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

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

📘 Get NCERT Solutions for Class 11 Chemistry Chapter 4 – Chemical Bonding and Molecular Structure
with step-by-step explanations, solved NCERT questions, and practice MCQs.
These solutions are prepared according to the latest CBSE syllabus and are helpful for
Class 11 Board Exams, NEET, JEE and other competitive exams.
Important topics covered in this chapter include
Kossel–Lewis Approach, Ionic Bond, Covalent Bond, VSEPR Theory, Valence Bond Theory, Hybridization and Molecular Geometry.
All concepts are explained in simple language so students can easily understand the formation and structure of molecules. ✅

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

Showing all questions

Q.1:
Explain the formation
of a
chemical bond
.

Answer
✅ Most elements do not exist as
independent atoms; they combine to form
molecules.
✅ The attractive force that holds atoms together in a molecule is called a
chemical bond.

✅ According to the
Kossel–Lewis theory, chemical bonds are formed mainly by:
1️⃣
Transfer of valence electrons
2️⃣
Sharing of valence electrons

✅ Through these processes, atoms try to achieve a stable
octet electronic configuration.

Explanation (Step by Step)
Step 1: Relationship between matter, atoms and molecules
👉 In most substances, atoms do not remain alone but combine to form
stable groups called molecules.
👉 The force that holds these atoms together is known as a
chemical bond.

Step 2: Definition of a chemical bond
A chemical bond is the
attractive force that holds atoms or ions together in a
chemical species.

Step 3: Kossel–Lewis concept
👉 Atoms form bonds in order to achieve
greater stability.
👉 This stability is usually obtained by achieving the
octet configuration of noble gases.
👉 This can occur either by:
•
transfer of electrons
•
sharing of electrons

Step 4: Example (Formation of NaCl)
👉 Sodium (Na) loses one electron to form
Na⁺.
👉 Chlorine (Cl) gains one electron to form
Cl⁻.
👉 The electrostatic attraction between
Na⁺ and
Cl⁻
forms an
ionic (electrovalent) bond.

Did You Know?
👉
Noble gases rarely form chemical bonds because their
outermost shell is already stable.

👉 The formation of chemical bonds usually results in
lower energy and
greater stability.

👉 When electrons are transferred, an
ionic bond is formed, and when electrons are shared, a
covalent bond is formed.

Q.2: Write the
Lewis dot symbols
for atoms of the following elements:
Mg, Na, B, O, N, Br
.

Answer
✅ In a
Lewis dot symbol, only the
valence electrons are represented by dots (•).

• Mg (2,8,2) →
2 valence electrons
• Na (2,8,1) →
1 valence electron
• B (2,3) →
3 valence electrons
• O (2,6) →
6 valence electrons
• N (2,5) →
5 valence electrons
• Br (2,8,18,7) →
7 valence electrons

Lewis Dot Symbols:
• Mg : •Mg•
• Na : Na•
• B : •B••
• O : :O:•• (total 6 dots)
• N : :N••• (total 5 dots)
• Br : :Br:••• (total 7 dots)

Note:
The positions of dots (top, bottom, left, or right) may vary, but the
total number of dots must equal the number of valence electrons.

Explanation (Step by Step)
Step 1: What is a Lewis dot symbol?
👉 A
Lewis dot symbol represents the
valence electrons of an atom by dots placed around its chemical symbol.

Step 2: How to determine Lewis dot symbols?
1️⃣ Write the
electronic configuration of the element.
2️⃣ Identify the electrons in the
outermost shell (valence electrons).
3️⃣ Represent these electrons as
dots around the element symbol.

Step 3: Applying to the given elements
• Mg (12): 2,8,2 →
2 valence electrons
• Na (11): 2,8,1 →
1 valence electron
• B (5): 2,3 →
3 valence electrons
• O (8): 2,6 →
6 valence electrons
• N (7): 2,5 →
5 valence electrons
• Br (35): 2,8,18,7 →
7 valence electrons

Did You Know?
👉 For most
main group elements, the number of
valence electrons is related to their
group number in the periodic table.

👉 Lewis dot symbols are extremely useful for understanding
chemical bonding, especially
ionic bonds and
covalent bonds.

👉 Elements like
oxygen (6 valence electrons) and
bromine (7 valence electrons)
tend to gain or share electrons to achieve a
stable octet configuration.

