How to Use This Page
This page provides complete exercise solutions for Chapter 10 — Chemical Bonding of the Samacheer Kalvi Class 11 Chemistry textbook. Work through each section:
1. Key Concepts — revise definitions, theories, and geometries. 2. MCQs through Match the Following — test your conceptual understanding. 3. Brief and Detailed Answers — practise writing structured explanations. 4. VSEPR and Hybridisation — master geometry predictions and orbital diagrams. 5. Common Mistakes & Exam Checklist — review before your exam.
Key Concepts (Quick Recap)
| Concept | Description |
|---|
| Ionic Bond | Complete transfer of electrons from metal to non-metal; held by electrostatic attraction |
| Covalent Bond | Sharing of electron pairs between atoms; sigma (σ) and pi (π) bonds |
| Coordinate Bond | Shared pair donated by one atom; donor has a lone pair, acceptor has empty orbital |
| Electronegativity | Ability of an atom to attract shared electrons; increases across a period, decreases down a group |
| Lattice Energy | Energy released when gaseous ions form 1 mole of ionic solid; larger for small ions with high charges |
| Bond Length | Average distance between nuclei of two bonded atoms; single > double > triple |
| Bond Energy | Energy required to break 1 mole of bonds; triple > double > single |
| VSEPR Theory | Electron pairs around central atom arrange to minimise repulsion; determines geometry |
| Hybridisation | Mixing of atomic orbitals to form equivalent hybrid orbitals for stronger bonding |
| Molecular Orbital Theory | Atomic orbitals combine to form bonding and antibonding molecular orbitals |
| Hydrogen Bond | Weak bond between H (bonded to F, O, or N) and another electronegative atom |
| Metallic Bond | Positive metal ions in a sea of delocalised electrons |
Hybridisation Summary Table:
| Hybridisation | Orbitals Mixed | Geometry | Bond Angle | Examples |
|---|
| sp | 1s + 1p | Linear | 180° | BeCl₂, C₂H₂, CO₂ |
| sp² | 1s + 2p | Trigonal planar | 120° | BF₃, C₂H₄, SO₃ |
| sp³ | 1s + 3p | Tetrahedral | 109.5° | CH₄, NH₃, H₂O |
| sp³d | 1s + 3p + 1d | Trigonal bipyramidal | 90°, 120° | PCl₅ |
| sp³d² | 1s + 3p + 2d | Octahedral | 90° | SF₆ |
VSEPR Geometry Summary:
| Electron Pairs | Bonding Pairs | Lone Pairs | Geometry | Example |
|---|
| 2 | 2 | 0 | Linear | BeCl₂ |
| 3 | 3 | 0 | Trigonal planar | BF₃ |
| 3 | 2 | 1 | Bent (V-shape) | SO₂ |
| 4 | 4 | 0 | Tetrahedral | CH₄ |
| 4 | 3 | 1 | Pyramidal | NH₃ |
| 4 | 2 | 2 | Bent (V-shape) | H₂O |
| 5 | 5 | 0 | Trigonal bipyramidal | PCl₅ |
| 6 | 6 | 0 | Octahedral | SF₆ |
I. Choose the Best Answer (MCQs)
1. The geometry of NH₃ according to VSEPR theory is:
Answer
(a) Tetrahedral (b) Trigonal pyramidal (c) Trigonal planar (d) Linear
Answer(b) Nitrogen in NH₃ has 4 electron pairs (3 bonding + 1 lone). The electron pair geometry is tetrahedral, but the molecular geometry (considering only bonded atoms) is trigonal pyramidal. The lone pair pushes the bonding pairs closer, reducing the bond angle from 109.5° to about 107°.
2. The hybridisation of carbon in ethylene (C₂H₄) is:
Answer
(a) sp (b) sp² (c) sp³ (d) sp³d
Answer(b) In ethylene, each carbon is bonded to 2 H atoms and 1 C atom with a double bond. Carbon forms 3 sigma bonds (2 C-H + 1 C-C) and 1 pi bond (C=C). Three sigma bonds require sp² hybridisation (3 hybrid orbitals). The remaining unhybridised p orbital forms the pi bond.
