Chemical Bonding and Molecular Structure — NCERT Class 11 Chemistry

Q: What are the different types of hybridization?

The main types are: sp (linear, 180°, e.g., BeCl₂, C₂H₂), sp² (trigonal planar, 120°, e.g., BF₃, C₂H₄), sp³ (tetrahedral, 109.5°, e.g., CH₄, NH₃, H₂O), sp³d (trigonal bipyramidal, e.g., PCl₅), sp³d² (octahedral, e.g., SF₆). The hybridization type determines the molecular geometry.

Q: What is VSEPR theory and how does it predict molecular shapes?

VSEPR (Valence Shell Electron Pair Repulsion) theory states that electron pairs (bonding and lone) around a central atom arrange themselves to minimise repulsion. Lone pair–lone pair repulsion > lone pair–bond pair repulsion > bond pair–bond pair repulsion. Based on the number of electron pairs, VSEPR predicts shapes: 2 pairs = linear, 3 = trigonal planar, 4 = tetrahedral, 5 = trigonal bipyramidal, 6 = octahedral.

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This page provides complete, step-by-step NCERT solutions for Class 11 Chemistry Chapter 4 — Chemical Bonding and Molecular Structure. Every intext question and chapter-end exercise question is answered with clear diagrams and explanations. Use the section headings to find the question you need.

Key Concepts (Quick Recap)

Chemical bond: Force of attraction holding atoms together in a molecule
Kossel-Lewis approach: Atoms tend to achieve noble gas configuration (octet) by gaining, losing, or sharing electrons
Lewis symbols: Valence electrons shown as dots around the atomic symbol
Ionic bond: Electrostatic attraction between cation and anion (electron transfer); formed between metals and non-metals
Covalent bond: Sharing of electron pairs between atoms (non-metals)
Formal charge: FC = (valence e⁻) − (non-bonding e⁻) − ½(bonding e⁻); helps determine the most stable Lewis structure
Bond parameters: Bond length, bond angle, bond enthalpy, bond order
VSEPR theory: Electron pairs repel each other; arrangement minimises repulsion
Hybridization: Mixing of atomic orbitals to form equivalent hybrid orbitals
sp → linear (180°)
sp² → trigonal planar (120°)
sp³ → tetrahedral (109.5°)
sp³d → trigonal bipyramidal
sp³d² → octahedral
Molecular orbital theory (MO): Orbitals are formed by linear combination of atomic orbitals (LCAO); bonding MOs (lower energy) and antibonding MOs (higher energy)
Bond order: (N_b − N_a)/2 where N_b = electrons in bonding MOs, N_a = electrons in antibonding MOs
Hydrogen bonding: H bonded to F, O, or N interacts with lone pair on another electronegative atom; intermolecular (between molecules) or intramolecular (within molecule)

Intext Questions

Page 72–73 (Kossel-Lewis Approach, Lewis Symbols)

Q4.1: Explain the formation of ionic bond in NaCl.

AnswerSodium (Na, Z = 11) has electronic configuration [Ne]3s¹. It loses one electron to form Na⁺: Na → Na⁺ + e⁻ (IE = 496 kJ/mol). Chlorine (Cl, Z = 17) has configuration [Ne]3s²3p⁵. It gains one electron to form Cl⁻: Cl + e⁻ → Cl⁻ (EGE = −349 kJ/mol). The resulting Na⁺ and Cl⁻ ions are held together by strong electrostatic attraction, forming the ionic bond in NaCl.

Q4.2: Write the Lewis dot structure for: (i) H₂O, (ii) CO₂, (iii) N₂.

Answer

(i) H₂O: O has 6 valence electrons, each H has 1. O shares 2 electrons with two H atoms. Structure: H—Ö—H (2 lone pairs on O).
(ii) CO₂: C has 4 valence electrons, O has 6. C forms double bonds with each O. Structure: O═C═O (2 lone pairs on each O).
(iii) N₂: Each N has 5 valence electrons. They share 3 pairs forming a triple bond. Structure: N≡N (1 lone pair on each N).

