Important Questions Class 11 Chemistry Chapter 4 Chemical Bonding and Molecular Structure

Chemical bonding and molecular structure explains how atoms combine to form stable molecules through ionic bonds, covalent bonds, orbital overlap, hybridisation, and molecular orbitals. Important questions class 11 chemistry chapter 4 cover Lewis structures, VSEPR theory, valence bond theory, hybridisation, MOT, dipole moment, bond parameters, and hydrogen bonding.

A molecule’s shape decides many of its properties. Water is bent, carbon dioxide is linear, ammonia is pyramidal, and oxygen is paramagnetic because bonding is not only about sharing electrons.

Class 11 Chemistry Chapter 4 helps students understand why molecules form, how their shapes are predicted, and how bond order affects stability. Once students connect electron pairs, lone pairs, overlap, and molecular orbitals, VSEPR, hybridisation, and MOT questions become much easier.

Key Takeaways

Class 11 Chemistry Units at a Glance

Important Topics in Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure has many subtopics, but most exam questions come from shape prediction, hybridisation, resonance, dipole moment, and MOT.

1. Kössel-Lewis approach and octet rule
2. Ionic and covalent bond formation
3. Lewis structures and formal charge
4. Resonance and resonance hybrid
5. Bond length, bond angle, bond enthalpy, and bond order
6. Bond polarity and dipole moment
7. VSEPR theory and molecular shapes
8. Valence bond theory and orbital overlap
9. Hybridisation: sp, sp², sp³, sp³d, sp³d²
10. Sigma and pi bonds
11. Molecular orbital theory
12. Bonding in H₂, He₂, Li₂, C₂, N₂, O₂
13. Intermolecular and intramolecular hydrogen bonding

Important Questions of Chemical Bonding and Molecular Structure

Important questions of chemical bonding and molecular structure should be revised by concept.

Shape questions need VSEPR. Hybridisation questions need steric number. MOT questions need bond order. Resonance questions need equivalent structures and formal charge.

Important Questions Class 11 Chemistry Chapter 4 with Answers

Important questions class 11 chemistry chapter 4 are arranged from basic definitions to advanced MOT and PYQ-style questions.

These chemical bonding and molecular structure class 11 important questions help students practise theory, diagrams, reasoning, and formula-based answers together.

Very Short Answer Questions from Chemical Bonding and Molecular Structure

Q1. Define a chemical bond.
Ans. A chemical bond is the attractive force that holds atoms, ions, or other particles together in a chemical species.

Q2. What is the octet rule?
Ans. The octet rule states that atoms combine by losing, gaining, or sharing electrons to complete eight electrons in their valence shell.

Q3. Define bond order.
Ans. Bond order is the number of bonds between two atoms.

In molecular orbital theory:

Bond order = 1/2 × (number of bonding electrons - number of antibonding electrons)

Q4. Define dipole moment.
Ans. Dipole moment is the product of charge and distance between the centres of positive and negative charge.

Dipole moment = charge × distance

It is expressed in Debye.

Q5. What is a sigma bond?
Ans. A sigma bond is formed by end-to-end overlap of atomic orbitals along the internuclear axis.

Q6. What is a pi bond?
Ans. A pi bond is formed by sideways overlap of p orbitals above and below the internuclear axis.

Q7. Define hybridisation.
Ans. Hybridisation is the mixing of atomic orbitals of comparable energy to form equivalent hybrid orbitals of the same shape and energy.

Q8. Why is the dipole moment of CO₂ zero?
Ans. CO₂ has a linear shape.

The two C=O bond dipoles are equal and opposite. They cancel each other, so net dipole moment is zero.

Q9. Define bond length.
Ans. Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule.

Q10. What are the two types of hydrogen bonding?
Ans. The two types are intermolecular hydrogen bonding and intramolecular hydrogen bonding.

Q11. Which has a higher boiling point: H₂O or H₂S?
Ans. H₂O has a higher boiling point.

Water molecules form strong intermolecular hydrogen bonds. H₂S does not form such strong hydrogen bonding.

Q12. Define bond enthalpy.
Ans. Bond enthalpy is the energy required to break one mole of a particular bond in gaseous molecules.

Its unit is kJ mol⁻¹.

