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State Board · Class 11 · Chemistry

Lesson 5: Chemical Bonding

7 concepts · 56 practice questions with answers · Free to read

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Kossel–Lewis approach, ionic bond and lattice enthalpy

Link to Std 9–10 : In Std 9–10 you saw NaCl form from Na and Cl; here we explain it with energy accounting.

From daily life

If one friend has extra snacks and the other too few, one gives and one takes and both are content. Atoms stabilise by giving and taking electrons.

Simple definition

Octet rule: an atom becomes stable with 8 electrons in its outer shell (2 for H, Li). Ionic bond: electrostatic attraction between cation and anion. Lattice enthalpy: energy needed to separate 1 mole of an ionic crystal into gaseous ions.

Ionic bondNa2, 8, 1Cl2, 8, 7e⁻→Na⁺2, 8Cl⁻2, 8, 8Ionic bond: Na gives an electron, Cl takes itBorn–Haber (kJ/mol): Na(s) → Na(g) +108 · IE₁ +496½Cl₂ → Cl +122 · electron gain −349 · lattice −788ΔfH = 108 + 496 + 122 − 349 − 788 = −411 kJ/mol

In the figure : Na → Na⁺, Cl → Cl⁻ and the Born–Haber sum.

Rule 1: When an ionic bond forms — Low ionisation enthalpy (metal) + high electron gain (non-metal) + large lattice enthalpy.

  • Easy : Na (2,8,1) → Na⁺ (2,8); Cl (2,8,7) → Cl⁻ (2,8,8).
  • Medium : Mg + O → Mg²⁺ O²⁻ (2 e⁻ transferred).
  • Exam-type : Ca + 2F → CaF₂ (1 e⁻ to each F).

Rule 2: Factors affecting lattice enthalpy — Lattice enthalpy ∝ (q₁ q₂) ÷ r. Higher charges and smaller ions ⇒ larger.

  • Easy : NaCl ≈ 788 kJ/mol; KCl ≈ 715 (K⁺ is larger).
  • Medium : MgO ≈ 3791 kJ/mol, much more than NaCl (2+/2−).
  • Exam-type : So MgO melts higher (2852°C) than NaCl (801°C).

Rule 3: Properties of ionic compounds — Hard, brittle, high melting; conduct only when molten or dissolved; dissolve in polar solvents.

  • Easy : Solid NaCl does not conduct.
  • Medium : Molten NaCl conducts (ions are free).
  • Exam-type : A hammer blow shatters the crystal (like charges come together).

Derivation: lattice enthalpy from the Born–Haber cycle

  • Step 1: Na(s) → Na(g) +108; Na(g) → Na⁺ + e⁻ +496; ½Cl₂ → Cl +122; Cl + e⁻ → Cl⁻ −349.
  • Step 2: Na⁺(g) + Cl⁻(g) → NaCl(s) releases the lattice enthalpy U.
  • Step 3: Hess's law: Δ_fH = 108 + 496 + 122 − 349 + U = −411 ⇒ U = −411 − 377 = −788 kJ/mol.

CET level (MCQ)

  • CET 1) Highest lattice enthalpy? 1) NaCl 2) KCl 3) MgO 4) NaBr — Answer: 3). (2+/2− charges and small ions.)
  • CET 2) An ion that obeys the octet rule? 1) Na⁺ 2) Fe³⁺ 3) Zn²⁺ 4) Cu²⁺ — Answer: 1). (transition-metal ions do not.)

Real-life use: Salt solution conducts electricity; MgO is used in furnace bricks because of its very high melting point.

Remember

Lattice enthalpy ∝ q₁q₂ ÷ r; an ionic compound conducts only when molten or dissolved.

Common mistake

Writing that an ionic solid conducts electricity.

CET trap

Stability is decided by the lattice enthalpy; forming Na⁺ costs energy but the lattice energy repays it.

Shortcut

Memory clue: "charge up, size down ⇒ lattice up".