Q4.3: What is the octet rule? Are there exceptions?

AnswerThe octet rule states that atoms tend to gain, lose, or share electrons to achieve 8 electrons in their valence shell (noble gas configuration). Exceptions include: (1) Incomplete octet — BeCl₂ (4 e⁻ around Be), BF₃ (6 e⁻ around B). (2) Expanded octet — PCl₅ (10 e⁻ around P), SF₆ (12 e⁻ around S). (3) Odd electron molecules — NO (11 e⁻ total).

Page 78–80 (Ionic Bond, Lattice Energy)

Q4.4: Define lattice energy. How does it affect the stability of ionic compounds?

AnswerLattice energy is the energy released when one mole of an ionic compound is formed from its gaseous ions. For example, Na⁺(g) + Cl⁻(g) → NaCl(s) + energy. Higher lattice energy means stronger electrostatic forces between ions, resulting in greater stability, higher melting point, and lower solubility. Lattice energy increases with: (1) higher ionic charge, (2) smaller ionic radii.

Q4.5: Why is NaCl more stable than Na⁺ and Cl⁻ ions existing separately?

AnswerThe lattice energy released during the formation of NaCl solid from gaseous Na⁺ and Cl⁻ ions (≈ −788 kJ/mol) is much greater than the energy required to form these ions (IE of Na − EGE of Cl ≈ 496 − 349 = +147 kJ/mol). The net energy change is favourable, making NaCl more stable than separated ions.

Page 85–88 (VSEPR Theory, Shapes of Molecules)

Q4.6: Predict the shape of: (i) BeCl₂, (ii) BF₃, (iii) NH₃, (iv) H₂O.

Answer

(i) BeCl₂: Be has 2 bond pairs, 0 lone pairs. Electron pairs = 2. Shape = linear (180°).
(ii) BF₃: B has 3 bond pairs, 0 lone pairs. Electron pairs = 3. Shape = trigonal planar (120°).
(iii) NH₃: N has 3 bond pairs, 1 lone pair. Electron pairs = 4. Shape = trigonal pyramidal (107°).
(iv) H₂O: O has 2 bond pairs, 2 lone pairs. Electron pairs = 4. Shape = bent/V-shaped (104.5°).

Q4.7: Why is the bond angle in H₂O less than in NH₃?

AnswerIn H₂O, O has 2 lone pairs which exert greater repulsion than the 1 lone pair in NH₃. Lone pair–lone pair repulsion > lone pair–bond pair repulsion. This pushes the O—H bonds closer together, reducing the bond angle in H₂O (104.5°) more than in NH₃ (107°).

Page 93–95 (Hybridization, MO Theory Basics)

Q4.8: What is hybridization? Explain with an example.

AnswerHybridization is the concept of mixing atomic orbitals of similar energies on the same atom to form an equal number of equivalent hybrid orbitals. For example, in CH₄, one 2s and three 2p orbitals of C mix to form four sp³ hybrid orbitals, each with 25% s-character and 75% p-character, directed towards the corners of a tetrahedron at 109.5°.

Q4.9: Explain the formation of a sigma (σ) bond and a pi (π) bond.

AnswerA σ bond is formed by head-on (axial) overlap of atomic orbitals along the internuclear axis. It is stronger and allows free rotation. Examples: s-s overlap (H₂), s-p overlap (HCl), p-p overlap (Cl₂). A π bond is formed by sidewise (lateral) overlap of p-orbitals perpendicular to the internuclear axis. It is weaker and restricts rotation. A double bond has one σ + one π; a triple bond has one σ + two π bonds.

Q4.10: Write the MO configuration of O₂ and explain its paramagnetic nature.

AnswerO₂ (16 electrons): KK(σ2s)²(σ2s)²(σ2p_z)²(π2p_x)²(π2p_y)²(π2p_x)¹(π2p_y)¹. Bond order = (10 − 6)/2 = 2. The two unpaired electrons in π2p orbitals make O₂ paramagnetic.