Q.3: Write the
Lewis symbols
for the following
atoms and ions
:
S and S²⁻; Al and Al³⁺; H and H⁻

Answer
Lewis symbols:

• S : :S:•• (6 valence electrons)
• S²⁻ : [:S:]²⁻ (8 electrons, octet complete)

• Al : Al••• (3 valence electrons)
• Al³⁺ : [Al]³⁺ (no valence electrons shown)

• H : H• (1 valence electron)
• H⁻ : [:H:]⁻ (2 electrons, duplet complete)

Explanation (Step by Step)
Step 1: What are Lewis symbols?
👉 Lewis symbols represent the
valence electrons of an atom using dots around the element symbol.

Step 2: Determine valence electrons
• S → Group 16 →
6 valence electrons
• Al → Group 13 →
3 valence electrons
• H → Group 1 →
1 valence electron

Step 3: Effect of ion formation
👉 When atoms gain or lose electrons, their Lewis symbols change.
•
S²⁻ gains 2 electrons → total 8 electrons (octet).
•
Al³⁺ loses 3 electrons → no valence electrons shown.
•
H⁻ gains 1 electron → 2 electrons (duplet configuration).

Did You Know?
👉 Atoms generally gain or lose electrons to achieve a
stable noble gas configuration.

👉 Most atoms follow the
octet rule, while hydrogen follows the
duplet rule.

👉 Lewis symbols are extremely useful for understanding
ionic and covalent bonding.

Q.4: Draw the
Lewis structures
for the following
molecules and ions
:
H₂S, SiCl₄, BeF₂, CO₃²⁻, HCOOH

Answer
Lewis Structures:

1️⃣
H₂S
H – S – H

(Sulfur has two lone pairs)

2️⃣
SiCl₄
Cl – Si – Cl

|
Cl Cl

(Si forms four single bonds with Cl atoms)

3️⃣
BeF₂
F – Be – F

(Linear molecule; Be forms two bonds)

4️⃣
CO₃²⁻
O – C = O
|
O⁻

(Resonance occurs among three equivalent structures)

5️⃣
HCOOH (Formic Acid)
H – C = O
|
O – H

Explanation (Step by Step)
Step 1: Count valence electrons
👉 Determine the total number of
valence electrons of all atoms.

Example:
• H₂S → S = 6, H = 1×2 → total = 8 electrons
• CO₃²⁻ → C = 4, O = 6×3 + 2 extra electrons → total = 24 electrons

Step 2: Choose the central atom
👉 Usually the
least electronegative atom becomes the central atom.
Example:
S in H₂S, Si in SiCl₄, C in CO₃²⁻.

Step 3: Form single bonds
👉 Connect surrounding atoms with
single bonds first.

Step 4: Complete octet
👉 Add
lone pairs to complete the octet rule.
👉 In some molecules like BeF₂, the central atom may have
incomplete octet.

Step 5: Consider resonance
👉 Some molecules like
CO₃²⁻ show
resonance structures.

Did You Know?
👉
BeF₂ is an example of a molecule with an
incomplete octet.

👉
CO₃²⁻ shows
three resonance structures, making all C–O bonds equivalent.

👉 Lewis structures help in predicting
molecular shape and bonding.

Q.5: Define the
octet rule
and write its
significance and limitations
.

Answer
✅ According to the
octet rule, atoms tend to form chemical bonds in such a way that they obtain
eight electrons in their outermost shell
(or two electrons in the case of helium), thereby achieving a stable
noble gas electronic configuration.

✅ This rule was proposed by
Kossel and Lewis (1916)
based on the stability of noble gases.

✅ According to this concept, chemical bonds are formed mainly by:
1️⃣
transfer of electrons (ionic bond)
2️⃣
sharing of electrons (covalent bond)

Explanation (Step by Step)
Step 1: Basis of the octet rule
👉
Noble gases are highly stable because their
outermost shell is completely filled.
👉 Helium has
2 electrons (duplet), while most other noble gases have
8 electrons (octet) in the outer shell.
👉 Other atoms form chemical bonds to achieve this
stable configuration.

Step 2: How is the octet achieved?
(a) By transfer of electrons
Example: Formation of sodium chloride
Na → Na⁺ + e⁻
Cl + e⁻ → Cl⁻
Na⁺ + Cl⁻ → NaCl
👉 Both atoms achieve a
noble gas configuration.

(b) By sharing of electrons
👉 Two atoms share one or more pairs of electrons to obtain
stable octet configuration,
as seen in many
covalent molecules.

