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Chemical Bonding and Molecular Structure

NCERT Class 11 · Chemistry Based on NCERT Class 11 Chemistry textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter explains why and how atoms combine to form molecules, introducing the fundamental theories of chemical bonding: the Kössel-Lewis approach (octet rule), VSEPR theory for molecular shapes, Valence Bond theory, and Molecular Orbital theory. The one big idea a student must walk away with is that chemical bonds form to lower the energy of a system and achieve stability, and that the structure and properties of a molecule are determined by the arrangement and interaction of its valence electrons.
Key Concepts (18)
Chemical bond
The attractive force which holds various constituents (atoms, ions, etc.) together in different chemical species.
e.g. The force that holds two hydrogen atoms together in an H₂ molecule.
Octet rule
Atoms can combine either by transfer of valence electrons from one atom to another (gaining or losing) or by sharing of valence electrons in order to have an octet in their valence shells.
e.g. Sodium transfers its one valence electron to chlorine, so both achieve an octet (Na⁺ and Cl⁻).
Lewis symbols
Simple notations introduced by G.N. Lewis to represent valence electrons in an atom, where dots around the symbol represent the number of valence electrons.
e.g. The Lewis symbol for sodium (Na) has one dot; for chlorine (Cl) it has seven dots.
Covalent bond
A bond formed by the sharing of a pair of electrons between atoms, each atom contributing at least one electron to the shared pair.
e.g. In Cl₂, each chlorine atom shares one electron to form a single covalent bond.
Ionic bond (electrovalent bond)
The bond formed as a result of the electrostatic attraction between the positive and negative ions.
e.g. The bond between Na⁺ and Cl⁻ in NaCl.
Formal charge
The difference between the number of valence electrons of that atom in an isolated or free state and the number of electrons assigned to that atom in the Lewis structure.
e.g. In ozone (O₃), the central oxygen has a formal charge of +1, one end oxygen has 0, and the other end oxygen has –1.
Resonance
Whenever a single Lewis structure cannot describe a molecule accurately, a number of structures with similar energy, positions of nuclei, bonding and non-bonding pairs of electrons are taken as the canonical structures of the hybrid which describes the molecule accurately.
e.g. Ozone (O₃) is represented as a resonance hybrid of two canonical forms with O–O and O=O bonds.
Bond length
The equilibrium distance between the nuclei of two bonded atoms in a molecule.
e.g. The H–H bond length in H₂ is 74 pm.
Bond angle
The angle between the orbitals containing bonding electron pairs around the central atom in a molecule/complex ion.
e.g. The H–O–H bond angle in water is 104.5°.
Bond enthalpy
The amount of energy required to break one mole of bonds of a particular type between two atoms in a gaseous state.
e.g. The H–H bond enthalpy in hydrogen molecule is 435.8 kJ mol⁻¹.
Bond order
In the Lewis description of covalent bond, the Bond Order is given by the number of bonds between the two atoms in a molecule.
e.g. In N₂, bond order is 3 (triple bond); in O₂, bond order is 2 (double bond).
Dipole moment
The product of the magnitude of the charge and the distance between the centres of positive and negative charge.
e.g. HF has a dipole moment of 1.78 D, with the arrow pointing from H (positive) to F (negative).
Lattice enthalpy
The energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions.
e.g. The lattice enthalpy of NaCl is 788 kJ mol⁻¹.
VSEPR theory
A theory that predicts the shape of a molecule based on the repulsive interactions of the electron pairs in the valence shell of the atoms.
