Class 11 Chemistry

Chapter 4: Chemical Bonding and Molecular Structure

Important Practice Questions (MCQ, Fill in the Blanks, True/False, Match the Following & Very Short Questions)

A. Multiple Choice Questions (MCQs)

1. The force that holds atoms together in a molecule is called:

A) Nuclear force
B) Chemical bond
C) Gravitational force
D) Magnetic force

Answer: B) Chemical bond

2. The tendency of atoms to achieve noble gas configuration is explained by:

A) VSEPR theory
B) Octet rule
C) MOT
D) Hybridisation

Answer: B) Octet rule

3. The maximum number of electrons present in the valence shell according to octet rule is:

A) 2
B) 4
C) 8
D) 18

Answer: C) 8

4. Which molecule follows duplet rule?

A) CH₄
B) H₂
C) NH₃
D) H₂O

Answer: B) H₂

5. The bond formed by transfer of electrons is:

A) Covalent bond
B) Ionic bond
C) Hydrogen bond
D) Metallic bond

Answer: B) Ionic bond

6. NaCl contains:

A) Covalent bond
B) Ionic bond
C) Hydrogen bond
D) Coordinate bond

Answer: B) Ionic bond

7. The number of electrons shared in a single covalent bond is:

A) 1
B) 2
C) 4
D) 6

Answer: B) 2

8. Which molecule contains a triple bond?

A) O₂
B) N₂
C) H₂
D) Cl₂

Answer: B) N₂

9. Bond order of nitrogen molecule (N₂) is:

A) 1
B) 2
C) 3
D) 4

Answer: C) 3

10. The strongest bond among the following is:

A) C–C
B) C=C
C) C≡C
D) C–H

Answer: C) C≡C

Lewis Structure & Formal Charge

11. Lewis symbols represent:

A) Atomic mass
B) Valence electrons
C) Neutrons
D) Protons

Answer: B) Valence electrons

12. Formal charge is calculated by:

A) Atomic number
B) Mass number
C) Valence electrons and bonding electrons
D) Number of neutrons

Answer: C

13. Resonance structures have:

A) Different atoms arrangement
B) Same molecular formula but different atoms
C) Same arrangement of atoms but different electron arrangement
D) Different molecular formula

Answer: C

14. Resonance increases:

A) Energy
B) Instability
C) Stability
D) Bond length

Answer: C

Bond Polarity & Dipole Moment

15. Dipole moment is represented by:

A) μ
B) λ
C) α
D) β

Answer: A) μ

16. Dipole moment formula is:

A) μ = q/r
B) μ = q × r
C) μ = q + r
D) μ = q – r

Answer: B

17. Which molecule has zero dipole moment?

A) H₂O
B) NH₃
C) CO₂
D) HF

Answer: C) CO₂

18. The shape of CO₂ is:

A) Bent
B) Linear
C) Pyramidal
D) Tetrahedral

Answer: B) Linear

VSEPR Theory

19. VSEPR theory is based on:

A) Attraction between electrons
B) Repulsion between electron pairs
C) Nuclear forces
D) Proton movement

Answer: B

20. The order of repulsion is:

A) BP-BP > LP-BP > LP-LP
B) LP-LP > LP-BP > BP-BP
C) LP-BP > BP-BP > LP-LP
D) All equal

Answer: B

21. Shape of CH₄ is:

A) Linear
B) Bent
C) Tetrahedral
D) Pyramidal

Answer: C

22. Bond angle in methane is:

A) 90°
B) 107°
C) 109.5°
D) 120°

Answer: C

23. Shape of NH₃ is:

A) Linear
B) Trigonal pyramidal
C) Trigonal planar
D) Octahedral

Answer: B

24. Bond angle in water molecule is:

A) 180°
B) 120°
C) 109.5°
D) 104.5°

Answer: D

Hybridisation

25. Hybridisation of carbon in methane is:

A) sp
B) sp²
C) sp³
D) sp³d

Answer: C

26. Hybridisation in BeCl₂ is:

A) sp
B) sp²
C) sp³
D) sp³d

Answer: A

27. BF₃ has:

A) sp hybridisation
B) sp² hybridisation
C) sp³ hybridisation
D) sp³d hybridisation

