Match List – I with List – II (Chemical Bonding) List – I List – II (A) [PtCl 4 ] 2 - (I) sp 3 d (B) BrF 5 (II) d 2 sp 3 (C) PCl 5 (III) dsp 2 (D) [Co(NH 3 ) 6 ] 3+ (iv) sp 3 d 2 Choose the most appropriate answer from the options given below :

(A) – (III), (B) – (IV), (C) – (I), (D) – (II)

(A) – (II) – (B) – (IV), (C) – (I), (D) – (III)

(A) – (III), (B) – (I), (C) – (IV), (D) – (II)

(A) – (II), (B) – (I), (C) – (IV), (D) – (III)

NCERT Solutions for Class 11 Chemistry Chapter 4 – Chemical Bonding and Molecular Structure

Q: In Chemical bonding, what are the four types of bonds?

In Chemistry, the bonds are of four types - Covalent bond, Ionic bond, Hydrogen bond, and Metallic bond.

Q: Explain the concept of chemical bonding.

It is the bond that holds the ions, atoms or molecules to form stable substances, compounds and larger structures.

Q: According to bonding chemistry, what are different types of chemical bonds?

The chemical bonds can be primary or strong bonds such as ionic, covalent, and metallic bonds. It can also be the secondary or weak bonds like hydrogen bonding, London dispersion force, and dipole interactions.

Q: Mention three types of chemical bonds.

The three types of chemical bonds are - ionic, metallic and covalent bonds. When the electrons transfer between the atoms, they form the Ionic bonds by producing charged ions that are attracted to each other. When atoms share electrons, covalent bonds are created. When metal atoms share a sea of delocalized electrons, metallic bonds get created.

Q: In Chemical Bonding and Molecular Structure, which are three types of chemical laws?

The three fundamental laws of chemistry are - Law of Definite Proportions, Law of Conservation of Mass, and Law of Multiple Proportions.

Q: 4.4. Draw the Lewis structures for the following molecules and ions: H2S, SiCl4, BeF2, CO32-, HCOOH.

4.4. The properties of molecules and compounds are determined by their molecular structure, which is the arrangement of the atoms in space. The molecular structure is determined by the type of bond and the number and arrangement of atoms in the molecule. The shape of the molecule affects its polarity, reactivity, and physical properties such as boiling point, melting point, and solubility. The valence shell electron pair repulsion (VSEPR) theory is a model used to predict the molecular structure of covalent molecules based on the repulsion between valence electron pairs. This theory states that the electron pairs in the valence shell of an atom repel each other and try to occupy positions that minimise repulsion. The VSEPR theory is used to predict the geometry and shape of molecules, such as linear, trigonal planar, tetrahedral, and octahedral. In summary, chemical bonding and molecular structure are important concepts in chemistry that help us understand the behaviour and properties of molecules and compounds. The type of bond and molecular structure determine the properties of a substance, including its reactivity, polarity, and physical properties. The VSEPR theory is a useful tool for predicting the molecular structure of covalent molecules based on the repulsion between valence electron pairs.

Q: 4.14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions (a) K and S (b) Ca and O (c) Al and N.

4.14. Kindly go through the solution

Q: 4.7. Discuss the shape of the following molecules using the VSEPR model: BeCl2, BCl3, SiCl4, AsF5, H2S, PH3.

4.7 Molecule Number of electron pairs around central atom Type Molecular geometry Bond angles BeCl2? 2 AB2 Linear 180o BCl3? 3 AB3 trigonal planar 120o SiCl4? 4 AB4 tetrahedral 109.5o AsF5? 5 AB5 trigonal bipyramidal three 120 o, two 90 o H2?S 6 AB2L2 V-shaped/bent 92 o PH3? 5 AB3L trigonal pyramidal 93.5o

Q: 4.6. Write the favourable factors for the formation of ionic bond.

4.6. The favourable factors for the formation of ionic bond are: 1. Low ionization enthalpy of metal atoms. 2. High electron gain enthalpy of non-metal atoms. High lattice enthalpy of compound formed.