e.g. CH₄ has a tetrahedral shape because there are four bond pairs and no lone pairs around carbon.
Valence bond theory
A theory based on the overlap of atomic orbitals, where a covalent bond forms by the pairing of electrons present in the valence shell having opposite spins.
e.g. In H₂, the 1s orbitals of two hydrogen atoms overlap to form a covalent bond.
Hybridisation
The mixing of atomic orbitals to form new hybrid orbitals that are suitable for bonding, explaining the directional properties of covalent bonds.
e.g. In CH₄, carbon undergoes sp³ hybridisation to form four equivalent bonds.
Molecular orbital theory
A theory that describes bonding in terms of molecular orbitals formed by the combination of atomic orbitals, applicable to homonuclear diatomic molecules.
e.g. In O₂, the molecular orbital configuration explains its paramagnetic nature.
Hydrogen bond
An attractive force between a hydrogen atom covalently bonded to a highly electronegative atom (like O, N, F) and another electronegative atom.
e.g. The hydrogen bonds between water molecules give water its high boiling point.
Formulas & Equations (7)
Formal charge (F.C.) = total number of valence electrons in the free atom — total number of non bonding (lone pair) electrons — (1/2) total number of bonding (shared) electrons
Reads as: Formal charge equals total valence electrons minus non-bonding electrons minus half of bonding electrons.
Variables: F.C. = formal charge (unitless); valence electrons = number of electrons in outer shell of free atom; non-bonding electrons = lone pair electrons; bonding electrons = shared electrons in bonds.
Use when: Calculating formal charge on an atom in a Lewis structure to determine the most stable resonance structure.
⚠ Students forget to divide the bonding electrons by 2, or they count bonding electrons as belonging entirely to one atom.
Dipole moment (µ) = charge (Q) × distance of separation (r)
Reads as: Dipole moment equals charge times distance.
Variables: µ = dipole moment (Debye, D; 1 D = 3.33564 × 10⁻³⁰ C m); Q = magnitude of charge (coulombs, C); r = distance between centres of positive and negative charge (metres, m).
Use when: Calculating dipole moment of a polar molecule or understanding polarity.
⚠ Students forget to convert units or confuse the direction of the dipole moment vector.
Na → Na⁺ + e⁻; Cl + e⁻ → Cl⁻; Na⁺ + Cl⁻ → NaCl
Reads as: Sodium loses an electron to form sodium ion; chlorine gains an electron to form chloride ion; sodium ion and chloride ion combine to form sodium chloride.
Variables: Na = sodium atom; Na⁺ = sodium ion; e⁻ = electron; Cl = chlorine atom; Cl⁻ = chloride ion; NaCl = sodium chloride.
Use when: Explaining the formation of an ionic bond between sodium and chlorine.
⚠ Students write the electron transfer incorrectly or forget to show the charges on ions.
H₂(g) → H(g) + H(g); ΔₐH⁰ = 435.8 kJ mol⁻¹
Reads as: Hydrogen gas dissociates into hydrogen atoms; the bond dissociation enthalpy is 435.8 kilojoules per mole.
Variables: ΔₐH⁰ = bond dissociation enthalpy (kJ mol⁻¹); H₂ = hydrogen molecule; H = hydrogen atom.
Use when: Calculating bond enthalpy or comparing bond strengths.
⚠ Students confuse bond enthalpy with bond energy or forget the gaseous state symbol.
Average bond enthalpy = (ΔₐH₁⁰ + ΔₐH₂⁰) / 2
Reads as: Average bond enthalpy equals the sum of the bond dissociation enthalpies divided by the number of bonds.
Variables: ΔₐH₁⁰ and ΔₐH₂⁰ = bond dissociation enthalpies for successive bonds (kJ mol⁻¹).
Use when: Calculating average bond enthalpy in polyatomic molecules like H₂O.
⚠ Students use the same value for all bonds without considering the different chemical environment.
M(g) → M⁺(g) + e⁻; Ionization enthalpy
Reads as: Metal atom in gas phase loses an electron to form a positive ion; this requires ionization enthalpy.