Lewis Structure, Octet Rule and Formal Charge Questions

Lewis structures show valence electrons, bond pairs, lone pairs, and formal charges.

For polyatomic ions, always count total valence electrons first, including charge.

Chemical Bonding and Molecular Structure Question Answer

Q13. Write the Lewis dot structure of CO₂.
Ans. Carbon has 4 valence electrons.

Each oxygen has 6 valence electrons.

Total valence electrons = 4 + 6 + 6 = 16

Carbon forms double bonds with both oxygen atoms.

Structure: O=C=O

Each oxygen has two lone pairs. All atoms complete their octet.

Q14. Draw the Lewis structure of NO₂⁻.
Ans. Total valence electrons:

Nitrogen = 5
Two oxygen atoms = 12
One extra electron due to negative charge = 1

Total = 18 electrons

Nitrogen is the central atom. One N=O double bond and one N-O single bond are formed.

The ion has two resonance structures because the double bond can be on either oxygen.

Q15. Calculate the formal charge on the central oxygen atom in O₃.
Ans. Formal charge = Valence electrons - Non-bonding electrons - 1/2 bonding electrons

For central oxygen in O₃:

Valence electrons = 6

Non-bonding electrons = 2

Bonding electrons = 6

Formal charge = 6 - 2 - 1/2(6)

= 6 - 2 - 3

= +1

So, formal charge on central oxygen is +1.

Q16. Write the significance and limitations of the octet rule.
Ans. The octet rule explains bond formation in many ionic and covalent compounds.

It also helps explain the structures of many simple molecules.

Limitations:

1. It fails for incomplete octet molecules like BeH₂, BF₃, and AlCl₃.
2. It fails for odd-electron molecules like NO and NO₂.
3. It fails for expanded octet molecules like PF₅ and SF₆.
4. It does not explain molecular shape.
5. It does not explain relative stability.

Q17. Why cannot CO₃²⁻ be represented by a single Lewis structure?
Ans. A single Lewis structure would show one C=O double bond and two C-O single bonds.

This suggests unequal C-O bond lengths.

Experimentally, all three C-O bonds in CO₃²⁻ are equal.

So, CO₃²⁻ is represented as a resonance hybrid of three equivalent structures.

Q18. Write the resonance structures of O₃.
Ans. Ozone has two resonance structures.

In one structure, the double bond is between the left oxygen and central oxygen.

In the other structure, the double bond is between the central oxygen and right oxygen.

The actual ozone molecule is a resonance hybrid.

Both O-O bonds have equal bond length, intermediate between single and double bond length.

VSEPR Theory and Molecular Shape Questions

VSEPR theory predicts molecular shape from repulsions between electron pairs.

Use this repulsion order:

Lone pair-lone pair > lone pair-bond pair > bond pair-bond pair

Chemical Bonding and Molecular Structure Class 11 Question Answer

Q19. Predict the shape of NH₃ using VSEPR theory.
Ans. Nitrogen in NH₃ has three bond pairs and one lone pair.

The four electron pairs arrange themselves tetrahedrally.

Because one lone pair is present, the molecular shape becomes trigonal pyramidal.

The H-N-H bond angle reduces from 109.5° to about 107°.

Q20. Explain the non-linear shape of H₂S using VSEPR theory.
Ans. Sulphur in H₂S has two bond pairs and two lone pairs.

The electron pair geometry is tetrahedral.

Due to lone pair-lone pair and lone pair-bond pair repulsions, the molecule becomes bent or V-shaped.

Q21. Although NH₃ and H₂O both have distorted tetrahedral geometry, the bond angle in H₂O is less than in NH₃. Explain.
Ans. NH₃ has one lone pair and three bond pairs.

H₂O has two lone pairs and two bond pairs.

Lone pair-lone pair repulsion is greater than lone pair-bond pair repulsion.

Since H₂O has two lone pairs, its bond angle is compressed more.

So, H₂O has bond angle 104.5°, while NH₃ has about 107°.

Q22. Discuss the shape of BeCl₂, BCl₃, and SiCl₄ using VSEPR theory.
Ans.

Q23. Explain why PCl₅ is trigonal bipyramidal while IF₅ is square pyramidal.
Ans. PCl₅ has five bond pairs and no lone pair on phosphorus.

So, it has trigonal bipyramidal geometry.

