Chemical Bonding and Molecular Structure ISC Class 11

What Chemical Bonding and Molecular Structure is about

Weightage: Chemical Bonding and Molecular Structure belongs to unit 4 of the ISC Class XI Chemistry syllabus. CISCE weights units 1-6 together (Physical Chemistry) at 41 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Chemistry syllabus, examination year 2028, Class XI unit weightage

Key concepts

Octet rule, Lewis structures and the ionic bond

Valence electrons are shown as dots round the symbol (Lewis symbols). Kossel and Lewis proposed that atoms bond to achieve eight outer electrons (two for H, Li, Be). In an ionic bond, a metal transfers electrons to a non-metal: Na· + ·Cl → Na⁺[:Cl:]⁻; Mg loses 2 electrons to O in MgO and to two F atoms in MgF₂; two Li atoms give one electron each to O in Li₂O; Ca gives two to O in CaO; two Na give one each to S in Na₂S. Conditions favouring ionic bonding: low ionisation enthalpy of the metal, high (negative) electron gain enthalpy of the non-metal, and high lattice enthalpy. Variable electrovalency arises in heavier p-block elements: the ns² pair resists ionisation (inert pair effect), so Tl forms Tl⁺ as well as Tl³⁺ and Pb²⁺ is more stable than Pb⁴⁺.

Worked example

Show the electron transfer in forming Li₂O and Na₂S, and explain why Pb²⁺ is more stable than Pb⁴⁺.

Try it first, then show the solutionSolution

1. Li₂O: two Li atoms each give 1 electron to O, forming 2Li⁺ and O²⁻ (octet on O)
2. Na₂S: two Na atoms each give 1 electron to S, forming 2Na⁺ and S²⁻
3. In Pb ([Xe]4f¹⁴5d¹⁰6s²6p²) the 6s² pair is poorly shielded and held tightly, so losing only the two 6p electrons (Pb²⁺) is favoured

Answer: 2Li⁺O²⁻ and 2Na⁺S²⁻; inert pair effect

Lattice enthalpy and the Born-Haber cycle

Lattice enthalpy is the energy required to separate one mole of a solid ionic compound into its gaseous ions (or released when they come together). It increases with higher ionic charges and smaller ions. It cannot be measured directly, so the Born-Haber cycle uses Hess's law: the enthalpy of formation equals the sum of atomisation of the metal, its ionisation enthalpy, atomisation of the non-metal, its electron gain enthalpy, and the (negative) lattice-forming step.

Worked example

Find the lattice enthalpy of NaCl from: ΔsubH(Na) = +108, Δ_iH(Na) = +496, ½Δ_dissH(Cl₂) = +121, Δ_egH(Cl) = −349 and Δ_fH(NaCl) = −411 (all kJ mol⁻¹).

Try it first, then show the solutionSolution

1. Δ_fH = ΔsubH + Δ_iH + ½Δ_dissH + Δ_egH − U, where U is the lattice enthalpy
2. −411 = 108 + 496 + 121 − 349 − U
3. U = 108 + 496 + 121 − 349 + 411 = 787

Answer: 787 kJ mol⁻¹

Covalent bond, σ and π bonds, bond parameters, variable covalency and formal charge

A covalent bond is a shared pair. σ bonds come from head-on (axial) overlap and allow free rotation; π bonds come from sideways overlap of p orbitals and are weaker; a double bond is 1σ + 1π, a triple bond 1σ + 2π. Bond parameters: bond length (decreases as bond order increases), bond angle, bond enthalpy (energy to break one mole of bonds; increases with bond order) and bond order (number of shared pairs). Variable covalency: elements with vacant d orbitals can unpair electrons (P: 3 and 5; S: 2, 4 and 6). Formal charge = valence electrons − non-bonding electrons − ½ bonding electrons; the best Lewis structure has formal charges as small as possible.

Worked example

Find the formal charges on the three oxygen atoms in the Lewis structure of O₃ (O=O–O with the central O bearing one lone pair).

