Chemical Bonding and Molecular Structure Class 11: VSEPR, Hybridisation and MO Theory

What Chemical Bonding and Molecular Structure is about

Weightage: This chapter is Unit IV, Chemical Bonding and Molecular Structure, which the syllabus gives 7 of the 70 theory marks in the school annual exam.Source: CBSE Chemistry (043) syllabus 2026-27, Class XI course structure, Unit IV Chemical Bonding and Molecular Structure

Key concepts

Kössel-Lewis approach: octet rule, Lewis structures and formal charge

NCERT section 4.1 Kössel-Lewis Approach to Chemical Bonding (4.1.1 to 4.1.5)

Lewis symbols show the valence electrons of an atom as dots around its symbol (for example, N with five dots). The number of dots gives the group valence.

Kössel and Lewis (1916) proposed that atoms combine to attain the stable octet (eight electrons in the valence shell, like the noble gases; two for H, like He):

• Electropositive and electronegative atoms do this by transferring electrons, giving ions held by electrostatic attraction (electrovalent bond).
• Atoms of similar electronegativity share electron pairs, giving a covalent bond. Sharing one, two or three pairs gives single, double or triple bonds (Cl–Cl, O=C=O, N≡N).

Drawing a Lewis structure:
1. Add the valence electrons of all atoms; add one for each negative charge and subtract one for each positive charge.
2. Put the least electronegative atom in the centre (H and F are always terminal).
3. Join atoms with single bonds, complete the octets of the outer atoms, and put the rest on the central atom.
4. If the central atom lacks an octet, turn lone pairs on outer atoms into multiple bonds.

Formal charge helps pick the best structure:
formal charge = (valence electrons in free atom) − (non-bonding electrons) − ½(bonding electrons).
The preferred structure has the smallest formal charges, with any negative charge on the more electronegative atom. Formal charges are bookkeeping; they are not real charges.

Limitations of the octet rule:

• Incomplete octet: BeH₂ (4 electrons on Be), BF₃ and BCl₃ (6 on B), LiCl.
• Odd-electron molecules: NO and NO₂ cannot give every atom an octet.
• Expanded octet: elements of period 3 and beyond can use d orbitals: PCl₅ (10), SF₆ (12), H₂SO₄ (12 around S).
• Noble gases form compounds (XeF₂, XeF₄, XeOF₄); the rule does not explain shape or energy of molecules.

Worked example

Draw the Lewis structure of ozone, O₃ (with one O=O and one O–O bond), and calculate the formal charge on each oxygen atom. (3 marks)

Try it first, then show the solutionSolution

1. Valence electrons = 3 × 6 = 18. Structure: O=O–O with the central O carrying one lone pair, the double-bonded end O two lone pairs and the single-bonded end O three lone pairs. (1)
2. Central O: 6 − 2 − ½(6) = +1. Double-bonded end O: 6 − 4 − ½(4) = 0. (1)
3. Single-bonded end O: 6 − 6 − ½(2) = −1. The charges add up to 0, as they must for a neutral molecule. (1)

Answer: Central O +1; double-bonded O 0; single-bonded O −1

Ionic bond and lattice enthalpy

NCERT section 4.2 Ionic or Electrovalent Bond (4.2.1 Lattice Enthalpy)

An ionic bond forms when one atom loses electrons (forming a cation) and another gains them (forming an anion), and the ions attract each other. Its formation is favoured by:

• low ionisation enthalpy of the metal,
• high negative electron gain enthalpy of the non-metal,
• high lattice enthalpy of the solid.

In the gas phase, forming Na⁺ and Cl⁻ from atoms actually costs energy (ionisation of Na needs 495.8 kJ mol⁻¹; electron gain by Cl releases only 348.7 kJ mol⁻¹). The compound is stable because the ions pack into a three-dimensional crystal lattice, and the energy released then is large.

Lattice enthalpy is the energy needed to separate one mole of a solid ionic compound completely into gaseous ions. For NaCl it is 788 kJ mol⁻¹. The larger the charges and the smaller the ions, the higher the lattice enthalpy.

Ionic compounds do not form discrete molecules; each ion is surrounded by ions of opposite charge (in NaCl, six of each), so we speak of a formula unit. Most ionic compounds have cations from metals and anions from non-metals; the ammonium ion is an exception made only of non-metals.

