone pairs sit in hybrids: NH₃ is sp³ but pyramidal; H₂O is sp³ but bent; SO₂ is sp² but bent.

Steric-Number Shortcut (fastest exam method)

$\text{SN} = \sigma\text{-bonds} + \text{lone pairs on central atom}$ → SN 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d². Carbon quick-read: only single bonds → sp³; one double → sp²; one triple or two doubles → sp. s-character controls properties: more s-character (sp 50% > sp² 33% > sp³ 25%) ⟹ shorter/stronger bonds and higher electronegativity of that carbon.

6.2 Classic Walkthroughs (Board Favourites)

Hybridization Orbital Energy Box Diagrams — sp, sp², sp³, sp³d, sp³d²

sp (Be in BeCl₂) Before: ↑↓ 2s × 2p 2p → After: ↑ sp ↑ sp p (unhyb) p (unhyb) 2 sp hybrids → Linear 180° | BeCl₂, C₂H₂ sp² (B in BF₃) Before: ↑↓ 2s ↑ 2px 2py 2pz → ↑ sp² ↑ sp² ↑ sp² pz (π) 3 sp² hybrids → Trigonal Planar 120° + 1 unhyb p for π | BF₃, C₂H₄ sp³ (C in CH₄) Before: ↑↓ 2s ↑ ↑ ↑ 2px 2py 2pz → ↑ sp³ ↑ sp³ ↑ sp³ ↑ sp³ 4 sp³ hybrids → Tetrahedral 109.5° | CH₄, NH₃, H₂O sp³d (P in PCl₅) 5 hybrid orbitals = 3 equatorial (120°) + 2 axial (90°) Trigonal Bipyramidal │ d orbital from 3d sub-shell Axial bonds LONGER than equatorial (3 vs 2 repulsions) sp³d² (S in SF₆) 6 equivalent hybrid orbitals pointing at octahedron vertices Octahedral │ all bond angles 90° All bonds equivalent │ SF₆ is exceptionally inert SN → Hybridisation Quick Key SN 2 → sp │ SN 3 → sp² │ SN 4 → sp³ SN 5 → sp³d │ SN 6 → sp³d² SN = (number of σ bonds) + (lone pairs on central atom) π bonds always use UNHYBRIDISED p orbitals

Hybridisation Decision Flowchart (Steric Number Method)

Step 1 Pick CENTRAL atom (least electroneg.) Step 2 Count σ-bonds (each bond = 1σ) Step 3 Count lone pairs (lp) on central atom SN = σ + lp Steric Number SN 2 → sp → Linear 180° BeCl₂, CO₂, C₂H₂, HCN, NO₂⁺ SN 3 → sp² → Trig. Planar 120° BF₃, SO₃; SO₂/O₃ bent if 1 lp SN 4 → sp³ → Tetrahedral 109.5° CH₄; NH₃ pyramidal; H₂O bent SN 5 → sp³d → Trig. Bipyramidal PCl₅; SF₄ see-saw; ClF₃ T; XeF₂ linear SN 6 → sp³d² → Octahedral 90° SF₆; BrF₅ sq. pyramidal; XeF₄ sq. planar Key Rules to Remember: • π bonds always use UNHYBRIDISED p orbitals (NOT counted in SN) • Lone pairs sit in hybrids → shape ≠ geometry (NH₃: sp³ but pyramidal) • More s-character → shorter bond, higher EN: sp (50%) > sp² (33%) > sp³ (25%)
✍ IN-TEXT PRACTICE 6.1 — NEET / JEE (Hybridisation spotting)

Q. State the hybridisation of the central atom in: (a) SO₄²⁻, (b) NH₄⁺, (c) XeF₂, (d) each carbon of CH₃–CH=CH₂.

(a) S: 4σ + 0 lp → sp³; (b) N: 4σ + 0 lp → sp³; (c) Xe: 2σ + 3 lp → SN 5 → sp³d (linear).
(d) CH₃ carbon: sp³; both alkene carbons (one double bond each): sp².

7. Molecular Orbital Theory (MOT)

Molecular Orbital Theory (F. Hund & R.S. Mulliken, 1932) considers that atomic orbitals combine to form molecular orbitals (MOs) spread over all nuclei in the molecule. Electrons occupy MOs following the Aufbau principle, Pauli exclusion principle, and Hund's rule.