3. Which of the following has the highest lattice energy?
Answer
(a) NaCl (b) NaBr (c) NaF (d) NaI
Answer(c) Lattice energy increases with decreasing ionic size and increasing ionic charge. F⁻ is the smallest anion among the halides listed, so NaF has the shortest interionic distance and the highest lattice energy.
4. The number of sigma and pi bonds in N₂ are:
Answer
(a) 1 sigma, 2 pi (b) 2 sigma, 1 pi (c) 3 sigma, 0 pi (d) 1 sigma, 1 pi
Answer(a) The nitrogen molecule has a triple bond: N≡N. A triple bond consists of 1 sigma bond and 2 pi bonds. So N₂ has 1 sigma bond and 2 pi bonds.
5. The bond order of O₂ according to molecular orbital theory is:
Answer
(a) 1 (b) 2 (c) 2.5 (d) 3
Answer(b) O₂ has 16 electrons. The molecular orbital configuration fills bonding and antibonding orbitals. Bond order = (Nb − Na)/2 = (10 − 6)/2 = 2. This explains the double bond character and the paramagnetic nature of O₂ (due to 2 unpaired electrons in π* orbitals).
6. Which of the following molecules is polar?
Answer
(a) CCl₄ (b) BF₃ (c) NH₃ (d) CO₂
Answer(c) NH₃ is polar because its trigonal pyramidal geometry (due to the lone pair on N) means the individual bond dipoles do not cancel. CCl₄ (tetrahedral, symmetric), BF₃ (trigonal planar, symmetric), and CO₂ (linear, symmetric) all have their bond dipoles cancelling out.
7. The type of hybridisation in SF₆ is:
Answer
(a) sp³ (b) sp³d (c) sp³d² (d) sp³d³
Answer(c) Sulphur in SF₆ is bonded to 6 fluorine atoms with no lone pairs. This requires 6 hybrid orbitals, formed by mixing one s, three p, and two d orbitals → sp³d² hybridisation. The geometry is octahedral with 90° bond angles.
II. Fill in the Blanks
1. The ionic bond is formed by the complete transfer of electrons from a metal to a non-metal. 2. A covalent bond is formed by the sharing of electron pairs between atoms. 3. A coordinate bond is a type of covalent bond in which the shared pair is donated by one atom only. 4. The bond angle in a perfect tetrahedral molecule is 109.5°. 5. In VSEPR theory, lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion, which is greater than bond pair-bond pair repulsion. 6. The hybridisation of carbon in methane (CH₄) is sp³. 7. The number of lone pairs on oxygen in H₂O is 2. 8. The bond order of H₂ according to MOT is 1. 9. Hydrogen bonding occurs when hydrogen is bonded to F, O, or N (electronegative atoms). 10. The shape of BF₃ is trigonal planar. 11. In a sigma bond, the electron density is concentrated along the internuclear axis. 12. In a pi bond, the electron density is concentrated above and below the internuclear axis. 13. The hybridisation of Beryllium in BeCl₂ is sp. 14. The strongest bond among single, double, and triple bonds is the triple bond. 15. Metallic bonding is described as positive ions in a sea of delocalised electrons.
III. True or False (with Corrections)
1. Lattice energy increases with increasing ionic size.
False. Lattice energy increases with DECREASING ionic size (and increasing ionic charge). Smaller ions can get closer together, resulting in stronger electrostatic attraction and higher lattice energy.
2. A molecule with polar bonds must be polar.
False. A molecule with polar bonds may be polar OR nonpolar depending on its geometry. If the molecule has symmetric geometry (e.g., CO₂, CCl₄, BF₃), the bond dipoles cancel and the molecule is nonpolar. Polarity depends on both bond polarity and molecular shape.
3. In sp hybridisation, two hybrid orbitals are formed.
True. sp hybridisation involves mixing 1 s orbital and 1 p orbital to form 2 equivalent sp hybrid orbitals, oriented 180° apart (linear geometry). The remaining 2 unhybridised p orbitals are available for pi bonding.
4. The bond angle in H₂O (104.5°) is less than in NH₃ (107°) because oxygen has more lone pairs.
True. Oxygen in H₂O has 2 lone pairs while nitrogen in NH₃ has 1 lone pair. More lone pairs cause greater repulsion and a smaller bond angle. H₂O: 104.5° < NH₃: 107° < CH₄: 109.5°.