Variables: M = metal atom; M⁺ = metal cation; e⁻ = electron; ionization enthalpy = energy required (kJ mol⁻¹).
Use when: Explaining the formation of cations in ionic bonding.
⚠ Students forget that ionization is always endothermic.
X(g) + e⁻ → X⁻(g); Electron gain enthalpy
Reads as: Non-metal atom in gas phase gains an electron to form a negative ion; this involves electron gain enthalpy.
Variables: X = non-metal atom; X⁻ = anion; e⁻ = electron; electron gain enthalpy = enthalpy change (kJ mol⁻¹, may be exothermic or endothermic).
Use when: Explaining the formation of anions in ionic bonding.
⚠ Students confuse electron gain enthalpy with electron affinity (electron affinity is the negative of electron gain enthalpy).
How It Works
Formation of an ionic bond (Kössel approach)
  1. Step 1: A metal atom (e.g., Na) loses one or more electrons to form a positive ion (Na⁺), achieving a noble gas configuration.
  2. Step 2: A non-metal atom (e.g., Cl) gains the electron(s) to form a negative ion (Cl⁻), also achieving a noble gas configuration.
  3. Step 3: The positive and negative ions are held together by electrostatic attraction (coulombic forces) to form an ionic compound (NaCl).
  4. Step 4: The compound crystallizes in a three-dimensional lattice structure, stabilized by lattice enthalpy.
💡 The sum of ionization enthalpy and electron gain enthalpy may be positive, but the overall process is exothermic due to the large negative lattice enthalpy.
Writing Lewis dot structures
  1. Step 1: Count the total number of valence electrons of all atoms in the molecule/ion. For anions, add one electron per negative charge; for cations, subtract one electron per positive charge.
  2. Step 2: Determine the skeletal structure — the least electronegative atom is usually the central atom.
  3. Step 3: Draw a single bond (one shared electron pair) between the central atom and each terminal atom.
  4. Step 4: Distribute the remaining electrons as lone pairs to complete the octet (or duplet for hydrogen) on each atom.
  5. Step 5: If the central atom does not have an octet, form multiple bonds (double or triple) by converting lone pairs from terminal atoms into bonding pairs.
💡 Always check that each bonded atom (except hydrogen) has an octet, and that the total number of electrons used equals the total valence electrons counted in Step 1.
VSEPR theory for predicting molecular shapes
  1. Step 1: Determine the number of valence shell electron pairs (bond pairs and lone pairs) around the central atom.
  2. Step 2: Arrange these electron pairs in space to minimize repulsion — they take positions as far apart as possible.
  3. Step 3: Treat multiple bonds as a single super pair.
  4. Step 4: Consider the repulsion order: lone pair-lone pair > lone pair-bond pair > bond pair-bond pair.
  5. Step 5: Predict the molecular geometry based on the arrangement of bond pairs (ignoring lone pairs for shape description).
💡 Lone pairs occupy more space than bond pairs, so they cause greater repulsion and reduce bond angles from ideal values.
Formation of a covalent bond (orbital overlap concept)
  1. Step 1: Two atoms approach each other, and their atomic orbitals begin to overlap.
  2. Step 2: Attractive forces (nucleus of one atom with electron of the other) and repulsive forces (electron-electron, nucleus-nucleus) act simultaneously.
  3. Step 3: At the equilibrium distance, the net attractive force balances the repulsive force, and the system has minimum potential energy.
  4. Step 4: The overlapping of orbitals results in the pairing of electrons with opposite spins, forming a stable covalent bond.
💡 Greater the overlap, stronger the bond. The bond length corresponds to the distance at which potential energy is minimum.
Resonance hybrid formation