Linear Combination of Atomic Orbitals (LCAO)

LCAO Combination: Formation of Bonding ($\sigma$) and Antibonding ($\sigma^*$) Molecular Orbitals

$\psi_A$ + $\psi_B$ $\sigma 1s$ Bonding MO ($\psi = \psi_A + \psi_B$) Lower energy, high electron density between nuclei $\psi_A$ − $\psi_B$ + − $\sigma^* 1s$ Antibonding MO ($\psi^* = \psi_A - \psi_B$) Higher energy, nodal plane (zero density) at centre

7.1 The Two Energy Sequences (Board & Competitive Exam Core)

⭐ THE TWO ENERGY SEQUENCES IN MOT

1. For $Z \le 7$ ($14$ or fewer valence/total electrons: $\mathrm{Li_2, Be_2, B_2, C_2, N_2}$): Due to $2s-2p$ mixing, $\sigma 2p_z$ is pushed to a higher energy than the degenerate $\pi 2p_x, \pi 2p_y$ pair:

$$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < (\pi 2p_x = \pi 2p_y) < \sigma 2p_z < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$

2. For $Z > 7$ (more than $14$ electrons: $\mathrm{O_2, F_2, Ne_2}$): Negligible $2s-2p$ mixing $\implies \sigma 2p_z$ is lower in energy than $\pi 2p$:

$$\sigma 1s < \sigma^* 1s < \sigma 2s < \sigma^* 2s < \sigma 2p_z < (\pi 2p_x = \pi 2p_y) < (\pi^* 2p_x = \pi^* 2p_y) < \sigma^* 2p_z$$

MO Energy Level Diagrams: $Z \le 7$ ($\mathrm{N_2}$, with $2s-2p$ mixing) vs $Z > 7$ ($\mathrm{O_2}$, paramagnetism)

$\mathrm{N_2}$ ($14\text{ }e^-$): $2s-2p$ Mixing Active Energy (E) ↑ $\sigma^* 2p_z$ $\pi^* 2p_x = \pi^* 2p_y$ ↿⇂ $\sigma 2p_z$ ↿⇂ ↿⇂ $\pi 2p_x = \pi 2p_y$ ↿⇂ $\sigma^* 2s$ ↿⇂ $\sigma 2s$ $\text{Bond Order} = \frac{10 - 4}{2} = \mathbf{3}$ (Diamagnetic) $\mathrm{O_2}$ ($16\text{ }e^-$): Paramagnetic Ground State Energy (E) ↑ $\sigma^* 2p_z$ ↿ ↿ $\pi^* 2p_x^1 = \pi^* 2p_y^1$ (2 Unpaired $e^-$) ↿⇂ ↿⇂ $\pi 2p_x = \pi 2p_y$ ↿⇂ $\sigma 2p_z$ ↿⇂ $\sigma^* 2s$ ↿⇂ $\sigma 2s$ $\text{Bond Order} = \frac{10 - 6}{2} = \mathbf{2}$ (Paramagnetic!)

7.2 Homonuclear Diatomics Series & Key NCERT Cases

Species Total $e^-$ MO Electronic Configuration $N_b$ $N_a$ Bond Order Magnetic Character
$\mathrm{H_2}$ 2 $\sigma 1s^2$ 2 0 1 Diamagnetic
$\mathrm{He_2}$ 4 $\sigma 1s^2 \, \sigma^* 1s^2$ 2 2 0 Does not exist
$\mathrm{Li_2}$ 6 $\mathrm{KK} \, \sigma 2s^2$ 4 2 1 Diamagnetic
$\mathrm{Be_2}$ 8 $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2$ 4 4 0 Does not exist
$\mathrm{B_2}$ 10 $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2 \, (\pi 2p_x^1 = \pi 2p_y^1)$ 6 4 1 Paramagnetic (2 unpaired $e^-$)
$\mathrm{C_2}$ 12 $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2 \, (\pi 2p_x^2 = \pi 2p_y^2)$ 8 4 2 Diamagnetic (Both bonds are $\pi$ bonds!)
$\mathrm{N_2}$ 14 $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2 \, (\pi 2p_x^2 = \pi 2p_y^2) \, \sigma 2p_z^2$ 10 4 3 Diamagnetic (1 $\sigma$ + 2 $\pi$)
$\mathrm{O_2}$ 16 $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2 \, \sigma 2p_z^2 \, (\pi 2p_x^2 = \pi 2p_y^2) \, (\pi^* 2p_x^1 = \pi^* 2p_y^1)$ 10 6 2 Paramagnetic (2 unpaired $e^-$)
$\mathrm{F_2}$ 18 $\mathrm{KK} \, \sigma 2s^2 \, \sigma^* 2s^2 \, \sigma 2p_z^2 \, (\pi 2p_x^2 = \pi 2p_y^2) \, (\pi^* 2p_x^2 = \pi^* 2p_y^2)$ 10 8 1 Diamagnetic
$\mathrm{Ne_2}$ 20 $\dots \, \sigma^* 2p_z^2$ 10 10 0 Does not exist
NCERT Highlights & Exam Exceptions in MOT

