Chemistry Teacher IQ | CBSE, NEET & JEE Chemistry Notes, MCQs & Questions

 - Dr.Sanjaykumar Pawar  

CBSE Chemistry Notes, MCQs & Questions | Chemistry Teacher IQ 

Chemistry Teacher IQ – Complete Chemistry Learning Resources for CBSE, NEET & JEE Aspirants

⚛️ CHEMICAL BONDING

Topper's Detailed Notes

VSEPR Theory • Molecular Geometry • Bond Angles • Electronegativity • Bond Dipoles • Molecular Dipole Moment

Class 11/12 • NEET • JEE Foundation

1. Why Do Atoms Form Chemical Bonds?

Atoms form chemical bonds because a bonded system can achieve a more stable electronic arrangement. The electrons involved in bonding are mainly the valence electrons.

For many main-group elements, stability is associated with achieving a noble-gas-like valence-shell configuration. This is commonly described using the octet rule, although the octet rule has important exceptions.

Core idea:

Chemical bonding is fundamentally an interaction involving electrons and nuclei. The arrangement of these interactions determines:

• Bond length
• Bond strength
• Bond angle
• Molecular geometry
• Polarity
• Physical and chemical properties

2. Lewis Structures — The Starting Point

Before applying VSEPR theory, we generally need a reasonable Lewis structure. A Lewis structure shows valence electrons as bonding pairs and lone pairs.

Example: H₂O

H — O — H

Oxygen also has two lone pairs:

•• O ••

The oxygen atom has:

• 2 O–H bonding pairs
• 2 lone pairs

Therefore, oxygen has 4 electron domains.

Electron domains = Bonding domains + Lone-pair domains

3. VSEPR Theory

VSEPR means:

Valence Shell Electron Pair Repulsion

The theory states that electron domains around a central atom arrange themselves so that their mutual repulsions are minimized.

Why?

Electrons are negatively charged. Therefore, electron-rich regions repel each other. The molecule adopts an arrangement that provides the most favorable spatial separation of these regions.

Electron domains

↓

Repel one another

↓

Move as far apart as possible

↓

Three-dimensional arrangement

↓

Molecular geometry

4. What Is an Electron Domain?

An electron domain is a region of electron density around the central atom.

Feature	Electron domain?
Single bond	Yes — one domain
Double bond	Yes — one domain
Triple bond	Yes — one domain
Lone pair	Yes — one domain

VERY IMPORTANT EXAM POINT

A double or triple bond contains multiple shared electron pairs, but in VSEPR it is treated as one electron domain.

5. Relative Strength of Electron-Pair Repulsions

Electron domains do not repel with exactly equal strength.

Lone Pair–Lone Pair > Lone Pair–Bond Pair > Bond Pair–Bond Pair

Why are lone pairs stronger?

A bonding pair is attracted by two nuclei because it lies between two atoms. A lone pair is associated primarily with one central atom and occupies more space around that atom. Therefore, its electron density causes stronger repulsion.

Strongest

LP ↔ LP

↓

LP ↔ BP

↓

BP ↔ BP

Weakest

6. Electron Geometry vs Molecular Geometry

These two terms are frequently confused.

Electron Geometry	Molecular Geometry
Considers all electron domains.	Describes the arrangement of atoms around the central atom.
Lone pairs are included.	Lone pairs are not named as atoms in the molecular shape.

Example: NH₃

Nitrogen has:

• 3 bonding pairs
• 1 lone pair

Total = 4 electron domains.

Four electron domains give a tetrahedral electron geometry. However, one position is occupied by a lone pair, leaving three atoms. Thus the molecular geometry is trigonal pyramidal.

Remember:
NH₃ → Electron geometry = tetrahedral
NH₃ → Molecular geometry = trigonal pyramidal

7. Steric Number

A convenient way to organize VSEPR problems is to use the steric number.

Steric Number = Number of σ bonds + Number of lone pairs on central atom

Multiple bonds still count as one σ-bond domain.

