Chemical Bonding MCQs – VSEPR and Hybridisation Class 11
Chemical Bonding MCQs – VSEPR & Hybridisation (Class 11 Chemistry)
Course: CBSE Class 11 Chemistry – MCQs with Answers and Explanations
Section: Inorganic Chemistry – MCQ Titles
The following 50 Multiple Choice Questions (MCQs) are strictly aligned with the NCERT chapter “Chemical Bonding and Molecular Structure”, with special focus on VSEPR theory and hybridisation. The questions are arranged section-wise and include clear, concept-clearing explanations, making them ideal for CBSE Class 11 board exams and concept revision.
Section A: Basics of Chemical Bonding
1. Chemical bonding refers to
A. attraction between nuclei
B. force holding atoms together in molecules
C. repulsion between electrons
D. breaking of atoms
Answer: B
Explanation: Chemical bonds keep atoms together to form molecules or ions.
2. A chemical bond is formed due to
A. decrease in potential energy
B. increase in kinetic energy
C. loss of mass
D. increase in entropy only
Answer: A
Explanation: Bond formation lowers the overall potential energy of the system.
3. Which bond involves sharing of electrons?
A. Ionic bond
B. Metallic bond
C. Covalent bond
D. Hydrogen bond
Answer: C
Explanation: Covalent bonding involves sharing of electron pairs.
Section B: Lewis Structures and Octet Rule
4. The octet rule states that atoms tend to achieve
A. 2 electrons
B. 6 electrons
C. 8 electrons
D. 10 electrons
Answer: C
Explanation: Atoms become stable by attaining noble gas configuration.
5. Which element does NOT follow the octet rule?
A. Carbon
B. Oxygen
C. Neon
D. Hydrogen
Answer: D
Explanation: Hydrogen follows the duplet rule (2 electrons).
6. Expanded octet is shown by elements of
A. 1st period
B. 2nd period
C. 3rd period and beyond
D. noble gases only
Answer: C
Explanation: Availability of d-orbitals allows expanded octet.
Section C: VSEPR Theory
7. VSEPR theory is used to predict
A. bond length
B. bond energy
C. molecular shape
D. bond polarity
Answer: C
Explanation: VSEPR explains molecular geometry based on electron pair repulsions.
8. According to VSEPR theory, electron pairs repel each other because they
A. are negatively charged
B. are positively charged
C. have mass
D. are neutral
Answer: A
Explanation: Electron pairs are negatively charged and repel each other.
9. Which repulsion is strongest?
A. Bond pair–bond pair
B. Lone pair–bond pair
C. Lone pair–lone pair
D. All are equal
Answer: C
Explanation: Lone pairs occupy more space, causing maximum repulsion.
10. Shape of BeCl₂ according to VSEPR theory is
A. bent
B. trigonal planar
C. linear
D. tetrahedral
Answer: C
Explanation: Two bond pairs and no lone pairs give linear shape.
Section D: Shapes of Molecules (VSEPR Based)
11. Shape of methane (CH₄) is
A. square planar
B. tetrahedral
C. trigonal planar
D. linear
Answer: B
Explanation: Four bond pairs around carbon arrange tetrahedrally.
12. Shape of ammonia (NH₃) is
A. tetrahedral
B. trigonal planar
C. trigonal pyramidal
D. bent
Answer: C
Explanation: One lone pair distorts tetrahedral geometry.
13. Shape of water molecule is
A. linear
B. bent
C. trigonal planar
D. tetrahedral
Answer: B
Explanation: Two lone pairs cause bent structure.
14. Which molecule is linear?
A. H₂O
B. NH₃
C. CO₂
D. CH₄
Answer: C
Explanation: Two bond pairs around carbon give linear geometry.
Section E: Bond Angle and Lone Pair Effect
15. Bond angle in CH₄ is
A. 90°
B. 104.5°
C. 109.5°
D. 120°
Answer: C
Explanation: Tetrahedral angle is 109.5°.
16. Bond angle in NH₃ is less than CH₄ due to
A. higher electronegativity
B. lone pair–bond pair repulsion
C. larger atomic size
D. hydrogen bonding
Answer: B
Explanation: Lone pair causes greater repulsion.
17. Bond angle in H₂O is
A. 109.5°
B. 107°
C. 104.5°
D. 90°
Answer: C
Explanation: Two lone pairs reduce bond angle significantly.
Section F: Concept of Hybridisation
18. Hybridisation is the mixing of
A. atomic orbitals of same energy
B. molecular orbitals
C. s and p orbitals only
D. d and f orbitals only
Answer: A
Explanation: Hybridisation involves mixing of atomic orbitals of similar energy.
19. Hybridisation of carbon in methane is
A. sp
B. sp²
C. sp³
D. dsp²
Answer: C
Explanation: Four equivalent sp³ orbitals are formed.
20. Number of hybrid orbitals formed in sp² hybridisation is
A. 2
B. 3
C. 4
D. 5
Answer: B
Explanation: One s and two p orbitals mix to form three sp² orbitals.
Section G: Types of Hybridisation and Shapes
21. sp hybridisation results in
A. linear geometry
B. trigonal planar geometry
C. tetrahedral geometry
D. square planar geometry
Answer: A
Explanation: Two sp orbitals arrange linearly at 180°.
22. Hybridisation of carbon in ethene (C₂H₄) is
A. sp
B. sp²
C. sp³
D. dsp
Answer: B
Explanation: Presence of double bond indicates sp² hybridisation.
23. Hybridisation of carbon in ethyne (C₂H₂) is
A. sp³
B. sp²
C. sp
D. dsp²
Answer: C
Explanation: Triple bond involves sp hybridisation.