Answer: B

28. Hybridisation of phosphorus in PCl₅ is:

A) sp²
B) sp³
C) sp³d
D) sp³d²

Answer: C

29. Hybridisation of sulphur in SF₆ is:

A) sp
B) sp²
C) sp³
D) sp³d²

Answer: D

Sigma and Pi Bonds

30. Sigma bond is formed by:

A) Sidewise overlap
B) Head-on overlap
C) No overlap
D) Electron transfer

Answer: B

31. Pi bond is formed by:

A) Head-on overlap
B) Sidewise overlap
C) Electron transfer
D) Ionic attraction

Answer: B

32. Double bond consists of:

A) Two sigma bonds
B) Two pi bonds
C) One sigma and one pi bond
D) Three sigma bonds

Answer: C

33. Triple bond contains:

A) 3σ
B) 2σ + 1π
C) 1σ + 2π
D) 3π

Answer: C

Molecular Orbital Theory

34. Bond order formula is:

A) (Nb + Na)/2
B) (Nb – Na)/2
C) Nb × Na
D) Nb/Na

Answer: B

35. O₂ is:

A) Diamagnetic
B) Paramagnetic
C) Non-magnetic
D) Ionic

Answer: B

36. Molecule with bond order zero:

A) H₂
B) O₂
C) He₂
D) N₂

Answer: C

Hydrogen Bonding

37. Hydrogen bonding occurs mainly with:

A) C, S, P
B) F, O, N
C) Na, Mg, Al
D) Cl, Br, I

Answer: B

38. Example of intramolecular hydrogen bonding:

A) Water
B) HF
C) o-nitrophenol
D) NH₃

Answer: C

B. Fill in the Blanks

Chemical bonds are formed to achieve ______ configuration.
Answer: noble gas
The outermost electrons are called ______ electrons.
Answer: valence
Ionic bond is formed by ______ of electrons.
Answer: transfer
Covalent bond is formed by ______ of electrons.
Answer: sharing
The shape of CH₄ is ______.
Answer: tetrahedral
The bond angle in CH₄ is ______.
Answer: 109.5°
The hybridisation of carbon in ethene is ______.
Answer: sp²
The hybridisation of carbon in ethyne is ______.
Answer: sp
CO₂ molecule has ______ dipole moment.
Answer: zero
The strongest type of overlap is ______ overlap.
Answer: sigma
A double bond contains one sigma and one ______ bond.
Answer: pi
Oxygen molecule is ______ due to unpaired electrons.
Answer: paramagnetic
Bond order of H₂ molecule is ______.
Answer: 1
SF₆ has ______ hybridisation.
Answer: sp³d²
Hydrogen bonding occurs between hydrogen and ______ electronegative atoms.
Answer: highly

C. True or False

Ionic bonds are formed by sharing of electrons.
❌ False
Covalent bonds are directional.
✅ True
CO₂ is a polar molecule.
❌ False
H₂O has bent geometry.
✅ True
Lone pair repulsion is greater than bond pair repulsion.
✅ True
Sigma bond is weaker than pi bond.
❌ False
O₂ is paramagnetic.
✅ True
CH₄ has sp³ hybridisation.
✅ True
BF₃ has trigonal planar shape.
✅ True
He₂ is stable.
❌ False

D. Match the Following

Column A	Column B
CH₄	Tetrahedral
BF₃	Trigonal planar
BeCl₂	Linear
NH₃	Pyramidal
H₂O	Bent
PCl₅	Trigonal bipyramidal
SF₆	Octahedral

E. Very Short Answer Questions

What is octet rule?
Answer: Atoms gain, lose or share electrons to achieve eight electrons in their valence shell.
What is bond order?
Answer: Number of electron pairs shared between two atoms.
Why is O₂ paramagnetic?
Answer: Due to presence of unpaired electrons.
What is hybridisation?
Answer: Mixing of atomic orbitals to form equivalent hybrid orbitals.
What is hydrogen bonding?
Answer: Attraction between H attached to F/O/N and another electronegative atom.