Worked example

Forming Na⁺(g) and Cl⁻(g) from Na(g) and Cl(g) costs a net 147 kJ mol⁻¹, yet NaCl is very stable. (a) Explain. (b) Which would you expect to have the higher lattice enthalpy, NaCl or MgO? Give a reason. (3 marks)

Try it first, then show the solutionSolution

1. Net cost: Na → Na⁺ needs 495.8 kJ mol⁻¹; Cl → Cl⁻ releases 348.7 kJ mol⁻¹; difference ≈ +147 kJ mol⁻¹. (1)
2. (a) When the gaseous ions come together to form the crystal lattice, 788 kJ mol⁻¹ of lattice energy is released, which more than outweighs this cost. (1)
3. (b) MgO: its ions carry charges of +2 and −2 (against +1 and −1 in NaCl) and are small, so the attraction and the lattice enthalpy are much greater. (1)

Answer: The large lattice enthalpy (788 kJ mol⁻¹) more than pays for ion formation; MgO has the higher lattice enthalpy because of doubled ionic charges.

Bond parameters and resonance

NCERT section 4.3 Bond Parameters (4.3.1 to 4.3.5)

• Bond length: equilibrium distance between the nuclei of two bonded atoms. Covalent radius = half the bond length for identical atoms. Bond length decreases as bond multiplicity increases: C–C 154 pm, C=C 134 pm, C≡C 120 pm.
• Bond angle: the angle between the orbitals holding bonding pairs around a central atom (H–O–H in water is 104.5°). It tells us the shape.
• Bond enthalpy: energy needed to break one mole of bonds of a type between two atoms in the gas phase (H–H 435.8 kJ mol⁻¹, O=O 498 kJ mol⁻¹, N≡N 946 kJ mol⁻¹). For polyatomic molecules, such as the two O–H bonds of water, the breaking energies differ, so we use the average bond enthalpy.
• Bond order: number of bonds between two atoms (H₂ 1, O₂ 2, N₂ 3). Isoelectronic species have the same bond order (N₂, CO and NO⁺ all have 3). Higher bond order means greater bond enthalpy and shorter bond length.

Resonance: when a single Lewis structure cannot explain the observed properties, the molecule is described by two or more canonical (resonance) structures; the real molecule is the resonance hybrid. In O₃, both O–O bonds are equal (128 pm), between a single (148 pm) and a double (121 pm) bond. Resonance stabilises the molecule: the hybrid is lower in energy than any single canonical form. The canonical forms are not real and the molecule does not switch between them. CO₃²⁻ (three equal C–O bonds) and CO₂ are other examples.

Worked example

Explain why all three C–O bonds in the carbonate ion are equal in length, and calculate the average C–O bond order. (2 marks)

Try it first, then show the solutionSolution

1. One Lewis structure would show one C=O and two C–O⁻ bonds of different lengths. In reality the double bond can be drawn in any of three positions, giving three equivalent canonical forms; the actual ion is their resonance hybrid, so all three bonds are identical. (1)
2. Total bonds = 4 (one double + two single) shared over 3 positions, so bond order = 4 ÷ 3 = 1.33. (1)

Answer: Resonance among three equivalent structures makes all C–O bonds equal; bond order = 1.33.

Polarity of bonds and dipole moment

NCERT section 4.3.6 Polarity of Bonds

In H₂ or Cl₂ the shared pair lies exactly midway: a non-polar covalent bond. When the atoms differ in electronegativity, as in HF, the shared pair moves towards the more electronegative atom, giving partial charges ($H^{\delta +}$–$F^{\delta -}$): a polar covalent bond.

Dipole moment (μ) measures polarity: $\mu =q\times d$ (charge × distance). It is a vector, drawn as an arrow from the positive to the negative end (chemists' convention; physicists draw it the other way). Unit: debye, 1 D = 3.33564 × 10⁻³⁰ C m.

For a molecule, the dipole moment is the vector sum of all bond dipoles (and lone-pair effects). So shape matters:

• CO₂ (linear) and BF₃ (trigonal planar) and CCl₄ (tetrahedral): bond dipoles cancel, μ = 0.
• H₂O (bent): μ = 1.85 D.
• NH₃ (1.47 D) vs NF₃ (0.24 D): in NH₃ the lone pair's dipole points the same way as the N–H bond dipoles and adds to them; in NF₃ the N–F bond dipoles point towards F, opposite to the lone pair, so they largely cancel.