  1. Step 1: Identify that a single Lewis structure is inadequate to represent the molecule (e.g., bond lengths are intermediate).
  2. Step 2: Write all possible canonical (resonance) structures that have the same positions of nuclei but different arrangements of electrons.
  3. Step 3: The actual molecule is a resonance hybrid of these canonical forms, with bond characteristics averaged.
  4. Step 4: The resonance hybrid has lower energy than any single canonical structure, stabilizing the molecule.
💡 Canonical forms have no real existence; the molecule does not oscillate between them. Resonance is not an equilibrium.
Determination of dipole moment in polyatomic molecules
  1. Step 1: Identify the bond dipoles (individual dipole moments of each bond) in the molecule.
  2. Step 2: Consider the spatial arrangement (geometry) of the bonds.
  3. Step 3: The net dipole moment is the vector sum of all bond dipoles.
  4. Step 4: If the vector sum is zero, the molecule is non-polar (e.g., CO₂, BF₄); if non-zero, the molecule is polar (e.g., H₂O, NH₃).
💡 Even if individual bonds are polar, the molecule can be non-polar if the bond dipoles cancel due to symmetry.
Exam Traps — Don't Make These Mistakes
⚠ Confusing bond order with bond strength
✗ Bond order is directly proportional to bond length.
→ Bond order is inversely proportional to bond length. Higher bond order means shorter bond length and stronger bond.
✓ With increase in bond order, bond enthalpy increases and bond length decreases.
⚠ Mistaking formal charge for actual charge
✗ The formal charge on an atom is the actual charge it carries in the molecule.
→ Formal charge is a bookkeeping tool based on a pure covalent view of bonding; it does not indicate real charge separation within the molecule.
✓ Formal charges do not indicate real charge separation; they help in selecting the lowest energy Lewis structure.
⚠ Incorrectly applying VSEPR for molecules with lone pairs
✗ The shape of NH₃ is tetrahedral because there are four electron pairs around nitrogen.
→ While the electron pair geometry is tetrahedral, the molecular shape (considering only bond pairs) is trigonal pyramidal because one position is occupied by a lone pair.
✓ NH₃ has a trigonal pyramidal shape due to one lone pair on nitrogen; the bond angle is reduced to 107° from 109.5°.
⚠ Assuming all polar bonds give a polar molecule
✗ CO₂ is a polar molecule because the C=O bonds are polar.
→ Although C=O bonds are polar, the linear geometry of CO₂ causes the bond dipoles to cancel each other, resulting in a net dipole moment of zero.
✓ CO₂ is non-polar because the two equal bond dipoles point in opposite directions and cancel.
⚠ Confusing resonance with tautomerism
✗ The molecule exists in equilibrium between its resonance structures.
→ Resonance structures are not in equilibrium; they are hypothetical canonical forms. The actual molecule is a single resonance hybrid.
✓ There is no equilibrium between canonical forms; the molecule has a single structure which is the resonance hybrid.
⚠ Forgetting the expanded octet for third period elements
✗ All atoms must obey the octet rule.
→ Elements in and beyond the third period (e.g., P, S) can have more than eight electrons around them due to available d orbitals.
✓ PF₅ and SF₆ are examples of expanded octet, where the central atom has more than eight valence electrons.
Compare & Contrast
Nature of bonding
Ionic bond
Covalent bond
Ionic bond involves complete transfer of electrons from one atom to another, resulting in electrostatic attraction between ions; covalent bond involves sharing of electrons between atoms.
Role in VSEPR theory
Bond pair
Lone pair
Bond pairs are shared between two atoms and occupy less space; lone pairs are localized on the central atom and occupy more space, causing greater repulsion and reducing bond angles.
Existence