Updated on Sep 18, 2025 12:42 IST

By Pallavi Pathak, Assistant Manager Content

NCERT Solutions for Class 11 Chemistry Chapter 4: Chemical Bonding and Molecular includes all the exercise questions given in the textbook. It discusses the attractive force that holds various constituents together in different chemical species. A chemical bond holds atoms and ions together in a molecule.
Chemical Bonding Class 11 covers the Valence Shell Electron Pair Repulsion (VSEPR) Theory, the Kössel-Lewis approach, the VSEPR model, Molecular Orbital (MO) Theory, and Valence Bond (VB) Theory. Chemical Bonding Class 11 NCERT Solutions is created by the subject matter experts. All the solutions are given in a step-by-step format. It is easy to understand and helps students build their problem-solving skills. The solutions are accurate and reliable for CBSE Board exam preparation and for preparing for various competitive exams, such as NEET and JEE Main.
Get free NCERT Solutions for Class 11 and 12 in Maths, Physics, and Chemistry, and chapter-wise PDFs with key concepts simplified for easy learning, available here - NCERT Solutions Class 11 and 12 for Maths, Physics, Chemistry.

Table of content

Chemical Bonding and Molecular Structure Question and Answers
Class 11 Chemistry Chemical Bonding and Molecular Structure: Key Topics, Weightage
Important Formulae of Chemical Bonding and Molecular Structure
Class 11 Chemistry Chemical Bonding and Molecular Structure Solution PDF: Download PDF for Free
Chemical Bonding and Molecular Structure - FAQs

Chemical Bonding and Molecular Structure Question and Answers

Students can find the NCERT solutions for Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure for the intext questions.

4.1. Explain the formation of a chemical bond.

Answer: Chemical bond is an attractive force which binds atoms, ions etc. together in a compound. According to Kossel and Lewis, atoms combine together in order to complete their respective octets so as to acquire the configuration of the nearest stable inert gas. This can occur in two ways; by transfer of one or more electrons from one atom to other or by sharing of electrons between two or more atoms. The chemical bond formed by sharing of electrons is called a covalent bond. In this process a chemical bond is formed

4.2. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.

Answer:

Element	Atomic number	Atomic arrangement	Lewis symbol
Mg	12	2,8,2
Na	11	2,8,1
B	5	2,3
O	8	2,6
N	7	2,5
Br	35	2,8,18,7

4.3. Write Lewis symbols for the following atoms and ions: S and S^2–; Al and Al³⁺; H and H^–. (Beginner)

Answer:

₁₆S = 2, 8, 6

₁₃Al = 2, 8, 3

₁H = 1

Lewis symbols are:

4.4. Draw the Lewis structures for the following molecules and ions:
H₂S, SiCl_4,BeF₂, CO₃^2-, HCOOH.

Answer:

The properties of molecules and compounds are determined by their molecular structure, which is the arrangement of the atoms in space. The molecular structure is determined by the type of bond and the number and arrangement of atoms in the molecule. The shape of the molecule affects its polarity, reactivity, and physical properties such as boiling point, melting point, and solubility.

The valence shell electron pair repulsion (VSEPR) theory is a model used to predict the molecular structure of covalent molecules based on the repulsion between valence electron pairs. This theory states that the electron pairs in the valence shell of an atom repel each other and try to occupy positions that minimise repulsion. The VSEPR theory is used to predict the geometry and shape of molecules, such as linear, trigonal planar, tetrahedral, and octahedral.

In summary, chemical bonding and molecular structure are important concepts in chemistry that help us understand the behaviour and properties of molecules and compounds. The type of bond and molecular structure determine the properties of a substance, including its reactivity, polarity, and physical properties. The VSEPR theory is a useful tool for predicting the molecular structure of covalent molecules based on the repulsion between valence electron pairs.

AsF₅	5	AB₅	trigonal bipyramidal	three 120^o, two 90^o
H₂S	6	AB₂L₂	V-shaped/bent	92^o
PH₃	5	AB₃L	trigonal pyramidal	93.5^o

Commonly asked questions

4.4. Draw the Lewis structures for the following molecules and ions:

H₂S, SiCl_4,BeF₂, CO₃^2-, HCOOH.

4.14. Use Lewis symbols to show electron transfer between the following atoms to form cations and anions (a) K and S (b) Ca and O (c) Al and N.

4.27. Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂molecules.

4.7. Discuss the shape of the following molecules using the VSEPR model:

BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃.