Even ionic bonds have some covalent character. Fajans' rules: covalent character increases when the cation is small and highly charged (more polarising), the anion is large (more polarisable), and the cation has a pseudo-noble-gas configuration (such as Cu⁺) rather than a noble gas one.

Worked example

BF₃ has polar B–F bonds but zero dipole moment, while NH₃ has a dipole moment of 1.47 D. Explain, with the shapes of the molecules. (2 marks)

Try it first, then show the solutionSolution

1. BF₃ is trigonal planar with bond angles of 120°; the three equal B–F bond dipoles cancel by symmetry, so μ = 0. (1)
2. NH₃ is trigonal pyramidal with a lone pair on N; the bond dipoles and the lone-pair effect add up to a resultant, so μ = 1.47 D. (1)

Answer: Symmetry cancels the bond dipoles in planar BF₃; pyramidal NH₃ has a net dipole.

VSEPR theory and shapes of molecules

NCERT section 4.4 The Valence Shell Electron Pair Repulsion (VSEPR) Theory

Valence Shell Electron Pair Repulsion (VSEPR) theory (Sidgwick and Powell; refined by Nyholm and Gillespie):
1. The shape depends on the number of electron pairs (bonded and lone) around the central atom.
2. Pairs repel each other and arrange themselves as far apart as possible.
3. A multiple bond counts as one 'super pair'.
4. Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair. Lone pairs are held only by one nucleus and spread out more, so they squeeze bond angles.

Shapes with no lone pairs (AB_x):

• 2 pairs: linear, 180° (BeCl₂)
• 3: trigonal planar, 120° (BF₃)
• 4: tetrahedral, 109.5° (CH₄)
• 5: trigonal bipyramidal, 120° and 90° (PCl₅)
• 6: octahedral, 90° (SF₆)

Shapes with lone pairs (AB_xE_y):

• 3 bp + 1 lp: trigonal pyramidal, 107° (NH₃)
• 2 bp + 2 lp: bent, 104.5° (H₂O)
• 2 bp + 1 lp: bent, below 120° (SO₂)
• 4 bp + 1 lp: see-saw (SF₄)
• 3 bp + 2 lp: T-shaped (ClF₃)
• 5 bp + 1 lp: square pyramidal (BrF₅)
• 4 bp + 2 lp: square planar (XeF₄)

In a trigonal bipyramid, lone pairs occupy equatorial positions, where they have fewer 90° repulsions.

Worked example

Predict the shapes of (a) ClF₃ and (b) XeF₄ using VSEPR theory. Show how you count electron pairs. (3 marks)

Try it first, then show the solutionSolution

1. (a) Cl has 7 valence electrons; 3 are used for 3 Cl–F bonds, leaving 4 = 2 lone pairs. Total 5 pairs → trigonal bipyramidal arrangement. (1)
2. The 2 lone pairs go to equatorial positions, so the atoms form a T shape. (½)
3. (b) Xe has 8 valence electrons; 4 bond with F, leaving 4 = 2 lone pairs. Total 6 pairs → octahedral arrangement. (1)
4. The 2 lone pairs sit opposite each other (180° apart, minimising lp–lp repulsion), so the four F atoms are square planar. (½)

Answer: (a) ClF₃: T-shaped (b) XeF₄: square planar

Valence bond theory: orbital overlap, sigma and pi bonds

NCERT section 4.5 Valence Bond Theory (4.5.1 to 4.5.5)

Valence bond theory (Heitler and London, developed by Pauling and Slater) explains a covalent bond as the overlap of half-filled atomic orbitals of two atoms, with the two electrons having opposite spins.

When two H atoms approach, new attractive forces (nucleus–electron of the other atom) and repulsive forces (nucleus–nucleus, electron–electron) appear. Attraction dominates until the energy reaches a minimum at 74 pm, the bond length of H₂; 435.8 kJ mol⁻¹ is released. Closer approach makes repulsion rise steeply.

Overlap concept: the greater the overlap, the stronger the bond. Overlap can be positive (same sign, bonding), negative (opposite sign) or zero (wrong orientation, no bond).

Types of covalent bonds:

• Sigma (σ) bond: end-to-end (head-on) overlap along the internuclear axis. Possible with s–s, s–p and p–p (axial) overlap. Strong, because overlap is large, and it allows free rotation.
• Pi (π) bond: sideways overlap of parallel p orbitals above and below the axis. Weaker than σ, forms only after a σ bond, and prevents rotation.

