Chemical Bonding and Molecular Structure
Welcome to HSLC Guru! This page presents detailed English-medium notes and question answers for Class 11 Chemistry Chapter 4 — Chemical Bonding and Molecular Structure following the ASSEB (Assam State School Education Board) syllabus. Chemical bonds are the invisible threads that hold atoms together to form the matter around us, from the salt in your kitchen to the DNA in your cells. This chapter explores the various theories that explain how and why atoms combine, including the Kossel-Lewis approach, valence bond theory, and molecular orbital theory. You will learn to predict the shapes of molecules using VSEPR theory, explain bonding through hybridisation, and understand special interactions like hydrogen bonding. Mastering this chapter will give you a powerful framework for understanding chemical reactivity, physical properties, and biological structure throughout your science studies.
Chapter Summary
A chemical bond is the attractive force that holds atoms or ions together in a molecule or crystal. The Kossel-Lewis approach proposed that atoms achieve a stable noble-gas configuration (octet) either by transferring electrons (ionic bond) or sharing electrons (covalent bond). Lewis introduced the dot symbol notation, where valence electrons are shown as dots around the element symbol. An ionic bond forms between a metal and non-metal through complete electron transfer, producing oppositely charged ions held by electrostatic attraction (e.g., NaCl, MgO). A covalent bond forms between non-metals through electron sharing (e.g., H₂, Cl₂, CH₄). The strength of an ionic crystal is measured by lattice enthalpy, the energy released when one mole of ionic compound forms from gaseous ions. The Born-Haber cycle uses Hess’s law to determine lattice enthalpy by combining sublimation, ionisation, dissociation, electron-gain, and formation enthalpies.
Lewis structures represent bonded electron pairs as lines and lone pairs as dots, helping visualise how atoms connect. The formal charge on an atom in a Lewis structure helps select the most stable resonance form. Resonance describes molecules whose true structure is a hybrid of two or more contributing structures (e.g., O₃, CO₃²⁻, benzene). Important bond parameters include bond length (distance between nuclei of bonded atoms), bond angle (angle between two bonds at a common atom), bond enthalpy (energy required to break one mole of bonds), and bond order (number of bonds between two atoms). Higher bond order means shorter and stronger bonds. VSEPR theory (Valence Shell Electron Pair Repulsion) predicts molecular geometry by minimising repulsion between electron pairs. Lone-pair repulsion is greater than bond-pair repulsion, distorting ideal angles in molecules like H₂O and NH₃.
Valence Bond Theory (VBT), proposed by Heitler and London and extended by Pauling, describes covalent bonding as overlap of half-filled atomic orbitals. Hybridisation is the mixing of atomic orbitals to form equivalent hybrid orbitals: sp (linear, 180°, BeCl₂), sp² (trigonal planar, 120°, BF₃), sp³ (tetrahedral, 109.5°, CH₄), sp³d (trigonal bipyramidal, PCl₅), and sp³d² (octahedral, SF₆). Sigma (σ) bonds form by head-on overlap and pi (π) bonds by sideways overlap of p-orbitals. Molecular Orbital Theory (MOT) treats electrons as belonging to the whole molecule, occupying bonding (lower energy) and antibonding (higher energy) molecular orbitals formed by linear combination of atomic orbitals (LCAO). The bond order from MOT predicts molecular stability and magnetic behaviour. For example, O₂ is paramagnetic because of two unpaired electrons in π* antibonding orbitals, a fact MOT explains but VBT cannot.
Hydrogen bonding is a special dipole-dipole interaction in which hydrogen attached to a highly electronegative atom (F, O, N) attracts another electronegative atom. Intermolecular hydrogen bonding occurs between separate molecules (water, ice, HF), while intramolecular hydrogen bonding occurs within the same molecule (o-nitrophenol, salicylaldehyde). Hydrogen bonds explain water’s high boiling point, ice floating on water, and the secondary structure of proteins and DNA. Polarity arises when bonded atoms have different electronegativities, creating partial charges. The dipole moment (μ = q × d, measured in Debye) is a vector quantity indicating the polarity of a molecule. Symmetric molecules like CO₂ and CCl₄ have zero net dipole moment despite polar bonds, while asymmetric molecules like H₂O and NH₃ have non-zero dipole moments. Together these concepts explain the diverse physical and chemical behaviour of substances.
$\text{Bond order} = \frac{N_b – N_a}{2}$
$\text{FC} = V – L – \frac{B}{2}$
$\text{SN} = \text{lone pairs} + \sigma\text{-bonds}$
1-Mark Questions and Answers
Q1. What is a chemical bond?