Steric Number	Basic Electron Geometry
2	Linear
3	Trigonal planar
4	Tetrahedral
5	Trigonal bipyramidal
6	Octahedral

8. Important VSEPR Geometries

AX₂ — Linear

X — A — X

Bond angle = 180°

Example: CO₂

AX₃ — Trigonal Planar

X

↙ A ↘

X    X

120°

Example: BF₃

AX₄ — Tetrahedral

X

↙ A ↘

X    X

Another X lies in 3-D space.

109.5°

Example: CH₄

AX₃E — Trigonal Pyramidal

••

A

↙ ↓ ↘

X X X

≈107°

Example: NH₃

AX₂E₂ — Bent

••   ••

A

↙   ↘

X    X

≈104.5°

Example: H₂O

9. The Most Important Bond-Angle Comparison

Molecule	Bond pairs	Lone pairs	Shape	Approx. angle
CH₄	4	0	Tetrahedral	109.5°
NH₃	3	1	Trigonal pyramidal	≈107°
H₂O	2	2	Bent	≈104.5°

CH₄

109.5°

↓

NH₃

≈107°

↓

H₂O

≈104.5°

Reason:

As lone-pair repulsion becomes more important, bonding pairs are pushed closer together. Therefore the bond angle decreases.

10. Interactive VSEPR Visualization

Use the interactive model to visualize how bonding pairs and lone pairs affect the three-dimensional arrangement.

11. Electronegativity

Electronegativity is the tendency of an atom in a chemical bond to attract the shared electron density toward itself.

It is important to understand that electronegativity is a property of an atom when it is participating in a bond, not simply an isolated-atom property.

Periodic Trend

Electronegativity increases →

⬆

Generally toward the top-right of the periodic table.

Fluorine has the highest electronegativity on the Pauling scale.

Across a Period

Electronegativity generally increases from left → right.

The effective nuclear charge increases, so the atom attracts shared electrons more strongly.

Down a Group

Electronegativity generally decreases from top → bottom.

Atomic size increases and the bonding electrons are farther from the nucleus.

12. Factors Affecting Electronegativity

1. Atomic size

Smaller atoms generally attract shared electrons more strongly.

2. Effective nuclear charge

Greater effective nuclear attraction generally increases electronegativity.

3. Shielding effect

Inner electrons partially shield valence electrons from the nucleus. Increasing shielding generally reduces the attraction experienced by bonding electrons.

Smaller atomic size + stronger effective nuclear attraction → higher electronegativity

13. Bond Polarity

Consider a covalent bond between two different atoms. If their electronegativities differ, the bonding electrons are attracted more strongly toward one atom.

Example: H–Cl

H^δ+ → Cl^δ−

Electron density is shifted toward chlorine.

The symbols δ+ and δ− represent partial charges, not full ionic charges.

Δχ = |χA − χB|

The larger the electronegativity difference, generally, the greater the bond polarity.

14. Interactive Bond Polarity

Change the atoms and observe how electronegativity difference changes electron-density distribution and bond polarity.

15. Bond Dipole Moment

A polar bond has a separation of positive and negative charge. This creates a bond dipole.

The dipole moment is represented by μ.

μ = q × r

where:

• μ = dipole moment
• q = magnitude of charge separation
• r = distance between the centers of positive and negative charge

For molecular chemistry, dipole moment is commonly expressed in Debye (D).

16. Bond Dipole vs Molecular Dipole

Bond Dipole	Molecular Dipole Moment
Associated with an individual polar bond.	Net dipole resulting from the entire molecular charge distribution.
Depends on bond polarity.	Depends on bond polarity and geometry.
One molecule may contain several bond dipoles.	The individual dipoles may reinforce or cancel each other.

TOPPER'S RULE:

Never decide whether a molecule is polar merely by checking whether it contains polar bonds.

You must also examine the three-dimensional molecular geometry.

17. Dipole Moment Is a Vector

Dipole moment has both magnitude and direction, so it behaves as a vector quantity.

Therefore, when a molecule contains several polar bonds, we cannot simply add their numerical values without considering direction.

Individual bond dipoles

↓

Vector addition

↓

Net molecular dipole

μ_molecule = Vector sum of all relevant bond dipoles

18. Why CO₂ Is Non-Polar

The C=O bonds in CO₂ are polar because oxygen is more electronegative than carbon.

However, CO₂ has a linear geometry.

