📌 Snapshot
- Builds the conceptual core for "why atoms combine and what shape the resulting species takes" — covers Kossel–Lewis (ionic + Lewis dot), octet rule and its three exceptions, ionic bond energetics (lattice enthalpy, Fajans rules), bond parameters (length, angle, enthalpy, order, resonance, polarity), VSEPR shapes, valence bond / hybridisation, molecular orbital theory for second-row diatomics, and hydrogen bonding.
- One of the two most heavily examined Class XI chemistry units in CUET — hybridisation type, VSEPR shape, MOT bond order/magnetic behaviour, and H-bonding-driven anomalies appear almost every year.
- Tables 4.1–4.8 and Figs. 4.1–4.22 are dense with data; CUET stems are usually mined directly from these.
- Acts as a prerequisite for organic chemistry (sp/sp²/sp³ in C), coordination chemistry (hybridisation, geometry), and physical chemistry (bond enthalpy, dipole moment).
📖 Detailed Notes
2.1 Core concepts
- Chemical bond is the attractive force holding constituent atoms/ions together in a molecule; Kossel and Lewis (1916) gave the first satisfactory electronic explanation based on noble-gas inertness (NCERT §4.1, p. 101).
- Lewis dot symbols show the valence electrons of an atom; group valence equals either the number of dots or 8 minus the number of dots (NCERT §4.1, p. 101).
- Kossel view of ionic bond: the highly electronegative halogens and electropositive alkali metals are separated by noble gases; transfer of electrons gives ions with ns²np⁶ configuration that are stabilised by electrostatic attraction — e.g. NaCl, CaF₂ (NCERT §4.1, p. 101).
- Octet rule (Kossel–Lewis, 1916): atoms combine by transfer or sharing of valence electrons to attain an octet in their valence shell (NCERT §4.1.1, p. 102).
- Covalent bond (Langmuir 1919): formed by sharing of an electron pair; sharing 1, 2, 3 pairs gives single, double (e.g. CO₂, C₂H₄), triple (e.g. N₂, C₂H₂) bonds (NCERT §4.1.2, p. 102).
- Writing Lewis structures: sum valence electrons (add for anions, subtract for cations); least electronegative atom is central (e.g. N in NF₃, C in CO₃²⁻); after single bonds, leftover electrons go as multiple bonds or lone pairs so every atom gets an octet (NCERT §4.1.3, p. 103).
- Formal charge = (valence e⁻ in free atom) − (non-bonding e⁻) − ½(bonding e⁻); used to pick the lowest-energy Lewis structure (smallest formal charges). For O₃, central O has F.C. = +1 and the terminal O atoms have 0 and −1 (NCERT §4.1.4, p. 104–105).
- Limitations of octet rule: (i) incomplete octet — LiCl, BeH₂, BCl₃, AlCl₃, BF₃; (ii) odd-electron molecules — NO, NO₂; (iii) expanded octet — PF₅, SF₆, H₂SO₄ (3d orbitals available from period 3 onwards). Octet rule is also silent on shape, molecular energy, and noble-gas compounds like XeF₂, KrF₂ (NCERT §4.1.5, p. 105–106).
- Ionic bond formation is favoured by low ionisation enthalpy of the metal, high (negative) electron gain enthalpy of the non-metal, and large lattice enthalpy. For NaCl, IE(Na) + ΔₑgH(Cl) = +147.1 kJ mol⁻¹ but lattice enthalpy = −788 kJ mol⁻¹, so the crystal is stable (NCERT §4.2, p. 106).
- Lattice enthalpy is the energy required to separate one mole of a solid ionic compound into gaseous ions; NaCl lattice enthalpy = 788 kJ mol⁻¹. It is the true measure of ionic-compound stability, not mere octet completion (NCERT §4.2.1, p. 107).
- Bond length is the equilibrium internuclear distance; covalent radius is half the distance between two like atoms covalently bonded. Typical values: H–H 74 pm, Cl–Cl 199 pm, N≡N 109 pm, O=O 121 pm (Tables 4.2, 4.3) (NCERT §4.3.1, p. 107–108).