Question Bank – Part 2

Assertion-Reason, Case-Based Questions, More MCQs & Numericals

A. Assertion–Reason Questions

Choose the correct option:

A) Both Assertion and Reason are true and Reason is the correct explanation.
B) Both Assertion and Reason are true but Reason is not the correct explanation.
C) Assertion is true but Reason is false.
D) Assertion is false but Reason is true.

1.

Assertion: NaCl has a high melting point.
Reason: NaCl has strong electrostatic attraction between ions.

Answer: A

2.

Assertion: CO₂ is a non-polar molecule.
Reason: The two C=O bond dipoles cancel each other.

Answer: A

3.

Assertion: NH₃ has a lower bond angle than CH₄.
Reason: Nitrogen has a lone pair of electrons.

Answer: A

4.

Assertion: H₂O is a polar molecule.
Reason: Water has a linear structure.

Answer: C

5.

Assertion: O₂ is paramagnetic.
Reason: O₂ contains two unpaired electrons.

Answer: A

6.

Assertion: Sigma bonds are stronger than pi bonds.
Reason: Sigma bonds are formed by head-on overlap.

Answer: A

7.

Assertion: BF₃ is non-polar.
Reason: BF₃ has trigonal planar geometry.

Answer: A

8.

Assertion: CH₄ has tetrahedral geometry.
Reason: Carbon undergoes sp³ hybridisation.

Answer: A

9.

Assertion: He₂ molecule does not exist.
Reason: Its bond order is zero.

Answer: A

10.

Assertion: Hydrogen bonding increases boiling point.
Reason: Hydrogen bonds increase intermolecular attraction.

Answer: A

B. More MCQs

1. Which element has the highest electronegativity?

A) Oxygen
B) Nitrogen
C) Fluorine
D) Chlorine

Answer: C) Fluorine

2. Which compound has maximum ionic character?

A) LiF
B) LiI
C) CsI
D) CsF

Answer: D) CsF

3. The central atom in NH₃ is:

A) Hydrogen
B) Nitrogen
C) Oxygen
D) Carbon

Answer: B

4. The number of lone pairs present in H₂O is:

A) 1
B) 2
C) 3
D) 4

Answer: B

5. Which molecule has a linear shape?

A) NH₃
B) H₂O
C) CO₂
D) CH₄

Answer: C

6. Which molecule has sp hybridisation?

A) CH₄
B) BeCl₂
C) BF₃
D) NH₃

Answer: B

7. Which molecule contains both sigma and pi bonds?

A) H₂
B) Cl₂
C) O₂
D) He₂

Answer: C

8. The bond angle in BF₃ is:

A) 90°
B) 109.5°
C) 120°
D) 180°

Answer: C

9. The hybridisation of nitrogen in NH₃ is:

A) sp
B) sp²
C) sp³
D) sp³d

Answer: C

10. Which molecule has zero dipole moment?

A) NH₃
B) H₂O
C) CO₂
D) HF

Answer: C

11. Lattice enthalpy depends on:

A) Size of ions
B) Charge of ions
C) Distance between ions
D) All of these

Answer: D

12. Which molecule has expanded octet?

A) BF₃
B) BeCl₂
C) SF₆
D) CH₄

Answer: C

13. The bond order of O₂ is:

A) 1
B) 2
C) 3
D) 4

Answer: B

14. The number of sigma bonds in ethane (C₂H₆) is:

A) 6
B) 7
C) 8
D) 9

Answer: B

Explanation:
C–C = 1σ
Six C–H = 6σ
Total = 7σ

15. The number of pi bonds in ethyne (C₂H₂) is:

A) 0
B) 1
C) 2
D) 3

Answer: C

C. Case-Based Questions

Case Study 1

A student studies methane molecule. Carbon has four valence electrons and combines with four hydrogen atoms.