O^δ− ← C^δ+ → O^δ−

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

Net μ = 0

Conclusion:

CO₂ contains polar bonds but is a non-polar molecule.

19. Why H₂O Is Polar

Each O–H bond is polar because oxygen is more electronegative than hydrogen.

But H₂O is bent, not linear.

   H^δ+

  ↘

O^δ−

  ↗

H^δ+

The two O–H bond dipoles do not cancel.

Net μ ≠ 0

Conclusion:

H₂O is a polar molecule.

20. Symmetry and Molecular Polarity

Symmetry is one of the most useful ideas for predicting molecular polarity.

Molecule	Shape	Bond polarity	Net dipole	Molecular nature
CO₂	Linear	Polar bonds	Zero	Non-polar
BF₃	Trigonal planar	Polar bonds	Zero	Non-polar
CH₄	Tetrahedral	Nearly/non-polar C–H bonds in basic treatment	Zero	Non-polar
NH₃	Trigonal pyramidal	Polar N–H bonds	Non-zero	Polar
H₂O	Bent	Polar O–H bonds	Non-zero	Polar

21. The Critical Difference: CO₂ vs H₂O

Property	CO₂	H₂O
Polar bonds?	Yes	Yes
Lone pairs on central atom?	0 on carbon	2 on oxygen
Shape	Linear	Bent
Dipoles cancel?	Yes	No
Molecular polarity	Non-polar	Polar

Exam lesson:

The presence of polar bonds is only the first step. The geometry decides whether the individual dipoles cancel or produce a net molecular dipole.

22. Complete Procedure for Solving VSEPR Questions

STEP 1
Draw Lewis structure.
↓
STEP 2
Identify central atom.
↓
STEP 3
Count σ-bond domains.
↓
STEP 4
Count lone pairs on central atom.
↓
STEP 5
Calculate steric number.
↓
STEP 6
Determine electron geometry.
↓
STEP 7
Determine molecular geometry.
↓
STEP 8
Estimate bond angle.
↓
STEP 9
Identify polar bonds.
↓
STEP 10
Use molecular geometry to determine dipole cancellation.

23. Solved Example 1 — CH₄

Step 1: Central atom

Carbon is the central atom.

Step 2: Bond pairs

Carbon forms four C–H single bonds.

Step 3: Lone pairs

Carbon has no lone pair in CH₄.

Step 4: Steric number

SN = 4 + 0 = 4

Step 5: Geometry

Electron geometry = tetrahedral.

Molecular geometry = tetrahedral.

Step 6: Bond angle

Ideal tetrahedral angle = 109.5°.

Step 7: Polarity

The tetrahedral arrangement is highly symmetrical, so the bond dipoles cancel in the idealized model.

CH₄ → non-polar molecule.

24. Solved Example 2 — NH₃

Step 1

Nitrogen is the central atom.

Step 2

There are 3 N–H bonding domains.

Step 3

There is 1 lone pair on nitrogen.

Step 4

SN = 3 + 1 = 4

Step 5

Electron geometry = tetrahedral.

Molecular geometry = trigonal pyramidal.

Step 6

Bond angle is approximately 107°.

Step 7

The N–H bonds are polar and their dipoles do not cancel because the molecule is not symmetrical in the same way as a tetrahedral molecule with four identical surrounding atoms.

NH₃ → polar molecule.

25. Solved Example 3 — H₂O

Step 1

Oxygen is the central atom.

Step 2

There are 2 O–H bonding domains.

Step 3

There are 2 lone pairs on oxygen.

Step 4

SN = 2 + 2 = 4

Step 5

Electron geometry = tetrahedral.

Molecular geometry = bent.

Step 6

Bond angle ≈ 104.5°.

Step 7

The O–H bond dipoles do not cancel because the molecule is bent.

H₂O → polar molecule.

26. Solved Example 4 — CO₂

Step 1

Carbon is the central atom.

Step 2

Carbon is connected to two oxygen atoms by double bonds.

Each double bond is treated as one electron domain in VSEPR.

Step 3

SN = 2 + 0 = 2

Step 4

Electron geometry = linear.

Molecular geometry = linear.

Step 5

Bond angle = 180°.

Step 6

The two C=O bond dipoles are equal and opposite.