- Bond angle is the angle between bonding orbitals around the central atom (e.g. 104.5° in H₂O) (NCERT §4.3.2, p. 108).
- Bond enthalpy is the energy to break one mole of a bond in the gas phase: H–H = 435.8, O=O = 498, N≡N = 946 kJ mol⁻¹. For polyatomics like H₂O, mean bond enthalpy = (502 + 427)/2 = 464.5 kJ mol⁻¹ (NCERT §4.3.3, p. 108–109).
- Bond order in Lewis theory = number of bonds between two atoms: H₂ = 1, O₂ = 2, N₂ = 3, CO = 3. Isoelectronic species have the same bond order (F₂ and O₂²⁻ → 1; N₂, CO, NO⁺ → 3). Higher bond order ⇒ higher bond enthalpy and shorter bond length (NCERT §4.3.4, p. 109).
- Resonance: when one Lewis structure cannot describe a species, several canonical forms with identical nuclear positions but different electron arrangements are drawn — their hybrid represents the real molecule. Examples: O₃ (both O–O bonds equal at 128 pm, between single 148 pm and double 121 pm), CO₃²⁻ (three equivalent C–O bonds), CO₂, benzene. The hybrid has lower energy than any canonical form; canonical forms have no real existence (NCERT §4.3.5, p. 109–110).
- Polarity / dipole moment: μ = Q × r, measured in Debye (1 D = 3.33564 × 10⁻³⁰ C m). HF is polar (μ = 1.78 D); H₂, BF₃, CO₂, CCl₄ have μ = 0 because bond dipoles cancel. H₂O is bent so μ = 1.85 D. NH₃ (1.47 D) > NF₃ (0.23 D) because in NH₃ the lone-pair dipole adds to the N–H bond dipoles whereas in NF₃ it opposes them (NCERT §4.3.6, p. 110–112).
- Fajans rules for partial covalent character of ionic bonds: small cation, large anion, high cation charge, and (n−1)d^n ns⁰ configuration all increase covalent character (NCERT §4.3.6, p. 112).
- VSEPR theory (Sidgwick–Powell 1940; Nyholm–Gillespie 1957): valence-shell electron pairs (bonded + lone) arrange to minimise repulsion. Repulsion order: lp–lp > lp–bp > bp–bp. A multiple bond is treated as one super-pair. Basic geometries: AB₂ linear, AB₃ trigonal planar, AB₄ tetrahedral, AB₅ trigonal bipyramidal, AB₆ octahedral (NCERT §4.4, Table 4.6, p. 112–114).
- With lone pairs (Tables 4.7, 4.8): AB₂E bent (e.g. SO₂, 119.5°), AB₃E trigonal pyramidal (NH₃, 107°), AB₂E₂ bent (H₂O, 104.5°), AB₄E see-saw (SF₄, lp at equatorial position), AB₃E₂ T-shape (ClF₃), AB₄E₂ square planar (XeF₄), AB₅E square pyramidal (BrF₅) (NCERT §4.4, Tables 4.7–4.8, p. 115–117).
- Valence Bond (VB) theory (Heitler–London 1927, Pauling): covalent bond forms when half-filled atomic orbitals with opposite-spin electrons overlap; greater overlap ⇒ stronger bond. For H₂, the potential-energy curve has a minimum at 74 pm corresponding to 435.8 kJ mol⁻¹ of bond enthalpy (NCERT §4.5, p. 117–118).
- σ vs π bond: σ bond by head-on (axial) overlap of s–s, s–p, p–p along internuclear axis; π bond by sidewise overlap of parallel p-orbitals, giving two electron clouds above and below the molecular plane. σ is stronger because of greater overlap. Multiple bonds = 1 σ + 1 (or 2) π (NCERT §4.5.4–4.5.5, p. 120).
- Hybridisation (Pauling): mixing of orbitals of slightly different energies to give equivalent hybrid orbitals. Types: sp (linear, 180°; BeCl₂), sp² (trigonal planar, 120°; BCl₃, C in C₂H₄), sp³ (tetrahedral, 109.5°; CH₄; also explains NH₃ at 107° and H₂O at 104.5° because of lone-pair repulsion), sp³d (trigonal bipyramidal; PCl₅), sp³d² (octahedral; SF₆), dsp² (square planar; [Ni(CN)₄]²⁻), sp³d² also gives square pyramidal BrF₅ (NCERT §4.6, p. 120–125).