Chapter-End Exercise

Q1. Explain the formation of a chemical bond.

AnswerA chemical bond is formed when atoms come close enough for their atomic orbitals to overlap, resulting in a decrease in energy (stabilisation). Atoms form bonds to achieve a stable electronic configuration (usually an octet). The bonded state has lower energy than the separated atoms, and this energy difference is the bond enthalpy.

Q2. Write Lewis dot symbols for: K, Ca, Al, N, O, F.

Answer

K: K̇ (1 valence electron)
Ca: Cä (2 valence electrons)
Al: Al⠇ (3 valence electrons)
N: ⠅N⠆ (5 valence electrons)
O: ⠇Ö⠇ (6 valence electrons)
F: ⠇F⠇⠇ (7 valence electrons)

Q3. Use the concept of formal charge to determine the most stable Lewis structure of CO₃²⁻.

AnswerThree resonance structures of CO₃²⁻ exist with one C=O double bond and two C—O single bonds. Formal charges:

All three resonance structures are equivalent, so the most stable structure is the resonance hybrid of all three, with each C—O bond having 1⅓ bond order.

Structure with C=O and two C—O⁻: C has FC = 4−0−4 = 0; double-bonded O has FC = 6−4−2 = 0; single-bonded O⁻ has FC = 6−6−1 = −1.
Total FC = 0 + 0 + (−1) + (−1) = −2 (matches the ion charge).

Q4. Differentiate between a sigma (σ) bond and a pi (π) bond.

Answer

Feature	σ Bond	π Bond
Overlap	Head-on (axial)	Sidewise (lateral)
Strength	Stronger	Weaker
Rotation	Free rotation possible	No free rotation
Shape	Cylindrical symmetry	Electron density above/below internuclear axis
Number	Only one between two atoms	One or two (in double/triple bonds)

Q5. Explain with examples: (i) coordinate bond, (ii) hydrogen bond.

Answer

(i) Coordinate (dative) bond: A covalent bond where both shared electrons come from one atom. Example: NH₃ + H⁺ → NH₄⁺. The lone pair on N is donated to H⁺ to form a coordinate bond (shown as →).
(ii) Hydrogen bond: When H is bonded to F, O, or N, it can form a weak bond with a lone pair on another electronegative atom. Example: In water, H of one molecule bonds to O of another (O—H···O). This is why H₂O is a liquid at room temperature despite its low molecular mass.

Q6. Predict the shapes of the following molecules using VSEPR theory: (i) BeCl₂, (ii) SF₆, (iii) NH₃, (iv) PCl₅.

Answer

(i) BeCl₂: 2 bond pairs, 0 lone pairs → Linear (180°)
(ii) SF₆: 6 bond pairs, 0 lone pairs → Octahedral (90°)
(iii) NH₃: 3 bond pairs, 1 lone pair → Trigonal pyramidal (107°)
(iv) PCl₅: 5 bond pairs, 0 lone pairs → Trigonal bipyramidal (120°, 90°)

Q7. What is meant by hybridization? What are the different types of hybridization?

AnswerHybridization is the mixing of atomic orbitals on the same atom to form new hybrid orbitals of equal energy. Types:

sp: 1 s + 1 p → 2 sp orbitals, linear (180°), e.g., BeCl₂, C₂H₂
sp²: 1 s + 2 p → 3 sp² orbitals, trigonal planar (120°), e.g., BF₃, C₂H₄
sp³: 1 s + 3 p → 4 sp³ orbitals, tetrahedral (109.5°), e.g., CH₄, NH₃, H₂O
sp³d: 1 s + 3 p + 1 d → 5 sp³d orbitals, trigonal bipyramidal, e.g., PCl₅
sp³d²: 1 s + 3 p + 2 d → 6 sp³d² orbitals, octahedral, e.g., SF₆

Q8. Describe the hybridization in: (i) CH₄, (ii) C₂H₄, (iii) C₂H₂.