CO₂ → non-polar molecule.

27. Common Student Mistakes

❌ Mistake 1: "Polar bonds always mean polar molecule."

Wrong. CO₂ has polar C=O bonds but zero net molecular dipole because of its linear symmetry.

❌ Mistake 2: Counting a double bond as two VSEPR domains.

Wrong. A double bond is treated as one electron domain.

❌ Mistake 3: Calling NH₃ tetrahedral molecular geometry.

Its electron geometry is tetrahedral, but its molecular geometry is trigonal pyramidal.

❌ Mistake 4: Ignoring lone pairs when predicting bond angle.

Lone pairs exert strong repulsion and can significantly compress bond angles.

❌ Mistake 5: Treating bond angle as a two-dimensional concept.

Molecular geometry is three-dimensional. Many molecules cannot be correctly understood from a flat drawing alone.

28. High-Yield Bond Angle Facts

Species	Shape	Approx. bond angle
CO₂	Linear	180°
BF₃	Trigonal planar	120°
CH₄	Tetrahedral	109.5°
NH₃	Trigonal pyramidal	≈107°
H₂O	Bent	≈104.5°

Quick memory sequence:

180° → 120° → 109.5° → 107° → 104.5°

Linear → Trigonal planar → Tetrahedral → NH₃ → H₂O

29. High-Yield Polarity Examples

Molecule	Shape	Polar bonds?	Net dipole?	Nature
CO₂	Linear	Yes	No	Non-polar
BF₃	Trigonal planar	Yes	No	Non-polar
CH₄	Tetrahedral	Very weak/approximately non-polar C–H bonds in basic treatment	No	Non-polar
NH₃	Trigonal pyramidal	Yes	Yes	Polar
H₂O	Bent	Yes	Yes	Polar

30. Topper's Concept Map 🧠

Electronegativity Difference

↓

Unequal sharing of electrons

↓

Bond Polarity

↓

But a molecule may contain several polar bonds

↓

VSEPR → Molecular Geometry

↓

Consider direction of every bond dipole

↓

Vector Cancellation / Addition

↓

Molecular Dipole Moment

↓

Polar or Non-polar Molecule

31. Topper's 10-Second Method

Whenever you see a question asking about shape, bond angle or polarity:

1. Draw the Lewis structure.
2. Find the central atom.
3. Count σ-bond domains.
4. Count lone pairs.
5. Determine steric number.
6. Find electron geometry.
7. Find molecular geometry.
8. Predict bond angle.
9. Identify polar bonds.
10. Check whether dipoles cancel.

Lewis Structure → VSEPR → Geometry → Bond Angle → Dipoles → Molecular Polarity

32. Advanced Exam Insight

VSEPR is a qualitative model. It is extremely useful for predicting molecular geometry and approximate bond angles, but real molecules can deviate from idealized angles because of several factors.

• Lone-pair repulsion
• Differences in ligand electronegativity
• Multiple bonds
• Bonding-pair distribution
• Substituent effects
• Electronic structure of the central atom

Important distinction:

VSEPR predicts geometry primarily from the distribution of electron domains. It does not mean that every molecule with the same electron-domain count has exactly the same bond angles.

33. Final Revision Sheet

Concept	Key Point
VSEPR	Electron domains arrange to minimize repulsions.
Strongest repulsion	LP–LP
Second strongest	LP–BP
Weakest	BP–BP
Linear	180°
Trigonal planar	120°
Tetrahedral	109.5°
NH₃	Trigonal pyramidal, ≈107°
H₂O	Bent, ≈104.5°
Electronegativity	Generally increases toward the top-right.
Fluorine	Highest electronegativity.
Bond dipole	Dipole associated with a polar bond.
Molecular dipole	Net vector result of molecular charge distribution.
CO₂	Polar bonds + linear geometry → non-polar molecule.
H₂O	Polar bonds + bent geometry → polar molecule.

34. Final Golden Rules ⭐

RULE 1: Electron pairs repel each other.

RULE 2: Lone pairs repel more strongly than bonding pairs.

RULE 3: More lone-pair repulsion generally means a smaller bond angle.

RULE 4: A multiple bond counts as one electron domain in basic VSEPR counting.