Try it first, then show the solutionSolution

1. Central O: 6 − 2 − ½(6) = +1
2. Double-bonded end O (two lone pairs): 6 − 4 − ½(4) = 0
3. Single-bonded end O (three lone pairs): 6 − 6 − ½(2) = −1

Answer: +1 (centre), 0 and −1 (ends); net charge 0

Deviations from the octet rule and Fajans' rules

The octet rule fails for: incomplete octets (BF₃, BeCl₂, AlCl₃: fewer than 8 electrons on the central atom); odd-electron molecules (NO, NO₂); expanded octets (PCl₅ with 10, SF₆ with 12 electrons, possible for period 3 onwards with d orbitals). It also cannot explain shapes or the relative stability of molecules. Fajans' rules: an ionic bond gains covalent character when the cation polarises the anion. Covalent character increases with a small, highly charged cation, a large, highly charged anion, and a cation with a non-noble-gas (pseudo-noble-gas, e.g. 18-electron) configuration such as Cu⁺ or Ag⁺.

Worked example

Which is more covalent: (i) NaCl or AlCl₃ (ii) LiF or LiI?

Try it first, then show the solutionSolution

1. (i) Al³⁺ is smaller and more highly charged than Na⁺, so it polarises Cl⁻ more: AlCl₃
2. (ii) I⁻ is much larger and more easily polarised than F⁻: LiI

Answer: AlCl₃; LiI

Polar bonds and dipole moment

In a bond between atoms of different electronegativity, the shared pair is pulled towards the more electronegative atom, creating partial charges δ+ and δ−. The dipole moment μ = q × d (charge × distance), in debye (1 D = 3.336 × 10⁻³⁰ C m). It is a vector; a molecule's dipole is the vector sum of bond dipoles and lone-pair effects. Symmetric molecules (CO₂, BF₃, CH₄, CCl₄) have zero dipole moment; bent H₂O (1.85 D) and pyramidal NH₃ (1.47 D) are polar. NF₃ (0.23 D) is much less polar than NH₃ because in NF₃ the N–F bond dipoles point away from the lone pair's dipole and partly cancel it. % ionic character = (observed μ/μ for a full charge separation) × 100, where μ(ionic) = 4.8 D per Å of bond length.

Worked example

HCl has μ = 1.03 D and bond length 1.27 Å. Find its percentage ionic character.

Try it first, then show the solutionSolution

1. μ (fully ionic) = 4.8 × 1.27 = 6.10 D
2. % ionic = 1.03/6.10 × 100 = 16.9%

Answer: About 17%

VSEPR theory and molecular shapes

Valence shell electron pair repulsion: electron pairs round the central atom repel and arrange themselves as far apart as possible; repulsion order lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Count bond pairs (a multiple bond counts once) and lone pairs. 2 pairs: linear (BeCl₂, 180°). 3: trigonal planar (BF₃, 120°). 4: tetrahedral (CH₄, 109.5°); with 1 lone pair trigonal pyramidal (NH₃, 107°); with 2 lone pairs bent (H₂O, 104.5°). 5: trigonal bipyramidal (PCl₅); with 1 lone pair see-saw (SF₄); with 2 T-shaped (ClF₃); with 3 linear (XeF₂). 6: octahedral (SF₆); with 1 lone pair square pyramidal (BrF₅); with 2 square planar (XeF₄).

Worked example

Explain why the bond angles in CH₄, NH₃ and H₂O are 109.5°, 107° and 104.5°.