Answer: A chemical bond is the attractive force that holds two or more atoms together in a molecule or compound, allowing them to achieve a more stable energy state.
Q2. State the octet rule.
Answer: The octet rule states that atoms tend to gain, lose, or share electrons so as to attain eight electrons in their outermost shell, achieving a stable noble-gas configuration.
Q3. Define lattice enthalpy.
Answer: Lattice enthalpy is the energy released when one mole of an ionic compound is formed from its gaseous ions under standard conditions.
Q4. What is bond length?
Answer: Bond length is the equilibrium distance between the nuclei of two covalently bonded atoms in a molecule, usually expressed in picometres (pm).
Q5. What is the geometry of a molecule with sp³ hybridisation?
Answer: A molecule with sp³ hybridisation has a tetrahedral geometry with bond angles of 109.5° (e.g., CH₄).
Q6. Define dipole moment.
Answer: Dipole moment is the product of the magnitude of charge and the distance between the centres of positive and negative charges, expressed in Debye units (D).
Q7. What type of bond exists in NaCl?
Answer: An ionic (electrovalent) bond exists in NaCl, formed by complete transfer of one electron from sodium to chlorine.
Q8. What is a sigma bond?
Answer: A sigma (σ) bond is a covalent bond formed by the head-on (axial) overlap of atomic orbitals along the internuclear axis.
Q9. Why is water a polar molecule?
Answer: Water is polar because of its bent shape and the higher electronegativity of oxygen compared to hydrogen, resulting in a non-zero net dipole moment.
Q10. What is hydrogen bonding?
Answer: Hydrogen bonding is a strong dipole-dipole attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O, N) and another electronegative atom in the same or a different molecule.
Q11. Define bond order according to MOT.
Answer: Bond order is half the difference between the number of electrons in bonding molecular orbitals (N_b) and antibonding molecular orbitals (N_a): Bond order = (N_b − N_a)/2.
Q12. What is electronegativity?
Answer: Electronegativity is the tendency of an atom in a chemical bond to attract the shared pair of electrons towards itself. Fluorine is the most electronegative element on the Pauling scale (4.0).
2 to 3-Mark Questions and Answers
Q1. Distinguish between ionic and covalent bonds.
Answer: An ionic bond is formed by complete transfer of electrons from one atom to another, producing oppositely charged ions held together by electrostatic forces. It typically forms between a metal and a non-metal (e.g., NaCl). A covalent bond is formed by sharing of electrons between two atoms, usually both non-metals (e.g., H₂, CH₄). Ionic compounds have high melting points, conduct electricity in molten or aqueous state, and are soluble in polar solvents. Covalent compounds generally have lower melting points, are poor conductors, and dissolve in non-polar solvents.
Q2. What are the postulates of VSEPR theory?
Answer: The main postulates of VSEPR theory are: (i) the shape of a molecule depends on the number of valence shell electron pairs around the central atom; (ii) electron pairs orient themselves to minimise mutual repulsion; (iii) the order of repulsion is lone pair-lone pair > lone pair-bond pair > bond pair-bond pair; (iv) multiple bonds are treated as a single electron pair for predicting geometry; and (v) lone pairs distort ideal bond angles.
Q3. Explain the formation of a covalent bond in H₂ using VBT.
Answer: According to Valence Bond Theory, when two hydrogen atoms approach, the half-filled 1s orbital of each atom overlaps along the internuclear axis. The electrons must have opposite spins (Pauli’s principle). This axial overlap creates a region of high electron density between the nuclei, forming a sigma (σ) bond. The energy of the system decreases until it reaches a minimum at the equilibrium bond length (74 pm), and bond enthalpy of about 436 kJ/mol is released.
Q4. Why is the bond angle in H₂O (104.5°) less than in CH₄ (109.5°)?
Answer: In CH₄, the central carbon atom has four bond pairs and no lone pairs, giving an ideal tetrahedral angle of 109.5°. In H₂O, the central oxygen atom has two bond pairs and two lone pairs. According to VSEPR theory, lone pair-lone pair and lone pair-bond pair repulsions are stronger than bond pair-bond pair repulsions. The two lone pairs push the O-H bonds closer together, reducing the H-O-H angle to 104.5°.
Q5. What is hybridisation? Give an example.