Step 3: Significance of the octet rule
✅ It explains the formation of many molecules such as
O₂, N₂, Cl₂, Br₂.
✅ It helps in understanding
ionic and covalent bonding.
✅ It is useful in explaining the structure of many
organic compounds.
✅ It works particularly well for
second period elements.

Step 4: Limitations of the octet rule
The octet rule is not universally applicable. Important exceptions include:

1) Incomplete Octet
👉 In some compounds, the central atom has fewer than
8 electrons.
Examples: BeH₂, BCl₃, AlCl₃

2) Odd Electron Molecules
👉 Some molecules contain an
odd number of electrons, so the octet cannot be completed for all atoms.
Examples: NO, NO₂

3) Expanded Octet
👉 Elements of the
third period and beyond can have more than
8 electrons in their valence shell.
Examples: PF₅, SF₆, H₂SO₄

Did You Know?
👉 The
octet rule is very useful but it
does not apply to every compound.

👉
Helium is stable with only
two electrons, so it follows the
duplet rule instead of the octet rule.

👉 In exams, when writing limitations of the octet rule, always mention these
three main exceptions:
• Incomplete octet
• Odd electron molecules
• Expanded octet

Q.6: Write the
favourable factors
for the formation of an
ionic bond
.

Answer
The following factors favour the formation of an
ionic bond:

✅
Low ionisation enthalpy
of the atom forming the cation.
✅
High (more negative) electron gain enthalpy
of the atom forming the anion.
✅
High lattice energy
of the resulting ionic crystal.

These factors together increase the
stability of ionic compounds.

Explanation (Step by Step)
Step 1: Effect of ionisation enthalpy
👉 In an ionic bond, one atom loses an electron to form a
cation.
👉 If the
ionisation enthalpy is low,
the electron can be removed easily.
👉 Therefore, atoms with low ionisation enthalpy favour the formation of
ionic bonds.

Example:
alkali metals and
alkaline earth metals easily form cations.

Step 2: Effect of electron gain enthalpy
👉 The second atom gains an electron to form an
anion.
👉 If the
electron gain enthalpy is highly negative,
the atom can accept electrons easily.
👉 This makes the formation of an ionic bond more favourable.

Example:
halogens (Group 17)
readily gain electrons.

Step 3: Effect of lattice energy
👉 When gaseous ions combine to form a
solid ionic crystal,
energy is released.
👉 This released energy is called
lattice energy.
👉 Greater lattice energy means
stronger ionic bonding and
greater stability.

A⁺(g) + B⁻(g) → A⁺B⁻(s) + U

Here
U represents lattice energy.

Step 4: Overall conclusion
👉 Ionic compounds are most stable when:
•
Ionisation enthalpy is low
•
Electron gain enthalpy is highly negative
•
Lattice energy is large

This means the overall process should be
energetically favourable.

Did You Know?
👉 The stability of an
ionic bond does not depend only on electron transfer but also strongly on
lattice energy.

👉 Ions with
smaller size and higher charge
attract each other more strongly, leading to
greater lattice energy.

👉 Easy exam trick to remember favourable factors:
Low IE + Highly negative EGE + High Lattice Energy → Ionic bond formation

Q.7: Discuss the
shape of the following molecules
using the
VSEPR model
:
BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃

Answer
According to the
VSEPR theory, the shape of a molecule depends on the
repulsion between electron pairs around the central atom.

Molecular Shapes:

•
BeCl₂ → Linear (Bond angle ≈ 180°)
•
BCl₃ → Trigonal planar (Bond angle ≈ 120°)
•
SiCl₄ → Tetrahedral (Bond angle ≈ 109.5°)
•
AsF₅ → Trigonal bipyramidal (Bond angles ≈ 90° and 120°)
•
H₂S → Bent or V-shaped (Bond angle ≈ 92°)
•
PH₃ → Trigonal pyramidal (Bond angle ≈ 93.5°)

Explanation (Step by Step)
Step 1: Principle of VSEPR theory
👉 According to the
Valence Shell Electron Pair Repulsion (VSEPR) theory,
electron pairs around a central atom arrange themselves as far apart as possible to minimize repulsion.

Step 2: Shapes of the given molecules

BeCl₂
👉 Central atom Be has
2 bonding pairs and no lone pairs.
👉 Geometry →
Linear (180°).

BCl₃
👉 B has
3 bonding pairs and no lone pairs.
👉 Geometry →
Trigonal planar (120°).

SiCl₄
👉 Si has
4 bonding pairs.
👉 Geometry →
Tetrahedral (109.5°).