Spot the difference

ItemNaClMgO
Charges1+, 1−2+, 2−
Lattice (kJ/mol)≈ 788≈ 3791
Melting point801°C2852°C

Practice questions

Easy
  1. 1State the octet rule.
  2. 2What is an ionic bond?
  3. 3How is MgO formed from Mg and O?
Medium
  1. 4Why is MgO's lattice enthalpy higher than NaCl's?
  2. 5Why does solid NaCl not conduct?
  3. 6Write the steps of the Born–Haber cycle for NaCl.
Challenging
  1. 7Find U if Δ_fH = −411 (with 108 + 496 + 122 − 349).
  2. 8Why is KCl's lattice enthalpy lower than NaCl's?
Show answers (try them yourself first)
  1. 1Atoms become stable with 8 outer electrons.
  2. 2Electrostatic attraction between cations and anions.
  3. 3Mg gives 2 e⁻, O takes them ⇒ Mg²⁺ O²⁻.
  4. 4Charges 2+/2− and smaller ions.
  5. 5Ions are locked in the lattice.
  6. 6Sublimation, IE, ½ dissociation, EA, lattice.
  7. 7U = −411 − 377 = −788 kJ/mol.
  8. 8K⁺ is larger ⇒ greater r.

Covalent bond and Lewis dot structures

From daily life

Two friends share one book; two atoms share an electron pair.

Simple definition

A covalent bond forms when two atoms share an electron pair: single (−), double (=), triple (≡). A Lewis structure shows valence electrons as dots/lines.

Lewis structuresH₂OOHHCO₂OCON₂NNNH₃NHHHLewis structure: 1) count total valence e⁻ 2) pick the central atom3) single bonds 4) octets 5) multiple bonds if shorte⁻ : H₂O 8 · CO₂ 16 · N₂ 10 · NH₃ 8

In the figure : Lewis structures of H₂O, CO₂, N₂ and NH₃.

Rule 1: Drawing a Lewis structure — Total valence e⁻ ⇒ skeleton ⇒ single bonds ⇒ octets on outer atoms ⇒ leftovers on the centre ⇒ multiple bonds if short.

  • Easy : H₂O: 2 + 6 = 8 e⁻; two lone pairs on O.
  • Medium : CO₂: 4 + 12 = 16 e⁻ ⇒ O=C=O.
  • Exam-type : N₂: 10 e⁻ ⇒ N≡N.

Rule 2: Bond order, length and enthalpy — Higher bond order ⇒ shorter bond and larger bond enthalpy.

  • Easy : C–C 154 pm, 347 kJ/mol.
  • Medium : C=C 134 pm, 614 kJ/mol.
  • Exam-type : C≡C 120 pm, 839 kJ/mol.

Rule 3: Coordinate (dative) bond — Both electrons come from one atom: NH₃ + H⁺ → NH₄⁺.

  • Easy : In NH₄⁺ all four N–H bonds are equal; one is dative.
  • Medium : H₃O⁺: H₂O + H⁺.
  • Exam-type : A dative bond can be drawn in O₃.

Example: Lewis structure of CO₃²⁻

  • Step 1: e⁻ = 4 + 3 × 6 + 2 = 24.
  • Step 2: C in the centre; three C–O single bonds (6 e⁻), 18 on O, total 24, but C has only 6 e⁻ ⇒ make one C=O.
  • Step 3: Two O now carry −1 each; total −2 ✓.

CET level (MCQ)

  • CET 1) σ and π bonds in N₂? 1) 3σ 2) 1σ + 2π 3) 2σ + 1π 4) 3π — Answer: 2).
  • CET 2) Shortest carbon–carbon bond? 1) C–C 2) C=C 3) C≡C 4) all equal — Answer: 3). (higher bond order ⇒ shorter.)

Real-life use: Air's N₂ is very stable because of the N≡N triple bond (945 kJ/mol), so it hardly reacts.

Remember

Higher bond order ⇒ shorter, stronger; single = 1σ, double = 1σ + 1π, triple = 1σ + 2π.

Common mistake

Forgetting the ion's charge when counting electrons (CO₃²⁻ adds 2).

CET trap

H is never the central atom; the least electronegative atom (not H) is central.

Shortcut

Memory clue: "count – pair up – octet".