Questions:

1. Hybridisation of carbon in methane is:

A) sp
B) sp²
C) sp³
D) sp³d

Answer: C

2. Shape of methane is:

A) Linear
B) Tetrahedral
C) Bent
D) Pyramidal

Answer: B

3. Bond angle in methane is:

Answer: 109.5°

Case Study 2

Oxygen molecule has 16 electrons. According to molecular orbital theory, electrons occupy bonding and antibonding orbitals.

Questions:

1. Oxygen molecule is:

A) Diamagnetic
B) Paramagnetic
C) Ionic
D) Non-polar only

Answer: B

2. Reason for paramagnetism is:

Answer: Presence of two unpaired electrons.

3. Bond order of O₂ is:

Answer: 2

D. Numerical Type Questions

1. Calculate bond order of H₂ molecule.

Given:

Bonding electrons = 2
Antibonding electrons = 0

Formula: $BO=\frac{N_b-N_a}{2}$ BO=2Nb−Na $BO=\frac{2-0}{2}=1$ BO=22−0=1

Answer: Bond order = 1

2. Calculate bond order of He₂.

Bonding electrons = 2
Antibonding electrons = 2 $BO=\frac{2-2}{2}=0$ BO=22−2=0

Answer: He₂ is unstable

3. Calculate formal charge on oxygen atom.

Formula: $FC=V-L-\frac{B}{2}$ FC=V−L−2B

Where:

V = Valence electrons
L = Lone pair electrons
B = Bonding electrons

4. A molecule has 3 bond pairs and 1 lone pair. Predict shape.

Electron pairs = 4

Hybridisation = sp³

Shape = Trigonal pyramidal

Example: NH₃

5. A molecule has 2 bond pairs and 2 lone pairs. Predict shape.

Hybridisation = sp³

Shape = Bent

Example: H₂O

E. One Word / One Line Questions

Theory explaining molecular shape?
VSEPR theory
Strongest covalent bond?
Sigma bond
Weakest covalent bond?
Pi bond
Hybridisation of BeCl₂?
sp
Hybridisation of BF₃?
sp²
Hybridisation of CH₄?
sp³
Hybridisation of PCl₅?
sp³d
Hybridisation of SF₆?
sp³d²
Magnetic nature of O₂?
Paramagnetic
Shape of CO₂?
Linear

F. Important Board Exam Questions

1. Explain octet rule with examples.

2. Draw Lewis structures of:

NH₃
H₂O
CO₂
BF₃

3. Explain VSEPR theory.

4. Explain hybridisation with examples.

5. Explain sigma and pi bonds.

6. Explain molecular orbital theory.

7. Why is O₂ paramagnetic?

8. Explain hydrogen bonding and its types.

9. Calculate bond order using MOT.

10. Explain factors affecting lattice enthalpy.

Question Bank – Part 3

HOTS Questions, Previous Year Pattern Questions, Diagram Practice & Advanced Revision Set

A. HOTS (Higher Order Thinking Skill) Questions

1. Why do atoms form chemical bonds?

Answer:
Atoms form chemical bonds to achieve a stable electronic configuration and lower their energy.

2. Why are noble gases chemically inactive?

Answer:
Noble gases have completely filled valence shells, making them highly stable and less reactive.

3. Why is NaCl solid but HCl is a gas at room temperature?

Answer:

NaCl has strong ionic attraction between ions.
HCl is a covalent molecule with weaker intermolecular forces.

4. Why does MgO have higher lattice enthalpy than NaCl?

Answer:
MgO contains Mg²⁺ and O²⁻ ions with higher charges compared to Na⁺ and Cl⁻, resulting in stronger attraction.