AsF₅
👉 As has
5 bonding pairs.
👉 Geometry →
Trigonal bipyramidal.

H₂S
👉 S has
2 bonding pairs and 2 lone pairs.
👉 Geometry →
Bent or V-shaped.

PH₃
👉 P has
3 bonding pairs and 1 lone pair.
👉 Geometry →
Trigonal pyramidal.

Did You Know?
👉 Lone pairs repel more strongly than bonding pairs, which is why bond angles often decrease when lone pairs are present.

👉 Example order of bond angles:
CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°)

👉 VSEPR theory is widely used to predict
molecular geometry and bond angles.

Q.8: Although the geometries of
NH₃
and
H₂O
molecules are
distorted tetrahedral
,
the bond angle in water is less than that of ammonia. Discuss.

Answer
✅ Both
NH₃
and
H₂O
have
tetrahedral electron pair geometry,
so their molecular shapes are considered
distorted tetrahedral.

✅ In
NH₃,
nitrogen has
one lone pair.
✅ In
H₂O,
oxygen has
two lone pairs.

✅ Since
lone pair–lone pair and lone pair–bond pair repulsions
are stronger, the bond pairs in water are pushed closer together.

Therefore:
• NH₃ bond angle ≈
107°
• H₂O bond angle ≈
104.5°

Explanation (Step by Step)
Step 1: Basis of VSEPR theory
👉 According to the
VSEPR theory,
electron pairs around the central atom repel each other and arrange themselves to minimize repulsion.

The order of repulsion is:
lone pair – lone pair > lone pair – bond pair > bond pair – bond pair

Step 2: Structure of NH₃
👉 In ammonia, the central atom
N has
4 electron pair regions:
• 3 bond pairs (N–H)
• 1 lone pair

👉 Because of lone pair–bond pair repulsion, the bond angle decreases from the ideal
109.5° to about
107°.

Step 3: Structure of H₂O
👉 In water, the central atom
O has
4 electron pair regions:
• 2 bond pairs (O–H)
• 2 lone pairs

👉 Because there are
two lone pairs,
repulsion is greater than in NH₃.
👉 This pushes the O–H bond pairs closer together and reduces the bond angle to about
104.5°.

Step 4: Conclusion
👉 Although both molecules have
tetrahedral electron geometry,
the greater number of
lone pairs in H₂O
increases repulsion and decreases the bond angle compared with NH₃.

Did You Know?
👉 The
ideal tetrahedral bond angle is
109.5°.

👉 Lone pairs occupy
more space
than bond pairs, so they push bonding atoms closer together.

👉 This gives the common bond angle order:
CH₄ (109.5°) > NH₃ (107°) > H₂O (104.5°)

Q.9: How do you express the
bond strength
in terms of
bond order
?

Answer
✅
Bond strength
is directly proportional to the
bond order.

✅ This means that as the
bond order increases,
the bond becomes
stronger.

✅ This happens because higher bond order usually results in greater
bond dissociation enthalpy.

Bond Strength ∝ Bond Order

Explanation (Step by Step)
Step 1: Basis of bond strength
👉 The strength of a bond depends on the
energy required to break the bond.
👉 This energy is known as
bond dissociation enthalpy.
👉 The greater this energy, the
stronger the bond.

Step 2: Effect of bond order
👉
Bond order represents the number of shared electron pairs between two atoms.
👉 When bond order increases, the
attraction between atoms increases.
👉 As a result, the bond becomes
shorter and stronger.

Step 3: Explanation with examples
• N₂ : Bond order =
3,
Bond enthalpy ≈
945 kJ mol⁻¹
• O₂ : Bond order =
2,
Bond enthalpy ≈
498 kJ mol⁻¹

👉 Since N₂ has a
higher bond order,
its bond is
stronger than that of O₂.

Step 4: Conclusion
👉 Therefore, bond strength increases with increase in bond order:

Bond Order ↑ → Bond Strength ↑ → Bond Dissociation Enthalpy ↑

Did You Know?
👉 Generally, bond strength follows the order:

Single bond < Double bond < Triple bond

👉 As bond order increases, the
bond length usually decreases.

👉 That is why molecules with
triple bonds such as
N₂
are extremely
stable and strong.

Q.10: Define the
bond length
.

Answer
✅ The
bond length is defined as the
equilibrium distance between the nuclei of two bonded atoms
in a molecule.

✅ It is usually expressed in
picometre (pm).