A single bond is one σ; a double bond is one σ + one π; a triple bond is one σ + two π.

VBT alone cannot explain why the four bonds of CH₄ are identical, or why water's angle is 104.5° instead of the 90° expected from p orbitals. Hybridisation fixes this.

Worked example

Count the σ and π bonds in (a) propyne, CH₃–C≡CH, and (b) acrylonitrile, CH₂=CH–C≡N. (2 marks)

Try it first, then show the solutionSolution

1. (a) Four C–H σ bonds (three in CH₃, one on the end carbon) + one C–C single σ + the C≡C (1 σ + 2 π) → 6 σ and 2 π. (1)
2. (b) Three C–H σ bonds; C=C has 1 σ + 1 π; C–C single 1 σ; C≡N has 1 σ + 2 π. Total σ = 3 + 1 + 1 + 1 = 6; π = 1 + 2 = 3. (1)

Answer: (a) 6 σ, 2 π (b) 6 σ, 3 π

Hybridisation

NCERT section 4.6 Hybridisation (4.6.1 to 4.6.3)

Hybridisation is the mixing of atomic orbitals of slightly different energies on the same atom to give an equal number of new, identical hybrid orbitals with equal energy and shape. Features:

• Only orbitals of similar energy mix (2s with 2p; 3s, 3p with 3d).
• The number of hybrid orbitals equals the number of orbitals mixed.
• Hybrid orbitals point in definite directions and form stronger σ bonds than pure orbitals.
• Unhybridised p orbitals can form π bonds.

Types:

• sp (one s + one p): linear, 180°, 50% s character. BeCl₂; each carbon in ethyne (with two unhybridised p orbitals forming two π bonds).
• sp² (one s + two p): trigonal planar, 120°. BCl₃; each carbon in ethene (one unhybridised p forms the π bond).
• sp³ (one s + three p): tetrahedral, 109.5°. CH₄, C₂H₆. In NH₃ and H₂O the central atom is also sp³, but lone pairs occupy one or two hybrid orbitals and compress the angle to 107° and 104.5°.
• sp³d (s + 3p + d_z²): trigonal bipyramidal. PCl₅: three equatorial P–Cl bonds (120°) and two longer axial bonds (90° to the plane), which explains why PCl₅ is reactive.
• sp³d² (s + 3p + d_z² + d_x²−y²): octahedral, 90°. SF₆.

Quick rule: hybridisation = number of σ bonds + number of lone pairs on the central atom (2 → sp, 3 → sp², 4 → sp³, 5 → sp³d, 6 → sp³d²).

Worked example

Give the hybridisation of the central atom and the shape of (a) SO₃ (b) SiCl₄ (c) ICl₄⁻. Why do all five atoms of ICl₄⁻ lie in one plane? (3 marks)

Try it first, then show the solutionSolution

1. (a) S in SO₃: 3 σ bonds (the π bonds do not count), 0 lone pairs → sp², trigonal planar. (½)
2. (b) Si: 4 σ bonds, 0 lone pairs → sp³, tetrahedral. (½)
3. (c) I has 7 valence electrons + 1 for the charge = 8; 4 are used in I–Cl bonds, leaving 2 lone pairs: 6 pairs → sp³d², octahedral arrangement. (1)
4. The two lone pairs take positions opposite each other (180° apart, which minimises lone pair–lone pair repulsion), so the four Cl atoms and I lie in one square plane. (1)

Answer: (a) sp², trigonal planar (b) sp³, tetrahedral (c) sp³d², square planar, because the two lone pairs sit trans to each other

Molecular orbital theory

NCERT section 4.7 Molecular Orbital Theory (4.7.1 to 4.7.5)

Molecular orbital (MO) theory (Hund and Mulliken) treats electrons as belonging to the whole molecule:

• Atomic orbitals of comparable energy and proper symmetry combine (LCAO: linear combination of atomic orbitals) to form the same number of molecular orbitals.
• Bonding MO: formed by addition (constructive interference); lower in energy, electron density concentrated between the nuclei. Antibonding MO (marked *): formed by subtraction; higher in energy, with a node between the nuclei.
• Conditions for combination: similar energy, same symmetry about the molecular axis, and maximum overlap.
• Types: σ (symmetrical about the axis, from s–s or axial p–p) and π (from sideways p–p).
• Electrons fill MOs by the aufbau principle, Pauli's principle and Hund's rule.