5. Why is BF₃ electron deficient?

Answer:
Boron has only six electrons around it after forming three B–F bonds, so it does not complete its octet.

6. Why does NH₃ have a smaller bond angle than CH₄?

Answer:
NH₃ contains one lone pair on nitrogen. Lone pair–bond pair repulsion reduces the bond angle from 109.5° to 107°.

7. Why does H₂O have a smaller bond angle than NH₃?

Answer:
H₂O contains two lone pairs, causing greater repulsion and reducing the bond angle to 104.5°.

8. Why is CO₂ non-polar while H₂O is polar?

Answer:

CO₂ has a linear shape, so bond dipoles cancel.
H₂O has a bent shape, so dipoles do not cancel.

9. Why is N₂ more stable than O₂?

Answer:
N₂ has bond order 3, while O₂ has bond order 2. Higher bond order makes N₂ stronger and more stable.

10. Why does oxygen show paramagnetism?

Answer:
O₂ contains two unpaired electrons in antibonding molecular orbitals.

B. Important Structure-Based Questions

Draw Lewis Structure of H₂O

Steps:

Oxygen has 6 valence electrons.
Two hydrogen atoms contribute 2 electrons.
Total electrons = 8.

Structure:

   ..
H : O : H
   ..

Shape:

Bent

Hybridisation:

sp³

Draw Lewis Structure of NH₃

Nitrogen:

5 valence electrons

Three H atoms:

3 electrons

Total:

8 electrons

Structure:

    ..
H — N — H
    |
    H

Shape:

Trigonal pyramidal

Draw Lewis Structure of CO₂

Structure:

O = C = O

Shape:

Linear

Bond angle:

180°

C. Reason Based Questions

1. Why is Fajan’s rule important?

Answer:
It predicts the covalent character present in ionic compounds.

2. Why does smaller cation increase covalent character?

Answer:
Small cations have greater polarising power and distort the electron cloud of anions.

3. Why is LiI more covalent than LiF?

Answer:
I⁻ is larger and more easily polarised than F⁻.

4. Why are ionic compounds soluble in water?

Answer:
Water molecules are polar and attract ions, helping dissolve ionic compounds.

5. Why do ionic compounds conduct electricity only in molten state?

Answer:
In molten state ions become free to move and carry electric current.

D. Assertion Reason Practice – Advanced

1.

Assertion: Bond length decreases as bond order increases.

Reason: More shared electron pairs increase attraction between atoms.

Answer: A

2.

Assertion: H₂O has higher boiling point than H₂S.

Reason: H₂O forms strong hydrogen bonds.

Answer: A

3.

Assertion: CO₂ has zero dipole moment.

Reason: Carbon dioxide has a symmetrical linear structure.

Answer: A

4.

Assertion: PCl₅ has trigonal bipyramidal geometry.

Reason: Phosphorus undergoes sp³d hybridisation.

Answer: A

5.

Assertion: Sigma bond allows free rotation.

Reason: Electron density lies along internuclear axis.

Answer: A

E. Multiple Choice Questions – Advanced Level

1. Which molecule violates octet rule?

A) CH₄
B) NH₃
C) BF₃
D) H₂O

Answer: C

2. Which has maximum bond angle?

A) NH₃
B) H₂O
C) CH₄
D) BF₃

Answer: D (120°)

3. Which molecule has maximum bond order?

A) O₂
B) N₂
C) F₂
D) H₂

Answer: B

4. Which molecule contains only sigma bonds?

A) C₂H₄
B) C₂H₂
C) CH₄
D) CO₂

Answer: C

5. Which molecule contains 1 sigma and 2 pi bonds?

A) O₂
B) N₂
C) C₂H₂
D) H₂

Answer: C

6. Which molecule is diamagnetic?

A) O₂
B) NO
C) N₂
D) B₂

Answer: C

7. Hybridisation of central atom in XeF₄ is:

A) sp³
B) sp³d
C) sp³d²
D) sp²

Answer: C

8. The shape of SF₄ is:

A) Linear
B) See-saw
C) Square planar
D) Tetrahedral

Answer: B

F. Important Numericals

1. Calculate bond order of N₂.

Electronic configuration:

Bonding electrons = 10
Antibonding electrons = 4 $BO=\frac{10-4}{2}$ BO=210−4 $BO=3$ BO=3

Answer: Bond order = 3

2. Calculate bond order of O₂⁺.

O₂ bond order = 2

Removing one electron from antibonding orbital increases bond order.

Answer: 2.5

3. Calculate bond order of O₂⁻.

Adding one electron to antibonding orbital decreases bond order.

Answer: 1.5

G. Final Chapter Memory Tricks

Shapes:

Be → Straight

BeCl₂ → Linear

B → Flat

BF₃ → Trigonal planar

C → Complete tetrahedron

CH₄ → Tetrahedral

N → One lone pair

NH₃ → Pyramidal

O → Two lone pairs

H₂O → Bent

Hybridisation Trick:

Number of electron regions:

2 → sp
3 → sp²
4 → sp³
5 → sp³d
6 → sp³d²

Bond Trick:

Single bond:

Double bond:

1σ + 1π

Triple bond:

1σ + 2π

Class 11 Chemistry

Chapter 4: Chemical Bonding and Molecular Structure

Question Bank – Part 4

Final Important Question Set (CBSE/Board + Competitive Level)

A. Very Important MCQs (Revision Set)

1. The bond formed by mutual sharing of electrons is:

A) Ionic bond
B) Covalent bond
C) Hydrogen bond
D) Metallic bond

Answer: B

2. The electronic configuration of noble gases is stable because they have:

A) High atomic mass
B) Complete valence shell
C) Large size
D) High density

Answer: B

3. The number of valence electrons in chlorine is:

A) 5
B) 6
C) 7
D) 8

Answer: C

4. Which compound has maximum ionic character?

A) LiF
B) LiCl
C) LiBr
D) LiI

Answer: A

5. Which of the following is electron deficient?

A) CH₄
B) NH₃
C) BF₃
D) H₂O

Answer: C

6. Which molecule has expanded octet?

A) BF₃
B) CH₄
C) PCl₅
D) NH₃

Answer: C

7. The geometry of NH₃ is:

A) Linear
B) Trigonal planar
C) Trigonal pyramidal
D) Tetrahedral

Answer: C

8. The geometry of XeF₄ is:

A) Tetrahedral
B) Square planar
C) Linear
D) Trigonal planar

Answer: B

9. The bond angle of methane is:

A) 90°
B) 104.5°
C) 107°
D) 109.5°

Answer: D

10. The strongest hydrogen bonding occurs in:

A) HCl
B) HF
C) HBr
D) HI

Answer: B

11. Which molecule has no lone pair on central atom?

A) NH₃
B) H₂O
C) CH₄
D) PCl₃

Answer: C

12. The number of lone pairs in NH₃ is:

A) 0
B) 1
C) 2
D) 3

Answer: B

13. Which molecule has highest bond order?

A) O₂
B) F₂
C) N₂
D) H₂

Answer: C

14. Paramagnetic molecule is:

A) N₂
B) H₂
C) O₂
D) CO

Answer: C

15. Hybridisation of carbon in CO₂ is:

A) sp
B) sp²
C) sp³
D) sp³d

Answer: A

B. Fill in the Blanks (Important)

Chemical bonding helps atoms achieve ______ configuration.
Answer: stable
The electrons involved in bonding are called ______ electrons.
Answer: valence
Ionic bond is also called ______ bond.
Answer: electrovalent
Covalent bond involves ______ of electrons.
Answer: sharing
Lewis structures represent ______ electrons.
Answer: valence
The shape of BF₃ is ______.
Answer: trigonal planar
The shape of CH₄ is ______.
Answer: tetrahedral
The hybridisation of NH₃ is ______.
Answer: sp³
The bond angle in CO₂ is ______.
Answer: 180°
The bond angle in BF₃ is ______.