RULE 5: Electronegativity generally increases toward the top-right of the periodic table.

RULE 6: A difference in electronegativity can produce a polar bond.

RULE 7: Polar bonds do NOT automatically make a polar molecule.

RULE 8: Molecular geometry determines whether bond dipoles cancel.

RULE 9: CO₂ is the classic example of polar bonds but a non-polar molecule.

RULE 10: H₂O is the classic example of polar bonds plus bent geometry giving a polar molecule.

🔥 MASTER FORMULA FOR POLARITY

Bond Polarity + Molecular Geometry → Molecular Dipole Moment

Master the geometry first. Then master the dipoles.

⚛️ Chemical Bonding

CBSE Class 11 Question Bank

VSEPR • Bond Angles • Molecular Geometry • Electronegativity • Dipoles • Molecular Polarity

SECTION A — Multiple Choice Questions (MCQs)

Q1. VSEPR theory is based on the principle that:

A. Nuclei repel each other only

B. Electron pairs repel each other and arrange themselves to minimize repulsion

C. Protons attract electron pairs

D. Electrons always remain stationary

Answer: B
Electron domains around a central atom arrange themselves as far apart as possible to minimize repulsion.

Q2. Which type of repulsion is strongest according to VSEPR theory?

A. Bond pair–bond pair

B. Lone pair–bond pair

C. Lone pair–lone pair

D. Nucleus–nucleus

Answer: C

Q3. The ideal bond angle in a tetrahedral molecule is:

A. 90°

B. 104.5°

C. 109.5°

D. 120°

Answer: C — 109.5°

Q4. The molecular geometry of NH₃ is:

A. Linear

B. Trigonal planar

C. Trigonal pyramidal

D. Tetrahedral

Answer: C — Trigonal pyramidal

Q5. The molecular geometry of H₂O is:

A. Linear

B. Bent

C. Trigonal planar

D. Tetrahedral

Answer: B — Bent

Q6. The approximate H–O–H bond angle in water is:

A. 180°

B. 120°

C. 109.5°

D. 104.5°

Answer: D — 104.5°

Q7. The approximate H–N–H bond angle in NH₃ is:

A. 90°

B. 104.5°

C. 107°

D. 120°

Answer: C — approximately 107°

Q8. Which molecule has a linear geometry?

A. H₂O

B. NH₃

C. CO₂

D. CH₄

Answer: C — CO₂

Q9. In VSEPR theory, a double bond is counted as:

A. One electron domain

B. Two electron domains

C. Three electron domains

D. Four electron domains

Answer: A

Q10. The most electronegative element is:

A. O

B. Cl

C. F

D. N

Answer: C — Fluorine

Q11. Electronegativity generally increases:

A. From right to left across a period

B. From left to right across a period

C. Down a group

D. Randomly

Answer: B

Q12. Which molecule has polar bonds but zero net molecular dipole moment?

A. H₂O

B. NH₃

C. CO₂

D. HCl

Answer: C — CO₂

Q13. Which molecule is polar?

A. CO₂

B. BF₃

C. H₂O

D. CCl₄

Answer: C — H₂O

Q14. The dipole moment is commonly represented by:

A. α

B. μ

C. β

D. λ

Answer: B — μ

Q15. The relationship between dipole moment, charge and separation is:

A. μ = q/r

B. μ = qr

C. μ = r/q

D. μ = q + r

Answer: B — μ = q × r

SECTION B — Very Short Answer Questions

Q1. What does VSEPR stand for?

Valence Shell Electron Pair Repulsion.

Q2. What is an electron domain?

A region of electron density around the central atom.

Q3. Which repels more strongly: a lone pair or a bonding pair?

A lone pair generally produces stronger repulsion.

Q4. What is the ideal bond angle in a linear arrangement?

180°.

Q5. What is the ideal bond angle in trigonal planar geometry?

120°.

Q6. What is the ideal tetrahedral bond angle?

109.5°.

Q7. What is the molecular shape of NH₃?

Trigonal pyramidal.

Q8. What is the molecular shape of H₂O?

Bent or angular.

Q9. Which element has the highest electronegativity?

Fluorine.

Q10. What is a bond dipole?

A separation of partial positive and partial negative charge within a polar bond.