Energy order:

• For O₂ and F₂ (and Ne₂): σ1s < σ*1s < σ2s < σ*2s < σ2p_z < (π2pₓ = π2p_y) < (π*2pₓ = π*2p_y) < σ*2p_z.
• For Li₂, Be₂, B₂, C₂ and N₂: σ1s < σ*1s < σ2s < σ*2s < (π2pₓ = π2p_y) < σ2p_z < (π*2pₓ = π*2p_y) < σ*2p_z.

Bond order = ½(N_b − N_a), where N_b and N_a are electrons in bonding and antibonding MOs.

• Bond order > 0: the molecule exists; zero or negative: it does not (He₂, Be₂, Ne₂).
• Higher bond order: greater stability, greater bond enthalpy, shorter bond.
• Magnetic nature: any unpaired electron makes the species paramagnetic (attracted by a magnetic field); all paired makes it diamagnetic.

Worked example

Calcium carbide contains the C₂²⁻ ion. Write its MO configuration, find its bond order and magnetic nature, and name a neutral molecule isoelectronic with it. (3 marks)

Try it first, then show the solutionSolution

1. C₂²⁻ has 2 × 6 + 2 = 14 electrons, so the N₂-type order applies: σ1s² σ*1s² σ2s² σ*2s² (π2pₓ² = π2p_y²) σ2p_z². (1)
2. N_b = 2 + 2 + 4 + 2 = 10; N_a = 2 + 2 = 4. Bond order = ½(10 − 4) = 3. (1)
3. All electrons are paired, so it is diamagnetic; it is isoelectronic with N₂ (and CO), which also has bond order 3. (1)

Answer: Bond order 3 (a triple bond, C≡C); diamagnetic; isoelectronic with N₂

Bonding in homonuclear diatomic molecules and hydrogen bonding

NCERT section 4.8 Bonding in Some Homonuclear Diatomic Molecules; 4.9 Hydrogen Bonding

Applying MO theory (using only the valence electrons is acceptable if the inner ones are noted as KK):

• H₂ (2 e): σ1s²; bond order 1; diamagnetic; bond enthalpy 438 kJ mol⁻¹.
• He₂ (4 e): σ1s² σ*1s²; bond order 0; does not exist.
• Li₂ (6 e): KK σ2s²; bond order 1; diamagnetic; exists in the vapour.
• C₂ (12 e): KK σ2s² σ*2s² (π2pₓ² = π2p_y²); bond order 2, made of two π bonds with no σ bond; diamagnetic.
• O₂ (16 e): KK σ2s² σ*2s² σ2p_z² (π2pₓ² = π2p_y²) (π*2pₓ¹ = π*2p_y¹); bond order 2; two unpaired electrons, so paramagnetic. This success of MO theory cannot be explained by the Lewis structure O=O.

For oxygen ions, removing or adding electrons in π* orbitals changes the bond order: O₂⁺ 2.5, O₂ 2, O₂⁻ 1.5, O₂²⁻ 1. Bond length rises in that order.

Hydrogen bonding: when hydrogen is bonded to a strongly electronegative atom (F, O or N), the H atom carries a significant δ+ charge and is attracted to a lone pair of another electronegative atom. This attraction is the hydrogen bond, shown by a dotted line ($H^{\delta +}-F^{\delta -}\cdots H^{\delta +}-F^{\delta -}$). It is much weaker than a covalent bond but stronger than ordinary van der Waals forces.

• Intermolecular (between molecules): HF, H₂O, NH₃, alcohols. It raises boiling points (water is a liquid while H₂S is a gas) and can make molecules associate.
• Intramolecular (within one molecule): o-nitrophenol, where the –OH hydrogen bonds to the neighbouring –NO₂ oxygen. This prevents association between molecules, so o-nitrophenol is more volatile than p-nitrophenol, which forms intermolecular hydrogen bonds.