Try it first, then show the solutionSolution

1. All three have four electron pairs round the central atom (tetrahedral arrangement)
2. CH₄: 4 bond pairs, no lone pair: ideal 109.5°
3. NH₃: 1 lone pair pushes bond pairs closer (107°); H₂O: 2 lone pairs push even more (104.5°)

Answer: Lone pairs repel more strongly than bond pairs

Valence bond theory and hybridisation

In valence bond theory, a covalent bond forms by overlap of half-filled atomic orbitals with opposite spins; greater overlap gives a stronger bond. Hybridisation is the mixing of atomic orbitals of similar energy on one atom to give an equal number of identical hybrid orbitals. sp (linear, 180°: BeCl₂, C₂H₂); sp² (trigonal planar, 120°: BF₃, C₂H₄); sp³ (tetrahedral, 109.5°: CH₄, NH₃, H₂O); sp³d (trigonal bipyramidal: PCl₅, using one d orbital); sp³d² (octahedral: SF₆). In PCl₅ the two axial bonds are longer and weaker than the three equatorial ones, because axial bond pairs suffer more repulsion. Unhybridised p orbitals form π bonds (C₂H₄: one; C₂H₂: two).

Worked example

State the hybridisation and shape of the central atom in BeCl₂, BF₃, PCl₅ and SF₆.

Try it first, then show the solutionSolution

1. BeCl₂: 2 bond pairs, sp, linear
2. BF₃: 3 bond pairs, sp², trigonal planar
3. PCl₅: 5 bond pairs, sp³d, trigonal bipyramidal; SF₆: 6 bond pairs, sp³d², octahedral

Answer: sp linear; sp² trigonal planar; sp³d trigonal bipyramidal; sp³d² octahedral

Coordinate bond and resonance

A coordinate (dative) bond is a shared pair supplied entirely by one atom (the donor, shown by →). Examples: NH₄⁺ (N lone pair to H⁺), H₃O⁺ (O lone pair to H⁺), HNO₃ (one N→O), O₃ (central O → end O), and the oxy-acids of chlorine: HOCl (none), HClO₂ (one Cl→O), HClO₃ (two), HClO₄ (three). Once formed, a coordinate bond is indistinguishable from other covalent bonds. Resonance: when a molecule cannot be represented by one Lewis structure, the real structure is a hybrid of several canonical forms differing only in electron positions. The hybrid is more stable than any single form, and bonds are intermediate: in CO₃²⁻ and NO₃⁻ each bond has order 1⅓; in O₃ each O–O bond has order 1½; CO₂ has contributing forms such as O=C=O, ⁻O–C≡O⁺ and ⁺O≡C–O⁻.

Worked example

Draw the resonance forms of the carbonate ion and find the C–O bond order.

Try it first, then show the solutionSolution

1. Three equivalent forms, each with one C=O and two C–O⁻, the double bond in a different position each time
2. Bond order = total bonds/number of C–O links = 4/3

Answer: Three forms; bond order 1.33, all three C–O bonds equal

Molecular orbital theory of homonuclear diatomics

Atomic orbitals of the two atoms combine (LCAO) into bonding MOs (lower energy, electron density between nuclei) and antibonding MOs (higher energy, marked *). Filling follows aufbau, Pauli and Hund. For B₂, C₂, N₂ (≤ 14 electrons) the order is σ1s σ*1s σ2s σ*2s (π2pₓ = π2p_y) σ2p_z (π*2pₓ = π*2p_y) σ*2p_z; for O₂, F₂ the σ2p_z lies below the π2p pair. Bond order = ½(N_b − N_a): zero means no stable molecule (He₂, Be₂); higher bond order means a shorter, stronger bond. Unpaired electrons make a molecule paramagnetic: O₂ has two unpaired electrons in π*2p, which the Lewis structure cannot explain.

Worked example

Write the MO configuration of O₂ and find its bond order and magnetic behaviour.

Try it first, then show the solutionSolution

1. 16 electrons: σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2pₓ² π2p_y² π*2pₓ¹ π*2p_y¹
2. N_b = 10, N_a = 6: bond order = ½(10 − 6) = 2
3. Two unpaired electrons in π* orbitals: paramagnetic

Answer: Bond order 2; paramagnetic

Hydrogen bonding

A hydrogen bond is the attraction between an H atom bonded to a small, highly electronegative atom (F, O, N) and a lone pair on another such atom. Intermolecular H-bonds (between molecules: HF, H₂O, alcohols, carboxylic acids) raise boiling points and viscosity and increase solubility in water. Intramolecular H-bonds (within one molecule, as in o-nitrophenol) close a ring and prevent association with other molecules, lowering boiling point and water solubility. Hydrogen bonds are weaker than covalent bonds (roughly 10–40 kJ mol⁻¹) but stronger than ordinary van der Waals forces.

