Chemical bonding is the fundamental concept that explains how atoms combine to form molecules and compounds.
In Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure, students learn the various types of bonds, such as ionic bonds, covalent bonds, and coordinate bonds, along with the theories that explain bonding behavior like the octet rule, VSEPR theory, and hybridization.
This chapter also introduces concepts such as bond parameters, molecular geometry, and polarity of molecules. Understanding these ideas is essential for grasping how substances behave chemically and physically. The knowledge gained in this chapter builds a strong base for advanced topics in organic, inorganic, and physical chemistry.
Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure
Q1. Explain the formation of a chemical bond.
Answer:
A chemical bond is formed when atoms combine to attain a stable electronic configuration, usually an octet in their valence shell. This can be achieved by sharing, gaining, or losing electrons. The force of attraction that holds these atoms together is called a chemical bond, and it leads to the formation of molecules or compounds.
Q2. Write Lewis dot symbols for atoms of the following elements: Be, Na, B, O, N, Br.
Answer:
- Be: Be••
- Na: Na•
- B: B•••
- O: ••O••
- N: ••N••
- Br: ••Br••
(Lewis symbols represent valence electrons around the symbol of the element.)
Q3. Write Lewis structures for the following molecules and ions: H₂S, SiCl₄, BeF₂, CO₃²⁻, HCOOH.
Answer:
- H₂S: H–S–H with two lone pairs on S
- SiCl₄: Central Si bonded to four Cl atoms (tetrahedral)
- BeF₂: Linear structure, F–Be–F
- CO₃²⁻: One C double bonded to O, two C–O single bonds with negative charges on the singly bonded O atoms; resonance present
- HCOOH: Formic acid → H–C(=O)–O–H
Q4. Draw the Lewis structure of the following molecules: HNO₃, NO₂, CO, NO, N₂.
Answer:
- HNO₃: H–O–N(=O)–O
- NO₂: Resonance structure with one double-bonded and one single-bonded oxygen
- CO: C≡O with a lone pair on each atom
- NO: N=O with one lone electron
- N₂: N≡N (triple bond between nitrogen atoms)
Q5. Define octet rule. Write its significance and limitations.
Answer:
Octet Rule: Atoms tend to gain, lose, or share electrons to attain 8 electrons in their valence shell.
Significance:
Explains stability of noble gases and formation of many molecules.
Limitations:
- Doesn’t apply to H (duplet), Be (4 electrons), B (6 electrons)
- Cannot explain expanded octet (e.g. PCl₅, SF₆)
- Fails for odd-electron molecules like NO
Q6. Write the favourable conditions for the formation of ionic bond.
Answer:
Favourable conditions for ionic bond formation:
- Low ionization enthalpy of metal atom (to lose electrons easily).
- High electron gain enthalpy of non-metal atom (to gain electrons easily).
- Large difference in electronegativity between the two atoms.
- High lattice energy to stabilize the ionic solid.
Q7. Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, CH₄, NH₃, H₂O.
Answer:
Using VSEPR theory:
- BeCl₂: Linear (2 bond pairs, 0 lone pair)
- BCl₃: Trigonal planar (3 bond pairs, 0 lone pair)
- CH₄: Tetrahedral (4 bond pairs, 0 lone pair)
- NH₃: Trigonal pyramidal (3 bond pairs, 1 lone pair)
- H₂O: Bent or V-shaped (2 bond pairs, 2 lone pairs)
Q8. Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer:
Both NH₃ and H₂O have sp³ hybridization and distorted tetrahedral shapes.
- In NH₃, one lone pair causes repulsion → bond angle ≈ 107°
- In H₂O, two lone pairs cause more repulsion → bond angle ≈ 104.5°
More lone pairs = more repulsion = smaller bond angle
Q9. How do lone pairs and bond pairs of electrons affect the shape of molecules?
Answer:
- Bond pairs repel each other and decide basic shape.
- Lone pairs repel more strongly than bond pairs.
More lone pairs distort the bond angles, changing the ideal shape (e.g. from tetrahedral to bent or pyramidal).
Q10. Distinguish between sigma (σ) and pi (π) bonds.
Answer:
| Sigma (σ) Bond | Pi (π) Bond |
|---|---|
| Formed by end-to-end overlap | Formed by sideways overlap |
| Stronger bond | Weaker bond |
| Present in all single bonds | Present in double/triple bonds |
| Electron density along internuclear axis | Electron density above and below axis |
Q11. Explain the formation of a double bond using the example of ethene (C₂H₄) molecule.
Answer:
In ethene (C₂H₄), each carbon atom is sp² hybridized, forming 3 sigma bonds (2 with H and 1 with C).
The unhybridized p-orbitals on each carbon atom overlap sideways to form a pi (π) bond, resulting in a double bond between the two carbon atoms.
So, the C=C bond in ethene consists of 1 sigma and 1 pi bond.
Q12. Explain the formation of a triple bond in N₂ molecule.
Answer:
Each nitrogen atom has 5 valence electrons. In N₂:
- One pair forms a sigma bond through head-on overlap of sp orbitals.
- The other two pairs form two pi bonds by sideways overlap of p orbitals.
Thus, a triple bond (1 sigma + 2 pi) exists between the two nitrogen atoms.
Q13. Draw the resonating structures of CO₃²⁻ and explain the delocalization of electrons.
Answer:
CO₃²⁻ (carbonate ion) has 3 equivalent resonance structures:
- One C=O double bond and two C–O single bonds with negative charges.