Spot the difference

Bondσ / πLength
Single1σLongest
Double1σ + 1πMedium
Triple1σ + 2πShortest

Practice questions

Easy
  1. 1What is a covalent bond?
  2. 2Total valence electrons of H₂O?
  3. 3How many σ and π bonds in N₂?
Medium
  1. 4Draw the Lewis structure of CO₂.
  2. 5How does NH₄⁺ have a coordinate bond?
  3. 6Order of lengths of C–C, C=C and C≡C.
Challenging
  1. 7Count the valence e⁻ of CO₃²⁻.
  2. 8Why is N₂ unreactive?
Show answers (try them yourself first)
  1. 1Sharing of an electron pair between two atoms.
  2. 22 + 6 = 8.
  3. 31σ + 2π.
  4. 4O=C=O with two lone pairs on each O.
  5. 5NH₃ donates its lone pair to H⁺.
  6. 6C–C > C=C > C≡C.
  7. 74 + 18 + 2 = 24.
  8. 8The N≡N bond is very strong (945 kJ/mol).

Formal charge, resonance and limits of the octet rule

From daily life

After sharing we check what each gets; formal charge shows how many electrons an atom 'owns'.

Simple definition

Formal charge FC = V − N − B/2. Resonance: several valid Lewis structures for one molecule; the real one is their hybrid. Octet limits: incomplete, odd-electron and expanded octets.

ResonanceOzone O₃: two equivalent resonance structuresO0O+1O−1↔O−1O+1O0Hybrid: both O–O bonds equal, bond order = 1.5Formal charge FC = V − N − B/2V = valence e⁻ · N = non-bonding e⁻ · B = bonding e⁻

In the figure : Two resonance structures of O₃.

Rule 1: Formal charge — The structure with FC nearest zero is best; negative FC belongs on the more electronegative atom.

  • Easy : Central O of O₃: 6 − 2 − 3 = +1.
  • Medium : Singly-bonded O: 6 − 6 − 1 = −1; doubly-bonded: 6 − 4 − 2 = 0.
  • Exam-type : N in NH₄⁺: 5 − 0 − 4 = +1.

Rule 2: Resonance — Resonance structures are not real; the true molecule is a hybrid with equal bonds and lower energy.

  • Easy : Bond order of O₃ = (2 + 1) ÷ 2 = 1.5.
  • Medium : C–O bond order in CO₃²⁻ = (2 + 1 + 1) ÷ 3 = 1.33.
  • Exam-type : All C–C bonds in benzene have order 1.5.

Rule 3: Exceptions to the octet rule — Incomplete: BeCl₂ (4), BF₃ (6). Odd electron: NO (11), NO₂ (17). Expanded: PCl₅ (10), SF₆ (12) — period 3 and beyond.

  • Easy : B in BF₃ has 6 e⁻.
  • Medium : NO has one unpaired e⁻ ⇒ paramagnetic.
  • Exam-type : S in SF₆ has 12 e⁻.

Example: FC sum for O₃

  • Step 1: +1 + 0 + (−1) = 0 ⇒ molecule is neutral ✓.
  • Step 2: The two structures are equivalent ⇒ both bonds equal at 1.5.
  • Step 3: Experimental O–O is 128 pm, between O–O (148) and O=O (121) ✓.

CET level (MCQ)

  • CET 1) FC formula? 1) V − N − B 2) V − N − B/2 3) V + N − B/2 4) N − V − B/2 — Answer: 2).
  • CET 2) Which violates the octet rule? 1) CH₄ 2) BF₃ 3) H₂O 4) NH₃ — Answer: 2).

Real-life use: The ozone layer's stability and benzene's special stability both come from resonance.

Remember

FC = V − N − B/2; resonance ⇒ equal bonds; exceptions: incomplete, odd, expanded.

Common mistake

Thinking resonance forms interconvert; the molecule is one stable hybrid.

CET trap

Bond order = total bonds ÷ number of positions (O₃: 3 ÷ 2); do not count only double bonds.

Shortcut

Memory clue: "V − N − half of B".