Q11. What is the SI unit of dipole moment?

Coulomb metre (C m).

Q12. What is the commonly used unit of molecular dipole moment in chemistry?

Debye (D).

Q13. Is CO₂ a polar molecule?

No. Its bond dipoles cancel because it is linear and symmetrical.

Q14. Is H₂O polar?

Yes. Its bent geometry prevents cancellation of the O–H bond dipoles.

Q15. How many electron domains are present around carbon in CO₂?

Two.

SECTION C — Short Answer Questions

Q1. Explain VSEPR theory.

VSEPR theory states that electron domains around a central atom repel one another and arrange themselves as far apart as possible. This arrangement minimizes electron-electron repulsion and determines the molecular geometry.

Q2. Why do lone pairs exert greater repulsion than bond pairs?

A lone pair is concentrated around one central atom and occupies a relatively larger region of space. A bonding pair is attracted by two nuclei and is distributed between two atoms. Hence lone-pair repulsion is generally stronger.

Q3. Why is the bond angle in NH₃ smaller than 109.5°?

NH₃ has four electron domains around nitrogen: three bond pairs and one lone pair. The lone pair repels the bonding pairs more strongly and compresses the H–N–H bond angle from the ideal tetrahedral value of 109.5° to approximately 107°.

Q4. Why is the bond angle in H₂O approximately 104.5°?

Oxygen in H₂O has four electron domains: two bonding pairs and two lone pairs. The stronger lone-pair repulsions compress the H–O–H angle to approximately 104.5°.

Q5. Differentiate between electron geometry and molecular geometry.

Electron geometry considers both bonding pairs and lone pairs. Molecular geometry describes the arrangement of atoms around the central atom, with lone pairs omitted from the shape name.

Q6. Explain the trend of electronegativity across a period.

Electronegativity generally increases from left to right across a period because effective nuclear charge increases and atomic size generally decreases. As a result, the atom attracts shared electrons more strongly.

Q7. Why is CO₂ non-polar although its C=O bonds are polar?

CO₂ is linear. The two equal C=O bond dipoles point in opposite directions and cancel each other. Therefore, the net molecular dipole moment is zero.

Q8. Why is H₂O polar?

The O–H bonds are polar and H₂O has a bent geometry. Therefore, the two bond dipoles do not cancel, giving H₂O a non-zero molecular dipole moment.

Q9. What is meant by electronegativity?

Electronegativity is the tendency of an atom in a chemical bond to attract the shared electron density toward itself.

Q10. Explain the difference between bond dipole and molecular dipole moment.

A bond dipole is associated with an individual polar bond. Molecular dipole moment is the net vector result of the dipoles and charge distribution of the entire molecule.

SECTION D — Long Answer Questions

Q1. Explain the shape and bond angle of CH₄, NH₃ and H₂O using VSEPR theory.

CH₄

Carbon has four bonding pairs and no lone pairs. There are four electron domains. They arrange themselves tetrahedrally.

Geometry: Tetrahedral

Bond angle: 109.5°

NH₃

Nitrogen has three bonding pairs and one lone pair. The four electron domains have tetrahedral electron geometry. Since one domain is a lone pair, the molecular geometry becomes trigonal pyramidal.

Bond angle: approximately 107°

H₂O

Oxygen has two bonding pairs and two lone pairs. The electron geometry is tetrahedral, but the molecular geometry is bent. Strong lone-pair repulsion compresses the H–O–H angle to approximately 104.5°.

Molecule	Bond pairs	Lone pairs	Shape	Angle
CH₄	4	0	Tetrahedral	109.5°
NH₃	3	1	Trigonal pyramidal	≈107°
H₂O	2	2	Bent	≈104.5°

Q2. Explain molecular polarity using CO₂ and H₂O as examples.

Molecular polarity depends on both bond polarity and molecular geometry.

CO₂

Each C=O bond is polar. However, CO₂ is linear and symmetrical. The two bond dipoles are equal and opposite, so they cancel.

Net dipole moment = 0

Therefore CO₂ is non-polar.

H₂O

The O–H bonds are polar. H₂O has a bent geometry because oxygen possesses two lone pairs. Therefore, the O–H bond dipoles do not cancel.