- PCl₅ geometry: three equatorial P–Cl bonds at 120°, two axial at 90° to the equatorial plane; axial bonds are longer/weaker due to greater repulsion from equatorial pairs, making PCl₅ reactive (NCERT §4.6.3, p. 125).
- Molecular Orbital (MO) theory (Hund–Mulliken 1932): atomic orbitals of comparable energy and same symmetry combine by LCAO to give equal numbers of bonding (lower energy, electron density between nuclei) and antibonding (higher energy, node between nuclei) MOs. Filling follows aufbau, Pauli, Hund (NCERT §4.7.1, p. 126).
- MO energy order: 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 Li₂, Be₂, B₂, C₂, N₂ — σ2p_z lies above (π2p_x = π2p_y) (NCERT §4.7.4, p. 128–129).
- Bond order = ½(N_b − N_a): positive ⇒ stable, zero/negative ⇒ unstable. H₂ b.o. = 1 (stable, diamagnetic); He₂ b.o. = 0 (does not exist); Li₂ b.o. = 1 (diamagnetic); C₂ b.o. = 2 (both bonds are π; diamagnetic); N₂ b.o. = 3; O₂ b.o. = 2 and is paramagnetic because of two unpaired electrons in π*2p_x, π*2p_y — a landmark success of MOT (NCERT §4.8, p. 129–131).
- Magnetic nature: all electrons paired ⇒ diamagnetic; one or more unpaired ⇒ paramagnetic (e.g. O₂, B₂) (NCERT §4.7.5, p. 129).
- Hydrogen bond: the attractive force between a hydrogen covalently bonded to a highly electronegative atom (F, O, N) of one molecule and an electronegative atom of another molecule; weaker than a covalent bond but stronger than van der Waals forces. Represented by a dotted line (NCERT §4.9, p. 131).
- Types of H-bond: (i) intermolecular — HF, H₂O, alcohols, ice; (ii) intramolecular — within the same molecule, e.g. o-nitrophenol where H sits between two oxygen atoms. H-bonding strongly affects structure and properties (e.g. anomalously high boiling point of water, lower density of ice than water) (NCERT §4.9.2, p. 132).
2.2 Definitions to memorise
| Term | Definition | Page |
|---|---|---|
| Chemical bond | Attractive force holding constituent atoms/ions together in a chemical species | 100 |
| Octet rule | Atoms combine by transfer/sharing of valence electrons to attain 8 electrons in the valence shell | 102 |
| Electrovalent (ionic) bond | Bond formed by electrostatic attraction between oppositely charged ions formed by electron transfer | 101–102 |
| Formal charge | (Valence e⁻ in free atom) − (lone-pair e⁻) − ½(bonding e⁻) | 104 |
| Lattice enthalpy | Energy required to completely separate one mole of a solid ionic compound into gaseous ions | 107 |
| Bond length | Equilibrium distance between the nuclei of two bonded atoms | 107 |
| Bond angle | Angle between orbitals containing bonding electron pairs around the central atom | 108 |
| Bond enthalpy | Energy required to break one mole of a particular type of bond in the gaseous state | 108 |
| Bond order (Lewis) | Number of bonds between two atoms in a molecule | 109 |
| Bond order (MOT) | ½ (N_b − N_a), where N_b, N_a are bonding and antibonding electrons | 129 |
| Resonance | Representation of a species by two or more canonical Lewis structures whose hybrid describes the actual species | 109 |
| Dipole moment (μ) | Product of magnitude of charge and distance between centres of positive and negative charge; μ = Q × r, unit Debye (1 D = 3.33564 × 10⁻³⁰ C m) | 110 |
| VSEPR theory | Shape determined by repulsion among valence-shell electron pairs (lp–lp > lp–bp > bp–bp) | 113 |
| Hybridisation | Intermixing of orbitals of slightly different energies to give a new set of equivalent orbitals | 120 |
| σ bond | Bond by head-on (axial) overlap along the internuclear axis | 120 |
| π bond | Bond by sidewise overlap of parallel p-orbitals; cloud lies above and below the molecular plane | 120 |
| Bonding MO | LCAO ψ_A + ψ_B; electron density between nuclei; lower energy than parent AO | 126 |
| Antibonding MO | LCAO ψ_A − ψ_B; nodal plane between nuclei; higher energy than parent AO | 126 |
| Hydrogen bond | Attractive force between H atom of one molecule and an F, O or N atom of another molecule | 131 |
2.3 Diagrams / processes to remember
- Fig. 4.1 — Bond length R = r_A + r_B (p. 107).