Spot the difference

ExceptionExamplee⁻
IncompleteBF₃6
Odd electronNO11
ExpandedSF₆12

Practice questions

Easy
  1. 1Write the formula of formal charge.
  2. 2What is resonance?
  3. 3How many electrons surround B in BF₃?
Medium
  1. 4Find the FC of the central O in O₃.
  2. 5What is the bond order of O₃?
  3. 6C–O bond order in CO₃²⁻?
Challenging
  1. 7Why is NO paramagnetic?
  2. 8How can P in PCl₅ have 10 electrons?
Show answers (try them yourself first)
  1. 1FC = V − N − B/2.
  2. 2Several valid structures; the real molecule is a hybrid.
  3. 36.
  4. 46 − 2 − 3 = +1.
  5. 51.5.
  6. 6(2 + 1 + 1) ÷ 3 = 1.33.
  7. 7It has one unpaired electron.
  8. 8P is in period 3 and uses vacant d orbitals for an expanded octet.

VSEPR theory and shapes of molecules

From daily life

Tie balloons together and each moves as far from the others as it can; electron pairs around a central atom do the same.

Simple definition

VSEPR: electron pairs (bonding and lone) around the central atom arrange themselves as far apart as possible. Repulsion: lp–lp > lp–bp > bp–bp.

VSEPRVSEPR: shapes and anglesLinear 180°BeCl₂Trigonal planar 120°BF₃Tetrahedral 109.5°CH₄Pyramidal 107°NH₃Bent 104.5°H₂OOctahedral 90°SF₆lp–lp > lp–bp > bp–bp (repulsion)

In the figure : Six main shapes and angles.

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VSEPR theory and shapes of molecules

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Rule 1: Shape from electron pairs — 2 pairs linear 180°; 3 trigonal planar 120°; 4 tetrahedral 109.5°; 5 TBP; 6 octahedral 90°.

  • Easy : BeCl₂: 2 bp ⇒ linear.
  • Medium : BF₃: 3 bp ⇒ trigonal planar.
  • Exam-type : CH₄: 4 bp ⇒ tetrahedral.

Rule 2: Effect of lone pairs — A lone pair takes more room ⇒ the bond angle shrinks.

  • Easy : NH₃: 3 bp + 1 lp ⇒ pyramidal 107°.
  • Medium : H₂O: 2 bp + 2 lp ⇒ bent 104.5°.
  • Exam-type : CH₄ 109.5° > NH₃ 107° > H₂O 104.5°.

Rule 3: Shapes with 5 and 6 pairs — 5 pairs (TBP): SF₄ see-saw, ClF₃ T-shaped, XeF₂ linear. 6 pairs: BrF₅ square pyramidal, XeF₄ square planar.

  • Easy : XeF₂: 5 pairs, 3 lp (equatorial) ⇒ linear.
  • Medium : XeF₄: 6 pairs, 2 lp (axial) ⇒ square planar.
  • Exam-type : Lone pairs take equatorial positions in TBP.

Method: finding the shape

  • Step 1: Count valence electrons on the central atom + outer atoms (adjust for charge).
  • Step 2: Total pairs = (V + M − C + A) ÷ 2; bonding pairs = number of outer atoms.
  • Step 3: Lone pairs = total − bonding ⇒ read the shape from the figure.

CET level (MCQ)

  • CET 1) Bond angle of H₂O? 1) 90° 2) 104.5° 3) 109.5° 4) 120° — Answer: 2).
  • CET 2) Shape of XeF₄? 1) tetrahedral 2) square planar 3) see-saw 4) octahedral — Answer: 2). (6 pairs, 2 lp.)

Real-life use: VSEPR shapes underlie ice's hexagonal structure, water's polarity and the 3-D fit of drugs.

Remember

lp–lp > lp–bp > bp–bp; more lone pairs ⇒ smaller angle; CH₄ > NH₃ > H₂O.

Common mistake

Counting lone pairs in the shape's name; the shape describes only the atoms.

CET trap

Electron-pair geometry and molecular shape differ: NH₃ has tetrahedral pair geometry but a pyramidal shape.

Shortcut

Memory clue: "A lone pair is a bulky friend who squeezes the others".