5. According to molecular orbital theory, O₂ is diamagnetic.
False. According to MOT, O₂ has 2 unpaired electrons in the π*2p orbitals, making it paramagnetic (attracted to a magnetic field), not diamagnetic. This is one of the great successes of MOT — it correctly predicts the paramagnetic nature of O₂, which valence bond theory could not explain.
IV. Match the Following
| Column A | Column B |
|---|
| 1. sp hybridisation | (a) Octahedral, 90° |
| 2. sp² hybridisation | (b) Tetrahedral, 109.5° |
| 3. sp³ hybridisation | (c) Linear, 180° |
| 4. sp³d hybridisation | (d) Trigonal bipyramidal, 90° and 120° |
| 5. sp³d² hybridisation | (e) Trigonal planar, 120° |
Answer
1 → (c) sp hybridisation: linear, 180° (e.g., BeCl₂, CO₂)
2 → (e) sp² hybridisation: trigonal planar, 120° (e.g., BF₃, C₂H₄)
3 → (b) sp³ hybridisation: tetrahedral, 109.5° (e.g., CH₄, NH₃)
4 → (d) sp³d hybridisation: trigonal bipyramidal, 90° and 120° (e.g., PCl₅)
5 → (a) sp³d² hybridisation: octahedral, 90° (e.g., SF₆)
V. Answer Briefly (2-mark)
1. Distinguish between sigma (σ) and pi (π) bonds.
Sigma (σ) bond: Pi (π) bond:
- Formed by head-on overlap - Formed by lateral (sideways) overlap
- Electron density along the axis - Electron density above and below the axis
- Stronger than pi bond - Weaker than sigma bond
- Free rotation possible - Free rotation NOT possible
- First bond between two atoms - Second and third bonds between two atoms
- Can exist independently - Always exists with a sigma bond
- Formed by s-s, s-p, p-p overlap - Formed only by p-p or p-d overlap
2. What is a coordinate bond? Give an example.
Answer
A coordinate bond (also called a dative bond) is a type of covalent bond in which both electrons of the shared pair come from one atom. The donor atom provides the lone pair and the acceptor atom provides the empty orbital.
Example: NH₃ → BF₃
NH₃ donates its lone pair on nitrogen to the empty orbital of boron in BF₃.
The product NH₃BF₃ has a coordinate bond: H₃N → BF₃
Another example: Formation of NH₄⁺ from NH₃ + H⁺
The lone pair on nitrogen in NH₃ is donated to H⁺ to form NH₄⁺.
3. Define lattice energy. What factors affect it?
Answer
Lattice energy is the energy released when 1 mole of an ionic solid is formed from its gaseous ions. It measures the strength of ionic bonding.
Na⁺(g) + Cl⁻(g) → NaCl(s) ΔH = −788 kJ/mol (lattice energy)
Factors affecting lattice energy:
1. Ionic charge: Higher charge → stronger attraction → higher lattice energy
(MgO > NaCl because Mg²⁺ and O²⁻ have higher charges)
2. Ionic size: Smaller ions → shorter distance → higher lattice energy
(NaF > NaCl > NaBr > NaI because F⁻ < Cl⁻ < Br⁻ < I⁻)
4. Why is H₂O a liquid while H₂S is a gas at room temperature?
Answer
Water (H₂O) is a liquid because it forms strong intermolecular hydrogen bonds between the H of one molecule and the O of another. Hydrogen bonds in water require significant energy to break, giving water a high boiling point.
H₂O: O is highly electronegative → strong H-bonds → liquid at RT
H₂S: S is less electronegative → weak dipole-dipole forces → gas at RT
Hydrogen bonding occurs only when H is bonded to F, O, or N.
Since S is not sufficiently electronegative, H₂S cannot form H-bonds.
VI. Answer in Detail (5-mark)
1. Explain VSEPR theory and predict the shapes of the following molecules: (i) BeCl₂, (ii) BF₃, (iii) CH₄, (iv) NH₃, (v) H₂O.