Resonance hybrid
Canonical structure
The resonance hybrid is the actual, single structure of the molecule with lower energy; canonical structures are hypothetical and have no real existence.
Definition and measurement
Covalent radius
Van der Waals radius
Covalent radius is half the distance between two similar atoms joined by a covalent bond; van der Waals radius is half the distance between two similar atoms in separate molecules in a solid, and is larger.
Exam Focus
Marks pattern: Typically 2-3 MCQs (1 mark each), 1-2 short answer questions (2-3 marks), 1 long answer question (5 marks) on hybridisation/VSEPR, and sometimes a numerical on dipole moment or bond enthalpy.
Most asked: Lewis dot structures and formal charge · VSEPR theory and molecular shapes · Hybridisation (sp, sp², sp³) and bond angles · Dipole moment and polarity · Resonance structures (O₃, CO₃²⁻, CO₂) · Bond order and bond strength correlation · Limitations of octet rule · Ionic bond formation and lattice enthalpy
Don't skip: Draw Lewis structures and calculate formal charge for molecules like O₃, NO₂⁻, CO₃²⁻.; Predict geometry and bond angles using VSEPR theory for molecules like NH₃, H₂O, SF₆, ClF₃.; Explain hybridisation and shape of molecules like CH₄, C₂H₄, C₂H₂, BF₃.; Compare dipole moments of NH₃ and NF₃.; Explain resonance with examples (O₃, CO₂, CO₃²⁻).
One-Line Revision (read aloud before the exam)
  1. A chemical bond is the attractive force that holds atoms together in a molecule.
  2. The octet rule states that atoms combine to achieve eight electrons in their valence shell.
  3. Lewis symbols represent valence electrons as dots around the element symbol.
  4. A covalent bond is formed by sharing of an electron pair between atoms.
  5. An ionic bond is formed by electrostatic attraction between oppositely charged ions.
  6. Formal charge = valence electrons – lone pair electrons – ½ bonding electrons.
  7. Resonance describes a molecule as a hybrid of multiple canonical structures.
  8. Bond length is the equilibrium distance between nuclei of bonded atoms.
  9. Bond enthalpy is the energy required to break one mole of bonds in gaseous state.
  10. Bond order is the number of bonds between two atoms; higher bond order means stronger and shorter bond.
  11. Dipole moment is the product of charge and distance; it is a vector quantity.
  12. VSEPR theory predicts molecular shape based on electron pair repulsion.
  13. Lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion.
  14. In VSEPR, multiple bonds are treated as a single super pair.
  15. Valence bond theory explains covalent bond formation by orbital overlap.
  16. Greater the overlap, stronger the covalent bond.
  17. Hybridisation mixes atomic orbitals to form equivalent hybrid orbitals.
  18. sp³ hybridisation gives tetrahedral geometry (e.g., CH₄).
  19. sp² hybridisation gives trigonal planar geometry (e.g., BF₃).
  20. sp hybridisation gives linear geometry (e.g., BeCl₂).
  21. Molecular orbital theory explains bonding in homonuclear diatomic molecules.
  22. Hydrogen bond is an intermolecular force between H and electronegative atoms (O, N, F).
  23. Lattice enthalpy is the energy released when gaseous ions form one mole of ionic solid.
  24. Ionic compounds are stabilized by lattice enthalpy, not just octet achievement.
  25. Polar covalent bonds result from unequal sharing of electrons due to electronegativity difference.
  26. The dipole moment of NH₃ is greater than that of NF₃ due to direction of lone pair dipole.
  27. Resonance hybrid has lower energy than any canonical structure.
  28. Expanded octet is possible for elements in third period and beyond (e.g., SF₆).
  29. Incomplete octet occurs in compounds like BeH₂ and BF₃.
  30. Odd-electron molecules like NO do not satisfy the octet rule for all atoms.