Spot the difference

Moleculebp + lpShape
CH₄4 + 0Tetrahedral
NH₃3 + 1Pyramidal
H₂O2 + 2Bent
XeF₄4 + 2Square planar

Practice questions

Easy
  1. 1State the basic idea of VSEPR.
  2. 2What is the angle in CH₄?
  3. 3State the shape of NH₃.
Medium
  1. 4Why is H₂O's angle less than CH₄'s?
  2. 5Shape and angle of SF₆.
  3. 6Why is XeF₂ linear?
Challenging
  1. 7Find the shape of ClF₃ (count the pairs).
  2. 8Compare the angles of NH₄⁺ and NH₃.
Show answers (try them yourself first)
  1. 1Electron pairs stay as far apart as possible.
  2. 2109.5°.
  3. 3Pyramidal (107°).
  4. 4Two lone pairs take more room.
  5. 5Octahedral, 90°.
  6. 63 lp equatorial ⇒ F axial at 180°.
  7. 75 pairs, 2 lp ⇒ T-shaped.
  8. 8NH₄⁺ 109.5° (no lp) > NH₃ 107°.

Valence bond theory and hybridisation

From daily life

The more two hands overlap the firmer the grip; the more orbitals overlap the stronger the bond.

Simple definition

Valence bond theory: bonds form by overlap of half-filled orbitals. σ: head-on overlap; π: sideways. Hybridisation: mixing of orbitals of nearby energy into equivalent new orbitals.

HybridisationHybridisation: shape and anglesp180° · linearBeCl₂, C₂H₂sp²120° · planarBF₃, C₂H₄sp³109.5° · tetrahedralCH₄, NH₃, H₂Os-character: sp 50% · sp² 33.3% · sp³ 25%sp³d (PCl₅, TBP) · sp³d² (SF₆, octahedral)sigma (σ): head-on overlap · pi (π): sideways overlap

In the figure : sp, sp², sp³ hybridisation and shape.

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Valence bond theory and hybridisation

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Rule 1: Finding hybridisation — H = ½ (V + M − C + A): 2 = sp, 3 = sp², 4 = sp³, 5 = sp³d, 6 = sp³d².

  • Easy : NH₃: ½(5 + 3) = 4 ⇒ sp³.
  • Medium : SO₂: ½(6 + 0) = 3 ⇒ sp².
  • Exam-type : SF₆: ½(6 + 6) = 6 ⇒ sp³d².

Rule 2: Hybridisation in carbon compounds — C₂H₆ sp³ (7σ); C₂H₄ sp² (5σ + 1π); C₂H₂ sp (3σ + 2π).

  • Easy : C₂H₄: 4 C–H + 1 C–C σ = 5σ, 1π.
  • Medium : C₂H₂: 2 C–H + 1 C–C σ = 3σ, 2π.
  • Exam-type : s-character sp 50% > sp² 33.3% > sp³ 25%; C–H acidity is highest for sp.

Rule 3: Comparing σ and π bonds — σ is strong and allows rotation; π is weaker, restricts rotation, with density off the axis.

  • Easy : Only one σ between two atoms.
  • Medium : A double bond is 1σ + 1π.
  • Exam-type : Rotation about C=C is blocked ⇒ cis/trans isomerism.

Derivation: sp³ hybridisation (CH₄)

  • Step 1: C is 2s² 2p² ⇒ promotion ⇒ 2s¹ 2p³ (four half-filled).
  • Step 2: One s + three p mix into four equivalent sp³ orbitals at 109.5°.
  • Step 3: Each sp³ overlaps an H 1s to give 4 σ bonds ⇒ tetrahedral.

CET level (MCQ)

  • CET 1) Hybridisation in C₂H₂? 1) sp 2) sp² 3) sp³ 4) sp³d — Answer: 1).
  • CET 2) σ : π in C₂H₄? 1) 4:2 2) 5:1 3) 3:2 4) 6:0 — Answer: 2).

Real-life use: Graphite (sp²) is slippery, diamond (sp³) is hard; benzene's flat ring is sp².

Remember

H = ½(V + M − C + A); sp 180°, sp² 120°, sp³ 109.5°; σ first, then π.

Common mistake

Calling NH₃ sp²; the lone pair counts in hybridisation.