Answer
VSEPR (Valence Shell Electron Pair Repulsion) theory predicts the shape of molecules based on the following postulates:
1. The shape of a molecule depends on the number of electron pairs (bonding + lone pairs) around the central atom. 2. Electron pairs repel each other and arrange themselves as far apart as possible to minimise repulsion. 3. The order of repulsion is: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. 4. Lone pairs occupy more space than bonding pairs, which reduces bond angles. 5. Only the positions of atoms (not lone pairs) determine the molecular geometry.
(i) BeCl₂ — Linear:
Central atom: Be
Valence electrons on Be: 2
Bonding pairs: 2 (Be-Cl bonds)
Lone pairs: 0
Total electron pairs: 2
Geometry: Linear
Bond angle: 180°
Hybridisation: sp
Cl — Be — Cl
(ii) BF₃ — Trigonal Planar:
Central atom: B
Valence electrons on B: 3
Bonding pairs: 3 (B-F bonds)
Lone pairs: 0
Total electron pairs: 3
Geometry: Trigonal planar
Bond angle: 120°
Hybridisation: sp²
F
/ \
F — B — F (all in one plane)
(iii) CH₄ — Tetrahedral:
Central atom: C
Valence electrons on C: 4
Bonding pairs: 4 (C-H bonds)
Lone pairs: 0
Total electron pairs: 4
Geometry: Tetrahedral
Bond angle: 109.5°
Hybridisation: sp³
H
|
H—C—H
|
H
(iv) NH₃ — Trigonal Pyramidal:
Central atom: N
Valence electrons on N: 5
Bonding pairs: 3 (N-H bonds)
Lone pairs: 1
Total electron pairs: 4
Electron pair geometry: Tetrahedral
Molecular geometry: Trigonal pyramidal
Bond angle: 107° (reduced from 109.5° due to lone pair repulsion)
Hybridisation: sp³
..
N — H
/ \
H H (lone pair at top)
The lone pair pushes the bonding pairs closer together, reducing the bond angle from 109.5° to 107°.
(v) H₂O — Bent (V-shaped):
Central atom: O
Valence electrons on O: 6
Bonding pairs: 2 (O-H bonds)
Lone pairs: 2
Total electron pairs: 4
Electron pair geometry: Tetrahedral
Molecular geometry: Bent (V-shaped)
Bond angle: 104.5° (further reduced by 2 lone pairs)
Hybridisation: sp³
.. ..
O — H
\
H (two lone pairs at top)
Two lone pairs repel more than one, reducing the bond angle
further from 109.5° to 104.5°.
Summary of bond angle trend due to lone pair repulsion:
CH₄ (0 lone pairs): 109.5°
NH₃ (1 lone pair): 107°
H₂O (2 lone pairs): 104.5°
More lone pairs → greater repulsion → smaller bond angle
2. Explain hybridisation with examples. Discuss sp, sp², and sp³ hybridisation in detail.
Answer
Hybridisation is the concept of mixing atomic orbitals of comparable energy on the same atom to form a new set of equivalent orbitals called hybrid orbitals. These hybrid orbitals are better oriented for overlap and form stronger bonds than pure atomic orbitals.
Key features of hybridisation:
1. Number of hybrid orbitals = number of atomic orbitals mixed
2. Hybrid orbitals are equivalent in energy and shape
3. They are oriented in specific directions (determining geometry)
4. The type of hybridisation determines the bond angle
5. Only orbitals of the same atom can hybridise
6. Hybridisation explains geometry that pure orbital overlap cannot
(i) sp Hybridisation (1 s + 1 p → 2 sp orbitals):
Process: 1 s orbital + 1 p orbital → 2 sp hybrid orbitals
Geometry: Linear
Bond angle: 180°
Example: BeCl₂
Be ground state: 1s² 2s²
Be excited state: 1s² 2s¹ 2p¹
Hybridisation: 2s + 2p → 2sp
Two sp orbitals overlap with Cl orbitals to form 2 sigma bonds.