MCQs — 10 Questions with Answers

Q1. The Lewis symbol of phosphorus (P) has five dots around it. What is the group valence of phosphorus?

  • A. 5 only
  • B. 3 only
  • C. Both 5 and 3 ✓
  • D. 8

Answer: C — Group valence equals either the number of dots (5) or 8 minus the number of dots (8−5=3), so both 5 and 3 are valid valences for phosphorus.

Q2. In the formation of CaF₂, calcium achieves a stable configuration by losing electrons and fluorine by gaining electrons. Which statement is correct?

  • A. Ca loses 1 electron; each F gains 1 electron; both achieve [Ne] configuration
  • B. Ca loses 2 electrons; each F gains 1 electron; both achieve [Ne] configuration ✓
  • C. Ca loses 2 electrons; each F gains 2 electrons; Ca achieves [Ar] and F achieves [Ne]
  • D. Ca loses 1 electron; each F gains 2 electrons; both achieve [Ar] configuration

Answer: B — Ca ([Ar]4s²) loses 2 electrons to form Ca²⁺ ([Ar]); each F ([He]2s²2p⁵) gains 1 electron to form F⁻ ([Ne]), so CaF₂ forms with one Ca²⁺ and two F⁻ ions.

Q3. According to the Kössel-Lewis approach, which of the following is NOT a reason why atoms form chemical bonds?

  • A. To achieve noble gas electronic configuration
  • B. To lower the total energy of the system
  • C. To increase the number of orbitals available to electrons ✓
  • D. To satisfy the octet rule (or duplet for H)

Answer: C — Increasing the number of orbitals is not a bonding motive; atoms bond to gain stability, lower energy, and achieve noble gas configurations.

Q4. Two atoms X and Y form a molecule X₂Y. X has 6 valence electrons and Y has 4 valence electrons. How many shared electron pairs are likely formed between X and each X-Y bond, assuming both atoms achieve octet or appropriate stable configuration?

  • A. Each X-Y bond has 1 shared pair; each X-X bond has 1 shared pair
  • B. Each X-Y bond has 2 shared pairs; X-X bond has 1 shared pair
  • C. Each X-Y bond has 1 shared pair; X-X bond has 2 shared pairs ✓
  • D. All bonds are single bonds with 1 shared pair each

Answer: C — For O₂S (analogy: X=O with 6 valence e⁻, Y=S with 4 valence e⁻), O atoms form a double bond (2 shared pairs) with S to both achieve octet, and the O-O bond would be single (1 pair) to satisfy the octet rule.

Q5. Which of the following compounds violates the octet rule?

  • A. NaCl
  • B. H₂O
  • C. PCl₅ ✓
  • D. CO₂

Answer: C — In PCl₅, phosphorus (group 15) has 5 valence electrons and forms 5 P-Cl bonds, resulting in 10 electrons around P, violating the octet rule; NaCl, H₂O, and CO₂ all obey the octet rule.

Q6. Which statement best explains why the Kössel-Lewis approach successfully explains ionic compound formation but has limitations for covalent compounds?

  • A. Ionic compounds always have exactly 8 electrons on each ion; covalent compounds may have different configurations
  • B. Ionic bonding involves complete electron transfer and clear ion formation; covalent bonding involves partial sharing and is not purely based on achieving a static octet ✓
  • C. Covalent compounds do not have noble gas configurations at all
  • D. Ionic compounds are always solid; covalent compounds are liquid or gas

Answer: B — The Kössel-Lewis model assumes electrons fully transfer (ionic) or achieve a fixed octet by sharing, but covalent bonding is dynamic and overlapping; some atoms achieve less than 8 electrons stably (e.g., BF₃ with 6 around B).

Q7. An element M in group 13 forms a compound with chlorine. Based on Lewis approach, which statement is correct?

  • A. M forms MCl₃ with electrovalence of +2 on M
  • B. M forms MCl₃ with electrovalence of +3 on M ✓
  • C. M forms MCl with electrovalence of +1 on M
  • D. M forms MCl₂ with electrovalence of +2 on M, because it has 2 valence electrons

Answer: B — Group 13 elements (like Al) have 3 valence electrons and lose all 3 to form covalent compounds (not purely ionic); MCl₃ forms with M having an oxidation state of +3 (electrovalence +3).

Q8. Assertion: The octet rule explains the formation of both ionic and covalent compounds. Reason: Both ionic and covalent bonding aim to achieve noble gas electronic configurations.

  • A. Both assertion and reason are correct, and reason is the correct explanation ✓
  • B. Both assertion and reason are correct, but reason is not the correct explanation
  • C. Assertion is correct, but reason is incorrect
  • D. Both assertion and reason are incorrect

Answer: A — The octet rule does explain both bonding types: ionic compounds achieve it through complete electron transfer, and covalent compounds achieve it through electron sharing; the underlying reason in both cases is attaining stable noble gas configurations.

Q9. Calculate the electrovalence of S in the compound H₂S, assuming sulfur achieves an octet by gaining electrons.

  • A. +2
  • B. −2 ✓
  • C. +1
  • D. −1

Answer: B — In H₂S, sulfur ([Ne]3s²3p⁴) gains 2 electrons to form S²⁻ and achieve [Ar]; the electrovalence is −2 (negative because it gains electrons).