Answer: Hybridisation is the process of intermixing of atomic orbitals of slightly different energies to produce an equal number of new hybrid orbitals of equivalent energy and shape. For example, in methane (CH₄), the 2s and three 2p orbitals of carbon mix to form four equivalent sp³ hybrid orbitals, each containing one electron. These overlap with the 1s orbitals of four hydrogen atoms forming four C-H sigma bonds in a tetrahedral arrangement.
Q6. Compare bonding and antibonding molecular orbitals.
Answer: A bonding molecular orbital (BMO) is formed by the additive combination of atomic orbitals (constructive interference). It has lower energy than the parent atomic orbitals and a high electron density between the nuclei, stabilising the molecule. An antibonding molecular orbital (ABMO) is formed by the subtractive combination (destructive interference). It has higher energy than the parent atomic orbitals, with a node between the nuclei. Electrons in BMO favour bonding, whereas electrons in ABMO oppose it. They are denoted σ/σ* and π/π*.
Q7. What is formal charge? Why is it useful?
Answer: Formal charge on an atom in a Lewis structure is calculated as: FC = (valence electrons in free atom) − (lone-pair electrons) − ½(bonding electrons). It helps in selecting the most stable Lewis structure: the structure with the smallest formal charges on atoms and with negative charges on more electronegative atoms is preferred. It is also useful for identifying which atom carries the positive or negative charge in a polyatomic ion.
Q8. Differentiate between sigma (σ) and pi (π) bonds.
Answer: A sigma bond is formed by head-on (axial) overlap of orbitals along the internuclear axis; it has cylindrical symmetry and is stronger. Examples: s-s, s-p, p-p (axial). A pi bond is formed by sideways (lateral) overlap of parallel p-orbitals perpendicular to the internuclear axis; it has electron density above and below the bond axis and is weaker. A single bond is one sigma; a double bond has one sigma and one pi; a triple bond has one sigma and two pi bonds.
5 to 7-Mark Questions and Answers
Q1. Describe the Born-Haber cycle for the formation of NaCl with a clear explanation of each step.
Answer: The Born-Haber cycle is an application of Hess’s law that calculates the lattice enthalpy of an ionic compound by combining various energy changes involved in its formation. For NaCl, the cycle involves the following steps: (i) Sublimation of solid sodium to gaseous sodium atoms, requiring sublimation enthalpy (ΔH_s = +108 kJ/mol); (ii) Ionisation of gaseous sodium atoms to Na⁺ ions, requiring ionisation enthalpy (IE = +495 kJ/mol); (iii) Dissociation of half a mole of Cl₂ molecules to gaseous Cl atoms, requiring half the bond dissociation enthalpy (½ ΔH_d = +121 kJ/mol); (iv) Electron gain by chlorine atoms forming Cl⁻ ions, releasing electron-gain enthalpy (ΔH_eg = -349 kJ/mol); (v) Combination of gaseous Na⁺ and Cl⁻ ions to form solid NaCl, releasing lattice enthalpy (U). The sum of all these steps equals the standard enthalpy of formation of NaCl (ΔH_f = -411 kJ/mol). Solving: U = ΔH_f − ΔH_s − IE − ½ΔH_d − ΔH_eg = -788 kJ/mol. The large negative value indicates the high stability of the NaCl crystal lattice.
Q2. Explain the shapes of NH₃, H₂O, BF₃, and PCl₅ using VSEPR theory.
Answer: (i) NH₃ — Nitrogen has 5 valence electrons; three are bonded to hydrogens and one lone pair remains. With three bond pairs and one lone pair, the geometry is tetrahedral but the shape is trigonal pyramidal with a bond angle of 107.5° due to lone pair-bond pair repulsion. (ii) H₂O — Oxygen has 6 valence electrons; two are bonded to hydrogens and two lone pairs remain. With two bond pairs and two lone pairs, the geometry is tetrahedral but the shape is bent with a bond angle of 104.5° due to greater lone pair-lone pair repulsion. (iii) BF₃ — Boron has 3 valence electrons, all forming bonds to fluorines, with no lone pair. With three bond pairs only, the shape is trigonal planar with bond angle 120° (sp² hybridisation). (iv) PCl₅ — Phosphorus has 5 valence electrons forming five bonds to chlorines, with no lone pair. The shape is trigonal bipyramidal (sp³d hybridisation) with three equatorial chlorines at 120° and two axial at 90° to the equatorial plane, giving bond angles of 90°, 120°, and 180°.
Q3. Discuss the molecular orbital diagrams of H₂, N₂, and O₂. Determine bond order and magnetic behaviour.