- Fig. 4.2 — Covalent radius (99 pm) vs van der Waals radius (180 pm) in Cl₂ (p. 107).
- Fig. 4.3 — Resonance hybrid of O₃ from two canonical forms (p. 109).
- Fig. 4.4 — Three canonical structures of CO₃²⁻ (p. 110); Fig. 4.5 — three canonical forms of CO₂ (p. 110).
- Table 4.5 — Dipole moments and geometries: HF 1.78 D linear, H₂O 1.85 D bent, NH₃ 1.47 D trigonal pyramidal, BF₃ 0 D trigonal planar, CO₂ 0 D linear, CCl₄ 0 D tetrahedral (p. 112).
- Tables 4.6–4.8 — VSEPR geometries (p. 114–117).
- Fig. 4.7 / 4.8 — Forces and potential-energy curve in H₂ formation (p. 118).
- Fig. 4.10–4.16 — Hybridisation diagrams for BeCl₂ (sp), BCl₃ (sp²), CH₄ (sp³), NH₃ and H₂O (sp³ with lone pairs), C₂H₄, C₂H₂ (p. 121–124).
- Fig. 4.17 — Trigonal bipyramidal PCl₅ (sp³d) with axial vs equatorial bonds (p. 125).
- Fig. 4.18 — Octahedral SF₆ (sp³d²) (p. 125).
- Fig. 4.20 — Bonding/antibonding MOs from 1s, 2p_z, 2p_x AOs (p. 128).
- Fig. 4.21 — MO occupancy table for B₂ through Ne₂ with bond order and magnetic property (p. 131).
- Fig. 4.22 — Intramolecular H-bond in o-nitrophenol (p. 132).
2.4 Common confusions / NTA trap points
- MOT energy order changes at O₂: for B₂, C₂, N₂ the (π2p) lies below σ2p_z; for O₂, F₂, Ne₂ the σ2p_z lies below (π2p). NTA often inverts this in distractors.
- NH₃ vs NF₃ dipole moments: N is more electronegative than H but less than F. Yet μ(NH₃) > μ(NF₃) because the lone-pair dipole adds in NH₃ but subtracts in NF₃. Easy trap.
- H₂O bent ≠ BeCl₂ linear even though both are AB₂ — H₂O has 2 lone pairs on O (AB₂E₂ → sp³ → bent), BeCl₂ has none (AB₂ → sp → linear).
- C₂ double bond is two π bonds (no σ from 2p_z because σ2p_z is empty in the (π2p)⁴ configuration), unlike most double bonds which are 1σ + 1π.
- Formal charge ≠ oxidation state ≠ real charge. It only helps in picking the best Lewis structure.
- Hybridisation does not require half-filled orbitals — even filled orbitals can hybridise (e.g. the lone-pair-bearing sp³ orbitals of O in H₂O).
- Bond order in resonance hybrid is fractional — in CO₃²⁻ each C–O bond order = 4/3 (1 σ + delocalised π over 3 oxygens).
- Ice less dense than water because tetrahedral H-bonding in ice creates empty cages; this is an H-bonding consequence often asked through MCQs on density/boiling-point anomalies.
🎯 Practice MCQs
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Q1. Which of the following pairs of molecules together contain examples of all three exceptions to the octet rule — incomplete octet, odd-electron, and expanded octet?
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Answer: B
BCl₃ shows an incomplete octet (B has only 6 electrons); NO is an odd-electron molecule; SF₆ shows an expanded octet (12 electrons on S). Option A is wrong because NH₃ obeys the octet rule.