CET trap

Hybridisation = bond pairs + lone pairs (not π bonds); do not count π bonds.

Shortcut

Memory clue: "2-3-4 : sp-sp²-sp³".

Spot the difference

HybridAngleExample
sp180°BeCl₂, C₂H₂
sp²120°BF₃, C₂H₄
sp³109.5°CH₄, NH₃
sp³d²90°SF₆

Practice questions

Easy
  1. 1State the difference between σ and π bonds.
  2. 2Hybridisation in CH₄?
  3. 3How many σ and π in C₂H₂?
Medium
  1. 4Find the hybridisation of NH₃ by the formula.
  2. 5What is the s-character in sp?
  3. 6Count σ and π in C₂H₄.
Challenging
  1. 7Hybridisation and shape of SF₆.
  2. 8Hybridisation and shape of CO₃²⁻.
Show answers (try them yourself first)
  1. 1σ is head-on, π sideways; σ is stronger.
  2. 2sp³.
  3. 33σ and 2π.
  4. 4½(5 + 3) = 4 ⇒ sp³.
  5. 550%.
  6. 65σ, 1π.
  7. 7sp³d², octahedral.
  8. 8½(4 + 0 + 2) = 3 ⇒ sp², trigonal planar.

Molecular orbital theory (MOT)

From daily life

Two waves can add (constructive) or cancel (destructive); atomic orbitals combine the same way into two molecular orbitals.

Simple definition

MOT: n atomic orbitals give n molecular orbitals — bonding (lower energy) and antibonding (higher). Bond order = (N_b − N_a) ÷ 2.

Molecular orbitalsMO diagram of O₂ (12 valence e⁻)σ*2p↑↑π*2p↑↓↑↓π2p↑↓σ2p↑↓σ*2s↑↓σ2sBO = (Nb − Na) ÷ 2O₂ : (8 − 4) ÷ 2 = 22 unpaired e⁻: paramagneticN₂ 3 · O₂⁺ 2.5 · O₂⁻ 1.5In B₂, C₂, N₂ π2p lies below σ2p (Z ≤ 7)

In the figure : MO diagram of O₂; two unpaired e⁻ in π*2p.

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Molecular orbital theory (MOT)

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Rule 1: Filling rules — Aufbau, Pauli, Hund. Order (O₂, F₂): σ2s < σ*2s < σ2p < π2p < π*2p < σ*2p; for Z ≤ 7 π2p < σ2p.

  • Easy : H₂: (σ1s)² ⇒ bond order 1.
  • Medium : He₂: (σ1s)²(σ*1s)² ⇒ 0 ⇒ does not exist.
  • Exam-type : N₂: 10 valence e⁻ ⇒ bond order 3.

Rule 2: Bond order and stability — Higher bond order ⇒ shorter, stronger, more stable; 0 ⇒ no molecule.

  • Easy : O₂⁺ (8 − 3) ÷ 2 = 2.5, O₂ 2, O₂⁻ 1.5, O₂²⁻ 1.
  • Medium : H₂⁺ has bond order 0.5 (exists).
  • Exam-type : Stability order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻.

Rule 3: Magnetic behaviour — Unpaired electrons ⇒ paramagnetic; all paired ⇒ diamagnetic.

  • Easy : O₂: two unpaired in π*2p ⇒ paramagnetic (liquid O₂ sticks to a magnet).
  • Medium : N₂ is diamagnetic.
  • Exam-type : B₂ is paramagnetic (two unpaired π e⁻); C₂ is diamagnetic.

Method: bond order of O₂

  • Step 1: 12 valence e⁻: σ2s², σ*2s², σ2p², π2p⁴, π*2p².
  • Step 2: N_b = 2 + 2 + 4 = 8; N_a = 2 + 2 = 4.
  • Step 3: Bond order = (8 − 4) ÷ 2 = 2; two unpaired in π*2p ⇒ paramagnetic.

CET level (MCQ)

  • CET 1) Bond order of O₂⁻? 1) 1 2) 1.5 3) 2 4) 2.5 — Answer: 2). (13 valence e⁻; N_b 8, N_a 5.)
  • CET 2) Which is paramagnetic? 1) N₂ 2) O₂ 3) C₂ 4) F₂ — Answer: 2).