Result: Cl — Be — Cl (linear, 180°)
Example: C₂H₂ (Acetylene/Ethyne)
Each C: 2s + 2p → 2sp orbitals + 2 unhybridised p orbitals
1 sp orbital forms C-H sigma bond
1 sp orbital forms C-C sigma bond
2 unhybridised p orbitals form 2 pi bonds (C≡C triple bond)
(ii) sp² Hybridisation (1 s + 2 p → 3 sp² orbitals):
Process: 1 s orbital + 2 p orbitals → 3 sp² hybrid orbitals
Geometry: Trigonal planar
Bond angle: 120°
1 unhybridised p orbital remains for pi bonding
Example: BF₃
B ground state: 1s² 2s² 2p¹
B excited state: 1s² 2s¹ 2px¹ 2py¹ 2pz⁰
Hybridisation: 2s + 2px + 2py → 3sp² orbitals
Three sp² orbitals overlap with F orbitals to form 3 sigma bonds.
Result: Trigonal planar, 120°
Example: C₂H₄ (Ethylene/Ethene)
Each C: 2s + 2px + 2py → 3sp² orbitals + 1 unhybridised pz
2 sp² orbitals form C-H sigma bonds
1 sp² orbital forms C-C sigma bond
1 unhybridised p orbital forms pi bond (C=C double bond)
(iii) sp³ Hybridisation (1 s + 3 p → 4 sp³ orbitals):
Process: 1 s orbital + 3 p orbitals → 4 sp³ hybrid orbitals
Geometry: Tetrahedral
Bond angle: 109.5° (reduced by lone pairs)
Example: CH₄ (Methane)
C ground state: 1s² 2s² 2px¹ 2py¹ 2pz⁰
C excited state: 1s² 2s¹ 2px¹ 2py¹ 2pz¹
Hybridisation: 2s + 2px + 2py + 2pz → 4sp³ orbitals
Four sp³ orbitals overlap with 4 H orbitals to form 4 sigma bonds.
Result: Tetrahedral, 109.5°
Example: NH₃ (Ammonia)
N: 1s² 2s² 2px¹ 2py¹ 2pz¹ → 4sp³ orbitals
3 sp³ orbitals form N-H bonds, 1 sp³ orbital holds lone pair
Result: Trigonal pyramidal, 107°
Example: H₂O (Water)
O: 1s² 2s² 2px² 2py¹ 2pz¹ → 4sp³ orbitals
2 sp³ orbitals form O-H bonds, 2 sp³ orbitals hold lone pairs
Result: Bent (V-shaped), 104.5°
Comparison Table:
| Feature | sp | sp² | sp³ |
|---|
| Orbitals mixed | 1s + 1p | 1s + 2p | 1s + 3p |
| Hybrid orbitals | 2 | 3 | 4 |
| Geometry | Linear | Trigonal planar | Tetrahedral |
| Bond angle | 180° | 120° | 109.5° |
| % s character | 50% | 33.3% | 25% |
| Examples | BeCl₂, CO₂, C₂H₂ | BF₃, C₂H₄, SO₃ | CH₄, NH₃, H₂O |
3. Explain the formation of an ionic bond using the Born-Haber cycle for NaCl. Describe the properties of ionic compounds.
Answer
An ionic bond is formed by the complete transfer of one or more electrons from a metal atom (low ionisation energy) to a non-metal atom (high electron affinity). The resulting oppositely charged ions are held together by strong electrostatic forces of attraction.
Formation of NaCl:
Step 1: Sublimation of sodium (solid to gas)
Na(s) → Na(g) ΔH_sub = +108.7 kJ/mol
Step 2: Ionisation of sodium (remove outermost electron)
Na(g) → Na⁺(g) + e⁻ ΔH_IE = +496 kJ/mol
Step 3: Dissociation of chlorine molecule
½Cl₂(g) → Cl(g) ΔH_D = +122 kJ/mol
Step 4: Electron affinity of chlorine (gain of electron)
Cl(g) + e⁻ → Cl⁻(g) ΔH_EA = −349 kJ/mol
Step 5: Lattice formation (ions combine to form crystal)
Na⁺(g) + Cl⁻(g) → NaCl(s) ΔH_Lattice = −788 kJ/mol
Overall: Na(s) + ½Cl₂(g) → NaCl(s) ΔH_f = −411.2 kJ/mol
By Hess's law (Born-Haber cycle):
ΔH_f = ΔH_sub + ΔH_IE + ΔH_D + ΔH_EA + ΔH_Lattice
−411.2 = 108.7 + 496 + 122 + (−349) + ΔH_Lattice
−411.2 = 377.7 + ΔH_Lattice
ΔH_Lattice = −788.9 kJ/mol
The Born-Haber Cycle is an application of Hess's law that allows us to calculate lattice energy when direct measurement is not possible. It relates the enthalpy of formation to sublimation energy, ionisation energy, bond dissociation energy, electron affinity, and lattice energy.