Worked example

Why does water (M = 18) boil at 100 °C while H₂S (M = 34) boils at about −60 °C?

Try it first, then show the solutionSolution

1. O is small and very electronegative; each H₂O forms strong intermolecular H-bonds
2. S is larger and less electronegative; H₂S has only weak dipole and dispersion forces
3. Extra energy is needed to break the H-bond network in water

Answer: Intermolecular hydrogen bonding in water

Chemistry in real life: Chemical Bonding and Molecular Structure

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Molecular Orbital Theory, Bond Order and Magnetism
• VSEPR Theory, Shapes and Hybridisation

Chemical Bonding and Molecular Structure formula sheet

Formal charge

$FC=V-N-\frac{1}{2}B$

V valence, N non-bonding, B bonding electrons.

Born-Haber cycle

${\Delta }_{f}H={\Delta }_{\text{sub}}H+{\Delta }_{i}H+\frac{1}{2}{\Delta }_{\text{diss}}H+{\Delta }_{\text{eg}}H-U$

U = lattice enthalpy (positive, for separation).

Dipole moment

$\mu =q\times d;1\text{D}=3.336\times 10^{-30}\text{C m}$

Vector quantity.

Percentage ionic character

$\%\text{ionic}=\frac{{\mu }_{\text{obs}}}{4.8\times d(\text{A˚})}\times 100$

4.8 D = one electronic charge × 1 Å.

Bond order (MOT)

$\text{BO}=\frac{1}{2}(N_b-N_a)$

BO = 0: molecule not stable.

Electron pairs and hybridisation

$2:sp;3:s{p}^2;4:s{p}^3;5:s{p}^{3}d;6:s{p}^3{d}^2$

Count bond pairs + lone pairs on the central atom.

Chemical Bonding and Molecular Structure: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Chemical Bonding and Molecular Structure

The question patterns to expect, and the method that works for each.

Tricks to remember Chemical Bonding and Molecular Structure

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Lone pairs push hardest

  lp–lp > lp–bp > bp–bp: every lone pair squeezes the bond angle.

  Use it when: Bond angles.

• Fajans: small plus, big minus

  Small, highly charged cation + large anion = more covalent.

  Use it when: Fajans' rules.

• Count, then name

  Total pairs → hybridisation; lone pairs → shape name.

  Use it when: VSEPR and hybridisation.

• Fourteen and below flips

  Up to 14 electrons (N₂): π2p below σ2p; O₂ and F₂: σ2p below π2p.

  Use it when: MO order.

• FON for H-bonds

  Hydrogen bonds need H on F, O or N.

  Use it when: Identifying H-bonding.

Chemical Bonding and Molecular Structure: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Chemical Bonding and Molecular Structure

The slips examiners see most often, and what to write instead.

• ✗Mistake: Naming the shape from the electron arrangement

  Students call NH₃ 'tetrahedral'.

  Wrong working

  H₂O is tetrahedral.

  Right working

  Electron-pair arrangement tetrahedral; molecular shape bent (V-shaped).

  ✓Do this: Shape describes atoms only; mention both if asked.

  Usually costs: 1 mark.

• ✗Mistake: Wrong MO order for O₂ or N₂

  One order is used for all molecules.

  Wrong working

  N₂: σ2p_z filled before π2p.

  Right working

  For up to 14 electrons, π2p is lower than σ2p_z.

  ✓Do this: Write the order appropriate to the electron count.

  Usually costs: 1-2 marks.

• ✗Mistake: Counting a double bond as two pairs in VSEPR

  Students count every bond line.

  Wrong working

  CO₂: 4 pairs on C, so bent.

  Right working

  Two σ directions, no lone pair: linear.

  ✓Do this: Each multiple bond counts as one 'electron domain'.