Q10. A compound XY₂ is formed where X is from group 16 and Y is from group 1. If both atoms follow the octet/duplet rule, how many total valence electrons are present in one molecule of XY₂?

  • A. 8
  • B. 10 ✓
  • C. 12
  • D. 14

Answer: B — X (group 16) has 6 valence electrons and needs 2 more for octet = 8; each Y (group 1) has 1 valence electron and achieves duplet with 2. Total = 6 (from X) + 1 (from first Y) + 1 (from second Y) = 8 in octet count, but counting all valence electrons: 6 + 1 + 1 = 8 for bonding; however, the compound has 6 (X) + 2 + 2 (Y₂) = 10 total valence electrons in the molecule.

Flashcards

What is a chemical bond?

The attractive force holding atoms, ions, or other constituents together in a chemical species.

Define the octet rule.

Atoms combine by transferring or sharing valence electrons to achieve eight electrons in their outer shell, like noble gases.

What are Lewis symbols and what do they show?

Simple notations representing an element's symbol with dots around it; the number of dots equals the number of valence electrons.

What is the relationship between group number and Lewis symbol dots for main-group elements?

The number of dots equals the group number (e.g., Group 17 halogens have 7 dots).

What is electrovalence?

The number of unit charges (positive or negative) on an ion formed by electron transfer.

How do Na and Cl achieve stable configurations in NaCl formation?

Na transfers one electron to Cl; Na becomes Na⁺ ([Ne]) and Cl becomes Cl⁻ ([Ar]).

What is the key difference between ionic and covalent bonding according to Lewis?

Ionic bonding involves transfer of electrons between atoms; covalent bonding involves sharing of electron pairs.

Name one limitation of the octet rule.

Compounds like PCl₅ and BF₃ have central atoms with fewer than eight valence electrons, violating the octet rule.

Who refined Lewis' theory in 1919 and introduced what key term?

Langmuir refined Lewis' theory and introduced the term 'covalent bond' to describe shared electron pairs.

How does Cl₂ formation demonstrate the covalent bond concept?

Each Cl atom ([Ne]3s²3p⁵) contributes one electron to form a shared pair, allowing both atoms to achieve the [Ar] octet.

Important Board Questions

Define chemical bonding. Why do atoms combine to form molecules? [2 marks]

Define as an attractive force between constituents. State that every system tends toward maximum stability by lowering energy, and bonding is nature's mechanism to achieve this stable state.

Explain the formation of NaCl using the Kössel-Lewis approach. Show how both sodium and chlorine achieve stable configurations and calculate their electrovalences. [5 marks]

Write electronic configurations: Na [Ne]3s¹ loses 1 e⁻ to form Na⁺[Ne]; Cl [Ne]3s²3p⁵ gains 1 e⁻ to form Cl⁻[Ar]. State electrovalence of Na is +1 (charge magnitude) and Cl is −1. Explain that electrostatic attraction stabilizes the ionic compound NaCl. Show both achieve noble gas octets.

State the octet rule and explain its limitations with specific examples. Why does the Kössel-Lewis model fail for some covalent compounds, and what does this reveal about chemical bonding? [6 marks]

Octet rule: atoms achieve 8 electrons in outer shell (2 for He/H). Limitations: PCl₅ (P has 10 e⁻), BF₃ (B has 6 e⁻), SF₆ (S has 12 e⁻), etc. Explain that the static octet model assumes bonding electrons arrange in a fixed cubical pattern (Lewis' model), but real covalent bonding involves dynamic electron sharing and molecular orbital overlap. This shows bonding is more complex than simple electron transfer/sharing for octet; it involves orbital interactions and variable electron densities.

Assertion-Reason Questions for Chemical Bonding and Molecular Structure

Q1. Assertion (A):

The octet rule states that atoms combine by transferring or sharing electrons to achieve a stable outer shell of 8 electrons.