4.6. Write the favourable factors for the formation of ionic bond.

4.10. Define the bond-length.

4.11. Explain the important aspects of resonance with reference to the CO₃^2-ion.

4.16. Write the significance/applications of dipole moment.

4.17. Define electronegativity. How does it differ from electron gain enthalpy?

4.25. Describe the change in hybridisation (if any) of the Al atom in the following reaction.

AlCl₃ + Cl^– ——> AlCl^4-

4.31. What do you understand by bond pairs and lone pairs of electrons? Illustrate by giving one example of each type.

4.33. Explain the formation of H₂ molecule on the basis of valence bond theory.

4.38. Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?

4.21. Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar with the four H atoms at the comers of the square and the C atom at its centre. Explain why CH₄ is not square planar?

4.22. Explain why BeH₂ molecule has a zero dipole moment although the Be—H bonds are polar.

4.26. Is there any change in the hybridisation of B and N atoms as a result of the following reaction?

BF₃ + NH₃ ——-> F₃B.NH₃

4.28. What is the total number of sigma and pi bonds in the following molecules?

(a) C₂H₂

(b) C₂H₄

4.29 Considering x-axis as the inter-nuclear axis which out of the following will not form a sigma bond and why?

(a) 1s and 1s

(b) 1s and 2p_x

(c) 2p_y and 2p_y

(d) 1s and 2s

4.2. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br.

4.3. Write Lewis symbols for the following atoms and ions: S and S^2–; Al and Al³⁺; H and H^–.

4.48. Which one is diamagnetic among NO+, NO and NO?

(a) NO⁺

(b) NO

(c) NO^–

(d) None of these

4.47. The axial overlap between the two orbitals leads to the formation of a:

(a) Sigma bond

(b) Pi bond

(c) Multiple Bond

(d) None of these

4.12. H₃PO₃can be represented by structures 1 and 2 shown below. Can these two structures be taken as the canonical forms of the resonance hybrid representing H₃PO₃? If not, give reasons for the same.

4.24. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.

4.35. Use molecular orbital theory to explain why the Be₂ molecule does not exist.

4.40. What is meant by the term bond order? Calculate the bond order of N₂, O₂, O₂⁺ and O₂.

4.42. Assertion: In BF₃, the dipole moment is zero

Reason: The three bond moments give anet sum of zero as the resultant of any two is equal and opposite to the third.

4.9. How do you express the bond strength in terms of bond order?

4.46. A co-ordinate bond is formed by:

(a) Sharing of electrons contributed by both the atoms

(b) Complete transfer of electrons

(c) Sharing of electrons contributed by one atom only

(d) None of these

4.53. What is meant by bond pairs of electrons?

4.1. Explain the formation of a chemical bond.

4.5. Define Octet rule. Write its significance and limitations.

4.8. Although geometries of NH₃and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.

4.32. Distinguish between a sigma bond and a pi bond.

4.34. Write the important conditions required for the linear combination of atomic orbitals to form molecular orbitals.

4.41. Assertion: The dipole moment in case of BeF₂ is zero.

Reason: There are two F atoms in a molecule of BeF₂.

4.15. Although both CO₂ and H₂O are tri atomic molecules, the shape of H₂O molecule is bent while that of CO₂ is linear. Explain this on the basis of dipole moment.

4.18. Explain with the help of suitable example polar covalent bond.

4.19. Arrange the bonds in order of increasing ionic character in the molecules: LiF, K₂O, N₂, SO₂ and ClF₃.

4.23. Which out of NH₃ and NF₃ has higher dipole moment and why?

4.36. Compare the relative stability of the following species and indicate their magnetic properties: O₂, O₂, O₂^– (Superoxide), O₂^2- (peroxide)

4.37. Write the significance of plus and minus sign in representing the orbitals.

4.39. Define hydrogen bonds. Is it weaker or stronger than the van der Waals forces?

4.43. Assertion: Sigma bond is weaker as compared to the pi bond.

Reason: The overlapping of orbitals takes place to a larger extent in sigma bond while the extent of overlapping occurs to a smaller extent in case of pi bond.

4.44. Assertion: A bonding molecular orbital always possesses lower energy than either of the atomic orbitals that have combined to form it.

Reason: Electrons placed in a bonding molecular orbital tend to destabilise a molecule.

4.45. Assertion: In PCl₅, the axial bonds are slightly longer and hence slightly weaker than the equatorial bonds.

Reason: The axial bond pairs suffer more repulsive interaction from the equatorial bond pairs.