Answer: (i) H₂ — Two electrons fill the σ1s bonding orbital. Configuration: (σ1s)². Bond order = (2-0)/2 = 1. Diamagnetic, stable single bond. (ii) N₂ — 14 electrons distributed as: (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(π2p_x)²(π2p_y)²(σ2p_z)². Bond order = (10-4)/2 = 3. Triple bond, diamagnetic, very stable. (iii) O₂ — 16 electrons distributed as: (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2p_z)²(π2p_x)²(π2p_y)²(π*2p_x)¹(π*2p_y)¹. Bond order = (10-6)/2 = 2. Double bond, paramagnetic due to two unpaired electrons in degenerate π* orbitals. MOT successfully explains the paramagnetism of O₂, which VBT cannot explain. The high bond order in N₂ accounts for its very high bond dissociation energy and chemical inertness.
Q4. What is hydrogen bonding? Distinguish between intermolecular and intramolecular hydrogen bonding with examples and effects.
Answer: Hydrogen bonding is a special type of dipole-dipole attraction between a hydrogen atom covalently bonded to a highly electronegative atom (F, O, N) and another electronegative atom carrying a lone pair. Although weaker than covalent bonds (typically 5-30 kJ/mol), it is stronger than ordinary van der Waals forces. (i) Intermolecular hydrogen bonding occurs between two or more molecules of the same or different compounds. Examples: H₂O, HF, NH₃, alcohols, carboxylic acids. It causes association of molecules, leading to abnormally high boiling points, viscosities, and surface tensions. The lower density of ice compared to water arises from a tetrahedral hydrogen-bonded network. (ii) Intramolecular hydrogen bonding occurs within a single molecule, forming a ring (chelation). Examples: o-nitrophenol, salicylaldehyde. It causes lower boiling and melting points compared to the corresponding para isomers, since intramolecular bonding cannot link molecules together. Hydrogen bonding is biologically vital — it stabilises the alpha-helix of proteins, base pairs in DNA, and the unique properties of water that support life.
Q5. Explain hybridisation in BeCl₂, BF₃, CH₄, PCl₅, and SF₆ with their geometries.
Answer: (i) BeCl₂ — Be has configuration 1s²2s². On excitation, one 2s electron moves to 2p giving 1s²2s¹2p¹. The 2s and one 2p orbital mix to form two sp hybrid orbitals at 180°. Each overlaps with a chlorine 3p orbital forming linear BeCl₂. (ii) BF₃ — B has configuration 1s²2s²2p¹. On excitation, configuration becomes 1s²2s¹2p². The 2s and two 2p orbitals form three sp² hybrid orbitals at 120° in a trigonal planar arrangement. (iii) CH₄ — C has configuration 1s²2s²2p². On excitation: 1s²2s¹2p³. The 2s and three 2p orbitals mix to form four sp³ hybrid orbitals directed towards the corners of a regular tetrahedron with bond angle 109.5°. (iv) PCl₅ — P uses 3s, 3p_x, 3p_y, 3p_z and 3d_z² forming five sp³d hybrid orbitals in trigonal bipyramidal geometry. Three equatorial bonds (120°) are shorter and stronger than two axial bonds (90°). (v) SF₆ — S uses 3s, three 3p, and two 3d orbitals (d_x²-y² and d_z²) forming six sp³d² hybrid orbitals in octahedral geometry with all bond angles 90°. These examples illustrate how hybridisation explains observed shapes and bond angles in covalent molecules.
Multiple Choice Questions (MCQ)
Q1. The bond present in NaCl is:
(a) Covalent (b) Ionic (c) Coordinate (d) Hydrogen
Answer: (b) Ionic
Q2. The shape of NH₃ molecule is:
(a) Linear (b) Trigonal planar (c) Trigonal pyramidal (d) Tetrahedral
Answer: (c) Trigonal pyramidal
Q3. The hybridisation in BF₃ is:
(a) sp (b) sp² (c) sp³ (d) sp³d
Answer: (b) sp²
Q4. Which of the following has the highest bond order?
(a) O₂ (b) N₂ (c) H₂ (d) F₂
Answer: (b) N₂
Q5. Which molecule has zero dipole moment?
(a) H₂O (b) NH₃ (c) CO₂ (d) HCl
Answer: (c) CO₂
Q6. The bond angle in CH₄ is:
(a) 90° (b) 104.5° (c) 107.5° (d) 109.5°
Answer: (d) 109.5°
Q7. Which of the following exhibits hydrogen bonding?