Q2. The formal charges on the central oxygen atom and the two terminal oxygen atoms in the conventional Lewis structure of the ozone molecule are, respectively:
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Answer: B
Using F.C. = (valence e⁻) − (non-bonding e⁻) − ½(bonding e⁻): central O = 6 − 2 − 3 = +1; doubly-bonded terminal O = 6 − 4 − 2 = 0; singly-bonded terminal O = 6 − 6 − 1 = −1.
Q3. Among the following, the molecule with the **largest** dipole moment is:
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Answer: C
H₂O is bent so the two O–H bond dipoles do not cancel, giving μ = 1.85 D. CO₂ (linear), BF₃ (trigonal planar) and CCl₄ (tetrahedral) are symmetric and the bond dipoles cancel exactly.
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Q4. The shapes of XeF₄, SF₄ and BrF₅ are, respectively:
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Answer: B
XeF₄ has 4 bond pairs + 2 lone pairs ⇒ square planar; SF₄ has 4 bp + 1 lp ⇒ see-saw with lp at equatorial position; BrF₅ has 5 bp + 1 lp ⇒ square pyramidal.
Q5. The bond angles in CH₄, NH₃ and H₂O follow the order:
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Answer: A
Lone pair–bond pair repulsion (and lp–lp in H₂O) is greater than bp–bp repulsion, compressing the bond angle from the ideal 109.5° as the number of lone pairs on the central atom increases.
Q6. The type of hybridisation present in PCl₅, SF₆ and BrF₅ is, respectively:
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Answer: A
PCl₅ uses one s + three p + one d ⇒ sp³d (trigonal bipyramidal). SF₆ uses one s + three p + two d ⇒ sp³d² (octahedral). BrF₅ has six electron pairs (5 bp + 1 lp) so it is also sp³d², giving a square pyramidal shape.
Q7. According to molecular orbital theory, the bond order and magnetic nature of the O₂ molecule are:
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Answer: B
O₂ has 16 electrons; 10 in bonding and 6 in antibonding MOs gives bond order 2. The two singly-occupied π* MOs make O₂ paramagnetic — a key triumph of MOT over Lewis theory.
Q8. Which one of the following statements about the He₂ molecule, on the basis of MOT, is correct?
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Answer: B
Equal numbers of bonding and antibonding electrons give zero bond order, so no net bonding influence — He₂ is not formed.
Q9. Which of the following sets contains species with **identical** bond order?
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Answer: A
N₂, CO and NO⁺ are all 14-electron isoelectronic species with bond order 3. F₂ (b.o. = 1) and O₂ (b.o. = 2) are not equal, ruling out B.
Q10. **Assertion (A):** The dipole moment of NH₃ is greater than that of NF₃, although fluorine is more electronegative than hydrogen. **Reason (R):** In NH₃ the lone-pair orbital dipole on nitrogen is in the same direction as the resultant N–H bond moments, whereas in NF₃ it opposes the resultant N–F bond moments.
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Answer: A
Measured values are μ(NH₃) = 4.90 × 10⁻³⁰ C m and μ(NF₃) = 0.80 × 10⁻³⁰ C m. The vector cancellation in NF₃ reduces its dipole moment despite F being more electronegative — R correctly explains A.
Q11. Match the molecules in List-I with their shapes in List-II. | List-I (Molecule) | List-II (Shape) | |---|---| | (P) BCl₃ | (1) Bent | | (Q) H₂O | (2) Trigonal pyramidal | | (R) NH₃ | (3) Trigonal planar | | (S) ClF₃ | (4) T-shape |
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Answer: A
BCl₃ has no lone pair on B (trigonal planar). H₂O (AB₂E₂) is bent. NH₃ (AB₃E) is trigonal pyramidal. ClF₃ (AB₃E₂) has two lone pairs at equatorial positions giving a T-shape.
Q12. Which of the following statements about hydrogen bonding is **incorrect**?
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Answer: D
Hydrogen bonds are weaker than covalent bonds; the hydrogen carries only a partial positive charge (δ+), not a full positive. Statements A, B and C all match the NCERT description.
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