8. Hydrogen Bonding

Definition & Condition

When H is covalently bonded to a small, highly electronegative atom (F, O, N only), the strongly δ+ hydrogen is attracted to a lone pair of another electronegative atom — a hydrogen bond (energy ~10–40 kJ/mol: much weaker than a covalent bond, much stronger than van der Waals).

Hydrogen Bonding — Intermolecular (HF & H₂O) and Intramolecular (o-Nitrophenol)

HF — Intermolecular H-bond Chain (Fig. 4.19) F δ− H δ+ H-bond F δ− H δ+ F H F H ··· Zigzag chain: F–H···F–H···F–H··· (strongest H-bond: F is most electronegative) H₂O — Intermolecular H-bonds (up to 4 per molecule) O H δ+ H δ+ 2 lp O H-bond O O O Each O can donate 2 H-bonds + accept 2 → tetrahedral network → ice open structure → lower density Intra- vs Intermolecular H-bonding (o- vs p-Nitrophenol) o-Nitrophenol (intramolecular) O H δ+ N O O δ− intramolecular H-bond (ring closure) Ring locked → no chains → LOWER b.p. (steam volatile) p-Nitrophenol (intermolecular) O H δ+ O H-bond (intermol.) NO₂ No ring lock → chains form → HIGHER b.p. (not steam volatile) Strength order: F–H···F > O–H···O > N–H···N | H-bond energy: ~10–40 kJ/mol (weak vs ~400 kJ/mol covalent) Boiling points: H₂O (100°C) > HF (19.5°C) > NH₃ (−33°C) — anomalously high due to H-bonding
Type Where Examples Consequence
Intermolecular Between different molecules H₂O, HF (zig-zag chains), NH₃, alcohols, DNA base pairs Raises b.p./m.p., association: H₂O liquid while H₂S is gas; ice less dense than water (open cage); carboxylic acids dimerise
Type Where Examples Consequence
Intramolecular Within ONE molecule (ring closure) o-nitrophenol, salicylaldehyde, o-fluorophenol LOWERS b.p. vs the para isomer (no chains form) — o-nitrophenol is steam-volatile, p-nitrophenol is not
✍ IN-TEXT PRACTICE 8.1 — NEET (H-bond reasoning)

Q. Explain: (a) H₂O is a liquid but H₂S a gas at room temperature; (b) o-nitrophenol has a lower boiling point than p-nitrophenol.

(a) O is small and highly electronegative → water forms extensive intermolecular H-bonds (up to 4 per molecule); S cannot → H₂S has only weak van der Waals forces.
(b) The ortho isomer's O–H locks onto the adjacent NO₂ within the molecule (intramolecular), preventing chain association; the para isomer H-bonds between molecules, so it needs more heat to separate.

9. Solved Examples — Every Question Type in the Chapter

One worked model of each distinct question type asked from this chapter, tagged by exam. Attempt each yourself first, then check.

✍ TYPE 1 — Boards · Define, State & Explain

Q. Define octet rule and list its limitations (3 marks). / Explain resonance with O₃ (3 marks). / State VSEPR postulates and predict the shape of H₂O (3 marks). / Why is O₂ paramagnetic? (2 marks — MOT only!)