(a) HCl (b) HBr (c) HF (d) HI
Answer: (c) HF
Q8. O₂ is paramagnetic because it has:
(a) Two unpaired electrons in π* orbitals (b) All paired electrons (c) Two unpaired electrons in σ orbital (d) Four unpaired electrons
Answer: (a) Two unpaired electrons in π* orbitals
Q9. The geometry of SF₆ is:
(a) Tetrahedral (b) Octahedral (c) Trigonal bipyramidal (d) Square planar
Answer: (b) Octahedral
Q10. The number of sigma and pi bonds in C₂H₄ are:
(a) 5σ, 1π (b) 4σ, 2π (c) 6σ, 0π (d) 3σ, 3π
Answer: (a) 5σ, 1π
Fill in the Blanks
Q1. An ionic bond is also called an ____________ bond.
Answer: electrovalent
Q2. The hybridisation in PCl₅ is ____________.
Answer: sp³d
Q3. The bond order of N₂ is ____________.
Answer: 3
Q4. Hydrogen bond is formed when H is bonded to ____________ , ____________ , or N.
Answer: F, O
Q5. The dipole moment of CO₂ is ____________.
Answer: zero
True or False
Q1. A sigma bond is stronger than a pi bond.
Answer: True
Q2. The shape of H₂O is linear.
Answer: False (it is bent/angular)
Q3. O₂ is diamagnetic according to MOT.
Answer: False (O₂ is paramagnetic)
Q4. Lattice enthalpy increases with the magnitude of the ionic charges.
Answer: True
Q5. Intramolecular hydrogen bonding raises the boiling point of a compound.
Answer: False (it lowers the boiling point compared to the para isomer)
Q6. The hybridisation in CH₄ is sp³.
Answer: True
Q7. Resonance structures differ only in the position of electrons, not atoms.
Answer: True
Glossary
| Term | Meaning |
|---|---|
| Chemical bond | Attractive force holding atoms together in a molecule or compound |
| Octet rule | Tendency of atoms to attain eight electrons in their valence shell |
| Ionic bond | Bond formed by complete transfer of electrons |
| Covalent bond | Bond formed by sharing of electron pairs |
| Lattice enthalpy | Energy released when one mole of ionic solid forms from gaseous ions |
| Born-Haber cycle | Thermochemical cycle for calculating lattice enthalpy |
| Lewis structure | Diagram showing valence electrons and bonds in a molecule |
| Formal charge | Hypothetical charge on an atom in a Lewis structure |
| Resonance | Description of a molecule by two or more contributing structures |
| Bond length | Equilibrium distance between nuclei of bonded atoms |
| Bond angle | Angle between two bonds at a central atom |
| Bond enthalpy | Energy required to break one mole of bonds in gas phase |
| Bond order | Number of bonds between two atoms (BMO − ABMO)/2 |
| VSEPR theory | Theory predicting molecular shape from electron pair repulsion |
| VBT | Valence bond theory based on overlap of half-filled atomic orbitals |
| Hybridisation | Mixing of atomic orbitals to form equivalent hybrid orbitals |
| sp hybridisation | Linear geometry, 180° (e.g., BeCl₂) |
| sp² hybridisation | Trigonal planar geometry, 120° (e.g., BF₃) |
| sp³ hybridisation | Tetrahedral geometry, 109.5° (e.g., CH₄) |
| sp³d hybridisation | Trigonal bipyramidal geometry (e.g., PCl₅) |
| sp³d² hybridisation | Octahedral geometry (e.g., SF₆) |
| Sigma bond | Covalent bond by axial overlap of orbitals |
| Pi bond | Covalent bond by sideways overlap of p-orbitals |
| Molecular orbital | Orbital extending over the entire molecule |
| Bonding MO | Lower-energy MO that stabilises the molecule |
| Antibonding MO | Higher-energy MO that destabilises the molecule |
| LCAO | Linear combination of atomic orbitals |
| Paramagnetic | Substance attracted by a magnetic field due to unpaired electrons |
| Diamagnetic | Substance weakly repelled by a magnetic field; all electrons paired |
| Hydrogen bond | Attraction between H and an electronegative atom (F, O, N) |
| Intermolecular H-bond | Hydrogen bonding between separate molecules |
| Intramolecular H-bond | Hydrogen bonding within the same molecule |
| Polarity | Unequal sharing of electrons creating partial charges |
| Dipole moment | Product of charge and distance, indicating bond polarity (Debye) |
| Electronegativity | Tendency of an atom to attract bonding electrons |