IF₅ has five bond pairs and one lone pair on iodine.

The six electron pairs arrange octahedrally, but one position is occupied by a lone pair.

So, IF₅ has square pyramidal shape.

Q24. Explain the shape of BrF₅.
Ans. Bromine in BrF₅ has five bond pairs and one lone pair.

The electron pair geometry is octahedral.

One lone pair occupies one position, so the molecular shape becomes square pyramidal.

The hybridisation is sp³d².

Hybridisation and Valence Bond Theory Questions

Hybridisation is one of the highest-weightage areas in Chapter 4.

For hybridisation questions, count sigma bonds and lone pairs around the central atom.

Class 11 Chemistry Chemical Bonding and Molecular Structure Important Questions

Q25. What is hybridisation? Describe the shapes of sp, sp², and sp³ hybrid orbitals with one example each.
Ans. Hybridisation is the mixing of atomic orbitals of similar energy to form equivalent hybrid orbitals.

sp hybridisation:

1. One s and one p orbital mix.
2. Two sp hybrid orbitals form.
3. Shape is linear.
4. Bond angle is 180°.
5. Example: BeCl₂.

sp² hybridisation:

1. One s and two p orbitals mix.
2. Three sp² hybrid orbitals form.
3. Shape is trigonal planar.
4. Bond angle is 120°.
5. Example: BCl₃.

sp³ hybridisation:

1. One s and three p orbitals mix.
2. Four sp³ hybrid orbitals form.
3. Shape is tetrahedral.
4. Bond angle is 109.5°.
5. Example: CH₄.

Q26. Explain the formation of CH₄, NH₃, and H₂O on the basis of sp³ hybridisation.
Ans. In CH₄, carbon undergoes sp³ hybridisation.

Four sp³ orbitals overlap with 1s orbitals of four hydrogen atoms.

All four bond pairs are equivalent.

Shape: tetrahedral
Bond angle: 109.5°

In NH₃, nitrogen undergoes sp³ hybridisation.

Three sp³ orbitals form N-H bonds. One sp³ orbital contains a lone pair.

Shape: trigonal pyramidal
Bond angle: 107°

In H₂O, oxygen undergoes sp³ hybridisation.

Two sp³ orbitals form O-H bonds. Two sp³ orbitals contain lone pairs.

Shape: bent
Bond angle: 104.5°

Q27. Describe the hybridisation in PCl₅ and explain why axial bonds are longer than equatorial bonds.
Ans. Phosphorus in PCl₅ undergoes sp³d hybridisation.

Five sp³d orbitals point toward the corners of a trigonal bipyramid.

There are three equatorial P-Cl bonds and two axial P-Cl bonds.

Axial bonds experience more repulsion because they are at 90° to three equatorial bonds.

This extra repulsion makes axial bonds slightly longer and weaker than equatorial bonds.

Q28. Is there any change in hybridisation of B and N in BF₃ + NH₃ → F₃B.NH₃?
Ans. In NH₃, nitrogen is sp³ hybridised.

It remains sp³ hybridised in the product.

In BF₃, boron is sp² hybridised.

After accepting the lone pair from nitrogen, boron becomes sp³ hybridised.

So, only boron’s hybridisation changes.

Q29. Which hybrid orbitals are used by carbon atoms in CH₃-CH₃, CH₃-CH=CH₂, and CH₃-CHO?
Ans.

CH₃-CH₃:

Both carbon atoms are sp³ hybridised.

CH₃-CH=CH₂:

The CH₃ carbon is sp³ hybridised.

The two double-bonded carbon atoms are sp² hybridised.

CH₃-CHO:

The CH₃ carbon is sp³ hybridised.

The carbonyl carbon is sp² hybridised.

Dipole Moment, Bond Polarity and Bond Parameters Questions

Dipole moment questions test both bond polarity and molecular shape.

A molecule with polar bonds can still have zero dipole moment if its geometry is symmetrical.

Q30. Explain why BeH₂ has zero dipole moment even though Be-H bonds are polar.
Ans. BeH₂ is a linear molecule.

The two Be-H bond dipoles are equal in magnitude and opposite in direction.

They cancel each other completely.

So, the net dipole moment is zero.

Q31. Which has a higher dipole moment: NH₃ or NF₃? Give a reason.
Ans. NH₃ has higher dipole moment than NF₃.