1 pm = 10⁻¹² m

Explanation (Step by Step)
Step 1: Basic meaning
👉 When two atoms form a
chemical bond,
a certain stable distance is established between their nuclei.
👉 This stable or
equilibrium distance
is called the
bond length.

Step 2: Unit of bond length
👉 Bond length is extremely small, so it is generally measured in
picometres (pm).

Step 3: In ionic compounds
👉 In ionic compounds, bond length is approximately equal to:
Radius of cation + Radius of anion

Step 4: In covalent compounds
👉 In covalent molecules, bond length is approximately equal to:
Sum of the covalent radii of the two bonded atoms

Did You Know?
👉 Generally, when
bond order increases,
the
bond length decreases.

👉 A
shorter bond
usually indicates a
stronger bond.

👉 Even between the same atoms, different types of bonds have different lengths:
Single bond > Double bond > Triple bond

Q.11: Explain the
important aspects of resonance
with reference to the
CO₃²⁻ ion
.

Answer
✅ The
carbonate ion (CO₃²⁻)
cannot be represented by a single Lewis structure.

✅ Instead, it is described by
three equivalent resonance structures
in which the double bond between carbon and oxygen is distributed among the three oxygen atoms.

✅ The actual structure is a
resonance hybrid, and all
C–O bonds are identical.

Explanation (Step by Step)
Step 1: Lewis structures of CO₃²⁻
👉 In one Lewis structure, carbon forms a
double bond with one oxygen
and single bonds with the other two oxygen atoms.
👉 However, the double bond can be placed with
any of the three oxygen atoms,
giving
three equivalent structures.

Step 2: Resonance concept
👉 The real structure of the carbonate ion is not any one of these structures but a
resonance hybrid of all three.
👉 In this hybrid structure, the electrons are
delocalized over all three C–O bonds.

Step 3: Consequences of resonance
👉 All three
C–O bonds have the same bond length.
👉 The bond order of each C–O bond is
1⅓.
👉 The ion is
more stable than any individual resonance structure.

Did You Know?
👉 Resonance explains why the three
C–O bonds in CO₃²⁻
are equal in length.

👉 The extra stability obtained due to electron delocalization is called
resonance stabilization.

👉 Many important molecules such as
benzene and nitrate ion
also show resonance.

Q.12: H₃PO₃ can be represented by structures (1) and (2) shown above.
Can these two structures be taken as the
canonical forms of the resonance hybrid
representing
H₃PO₃?
If not, give reasons.

Answer
❌
No, the two given structures
cannot be considered canonical (resonance) forms of
H₃PO₃.

👉 In resonance structures,
only electrons shift while the
positions of atoms remain the same.

👉 In the given structures, the
hydrogen atom changes its position
from oxygen to phosphorus.

👉 Therefore these two structures represent
different compounds, not resonance forms.

Explanation (Step by Step)
Step 1: Condition for resonance
👉 Resonance occurs when two or more structures differ only in the
distribution of electrons.
👉 The
positions of atoms must remain unchanged.

Step 2: Analysis of the given structures
👉 In structure (1), hydrogen is attached to
phosphorus.
👉 In structure (2), hydrogen is attached to
oxygen.

👉 This means the
position of hydrogen changes.

Step 3: Correct structure of H₃PO₃
👉 The correct structure contains:
• One
P=O double bond
• Two
P–OH groups
• One
P–H bond

👉 Therefore hydrogen is directly bonded to phosphorus in the actual molecule.

Did You Know?
👉
H₃PO₃ is called
phosphorous acid.

👉 It is actually
diprotic, not triprotic, because only
two hydrogens are attached to oxygen and can be ionized.

👉 The hydrogen attached to
phosphorus (P–H) does not show acidic behavior.

Q.13: Write the
resonance structures
for the following species:

SO₃, NO₂ and NO₃⁻

Answer
Resonance occurs when a molecule can be represented by
two or more Lewis structures that differ only in the
distribution of electrons.

1. SO₃
👉 Sulphur forms three equivalent resonance structures.
O=S–O ↔ O–S=O ↔ O–S–O

(The double bond shifts among the three oxygen atoms.)

2. NO₂
👉 Nitrogen dioxide has two resonance structures.
O=N–O ↔ O–N=O

(The double bond alternates between the two oxygen atoms.)

3. NO₃⁻
👉 Nitrate ion has three resonance structures.
O=N–O⁻ ↔ O⁻–N=O ↔ O–N=O⁻

(The double bond moves among the three oxygen atoms.)