Worked example

Compare the bond orders and magnetic behaviour of F₂, F₂⁺ and F₂⁻, and arrange them in order of increasing bond length. (3 marks)

Try it first, then show the solutionSolution

1. F₂ (18 e): KK σ2s² σ*2s² σ2p_z² (π2pₓ² = π2p_y²) (π*2pₓ² = π*2p_y²); N_b = 8, N_a = 6 counting only valence MOs → bond order 1; all paired, diamagnetic. (1)
2. F₂⁺ (17 e): one π* electron fewer → bond order ½(8 − 5) = 1.5, one unpaired electron, paramagnetic. F₂⁻ (19 e): the extra electron enters σ*2p_z → bond order ½(8 − 7) = 0.5, one unpaired electron, paramagnetic. (1)
3. Higher bond order means a shorter bond: F₂⁺ < F₂ < F₂⁻. (1)

Answer: Bond orders 1.5 (F₂⁺), 1 (F₂), 0.5 (F₂⁻); F₂ diamagnetic, both ions paramagnetic; bond length F₂⁺ < F₂ < F₂⁻

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

• MO Theory: Oxygen, Nitrogen and Related Diatomic Species
• VSEPR Theory, Shapes and Hybridisation

Chemical Bonding and Molecular Structure formula sheet

Formal charge

$FC=V-L-\frac{1}{2}S$

V = valence electrons of free atom, L = non-bonding electrons, S = bonding (shared) electrons.

Dipole moment

$\mu =q\times d$

1 D = 3.33564 × 10⁻³⁰ C m. A vector; for molecules add bond dipoles.

Steric number and hybridisation

$SN=\sigma \text{ bonds}+\text{lone pairs on central atom}$

2 sp, 3 sp², 4 sp³, 5 sp³d, 6 sp³d².

Lone pairs on central atom

$\text{lone pairs}=\frac{V-(\text{electrons used in bonds})\pm \text{charge}}{2}$

Add electrons for a negative charge, subtract for a positive one; count one electron per single bond to a monovalent atom.

Bond order (MO theory)

$B.O.=\frac{1}{2}(N_b-N_a)$

Zero or negative: the molecule does not exist.

Bond order from resonance

$B.O.=\frac{\text{total bonds between the atoms}}{\text{number of resonating positions}}$

CO₃²⁻: 4/3; NO₃⁻: 4/3; O₃: 3/2.

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.

• LP-LP > LP-BP > BP-BP

  Lone pairs are 'fat' because they belong to one nucleus only: two fat pairs push hardest.

  Use it when: Explaining angles: CH₄ 109.5°, NH₃ 107°, H₂O 104.5°.

• Count to hybridise

  σ bonds + lone pairs = 2, 3, 4, 5, 6 → sp, sp², sp³, sp³d, sp³d².

  Use it when: Any hybridisation question.

• 14 and below: pi before sigma

  For molecules with 14 or fewer electrons (up to N₂), π2p is below σ2p; above 14 (O₂, F₂), σ2p comes first.

  Use it when: Choosing the MO energy order.

• O₂ family ladder

  O₂²⁺ 3, O₂⁺ 2.5, O₂ 2, O₂⁻ 1.5, O₂²⁻ 1: each extra electron takes away half a bond.

  Use it when: Bond order and bond length questions.

• FON for hydrogen bonds

  Only H attached to F, O or N forms strong hydrogen bonds.

  Use it when: Spotting H-bonding molecules.

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 E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 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: Giving the electron-pair geometry instead of the shape

  Students stop after counting pairs.

  Wrong working

  NH₃ is tetrahedral.

  Right working

  NH₃ has a tetrahedral arrangement of pairs, but with one lone pair its shape is trigonal pyramidal.

  ✓Do this: Describe the shape by the positions of atoms only.

  Usually costs: 1 mark.

• ✗Mistake: Wrong MO energy order for N₂

  Students use the O₂ order for every molecule.

  Wrong working

  N₂: σ2p_z below π2p

  Right working

  N₂: π2pₓ = π2p_y below σ2p_z

  ✓Do this: Up to 14 electrons, π comes before σ2p_z.

  Usually costs: 1 mark (bond order still comes out right, but the configuration is marked wrong).

• ✗Mistake: Forgetting charge when counting electrons

  Ions are treated like neutral molecules.

  Wrong working

  NO₂⁻ has 5 + 12 = 17 electrons.

  Right working

  NO₂⁻ has 5 + 12 + 1 = 18 electrons.

  ✓Do this: Add one per negative charge, subtract one per positive charge.

  Usually costs: The whole Lewis structure.

• ✗Mistake: Saying resonance structures flip back and forth

  The double-headed arrow is misread.