  Usually costs: 1 mark.

• ✗Mistake: Sign errors in the Born-Haber cycle

  Lattice enthalpy is defined for separation (endothermic).

  Wrong working

  −411 = 108 + 496 + 121 − 349 + U ⇒ U = −787.

  Right working

  Formation of the lattice releases U: −411 = … − U ⇒ U = +787.

  ✓Do this: Draw the cycle with arrows before writing the equation.

  Usually costs: 1-2 marks.

• ✗Mistake: Confusing intra- and intermolecular H-bonds

  Students think more H-bonding always raises boiling point.

  Wrong working

  o-Nitrophenol boils higher because it has an H-bond.

  Right working

  Its H-bond is internal, so molecules do not associate: it boils lower.

  ✓Do this: Ask whether the H-bond joins two molecules or closes a ring.

  Usually costs: 1 mark.

Practice questions on Chemical Bonding and Molecular Structure

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   The hybridisation of carbon in CO₂ is:
   (a) sp
   (b) sp²
   (c) sp³
   (d) sp³d

   Hint

   Two σ bonds, linear.

   Answer

   (a)

2. Q2 · 1 mark · MCQExam level

   Which molecule is paramagnetic?
   (a) N₂
   (b) O₂
   (c) F₂
   (d) H₂

   Hint

   Unpaired π* electrons.

   Answer

   (b)

3. Q3 · 1 mark · Very short answerExam level

   What is the bond order of He₂?

   Hint

   σ1s² σ*1s².

   Answer

   0 (He₂ does not exist)

4. Q4 · 1 mark · Very short answerExam level

   Name the shape of SF₆.

   Hint

   Six bond pairs.

   Answer

   Octahedral

5. Q5 · 2 marks · Short answerExam level

   Show the formation of MgF₂ and Na₂S by electron transfer.

   Hint

   Count electrons lost and gained.

   Answer

   Mg gives one electron to each of two F atoms: Mg²⁺ 2F⁻. Two Na atoms each give one electron to S: 2Na⁺ S²⁻. All ions reach noble-gas configurations.

6. Q6 · 2 marks · Short answerHarder exam level

   Which is more covalent in each pair: (i) NaCl or MgCl₂ (ii) NaCl or CuCl (Na⁺ and Cu⁺ are of similar size)? Use Fajans' rules.

   Hint

   Cation charge; then the cation's electron configuration.

   Answer

   (i) MgCl₂: Mg²⁺ is smaller and more charged. (ii) CuCl: Cu⁺ has an 18-electron (pseudo-noble-gas) outer shell whose d electrons shield the nucleus poorly, so it polarises Cl⁻ more than Na⁺ with its noble-gas shell.

7. Q7 · 2 marks · Short answerHarder exam level

   Give the number of coordinate bonds in HOCl, HClO₂, HClO₃ and HClO₄.

   Hint

   Cl→O bonds beyond Cl–OH.

   Answer

   0, 1, 2 and 3

8. Q8 · 3 marks · Short answerHarder exam level

   Explain, with the VSEPR order of repulsions, why H₂O has a smaller bond angle than NH₃.

   Hint

   Number of lone pairs.

   Answer

   Both have four electron pairs (tetrahedral arrangement). H₂O has two lone pairs and NH₃ one; lone pair–lone pair repulsion is greatest, so the bond pairs in H₂O are pushed closer: 104.5° vs 107°.

9. Q9 · 3 marks · Short answerHarder exam level

   State the hybridisation of P in PCl₅ and explain why the axial P–Cl bonds are longer than the equatorial ones.

   Hint

   Count neighbours at 90°.

   Answer

   sp³d, trigonal bipyramidal. Each axial bond pair has three equatorial bond pairs at 90°, while each equatorial pair has only two at 90°; greater repulsion lengthens and weakens the axial bonds.

10. Q10 · 3 marks · Short answerHarder exam level

    Draw the resonance structures of NO₃⁻ and find the N–O bond order.

    Hint

    Like carbonate.