Real-life use: Liquid oxygen is pulled by a magnet; MOT explains this which Lewis structures cannot.

Remember

BO = (N_b − N_a)/2; π2p lies lower for Z ≤ 7; O₂ is paramagnetic; He₂ does not exist.

Common mistake

Using O₂'s order for N₂; in N₂ π2p is lower.

CET trap

An ion changes the electron count (O₂⁻ = 13); recount N_b and N_a.

Shortcut

Memory clue: "N₂ 3, O₂ 2, F₂ 1, Ne₂ 0" (bond order).

Spot the difference

SpeciesBOMagnetism
N₂3Dia
O₂2Para
O₂⁻1.5Para
F₂1Dia

Practice questions

Easy
  1. 1What are bonding and antibonding orbitals?
  2. 2Write the bond order formula.
  3. 3Bond order of H₂?
Medium
  1. 4Find the bond order of N₂.
  2. 5Why is O₂ paramagnetic?
  3. 6Why does He₂ not exist?
Challenging
  1. 7Bond orders and stability order of O₂⁺, O₂ and O₂⁻.
  2. 8Bond order and magnetic nature of F₂.
Show answers (try them yourself first)
  1. 1Bonding has lower energy; antibonding higher.
  2. 2(N_b − N_a) ÷ 2.
  3. 31.
  4. 4(8 − 2) ÷ 2 = 3.
  5. 5Two unpaired e⁻ in π*2p.
  6. 6Bond order 0.
  7. 72.5, 2, 1.5; O₂⁺ > O₂ > O₂⁻.
  8. 8(8 − 6) ÷ 2 = 1; diamagnetic.

Bond parameters, polarity, dipole moment and Fajans' rules

From daily life

In a tug of war the stronger side draws the rope closer; bonding electrons shift towards the more electronegative atom.

Simple definition

Bond parameters: bond length, angle, enthalpy and order. Dipole moment μ = q × d (unit D). Fajans' rules: the more a cation polarises an anion the more covalent character.

Dipole momentDipole moment μ = q × d (unit D; 1 D = 3.336 × 10⁻³⁰ C m)HClμδ+δ−HCl 1.03 DOCOμ = 0CO₂ 0 (symmetric)OHHμH₂O 1.85 D% ionic character = (μobserved ÷ μ100% ionic) × 100HCl: μionic = e × d = 6.1 D ⇒ 1.03 ÷ 6.1 ≈ 17 %Vector sum zero ⇒ non-polar: CO₂, BF₃, CH₄

In the figure : HCl polar, CO₂ non-polar, H₂O polar.

FajansFajans' rules: when does covalent character rise?Na⁺Cl⁻IonicAl³⁺Cl⁻Covalent (polarised)Small, highly charged cation + large, highly charged anion⇒ more polarisation ⇒ more covalent character (LiI, AlCl₃)

In the figure : Small, highly charged cation ⇒ more polarisation.

Rule 1: Dipole moment — μ = q × d; a vector. Symmetrical molecules have zero resultant.

  • Easy : HCl 1.03 D, HF 1.91 D, H₂O 1.85 D, NH₃ 1.47 D.
  • Medium : CO₂, BF₃, CH₄, CCl₄: μ = 0.
  • Exam-type : NF₃ (0.24 D) < NH₃ (1.47 D) because the N–F moments oppose the lone pair.

Rule 2: Percent ionic character — % = (μ_observed ÷ μ_ionic) × 100; μ_ionic = e × d.

  • Easy : HCl: d = 127 pm ⇒ μ_ionic = 1.602 × 10⁻¹⁹ × 127 × 10⁻¹² ÷ 3.336 × 10⁻³⁰ = 6.1 D.
  • Medium : Ionic character = 1.03 ÷ 6.1 ≈ 17 %.
  • Exam-type : H₂O: 1.85 D with d(O–H) = 96 pm ⇒ net polarity towards O.

Rule 3: Fajans' rules — Covalent character rises when: the cation is small and highly charged; the anion is large and highly charged; the cation has an 18-e⁻ (d) shell.