Where to answer from: Sections 1 & 1.3 (table), 3.1, 4 (postulates + AB₂E₂ logic), 7.2 (two unpaired π* electrons — VBT cannot explain it; MOT can: quote the electron configuration).
✍ TYPE 2 — NEET · Identify Shape + Hybridisation Together

Q. Match: SF₄, ClF₃, XeF₄, BrF₅ with their shapes and hybridisations.

SF₄: sp³d, see-saw (AB₄E). ClF₃: sp³d, T-shape (AB₃E₂). XeF₄: sp³d², square planar (AB₄E₂). BrF₅: sp³d², square pyramidal (AB₅E). Lone pairs equatorial in the sp³d cases.
✍ TYPE 3 — JEE Main · Bond-Angle Comparison

Q. Arrange in decreasing bond angle: CH₄, NH₃, H₂O, NH₄⁺, and separately NH₃ vs PH₃.

Lone pairs squeeze: NH₄⁺ = CH₄ (109.5°, 0 lp) > NH₃ (107°, 1 lp) > H₂O (104.5°, 2 lp).
Down the group: NH₃ (107°) > PH₃ (~93°) — bigger central atom, more p-character in bonds, lp spreads more.
✍ TYPE 4 — NEET · Dipole-Moment Reasoning

Q. BF₃ has zero dipole moment while NF₃ has a small one, and NH₃ a large one. Explain all three in two lines each.

BF₃: trigonal planar — three equal B–F vectors at 120° sum to zero.
NH₃ vs NF₃: both pyramidal; lone-pair moment ADDS to the N–H resultant (μ = 1.47 D) but OPPOSES the oppositely-directed N–F resultant (μ = 0.23 D).
✍ TYPE 5 — JEE Main · MOT Configuration → BO → Magnetism

Q. Write the MO configuration of N₂ and O₂⁻; give bond order and magnetic behaviour of each.

N₂ (14e⁻, mixed order): σ1s²σ*1s²σ2s²σ*2s² (π2p_x²=π2p_y²) σ2p_z² → BO = (10−4)/2 = 3, diamagnetic.
O₂⁻ (17e⁻, normal order): …σ2p_z² π2p⁴ π*2p³ → BO = (10−7)/2 = 1.5, paramagnetic (1 unpaired).
✍ TYPE 6 — JEE Advanced Level · Species That Do / Don't Exist

Q. Using MOT, decide which exist: He₂, He₂⁺, Be₂, H₂⁻.

He₂: BO 0 — no. He₂⁺ (3e⁻): BO = (2−1)/2 = 0.5 — exists (weakly). Be₂: BO 0 — no. H₂⁻ (3e⁻): BO 0.5 — exists but fragile. Rule: BO > 0 ⟹ existence possible.
✍ TYPE 7 — NEET · σ/π Count & Bond-Length Order

Q. Count σ and π bonds in CH₂=CH–C≡N, and order the three carbon–carbon/carbon–nitrogen bond lengths.

Count: σ: 3 C–H + C–C + C=C(1) + C≡N(1) = 6σ; π: 1 (C=C) + 2 (C≡N) = 3π.
Lengths: C≡N < C=C < C–C (higher bond order = shorter).
✍ TYPE 8 — JEE Main · Fajans / Covalent-Character Order

Q. Arrange in increasing covalent character: NaCl, MgCl₂, AlCl₃; and LiF, LiCl, LiBr, LiI.

Cation charge ↑, size ↓: NaCl < MgCl₂ < AlCl₃.
Anion size ↑ (polarisability): LiF < LiCl < LiBr < LiI.
✍ TYPE 9 — Boards / NEET · Boiling-Point Explanations (H-bond)

Q. Arrange and explain: HF, HCl, HBr, HI by boiling point.

HF is the outlier: strong intermolecular H-bonding lifts it to the top; the rest rise with molar mass (van der Waals): HCl < HBr < HI < HF.
✍ TYPE 10 — JEE Advanced Level · Isoelectronic + Odd Species Combo

Q. CO, CN⁻, NO⁺ and N₂ are isoelectronic. State their common bond order; then decide the bond order and magnetism of NO.

14-electron family: all have BO = 3 (diamagnetic).
NO (15e⁻): one electron enters π* → BO = (10−5)/2 = 2.5, paramagnetic; losing that electron gives NO⁺ with BO 3 — why NO⁺ is more stable than NO.

10. Common Mistakes & Misconceptions — Final Checklist

Night Before Exam

Read this the night before the exam:

  1. Formal charge ≠ real charge — it's a bookkeeping device; pick the structure with the LOWEST formal charges.
  2. For anions ADD electrons to the valence count, for cations SUBTRACT (CO₃²⁻ has 24, NH₄⁺ has 8).
  3. Octet exceptions: BeH₂/BF₃ (incomplete), NO/NO₂ (odd), PF₅/SF₆ (expanded), XeF₂ etc. (noble gas) — quote at least one of each.
  4. Resonance structures differ ONLY in electron positions; atoms never move; the hybrid is the single real molecule.
  5. Shape ≠ electron geometry when lone pairs exist: NH₃ is sp³/tetrahedral-geometry but PYRAMIDAL shape.
  6. Steric number counts a double or triple bond as ONE domain.
  7. In TBP, lone pairs sit EQUATORIAL; that's why ClF₃ is T-shaped and XeF₂ linear.
  8. μ = 0 does not mean nonpolar bonds — it means symmetric geometry (CO₂, BF₃, CH₄, CCl₄).
  9. NH₃ > NF₃ in dipole moment because the lone pair helps in NH₃ and fights in NF₃.
  10. π bonds use UNhybridised p orbitals; hybrids make σ bonds and hold lone pairs only.
  11. More s-character ⟹ shorter, stronger bond and higher electronegativity (sp > sp² > sp³).
  12. MOT energy order swaps at oxygen: up to N₂, π2p is BELOW σ2p_z; for O₂/F₂, σ2p_z is below.
  13. O₂'s paramagnetism is explained by MOT (2 unpaired π*), NOT by Lewis/VBT — say so explicitly.
  14. Removing an antibonding electron STRENGTHENS the bond (O₂ → O₂⁺); removing a bonding one WEAKENS it (N₂ → N₂⁺).
  15. H-bonding needs H attached to F, O, or N only; intramolecular H-bonding LOWERS boiling point (o-nitrophenol).

11. Rapid Revision — One-Page Fact & Formula Sheet

Concept Key result Note
Formal charge $FC = V - L - \tfrac12S$ Lowest FC = best structure
Octet exceptions BeH₂/BF₃; NO/NO₂; PF₅/SF₆; XeF₂ incomplete / odd / expanded / noble
Lattice enthalpy ∝ q₁q₂/r MgO ≫ NaCl; drives ionic bonding
Bond order vs length BO ↑ ⟹ length ↓, enthalpy ↑ C–C 154 > C=C 134 > C≡C 120 pm
Resonance BO total bonds / positions CO₃²⁻: 4/3; O₃: 1.5; C₆H₆: 1.5
Dipole moment $\mu = q\times d$ (debye) Vector sum; μ = 0 for CO₂, BF₃, CH₄, CCl₄
% ionic character $\frac{\mu_{obs}}{\mu_{ionic}}\times100$ μ_ionic = e·d
Fajans small cation + big anion + high charge ⟹ covalent LiI most covalent Li-halide
VSEPR repulsion lp–lp > lp–bp > bp–bp CH₄ 109.5° → NH₃ 107° → H₂O 104.5°
Key odd shapes SF₄ see-saw; ClF₃ T; XeF₂ linear; XeF₄ sq. planar; BrF₅ sq. pyramidal lp equatorial in TBP
Hybridisation ↔ SN 2 sp, 3 sp², 4 sp³, 5 sp³d, 6 sp³d² SN = σ + lp
s-character sp 50% > sp² 33% > sp³ 25% More s ⟹ shorter, stronger, more EN
σ vs π σ axial & strong; π lateral, needs σ first double = σ+π; triple = σ+2π
MOT bond order $BO = \frac{N_b - N_a}{2}$ BO 0 ⟹ doesn't exist (He₂, Be₂, Ne₂)
Energy-order swap ≤ N₂: π2p < σ2p_z; O₂/F₂: σ2p_z < π2p Makes B₂ paramagnetic, C₂'s bonds both π
Oxygen ladder O₂²⁻ 1 < O₂⁻ 1.5 < O₂ 2 < O₂⁺ 2.5 O₂: paramagnetic, 2 unpaired π*
Isoelectronic BO-3 club N₂, CO, CN⁻, NO⁺ NO itself: BO 2.5, paramagnetic
H-bond H on F/O/N; 10–40 kJ/mol b.p.: H₂O > HF > NH₃; intra lowers b.p.
Study Plan

How to use these notes: Day 1: Sections 1–3 (Lewis, ionic, bond parameters) — practise five formal-charge and two %-ionic problems; insert the NCERT resonance and dipole figures. Day 2: Section 4 + Fig A — write the 13-shape table from memory. Day 3: Sections 5–6 + Fig B and the NCERT hybridisation figures — do ten hybridisation spot-checks. Day 4: Sections 7–8 + Figs C & D and the NCERT MO diagrams — reproduce the N₂ and O₂ configurations unaided, then all ten Types of Section 9, followed by the mistakes checklist. Finish every session by writing the fact sheet from memory.

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