In NH₃, the resultant N-H bond dipoles and lone pair dipole act in the same direction.

In NF₃, the N-F bond dipoles oppose the lone pair dipole.

So, the net dipole moment of NF₃ is smaller.

Q32. What is the relationship between bond order, bond length, and bond enthalpy?
Ans. As bond order increases, bond length decreases.

As bond order increases, bond enthalpy increases.

A higher bond order means a stronger and shorter bond.

Example: N₂ has bond order 3, so it has very high bond enthalpy and short bond length.

Q33. Arrange N-H, F-H, C-H, and O-H bonds in increasing order of ionic character.
Ans. Ionic character depends on electronegativity difference.

Greater electronegativity difference means greater ionic character.

Order:

C-H < N-H < O-H < F-H

Molecular Orbital Theory Questions and Answers

Molecular orbital theory questions test bond order, magnetic nature, and stability.

For MOT questions, fill electrons correctly before calculating bond order.

Chemical Bonding and Molecular Structure Extra Questions

Q34. Write the electronic configuration of O₂ and explain its magnetic behaviour.
Ans. O₂ has 16 electrons.

Its molecular orbital configuration is:

σ1s² σ1s² σ2s² σ2s² σ2pz² π2px² = π2py² π2px¹ = π2py¹

Bond order = 1/2(10 - 6)

= 2

O₂ has two unpaired electrons in antibonding pi orbitals.

So, O₂ is paramagnetic.

Q35. Use molecular orbital theory to explain why He₂ does not exist.
Ans. Each helium atom has configuration 1s².

He₂ would have four electrons.

Molecular orbital configuration:

σ1s² σ*1s²

Bond order = 1/2(2 - 2)

= 0

Bond order zero means no net bond formation.

So, He₂ does not exist as a stable molecule.

Q36. Compare the bond energy and magnetic character of O₂⁺ and O₂⁻.
Ans. O₂⁺ has 15 electrons.

Bond order of O₂⁺ = 2.5

It has one unpaired electron, so it is paramagnetic.

O₂⁻ has 17 electrons.

Bond order of O₂⁻ = 1.5

It also has one unpaired electron, so it is paramagnetic.

O₂⁺ has higher bond order, so it has higher bond energy and greater stability.

Q37. Write the MO bond orders of N₂, N₂⁺, N₂⁻, and N₂²⁺. Arrange them in order of stability.
Ans.

Higher bond order means greater stability.

Order of stability:

N₂ > N₂⁺ = N₂⁻ > N₂²⁺

Q38. Write the conditions required for linear combination of atomic orbitals to form molecular orbitals.
Ans. Conditions for LCAO:

1. Atomic orbitals must have the same or nearly the same energy.
2. Atomic orbitals must have the same symmetry about the internuclear axis.
3. Atomic orbitals must overlap effectively.

Chemical Bonding and Molecular Structure Extra Questions with Answers

Chemical bonding and molecular structure extra questions usually combine VSEPR, hybridisation, hydrogen bonding, and MOT.

These questions help students practise reasoning beyond direct definitions.

Q39. Explain the formation of H₂ on the basis of valence bond theory.
Ans. When two hydrogen atoms approach each other, attractive and repulsive forces operate.

Attractive forces act between the nucleus of one atom and the electron of the other atom.

Repulsive forces act between two nuclei and between two electrons.

When attractive forces become greater than repulsive forces, potential energy decreases.

At a distance of 74 pm, minimum energy is reached.

The 1s orbitals overlap, electrons pair with opposite spins, and stable H₂ forms.

Q40. Classify H₂O, HOCl, BeCl₂, and Cl₂O as linear or non-linear.
Ans.

BeCl₂ is linear because Be has two bond pairs and no lone pair.

The others are bent due to lone pairs on the central atom.

Q41. Explain the difference between sigma and pi bonds.
Ans.

Q42. Elements X, Y, and Z have 4, 5, and 7 valence electrons respectively. Write molecular formulae for their hydrogen compounds. Which compound has the highest dipole moment?
Ans. Element X has 4 valence electrons.

Hydrogen compound = XH₄

Element Y has 5 valence electrons.

Hydrogen compound = H₃Y

Element Z has 7 valence electrons.