  • Easy : LiI is more covalent than other halides (large I⁻).
  • Medium : AlCl₃ is covalent (small Al³⁺, 3+); NaCl is ionic.
  • Exam-type : NaCl is more ionic than AgCl (Ag⁺ has d¹⁰).

Derivation: net dipole of H₂O

  • Step 1: Two O–H moments μ₁ = μ₂ = μ at angle θ = 104.5°.
  • Step 2: Resultant μ_R = 2μ cos(θ/2) = 2μ cos 52.25°.
  • Step 3: cos 52.25° ≈ 0.612 ⇒ μ_R ≈ 1.22 μ ⇒ non-zero, so H₂O is polar.

Formulas, units and dimensional formulae

ItemFormulaUnit
μq × dD (3.336 × 10⁻³⁰ C m)
Bond enthalpyEnergy to breakkJ mol⁻¹
Bond lengthInternuclear distancepm (10⁻¹² m)

CET level (MCQ)

  • CET 1) Which is non-polar? 1) H₂O 2) NH₃ 3) CCl₄ 4) CHCl₃ — Answer: 3). (symmetrical, resultant zero.)
  • CET 2) Most covalent character? 1) NaCl 2) MgCl₂ 3) AlCl₃ 4) KCl — Answer: 3). (small, 3+ Al³⁺.)

Real-life use: Water is polar so it dissolves salt but not oil; a microwave oven heats food by rotating polar water molecules.

Remember

μ = q × d; symmetrical ⇒ μ = 0; Fajans: small, highly charged cation + large anion ⇒ covalent.

Common mistake

Assuming polar bonds always make a polar molecule (CO₂ has polar bonds but is non-polar).

CET trap

Comparing NH₃ and NF₃: lone-pair and bond moments add in NH₃ but oppose in NF₃.

Shortcut

Memory clue: "small + highly charged = the puller; large anion = the pulled".

Spot the difference

ItemIonicCovalent
CationLarge, 1+Small, 3+
AnionSmallLarge
ExampleNaClAlCl₃

Practice questions

Easy
  1. 1What is the unit of dipole moment?
  2. 2Why is μ of CO₂ zero?
  3. 3State Fajans' rules.
Medium
  1. 4Find the % ionic character of HCl (1.03 D; μ_ionic = 6.1 D).
  2. 5Why is NF₃'s dipole less than NH₃'s?
  3. 6Which is more covalent, LiCl or NaCl?
Challenging
  1. 7Show that the net μ of H₂O is 2μ cos(θ/2).
  2. 8State the relation of bond order, length and enthalpy.
Show answers (try them yourself first)
  1. 1Debye (D).
  2. 2Two C=O moments are equal and opposite.
  3. 3Small, highly charged cation; large anion; 18-e⁻ cation ⇒ covalent.
  4. 41.03 ÷ 6.1 × 100 ≈ 17 %.
  5. 5In NF₃ the lone pair and N–F moments oppose.
  6. 6LiCl (small Li⁺).
  7. 7Resultant of two equal vectors = 2μ cos(θ/2).
  8. 8Bond order ↑ ⇒ length ↓, enthalpy ↑.

Quick revision of the whole lesson

ConceptKey formula / fact
Octet/ionicLattice enthalpy ∝ q₁q₂ ÷ r
Lewiscount e⁻ → pair → octet
FCV − N − B/2
VSEPRlp–lp > lp–bp > bp–bp ; 180 · 120 · 109.5 · 90
HybridisationH = ½(V + M − C + A)
MOTBO = (N_b − N_a)/2
Dipoleμ = q × d ; 1 D = 3.336 × 10⁻³⁰ C m
Fajanssmall, highly charged cation + large anion ⇒ covalent
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🎯 Most important points for the exam

  • Ionic bond and the Born–Haber cycle.
  • Lewis structures; bond order–length–enthalpy.
  • FC, resonance and octet exceptions.
  • VSEPR shapes and angles (CH₄ > NH₃ > H₂O).
  • Finding hybridisation; counting σ and π.
  • MOT: O₂ paramagnetic, N₂ BO 3.
  • Dipole moment and Fajans' rules.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.