24. Shape associated with sp² hybridisation is
A. linear
B. trigonal planar
C. tetrahedral
D. bent
Answer: B
Explanation: sp² orbitals lie in one plane at 120°.
Section H: Hybridisation of Central Atoms
25. Hybridisation of nitrogen in NH₃ is
A. sp
B. sp²
C. sp³
D. dsp
Answer: C
Explanation: Three bond pairs and one lone pair require sp³ hybridisation.
26. Hybridisation of oxygen in H₂O is
A. sp
B. sp²
C. sp³
D. dsp²
Answer: C
Explanation: Two bond pairs and two lone pairs → sp³.
27. Hybridisation of carbon in CO₂ is
A. sp³
B. sp²
C. sp
D. dsp
Answer: C
Explanation: Two regions of electron density → sp hybridisation.
Section I: Sigma and Pi Bonds
28. Sigma bond is formed by
A. sidewise overlap
B. end-to-end overlap
C. p–p overlap only
D. d–d overlap only
Answer: B
Explanation: Head-on overlap forms sigma bonds.
29. Pi bond is formed by
A. end-to-end overlap
B. s–s overlap
C. sidewise overlap
D. sp overlap
Answer: C
Explanation: Sidewise overlap of p orbitals forms π bonds.
30. Number of sigma bonds in ethene is
A. 3
B. 4
C. 5
D. 6
Answer: C
Explanation: Ethene has 5 sigma and 1 pi bond.
Section J: Relation Between Hybridisation and Geometry
31. Square planar geometry corresponds to
A. sp³
B. sp²
C. dsp²
D. sp
Answer: C
Explanation: dsp² hybridisation gives square planar shape.
32. Which molecule shows sp³ hybridisation but non-tetrahedral shape?
A. CH₄
B. NH₃
C. CCl₄
D. SiH₄
Answer: B
Explanation: Lone pair distorts geometry.
33. Which molecule has trigonal planar shape?
A. BF₃
B. NH₃
C. H₂O
D. PCl₅
Answer: A
Explanation: BF₃ has three bond pairs and no lone pair.
Section K: Conceptual NCERT-Based MCQs
34. Greater the s-character, the bond is
A. longer and weaker
B. shorter and stronger
C. non-polar
D. ionic
Answer: B
Explanation: s-orbitals are closer to nucleus, giving stronger bonds.
35. Order of increasing s-character is
A. sp < sp² < sp³
B. sp³ < sp² < sp
C. sp² < sp³ < sp
D. sp < sp³ < sp²
Answer: B
Explanation: sp has 50% s-character, sp² 33%, sp³ 25%.
36. Which hybridisation gives maximum bond angle?
A. sp³
B. sp²
C. sp
D. dsp²
Answer: C
Explanation: Linear geometry has 180° bond angle.
Section L: Application and Analytical MCQs
37. Which molecule has zero dipole moment?
A. NH₃
B. H₂O
C. CO₂
D. SO₂
Answer: C
Explanation: Linear and symmetrical molecule.
38. Which molecule is bent in shape?
A. CO₂
B. BF₃
C. SO₂
D. BeCl₂
Answer: C
Explanation: Lone pair on sulphur causes bent shape.
39. Hybridisation of boron in BF₃ is
A. sp
B. sp²
C. sp³
D. dsp²
Answer: B
Explanation: Three bond pairs → sp² hybridisation.
Section M: High-Order Thinking MCQs
40. Lone pairs reduce bond angle because they
A. occupy less space
B. attract bond pairs
C. repel bond pairs more strongly
D. form pi bonds
Answer: C
Explanation: Lone pairs exert stronger repulsion.
41. Which molecule violates octet rule?
A. NH₃
B. BF₃
C. H₂O
D. CH₄
Answer: B
Explanation: Boron has only six electrons.
42. Expanded octet is shown by
A. NH₃
B. H₂O
C. PCl₅
D. BF₃
Answer: C
Explanation: Phosphorus accommodates more than 8 electrons.
Section N: Final Conceptual MCQs
43. Hybridisation concept explains
A. molecular mass
B. molecular shape
C. atomic number
D. periodicity
Answer: B
Explanation: Hybridisation helps predict geometry.
44. Which molecule has trigonal bipyramidal geometry?
A. NH₃
B. BF₃
C. PCl₅
D. SF₄
Answer: C
Explanation: Five bond pairs around central atom.
45. Hybridisation of sulphur in SF₆ is
A. sp³
B. sp³d
C. sp³d²
D. dsp²
Answer: C
Explanation: Six equivalent orbitals are formed.
46. Which has the highest bond angle?
A. CH₄
B. NH₃
C. H₂O
D. CO₂
Answer: D
Explanation: Linear molecule has 180° bond angle.
47. Number of pi bonds in ethyne is
A. 1
B. 2
C. 3
D. 4
Answer: B
Explanation: One sigma and two pi bonds form triple bond.
48. Which bond is stronger?
A. single bond
B. double bond
C. triple bond
D. hydrogen bond
Answer: C
Explanation: Triple bond has maximum bond energy.
49. Which hybridisation gives tetrahedral shape?
A. sp
B. sp²
C. sp³
D. dsp²
Answer: C
Explanation: sp³ orbitals arrange tetrahedrally.
50. VSEPR theory mainly considers
A. nucleus–nucleus repulsion
B. electron–electron repulsion
C. proton–electron attraction
D. neutron interaction
Answer: B
Explanation: Electron pair repulsions determine shape.
✅ Completion Note
This completes a fully NCERT-aligned, CBSE-standard set of 50 MCQs on Chemical Bonding (VSEPR Theory & Hybridisation), ideal for concept clarity, structured revision, and board exam preparation.