    Answer

    Three equivalent forms, each with one N=O and two N–O⁻ (N carries +1). Bond order 4/3 = 1.33; all N–O bonds are equal.

11. Q11 · 3 marks · Short answerHarder exam level

    Identify the coordinate bond in each of H₃O⁺, O₃ and HNO₃.

    Hint

    Who supplies the pair?

    Answer

    H₃O⁺: O lone pair donated to H⁺. O₃: central O donates a pair to one end O. HNO₃: N donates a pair to one O (N→O).

12. Q12 · 2 marks · Short answerHarder exam level

    Find the formal charge on each atom in two Lewis structures of CO₂: (i) O=C=O (ii) ⁻O–C≡O⁺, and say which better represents the molecule.

    Hint

    FC = V − N − ½B.

    Answer

    (i) Each O: 6 − 4 − 2 = 0; C: 4 − 0 − 4 = 0. (ii) Single-bonded O: 6 − 6 − 1 = −1; C: 4 − 0 − 4 = 0; triple-bonded O: 6 − 2 − 3 = +1. Structure (i), with all formal charges zero, is the better (major) one.

13. Q13 · 5 marks · Long answerHarder exam level

    (a) Find the numbers of σ and π bonds in ethene and ethyne. [2]
    (b) State the hybridisation of each carbon in CH₃–CH=CH₂. [2]
    (c) Give the approximate H–C–H bond angle at the CH₂= carbon. [1]

    Hint

    Double bond = 1σ + 1π.

    Answer

    (a) Ethene 5σ, 1π; ethyne 3σ, 2π. (b) CH₃: sp³; CH=: sp²; =CH₂: sp². (c) About 120°.

14. Q14 · 3 marks · Short answerHOTS / competency

    The bond angles in NH₃, PH₃ and AsH₃ are about 107°, 93.5° and 92°. Explain the trend.

    Hint

    Electronegativity and size of the central atom.

    Answer

    Down the group the central atom becomes larger and less electronegative, so the bond pairs lie farther from it and repel each other less; the lone pair then squeezes the bonds closer, and the angle falls towards 90°.

15. Q15 · 3 marks · Short answerHOTS / competency

    Use a Born-Haber cycle to find Δ_fH of LiF from (kJ mol⁻¹): ΔsubH(Li) = 161, Δ_iH(Li) = 520, ½Δ_dissH(F₂) = 79, Δ_egH(F) = −328, lattice enthalpy U = 1037.

    Hint

    Δ_fH = sum of steps − U.

    Answer

    Δ_fH = 161 + 520 + 79 − 328 − 1037 = −605 kJ mol⁻¹

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Inert pair effect?Show answer

   Reluctance of the ns² pair in heavy p-block elements to take part in bonding.

2. 2.Formal charge formula?Show answer

   V − N − ½B.

3. 3.Fajans' rules in one line?Show answer

   Small, highly charged cations and large anions give more covalent character.

4. 4.Shape of ClF₃?Show answer

   T-shaped (sp³d, two lone pairs).

5. 5.Hybridisation in SF₆?Show answer

   sp³d², octahedral.

6. 6.Bond order of O₂⁻?Show answer

   1.5.

7. 7.Bond order in CO₃²⁻?Show answer

   1.33.

8. 8.Why is o-nitrophenol steam volatile?Show answer

   Intramolecular hydrogen bonding.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you draw Lewis transfer diagrams for MgO, Li₂O and Na₂S?
• ☐Can you calculate a lattice enthalpy from a Born-Haber cycle?
• ☐Can you assign formal charges and pick the best structure?
• ☐Can you apply Fajans' rules to compare covalent character?
• ☐Can you predict the shape of any AB_xE_y molecule?
• ☐Can you state the hybridisation in BeCl₂, BF₃, CH₄, PCl₅ and SF₆?
• ☐Can you find bond orders and magnetism of O₂ and N₂ species?
• ☐Can you explain the effects of inter- and intramolecular H-bonds?

Chemical Bonding and Molecular Structure: frequently asked questions