Properties of Ionic Compounds:
1. Physical state: Ionic compounds are hard, brittle crystalline solids at room temperature. The strong electrostatic forces create rigid 3D lattices.
2. Melting and boiling points: High, because large amounts of energy are needed to overcome the strong electrostatic forces between ions. E.g., NaCl melts at 801°C.
3. Electrical conductivity: Ionic compounds conduct electricity in the molten state and in aqueous solution (ions are free to move), but NOT in the solid state (ions are fixed in the lattice).
4. Solubility: Generally soluble in polar solvents (like water) because the polar solvent molecules can surround and separate the ions (hydration). Insoluble in nonpolar solvents.
5. Brittleness: When a force displaces one layer of ions relative to another, like-charged ions come close together. The resulting repulsion causes the crystal to shatter.
6. Crystal structure: Ionic compounds form ordered 3D crystal lattices. The arrangement depends on the relative sizes and charges of the ions (e.g., NaCl forms a face-centred cubic lattice).
7. Density: Generally higher density than covalent compounds because ions are packed closely together.
8. Directional nature: Ionic bonds are non-directional — the electrostatic attraction acts equally in all directions, unlike covalent bonds which have specific orientations.
Common Mistakes to Avoid
1. Confusing electron pair geometry with molecular geometry — electron pair geometry includes lone pairs; molecular geometry includes only bonded atoms. NH₃ has tetrahedral electron pair geometry but trigonal pyramidal molecular geometry.
2. Forgetting that hybridisation depends on the central atom only — count the number of electron pairs (bonding + lone pairs) around the CENTRAL atom to determine hybridisation, not all atoms in the molecule.
3. Confusing bond order with bond energy — bond order is the number of bonds between two atoms (1 for single, 2 for double, 3 for triple). Bond energy increases with bond order, but they are not the same thing.
4. Assuming all molecules with polar bonds are polar — molecular polarity depends on both bond polarity AND geometry. Symmetric molecules like CO₂, BF₃, and CCl₄ are nonpolar despite having polar bonds.
5. Confusing lattice energy with bond dissociation energy — lattice energy is for ionic compounds (gaseous ions → solid), bond dissociation energy is for covalent bonds (breaking a specific bond).
6. Writing hybridisation for atoms with only single bonds as sp — always count ALL electron pairs (bonding + lone pairs) around the central atom. Water has 4 electron pairs (2 bonding + 2 lone), so it is sp³, not sp.
7. Confusing coordinate bonds with ionic bonds — a coordinate bond is a type of covalent bond where both electrons come from one atom. An ionic bond involves complete electron transfer, not sharing.
8. Forgetting that hydrogen bonding is intermolecular, not intramolecular (usually) — hydrogen bonds are weak attractive forces BETWEEN molecules, not covalent bonds WITHIN molecules. They are much weaker than covalent bonds (10-40 kJ/mol vs 200-400 kJ/mol).
Exam Checklist
- [ ] Can you define ionic, covalent, and coordinate bonds with examples?
- [ ] Can you predict molecular geometry using VSEPR theory?
- [ ] Can you determine hybridisation (sp, sp², sp³, sp³d, sp³d²) from the number of electron pairs?
- [ ] Can you explain why bond angles decrease with increasing lone pairs (CH₄ → NH₃ → H₂O)?
- [ ] Can you distinguish between sigma and pi bonds?
- [ ] Can you explain the Born-Haber cycle and calculate lattice energy?
- [ ] Can you list the properties of ionic compounds and explain each?
- [ ] Can you explain hydrogen bonding and its effect on physical properties?
- [ ] Can you describe metallic bonding and its relation to metallic properties?
- [ ] Can you apply molecular orbital theory to determine bond order and magnetic nature of O₂, N₂, etc.?