Reason (R):

Noble gases with 8 electrons in their outer shells represent particularly stable electronic arrangements that other atoms try to achieve.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains why the octet rule exists. The stability of noble gas configurations with their filled valence shells (8 electrons) is the fundamental basis for the octet rule proposed by Kössel and Lewis.

Q2. Assertion (A):

Helium achieves stability with only 2 electrons in its valence shell, not 8 like other noble gases.

Reason (R):

Helium is a smaller atom and requires fewer electrons due to its lower atomic number compared to other noble gases.

Answer: C — A is true but R is false — The assertion is true—helium's first shell (n=1) can only accommodate a maximum of 2 electrons (1s²), unlike the second shell of neon which can hold 8. However, the reason is false. The fundamental cause is the orbital structure of shells (first shell has only s-orbital), not the size or atomic number of the element.

Q3. Assertion (A):

In the formation of CaF₂, calcium loses 2 electrons and fluorine gains 1 electron each, resulting in Ca²⁺ and two F⁻ ions.

Reason (R):

Both calcium and fluorine achieve the noble gas configuration of neon (1s² 2s² 2p⁶) through electron transfer.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both assertion and reason are true. However, the reason does NOT correctly explain the assertion. While Ca²⁺ achieves neon's configuration, each F⁻ achieves neon's configuration, but this is not the reason for electron transfer. The real reason is that calcium needs to lose 2 electrons to achieve argon's configuration (for Ca²⁺), and fluorine needs to gain 1 electron to achieve neon's configuration. The pairing (2 fluorines for 1 calcium) occurs because calcium loses 2 electrons total.

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Case-Based Questions (CBQ) for Chemical Bonding and Molecular Structure

Case Study:

A chemistry teacher conducted an experiment where students tested the conductivity of three substances: (1) solid sodium chloride, (2) molten sodium chloride at 800°C, and (3) aqueous solution of sodium chloride. The results showed that solid NaCl did not conduct electricity, while molten and aqueous NaCl conducted electricity strongly. The teacher explained this using the Kössel-Lewis theory: In solid NaCl, ions are fixed in crystal lattice positions, but in molten and aqueous states, ions are free to move. The formation of Na⁺ and Cl⁻ from neutral atoms occurred through electron transfer following the octet rule.

Q1. According to the Kössel-Lewis approach, why does solid NaCl not conduct electricity despite containing Na⁺ and Cl⁻ ions?

  • A. The ions are fixed in crystal lattice positions and cannot move freely ✓
  • B. The ions lose their charges in the solid state
  • C. Solid NaCl does not actually contain ions
  • D. The electrostatic attraction between ions prevents charge flow

Answer: A

Q2. The electronic configuration of Na⁺ formed in this experiment is [Ne]. Which of the following correctly represents this in terms of electron shells?

  • A. 1s² 2s² 2p⁶, achieving argon configuration
  • B. 1s² 2s² 2p⁶, the same as neon ✓
  • C. 1s² 2s² 2p⁵, one electron short of neon
  • D. 2s² 2p⁶ 3s¹, retaining one valence electron

Answer: B

Q3. Why does molten NaCl conduct electricity while solid NaCl does not, based on ionic bonding principles?

  • A. Melting creates new covalent bonds between ions
  • B. Ions gain mobility and can move freely in the molten state to transport charge ✓
  • C. Melting breaks the electrostatic attractions completely
  • D. The octet rule is satisfied only in molten state

Answer: B

Q4. In the formation of NaCl shown by the equation Na → Na⁺ + e⁻ and Cl + e⁻ → Cl⁻, which statement best reflects Kössel's observations about the stability of these ions?

  • A. Both ions achieve octet configurations resembling the nearest noble gas (Ne: 1s² 2s² 2p⁶) ✓
  • B. Na⁺ achieves helium configuration while Cl⁻ achieves argon configuration
  • C. The ions are stable due to their opposite charges, not their electron configurations
  • D. Only Cl⁻ achieves a stable noble gas configuration; Na⁺ remains unstable

Answer: A

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