Answer

(i) CH₄: C undergoes sp³ hybridization. One 2s and three 2p orbitals mix to form four sp³ hybrid orbitals, each overlapping with H 1s. Geometry: tetrahedral (109.5°).
(ii) C₂H₄: Each C undergoes sp² hybridization. Three sp² orbitals form σ bonds (2 C—H + 1 C—C). The remaining p orbital on each C forms a π bond. Geometry: trigonal planar (120°) around each C.
(iii) C₂H₂: Each C undergoes sp hybridization. Two sp orbitals form σ bonds (1 C—H + 1 C—C). Two remaining p orbitals on each C form two π bonds. Geometry: linear (180°).

Q9. Explain the formation of H₂ molecule using MO theory. What is its bond order?

AnswerH₂ has 2 electrons. MO configuration: (σ1s)². Bond order = (N_b − N_a)/2 = (2 − 0)/2 = 1. H₂ has a single bond, which is stable. The bonding MO (σ1s) is lower in energy than the individual 1s atomic orbitals.

Q10. Write the MO electronic configuration of N₂ and calculate its bond order.

AnswerN₂ has 14 electrons. MO configuration: (σ1s)²(σ1s)²(σ2s)²(σ2s)²(π2p_x)²(π2p_y)²(σ2p_z)². Bond order = (10 − 4)/2 = 3. This corresponds to the triple bond in N₂ (one σ + two π bonds). The high bond order explains the exceptional stability and low reactivity of nitrogen.

Q11. What is the bond order in O₂²⁻ (peroxide ion)?

AnswerO₂²⁻ has 18 electrons. MO configuration: KK(σ2s)²(σ2s)²(σ2p_z)²(π2p_x)²(π2p_y)²(π2p_x)²(π*2p_y)². Bond order = (10 − 8)/2 = 1. The peroxide ion has a single bond.

Q12. Compare the stability of O₂, O₂⁺, O₂⁻, and O₂²⁻ using bond order.

Answer

Species	Electrons	Bond Order	Stability
O₂⁺	15	(10−5)/2 = 2.5	Most stable
O₂	16	(10−6)/2 = 2	Stable
O₂⁻	17	(10−7)/2 = 1.5	Less stable
O₂²⁻	18	(10−8)/2 = 1	Least stable

Stability order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻.

Q13. Define dipole moment. What is its SI unit?

AnswerThe dipole moment (μ) is the product of the magnitude of charge (q) and the distance between the charges (d): μ = q × d. It measures the polarity of a bond or molecule. SI unit: coulomb-metre (C·m). Practical unit: Debye (D), where 1 D = 3.335 × 10⁻³⁰ C·m.

Q14. Both NH₃ and NF₃ have pyramidal shape with a lone pair on N. Why does NH₃ have a higher dipole moment than NF₃?

AnswerIn NH₃, the dipole moments of the three N—H bonds and the lone pair on N all point in the same direction (from H towards N and from N towards lone pair), adding up. In NF₃, the N—F bond dipoles point from N towards F (F is more electronegative), which opposes the lone pair dipole. The net dipole moment in NF₃ is the difference between the lone pair contribution and the bond dipoles, making it smaller than NH₃.

Q15. Explain why H₂O is a liquid while H₂S is a gas at room temperature.

AnswerIn H₂O, the O—H bonds create strong hydrogen bonding between molecules (O is small and highly electronegative). These intermolecular H-bonds require significant energy to break, keeping H₂O liquid. In H₂S, S is less electronegative and larger than O, so hydrogen bonding is negligible. Only weak van der Waals forces operate, making H₂S a gas at room temperature.

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Frequently Asked Questions

Ionic bonds involve the complete transfer of electrons from one atom (metal) to another (non-metal), forming oppositely charged ions that attract each other (e.g., NaCl). Covalent bonds involve the sharing of electron pairs between two non-metal atoms (e.g., H₂, H₂O). Ionic bonds are non-directional and form crystalline solids; covalent bonds are directional and can form molecules or network solids.