  ✓Do this: The real molecule is a single hybrid at all times; canonical forms are only pen-and-paper representations.

  Usually costs: 1 mark.

• ✗Mistake: Placing lone pairs axially in trigonal bipyramids

  Students put them anywhere.

  ✓Do this: Lone pairs go equatorial (fewer 90° repulsions): SF₄ see-saw, ClF₃ T-shaped, XeF₂ linear.

  Usually costs: 1 mark per shape.

• ✗Mistake: Counting a double bond as two σ bonds in hybridisation

  Students count all bonds.

  ✓Do this: Count only σ bonds (one per bonded atom) plus lone pairs.

  Usually costs: 1 mark.

NCERT exercise guide: Chemical Bonding and Molecular Structure

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

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 the central atom in XeF₂ is:
   (a) sp (b) sp³ (c) sp³d (d) sp³d²

   Hint

   2 bond pairs + 3 lone pairs.

   Answer

   (c) sp³d; the three lone pairs are equatorial, so the molecule is linear.

2. Q2 · 1 mark · MCQExam level

   Which molecule has zero dipole moment?
   (a) NH₃ (b) CHCl₃ (c) BF₃ (d) H₂O

   Hint

   Look for a symmetrical shape.

   Answer

   (c) BF₃

3. Q3 · 2 marks · Very short answerExam level

   Distinguish between a σ bond and a π bond (any two points). (2 marks)

   Hint

   Overlap type and strength.

   Answer

   σ: head-on overlap along the axis; π: sideways overlap above and below the axis (1). σ is stronger and allows free rotation; π is weaker, forms only after a σ bond and restricts rotation (1).

4. Q4 · 2 marks · Very short answerHarder exam level

   Write the resonance structures of CO₂ and explain why its C–O bond length (115 pm) lies between C=O (121 pm) and C≡O (110 pm). (2 marks)

   Hint

   Three canonical forms.

   Answer

   O=C=O ↔ ⁻O–C≡O⁺ ↔ ⁺O≡C–O⁻ (1). The real molecule is a hybrid of these, so each C–O bond is between a double and a triple bond (1).

5. Q5 · 2 marks · Very short answerHarder exam level

   Arrange the following in increasing order of bond angle and explain: H₂O, NH₃, CH₄, CO₂. (2 marks)

   Hint

   Count lone pairs.

   Answer

   H₂O (104.5°) < NH₃ (107°) < CH₄ (109.5°) < CO₂ (180°) (1). More lone pairs compress the angle; CO₂ is linear (sp) (1).

6. Q6 · 3 marks · Short answerHarder exam level

   Give the shape and hybridisation of (a) NH₄⁺ (b) BrF₅ (c) I₃⁻. (3 marks)

   Hint

   Count lone pairs including the charge.

   Answer

   (a) NH₄⁺: 4 bp, 0 lp, sp³, tetrahedral (1). (b) BrF₅: 5 bp, 1 lp, sp³d², square pyramidal (1). (c) I₃⁻: central I has 2 bp, 3 lp, sp³d, linear (1).

7. Q7 · 3 marks · Short answerHarder exam level

   Use MO theory to compare the stability of H₂⁺ and H₂⁻. Which is more stable and why? (3 marks)

   Hint

   Both have bond order 0.5.

   Answer

   H₂⁺: σ1s¹, bond order ½ (1). H₂⁻: σ1s² σ*1s¹, bond order ½ (1). Same bond order, but H₂⁺ is more stable because H₂⁻ has an electron in the antibonding orbital, which is more destabilising than a missing bonding electron (1).

8. Q8 · 3 marks · Short answerHarder exam level

   Suppose a student claims that because BF₃ has polar bonds and NH₃ has polar bonds, both must have similar dipole moments. Correct the student with reasons, and say which of the two can act as a Lewis base. (3 marks)

   Hint

   Shape and lone pair.

   Answer

   BF₃ is trigonal planar (sp²); its bond dipoles cancel, μ = 0 (1). NH₃ is pyramidal with a lone pair; μ = 1.47 D (1). NH₃ can donate its lone pair, so it is the Lewis base (BF₃ is a Lewis acid with an incomplete octet) (1).

9. Q9 · 2 marks · Very short answerHarder exam level

   Out of p-nitrophenol and o-nitrophenol, which has the higher boiling point? Explain with the type of hydrogen bonding. (2 marks)

   Hint

   Inter- versus intra-.