Answer: 120°
O₂ is ______ due to unpaired electrons.
Answer: paramagnetic
The strongest covalent bond is ______ bond.
Answer: sigma
Pi bond is formed by ______ overlap.
Answer: sidewise
Bond order of N₂ is ______.
Answer: 3
Hydrogen bonding occurs mainly with F, O and ______.
Answer: N

C. Match the Following

Column A	Column B
CH₄	sp³
BF₃	sp²
BeCl₂	sp
PCl₅	sp³d
SF₆	sp³d²
CO₂	Linear
H₂O	Bent
NH₃	Pyramidal

D. Short Answer Questions

1. Define chemical bond.

Answer:
Chemical bond is the attractive force that holds atoms together in a molecule or compound.

2. What is octet rule?

Answer:
Atoms gain, lose, or share electrons to achieve eight electrons in their outermost shell.

3. What is lattice enthalpy?

Answer:
Energy required to separate one mole of ionic solid into gaseous ions is called lattice enthalpy.

4. What is dipole moment?

Answer:
Dipole moment is the measure of polarity of a bond or molecule.

Formula: $\mu=q \times r$ μ=q×r

5. What is resonance?

Answer:
The phenomenon in which a molecule is represented by more than one Lewis structure due to electron delocalisation.

6. Why are ionic compounds hard?

Answer:
Because of strong electrostatic attraction between oppositely charged ions.

7. Why is CO₂ non-polar?

Answer:
Due to linear symmetrical structure, bond dipoles cancel each other.

8. Why is water polar?

Answer:
Because water has bent shape and its bond dipoles do not cancel.

9. What is hybridisation?

Answer:
Mixing of atomic orbitals to form equivalent hybrid orbitals.

10. Why is oxygen paramagnetic?

Answer:
Due to presence of two unpaired electrons in antibonding molecular orbitals.

E. Long Answer Questions (5 Marks)

1. Explain VSEPR theory.

Answer:

Main points:

Electron pairs repel each other.
They arrange themselves as far apart as possible.
Lone pairs have greater repulsion than bond pairs.
Molecular shape depends on electron pair arrangement.

Examples:

CH₄ → Tetrahedral
NH₃ → Pyramidal
H₂O → Bent

2. Explain hybridisation with examples.

Answer:

Hybridisation is mixing of atomic orbitals to form hybrid orbitals.

Examples:

Hybridisation	Example	Shape
sp	BeCl₂	Linear
sp²	BF₃	Trigonal planar
sp³	CH₄	Tetrahedral
sp³d	PCl₅	Trigonal bipyramidal
sp³d²	SF₆	Octahedral

3. Explain MOT and calculate bond order.

Formula: $BO=\frac{N_b-N_a}{2}$ BO=2Nb−Na

Where:

Nb = bonding electrons
Na = antibonding electrons

Higher bond order means:

Greater stability
Stronger bond
Shorter bond length

F. Important Competitive Questions

1. Which molecule has maximum bond angle?

A) NH₃
B) H₂O
C) CH₄
D) BF₃

Answer: BF₃ (120°)

2. Which molecule has highest polarity?

A) CO₂
B) BF₃
C) H₂O
D) CH₄

Answer: H₂O

3. Which species has bond order 2.5?

A) O₂
B) O₂⁺
C) O₂⁻
D) N₂

Answer: O₂⁺

4. Which molecule contains only sigma bonds?

A) C₂H₄
B) C₂H₂
C) CH₄
D) CO₂

Answer: CH₄

5. Which bond is strongest?

A) Single bond
B) Double bond
C) Triple bond
D) Hydrogen bond

Answer: Triple bond