4.49. Out of the following, intramolecular hydrogen bonding exists in:

(a) Water

(b) H₂S

(c) 4-nitrophenol

(d) 2-nitrophenol

4.50. Relation between bond order and stability of a molecule:

(a) Lower the bond order, greater is the stability

(b) Not related to each other

(c) Higher the bond order, greater is the stability

(d) None of the above

4.51. Write the type of hybridisation involved in CH₄, C₂H₄ and C₂H₂.

4.52. Predict the shapes of the following molecules using VSEPR theory?

(i) BeCl₂

(ii) SiCl₄

4.54. Why ethyl alcohol is completely miscible with water?

4.55. Which is more polar CO₂ or N₂O? Give reason.

4.56. Why N₂ is more stable than O₂? Explain on the basis of molecular orbital theory.

4.57. Out of bonding and antibonding molecular orbitals, which one has lower energy and which one has higher stability?

4.58. Name the two conditions which must be satisfied for hydrogen bonding to take place in a molecule.

4.59. What are Lewis structures? Write the Lewis structure of H₂, BeF₂ and H₂O.

4.60. Define Lattice energy. How is Lattice energy influenced by

(i) Charge on the ions

(ii) Size of the ions?

4.13. Write the resonance structures for SO₃, NO₂ and NO_3.

4.20 The skeletal structure of CH₃COOH as shown below is correct, but some of the bonds are shown incorrectly. Write the correct Lewis structure for acetic acid.

4.30. Which hybrid orbitals are used by carbon atoms in the following molecules?

(a) CH₃-CH₃ (b) CH₃-CH=CH₂ (c) CH₃-CH₂-OH (d) CH₃-CHO (e) CH₃COOH.

Class 11 Chemistry Chemical Bonding and Molecular Structure: Key Topics, Weightage

Class 11 Chapter 4 Chemical Bonding and Molecular Structure explores various key topics that help in understanding the chemical compound formation through various reaction mechanisms and molecular structures.

Before starting any chapter preparation, it is good to see the topics covered in the chapter for planned learning.

The following are the topics covered in this chapter:

Exercises	Topics Covered
4.1	Kossel-Lewis Approach to Chemical Bonding
4.2	Ionic or Electrovalent Bond
4.3	Bond Parameters
4.4	The Valence Shell Electron Pair Repulsion (VSEPR) Theory
4.5	Valence Bond Theory
4.6	Hybridisation
4.7	Molecular Orbital Theory
4.8	Bonding in Some Homonuclear Diatomic Molecules
4.9	Hydrogen Bonding

Chemical Bonding and Molecular Structure Weightage in NEET and JEE Main exam

Examination	Number of Questions	Weightage
NEET	2-3 questions	8-10%
JEE Main	2 questions	6-8%

Important Formulae of Chemical Bonding and Molecular Structure

Important Formulae of Chemical Bonding and Molecular Structure for CBSE and Competitive Exams

Formal Charge Calculation:
$\text{Formal Charge} = \left( \text{Valence electrons} \right) - \left( \text{Non-bonding electrons} \right) - \frac{1}{2} \left( \text{Bonding electrons} \right)$
Bond Order (MOT):
$\text{Bond Order} = \frac{(\text{Number of bonding electrons}) - (\text{Number of antibonding electrons})}{2}$
Dipole Moment:
$\mu = q \times d$
Hybridization Formula:
$Hybridization Number = \frac{1}{2} (Valence electrons of central atom + Number of monovalent atoms - Charge on cation + Charge on anion)$

Class 11 Chemistry Chemical Bonding and Molecular Structure Solution PDF: Download PDF for Free

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Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure NCERT Solution PDF: Download Free PDF

Chemical Bonding and Molecular Structure - FAQs

These are the frequently asked questions from the chemical bonding class 11:

Commonly asked questions

In Chemical bonding, what are the four types of bonds?

Explain the concept of chemical bonding.

According to bonding chemistry, what are different types of chemical bonds?

Mention three types of chemical bonds.

In Chemical Bonding and Molecular Structure, which are three types of chemical laws?

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NCERT Solutions for Class 11 Chemistry Chapter 4 – Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure Question and Answers

Commonly asked questions

Class 11 Chemistry Chemical Bonding and Molecular Structure: Key Topics, Weightage

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Important Formulae of Chemical Bonding and Molecular Structure

Important Formulae of Chemical Bonding and Molecular Structure for CBSE and Competitive Exams

Class 11 Chemistry Chemical Bonding and Molecular Structure Solution PDF: Download PDF for Free

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