Net dipole moment ≠ 0

Therefore H₂O is polar.

Q3. Discuss the periodic trend of electronegativity and explain the factors responsible for it.

Electronegativity generally increases from left to right across a period and decreases down a group.

Across a period

• Effective nuclear charge generally increases.
• Atomic radius generally decreases.
• The nucleus attracts shared electrons more strongly.
• Therefore electronegativity increases.

Down a group

• Number of electron shells increases.
• Atomic radius increases.
• Shielding effect increases.
• Bonding electrons are farther from the nucleus.
• Therefore electronegativity generally decreases.

SECTION E — Assertion & Reason

Choose the correct option:

A. Both Assertion and Reason are true, and Reason correctly explains Assertion.
B. Both Assertion and Reason are true, but Reason does not correctly explain Assertion.
C. Assertion is true, but Reason is false.
D. Assertion is false, but Reason is true.

Q1.

Assertion: The bond angle in NH₃ is smaller than 109.5°.

Reason: A lone pair repels bonding pairs more strongly than bonding pairs repel one another.

Answer: A

Q2.

Assertion: CO₂ has zero molecular dipole moment.

Reason: CO₂ contains no polar bonds.

Answer: C
CO₂ has polar C=O bonds, but the bond dipoles cancel because of its linear geometry.

Q3.

Assertion: H₂O is a polar molecule.

Reason: H₂O has a bent molecular geometry.

Answer: A

Q4.

Assertion: Electronegativity generally decreases down a group.

Reason: Atomic size and shielding generally increase down a group.

Answer: A

Q5.

Assertion: A double bond is counted as two electron domains in VSEPR theory.

Reason: A double bond contains two shared electron pairs.

Answer: D
The Reason is true, but the Assertion is false. A multiple bond is treated as one electron domain in basic VSEPR counting.

SECTION F — Fill in the Blanks

1. VSEPR stands for __________.
2. The strongest electron-pair repulsion is __________.
3. The ideal bond angle in a tetrahedral arrangement is __________.
4. NH₃ has __________ lone pair on nitrogen.
5. H₂O has __________ lone pairs on oxygen.
6. The molecular geometry of NH₃ is __________.
7. The molecular geometry of H₂O is __________.
8. CO₂ has a __________ molecular geometry.
9. The most electronegative element is __________.
10. Electronegativity generally increases from __________ to __________ across a period.
11. Dipole moment is represented by the symbol __________.
12. The formula for dipole moment is __________.
13. A molecule may contain polar bonds but still have zero __________ dipole moment.
14. CO₂ has a net dipole moment of __________.
15. H₂O has a __________ molecular dipole moment.

Answers:1. Valence Shell Electron Pair Repulsion
2. Lone pair–lone pair repulsion
3. 109.5°
4. One
5. Two
6. Trigonal pyramidal
7. Bent
8. Linear
9. Fluorine
10. Left, right
11. μ
12. μ = q × r
13. Molecular
14. Zero
15. Non-zero

SECTION G — Match the Columns

Column I	Column II
1. CO₂	A. Bent
2. BF₃	B. 109.5°
3. CH₄	C. Linear
4. NH₃	D. Trigonal planar
5. H₂O	E. Trigonal pyramidal

Answers:
1 → C
2 → D
3 → B
4 → E
5 → A

SECTION H — Statement-Based Questions

Q1. Consider the following statements:

I. Lone pairs generally occupy more space than bonding pairs.

II. Lone-pair repulsion is stronger than bond-pair repulsion.

III. Lone pairs generally increase the bond angle.

Which statements are correct?

A. I only

B. I and II only

C. II and III only

D. I, II and III

Answer: B

Q2. Consider:

I. CO₂ contains polar C=O bonds.

II. CO₂ is linear.

III. The bond dipoles cancel.

Which conclusion is correct?

A. CO₂ is polar.

B. CO₂ is non-polar.

C. CO₂ is ionic.

D. CO₂ has no C=O bond polarity.

Answer: B

Q3. Which statements about electronegativity are correct?

I. It generally increases across a period.

II. It generally decreases down a group.

III. Fluorine has the highest electronegativity.

Answer: All three statements are correct.