Hydrogen compound = HZ

HZ has the highest dipole moment because Z is likely a halogen and has high electronegativity.

Q43. Explain why intramolecular hydrogen bonding occurs in o-nitrophenol and not in p-nitrophenol.
Ans. In o-nitrophenol, the -OH group and -NO₂ group are close to each other.

So, hydrogen bonding can occur within the same molecule.

This is intramolecular hydrogen bonding.

In p-nitrophenol, the two groups are far apart.

So, hydrogen bonding occurs between different molecules.

This is intermolecular hydrogen bonding.

Chemical Bonding and Molecular Structure Previous Year Questions

Chemical bonding and molecular structure previous year questions often test shape, hybridisation, dipole moment, MOT, and bond order.

Practise these after finishing the concept-wise question bank.

Q44. Isostructural species have the same shape and hybridisation. Which pair is isostructural? (i) NF₃ and BF₃ (ii) BF₄⁻ and NH₄⁺ (iii) BCl₃ and BrCl₃ (iv) NH₃ and NO₃⁻
Ans. Correct answer: (ii) BF₄⁻ and NH₄⁺

Both have four bond pairs, sp³ hybridisation, and tetrahedral geometry.

Q45. Which species has the highest dipole moment? (i) CO₂ (ii) HI (iii) H₂O (iv) SO₂
Ans. Correct answer: (iii) H₂O

H₂O has a bent shape and two polar O-H bonds.

The resultant dipole moment is high.

CO₂ has zero dipole moment because it is linear.

Q46. Write the correct bond order sequence for O₂, O₂⁺, and O₂⁻.
Ans. Bond orders:

O₂⁻ = 1.5
O₂ = 2
O₂⁺ = 2.5

Increasing order of bond order:

O₂⁻ < O₂ < O₂⁺

Q47. Which species does not contain unpaired electrons? (i) N₂⁺ (ii) O₂ (iii) O₂²⁻ (iv) B₂
Ans. Correct answer: (iii) O₂²⁻

O₂²⁻ has all electrons paired in molecular orbitals.

So, it is diamagnetic.

Q48. What is the hybridisation of nitrogen in NO₂⁺, NO₃⁻, and NH₄⁺ respectively?
Ans. NO₂⁺ is linear, so nitrogen is sp hybridised.

NO₃⁻ is trigonal planar, so nitrogen is sp² hybridised.

NH₄⁺ is tetrahedral, so nitrogen is sp³ hybridised.

Answer: sp, sp², sp³

Assertion-Reason and MCQs from Chemical Bonding and Molecular Structure

MCQs and assertion-reason questions test precision. Check shape, lone pairs, and hybridisation before answering.

MCQs from Chemical Bonding and Molecular Structure

Q49. Which bond angle corresponds to sp² hybridisation? (i) 90° (ii) 120° (iii) 180° (iv) 109.5°
Ans. Correct answer: (ii) 120°

sp² hybrid orbitals are arranged in trigonal planar geometry.

Q50. Which of the following species has tetrahedral geometry? (i) BH₄⁻ (ii) NH₂⁻ (iii) CO₃²⁻ (iv) H₃O⁺
Ans. Correct answer: (i) BH₄⁻

BH₄⁻ has four bond pairs and no lone pair on boron.

So, it has tetrahedral geometry.

Q51. Assertion (A): The bond angle in H₂O is less than in NH₃. Reason (R): Lone pair-lone pair repulsion in H₂O is greater than lone pair-bond pair repulsion in NH₃.
Ans. Both assertion and reason are true, and the reason correctly explains the assertion.

H₂O has two lone pairs on oxygen, while NH₃ has one lone pair on nitrogen.

Greater lone pair repulsion in H₂O compresses the bond angle more.

Q52. Which statement is not correct from the viewpoint of molecular orbital theory? (i) Be₂ is not stable. (ii) He₂ is unstable but He₂⁺ can exist. (iii) Bond strength of N₂ is maximum among homonuclear diatomic molecules of the second period. (iv) In N₂, σ2pz is lower in energy than π2px and π2py.
Ans. Correct answer: (iv)

For B₂, C₂, and N₂, the π2px and π2py orbitals are lower in energy than σ2pz.

So, statement (iv) is incorrect.

Important Formulas from Chemical Bonding and Molecular Structure