   Answer

   p-Nitrophenol (1): it forms intermolecular hydrogen bonds that link molecules together; o-nitrophenol forms intramolecular hydrogen bonds, so its molecules stay separate and it is more volatile (1).

10. Q10 · 3 marks · Short answerHOTS / competency

    Suppose a chemist wants to know whether B₂ is paramagnetic. Write its MO configuration using the correct energy order, find the bond order and predict its magnetic nature. What would you predict if the O₂-type order were wrongly used? (3 marks)

    Hint

    B₂ has 10 electrons.

    Answer

    Correct order: σ1s² σ*1s² σ2s² σ*2s² (π2pₓ¹ = π2p_y¹); bond order ½(6 − 4) = 1; two unpaired electrons, paramagnetic (2). With the O₂-type order the last two electrons would pair in σ2p_z, predicting diamagnetism; experiment shows B₂ is paramagnetic, which supports the N₂-type order (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    (a) Explain with a diagram the formation of the π bond in ethene and why it prevents rotation. (b) Ethene has sp² carbons; give all bond angles. (c) In ethyne, how are the two π bonds arranged? (d) Which is shorter, the C–C bond in ethane, ethene or ethyne, and why? (5 marks)

    Hint

    Unhybridised p orbitals.

    Answer

    (a) Each sp² carbon keeps one unhybridised 2p orbital perpendicular to the molecular plane; these overlap sideways above and below the C–C axis to form the π bond (1). Rotating one CH₂ group would twist the p orbitals out of line and break the overlap, so rotation is restricted (1). (b) About 120° for H–C–H and H–C=C (all atoms in one plane) (1). (c) Each sp carbon has two unhybridised p orbitals (pₓ and p_y); they overlap to give two π bonds in perpendicular planes, forming a cylinder of electron density around the axis (1). (d) Ethyne (120 pm) < ethene (134 pm) < ethane (154 pm): more bonds and more s character pull the atoms closer (1).

12. Q12 · 1 mark · Assertion-reasonHOTS / competency

    Assertion (A): In PCl₅, all five P–Cl bonds are not equivalent.
    Reason (R): The axial bond pairs experience more repulsion from the equatorial bond pairs.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

    Hint

    Count 90° interactions.

    Answer

    (a)

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.Define lattice enthalpy.Show answer

   The energy needed to separate one mole of a solid ionic compound completely into gaseous ions.

2. 2.Give the formal charge formula.Show answer

   FC = valence electrons − non-bonding electrons − ½ bonding electrons.

3. 3.Name two molecules with an incomplete octet.Show answer

   BF₃ (or BCl₃) and BeH₂ (or BeCl₂, LiCl).

4. 4.What is the dipole moment of CO₂ and why?Show answer

   Zero, because it is linear and the two C=O bond dipoles cancel.

5. 5.What is the shape of SF₄?Show answer

   See-saw (sp³d, one equatorial lone pair).

6. 6.Which is stronger, σ or π bond?Show answer

   σ, because head-on overlap is greater.

7. 7.What is the hybridisation of carbon in ethene?Show answer

   sp² (trigonal planar, 120°).

8. 8.What is the bond order of He₂?Show answer

   Zero, so He₂ does not exist.

9. 9.Why is O₂ paramagnetic?Show answer

   It has two unpaired electrons in the π*2p orbitals.

10. 10.Why is o-nitrophenol more volatile than p-nitrophenol?Show answer

    It forms intramolecular hydrogen bonds, so molecules do not associate; p-nitrophenol forms intermolecular hydrogen bonds.

Quick revision checklist

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

• ☐Can you draw the Lewis structures of CO, HNO₃ and NO₂⁻ with formal charges?
• ☐Can you explain why NaCl is stable even though ion formation costs energy?
• ☐Can you explain the order of bond angles in CH₄, NH₃ and H₂O?
• ☐Can you predict the shapes of SF₄, ClF₃, XeF₂ and XeF₄?
• ☐Can you count σ and π bonds in any organic molecule?
• ☐Can you give hybridisation and shape of PCl₅ and SF₆ and explain the unequal bonds in PCl₅?
• ☐Can you write MO configurations of N₂ and O₂ and their ions with bond order?
• ☐Can you explain the two types of hydrogen bonds with examples?

Chemical Bonding and Molecular Structure: frequently asked questions