SECTION I — Case Study Based Questions

Case Study 1 — Comparing CH₄, NH₃ and H₂O

A student studies three molecules: CH₄, NH₃ and H₂O. Each central atom has four electron domains. However, the number of lone pairs is different. The student observes that their bond angles are not identical.

Q1. What is the electron geometry of all three molecules?

A. Linear

B. Trigonal planar

C. Tetrahedral

D. Octahedral

Answer: C

Q2. Which molecule has no lone pair on the central atom?

A. CH₄

B. NH₃

C. H₂O

D. Both NH₃ and H₂O

Answer: A

Q3. Which molecule has the smallest bond angle?

A. CH₄

B. NH₃

C. H₂O

D. All have the same angle

Answer: C

Q4. Arrange the approximate bond angles in decreasing order.

Answer: CH₄ (109.5°) > NH₃ (≈107°) > H₂O (≈104.5°).

Case Study 2 — CO₂ and H₂O

CO₂ and H₂O both contain bonds between atoms with different electronegativities. Nevertheless, CO₂ is non-polar while H₂O is polar.

Q1. Why are the C=O bonds polar?

Because oxygen is more electronegative than carbon, so the shared electron density is shifted toward oxygen.

Q2. Why does CO₂ have zero net dipole moment?

Its two equal C=O bond dipoles point in opposite directions in the linear molecule and cancel.

Q3. Why does H₂O have a non-zero dipole moment?

H₂O is bent, so its two O–H bond dipoles do not cancel.

Q4. What two factors must be considered to determine molecular polarity?

Bond polarity and molecular geometry.

SECTION J — Competency/HOTS Questions

Q1. A student says: "BF₃ is non-polar because B–F bonds are non-polar." Is the statement correct? Explain.

No. The B–F bonds are polar because fluorine is more electronegative than boron. However, BF₃ has a trigonal planar and symmetrical geometry. The three bond dipoles cancel each other vectorially. Therefore, BF₃ is a non-polar molecule despite having polar bonds.

Q2. Which has the greater bond angle: NH₃ or H₂O? Explain using VSEPR theory.

NH₃ has a bond angle of approximately 107°, while H₂O has approximately 104.5°.

NH₃ contains one lone pair, whereas H₂O contains two lone pairs on the central atom. The additional lone-pair repulsion in H₂O compresses its bond angle more strongly.

Therefore:

NH₃ > H₂O in bond angle.

Q3. A molecule contains polar bonds. Can its molecular dipole moment still be zero? Give an example.

Yes. If the molecule has a symmetrical geometry, individual bond dipoles can cancel.

Example: CO₂

The C=O bonds are polar, but the molecule is linear and the two dipoles are equal and opposite. Therefore, the net molecular dipole moment is zero.

SECTION K — Rapid Revision Questions

Question	Answer
VSEPR full form?	Valence Shell Electron Pair Repulsion
Strongest repulsion?	LP–LP
Linear angle?	180°
Trigonal planar angle?	120°
Tetrahedral angle?	109.5°
NH₃ shape?	Trigonal pyramidal
H₂O shape?	Bent
NH₃ angle?	≈107°
H₂O angle?	≈104.5°
CO₂ shape?	Linear
Most electronegative element?	Fluorine
Electronegativity across a period?	Generally increases
Electronegativity down a group?	Generally decreases
Dipole moment symbol?	μ
CO₂ polar or non-polar?	Non-polar
H₂O polar or non-polar?	Polar

⭐ CBSE Exam Master Tips

1. Always draw the Lewis structure before applying VSEPR.

2. Count electron domains, not individual electron pairs.

3. A double or triple bond counts as one electron domain.

4. Lone pairs cause stronger repulsion than bonding pairs.

5. Electron geometry and molecular geometry are not always the same.

6. Bond polarity depends mainly on electronegativity difference.

7. Molecular polarity depends on both bond polarity and geometry.

8. Symmetrical molecules can have polar bonds but zero net dipole.

9. CO₂ → polar bonds + linear geometry → non-polar molecule.

10. H₂O → polar bonds + bent geometry → polar molecule.

Lewis Structure → VSEPR → Geometry → Bond Angle → Bond Dipoles → Net Dipole