The double bond shifts among the three O atoms.
This delocalization of electrons results in equal bond length for all C–O bonds and adds stability to the ion.
Q14. Use Lewis symbols to show electron transfer between the following atoms to form ionic compounds:
(i) Ca and O
(ii) K and S
Answer:
(i) Ca (2 valence electrons) transfers 2 electrons to O (needs 2) → Ca²⁺ + O²⁻
(ii) 2 K atoms (each gives 1 electron) transfer electrons to S (needs 2) → 2K⁺ + S²⁻
Resulting compounds: CaO, K₂S
Q15. What is the coordinate bond? How is it different from a covalent bond?
Answer:
A coordinate bond is a type of covalent bond in which both electrons in the shared pair come from the same atom.
Difference:
| Covalent Bond | Coordinate Bond |
|---|---|
| Electrons shared by both atoms | Electrons donated by one atom only |
| Formed between similar or different atoms | Often formed with lone pairs (e.g. NH₄⁺) |
| Example: H–Cl | Example: NH₄⁺ (N → H⁺) |
Q16. Describe the molecular orbital theory of bonding in H₂ molecule.
Answer:
According to Molecular Orbital Theory:
- Atomic orbitals combine to form molecular orbitals: bonding (σ1s) and antibonding (σ*1s).
- In H₂, two 1s electrons occupy the bonding orbital (σ1s).
- Bond order = (Number of bonding electrons – Number of antibonding electrons) ÷ 2
= (2 – 0) ÷ 2 = 1
This indicates a stable single covalent bond in H₂.
Q17. What is meant by bond order? Calculate the bond order of: N₂, O₂, and F₂.
Answer:
Bond order = (Number of bonding electrons − Number of antibonding electrons) ÷ 2
- N₂ → BO = (10 − 4)/2 = 3
- O₂ → BO = (10 − 6)/2 = 2
- F₂ → BO = (8 − 6)/2 = 1
Higher bond order means stronger and shorter bond.
Q18. Which hybrid orbitals are used by carbon atoms in the following molecules?
(i) CH₃–CH₃
(ii) CH₃–CH=CH₂
(iii) CH₃–C≡CH
Answer:
(i) CH₃–CH₃ → All single bonds → sp³ hybridization
(ii) CH₃–CH=CH₂ → C=C double bond → involved C atoms are sp² hybridized
(iii) CH₃–C≡CH → C≡C triple bond → involved C atoms are sp hybridized
Q19. Describe the hybridization in the following molecules and shape:
(i) SF₆
(ii) PCl₅
(iii) BeCl₂
Answer:
(i) SF₆: sp³d² hybridization → Octahedral shape
(ii) PCl₅: sp³d hybridization → Trigonal bipyramidal shape
(iii) BeCl₂: sp hybridization → Linear shape
Q20. What do you understand by dipole moment? What does it signify?
Answer:
Dipole moment (μ) is the product of charge and distance of separation between charges in a polar molecule.
μ = q × d
Significance:
- Measures polarity of a bond or molecule
- Helps determine molecular geometry
- Molecules with dipole moment ≠ 0 are polar
Q21. Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Answer:
A hydrogen bond is a special type of dipole-dipole attraction between a hydrogen atom bonded to a highly electronegative atom (like N, O, or F) and another electronegative atom with a lone pair.
Strength:
Hydrogen bonds are stronger than van der Waals forces but weaker than covalent or ionic bonds.
Q22. What is resonance? Explain with an example.
Answer:
Resonance is a phenomenon where a molecule is best represented by two or more Lewis structures (resonance structures) that differ only in the position of electrons, not atoms.
Example:
Ozone (O₃) has two resonance structures:
O=O–O ↔ O–O=O
The actual structure is a resonance hybrid of both.
Q23. Explain the difference between polar covalent bond and non-polar covalent bond.
Answer:
| Polar Covalent Bond | Non-Polar Covalent Bond |
|---|---|
| Unequal sharing of electrons | Equal sharing of electrons |
| Occurs between atoms of different electronegativities | Occurs between identical or similar atoms |
| Creates dipole moment | No dipole moment |
| Example: H–Cl | Example: H–H, Cl–Cl |
Q24. Why does BF₃ have zero dipole moment though it has polar bonds?
Answer:
In BF₃, the molecule has a trigonal planar structure with 120° bond angles.
Though each B–F bond is polar, the dipoles cancel out due to symmetry.
Hence, the net dipole moment is zero.
For the official Class 10 Mathematics Solutions, you can visit:
- NCERT Textbooks (for Class 10):
Class-wise Solutions
Class 12:
Class 12 Physics – NCERT Solutions
Class 12 Chemistry – NCERT Solutions
Class 11:
- Class 11 Physics – NCERT Solutions
- Class 11 Chemistry – NCERT Solutions
- Class 11 Biology – NCERT Solutions
- Class 11 Math – NCERT Solutions
Class 10:
Class 9:
Class 8:
Class 7:
Class 6:
Subject-wise Solutions
Physics:
Chemistry:
Biology:
Math:
- Class 11 Math – NCERT Solutions
- Class 10 Math – NCERT Solutions
- Class 9 Math – NCERT Solutions
- Class 8 Math – NCERT Solutions
Science:
- Class 10 Science – NCERT Solutions
- Class 9 Science – NCERT Solutions
- Class 8 Science – Oxford Solutions
- Class 7 Science – Oxford Solutions
- Class 6 Science – Oxford Solutions