Explanation (Step by Step)
Step 1: What is resonance?
👉 Some molecules cannot be represented by a
single Lewis structure.
👉 Instead, they are described by multiple structures called
resonance structures.

Step 2: Resonance in SO₃
👉 Sulphur is bonded to three oxygen atoms.
👉 The
S=O double bond can appear with any oxygen atom.
👉 Therefore three equivalent resonance structures exist.
👉 As a result, all
S–O bonds become identical.

Step 3: Resonance in NO₂
👉 Nitrogen forms one double bond and one single bond with oxygen atoms.
👉 The
double bond shifts between the two oxygen atoms.
👉 Hence two resonance structures are possible.

Step 4: Resonance in NO₃⁻
👉 Nitrate ion contains three oxygen atoms bonded to nitrogen.
👉 One N=O double bond and two N–O single bonds are present in each structure.
👉 The double bond can shift among the
three oxygen atoms.
👉 Therefore three equivalent resonance structures are formed.

👉 Because of resonance, all
N–O bond lengths become equal.

Did You Know?
👉 The actual molecule is not any single resonance structure but a
resonance hybrid of all structures.

👉 Resonance increases the
stability of molecules because electrons become
delocalized.

👉 In
NO₃⁻, all N–O bonds have the same
bond length due to resonance.

Q.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

Answer
(a) K and S
K → K⁺ + e⁻
S + 2e⁻ → S²⁻

Lewis representation:
2K• + :S:•••• → 2K⁺ + [:S:]²⁻

(b) Ca and O
Ca → Ca²⁺ + 2e⁻
O + 2e⁻ → O²⁻

Lewis representation:
Ca•• + :O:•••• → Ca²⁺ + [:O:]²⁻

(c) Al and N
Al → Al³⁺ + 3e⁻
N + 3e⁻ → N³⁻

Lewis representation:
Al••• + :N:••• → Al³⁺ + [:N:]³⁻

Explanation (Step by Step)
Step 1: Determine valence electrons
👉 K → 1 valence electron
👉 Ca → 2 valence electrons
👉 Al → 3 valence electrons
👉 S → 6 valence electrons
👉 O → 6 valence electrons
👉 N → 5 valence electrons

Step 2: Electron transfer
👉 Metals lose electrons to form
cations.
👉 Non-metals gain electrons to form
anions.

Step 3: Formation of ionic compounds
• K₂S is formed from K⁺ and S²⁻.
• CaO is formed from Ca²⁺ and O²⁻.
• AlN is formed from Al³⁺ and N³⁻.

Did You Know?
👉 Ionic bonds are formed by
transfer of electrons from metals to non-metals.

👉 After gaining or losing electrons, atoms achieve a
stable noble gas configuration.

👉 Lewis symbols are very useful for understanding
ionic bond formation.

Q.15: Although both
CO₂ and H₂O
are
triatomic molecules
, the shape of
H₂O is
bent while that of
CO₂ is
linear.
Explain this on the basis of
dipole moment.

Answer
✅ In
CO₂, the two C–O bond dipoles are equal and opposite, so they cancel each other.
Therefore, the
net dipole moment is zero, indicating a
linear shape.

✅ In
H₂O, the O–H bond dipoles do not cancel because the molecule has a
bent geometry.
Hence, the molecule has a
non-zero dipole moment.

Explanation (Step by Step)
(i) Dipole moment concept
👉 Dipole moment depends on the
magnitude and direction of bond dipoles.
👉 If bond dipoles cancel each other, the molecule becomes
non-polar.

(ii) Case of CO₂
👉 CO₂ has a
linear geometry with a bond angle of
180°.
👉 The dipole moments of the two C=O bonds act in
opposite directions.
👉 They cancel each other, giving
net dipole moment = 0.

(iii) Case of H₂O
👉 H₂O has a
bent or V-shaped geometry with a bond angle of about
104.5°.
👉 The O–H bond dipoles are not opposite to each other.
👉 Therefore, they
do not cancel and a net dipole moment exists.

(iv) Conclusion
👉 CO₂ → Linear and non-polar (μ = 0)
👉 H₂O → Bent and polar (μ ≠ 0)

Did You Know?
👉 Dipole moment is an important tool to determine
molecular geometry.

👉 Even if bonds are polar, the molecule can be
non-polar if the dipoles cancel (as in CO₂).

👉 Water is highly polar due to its
bent structure and strong dipole moment.

Q.16: Write the
significance / applications
of
dipole moment
.

Answer
The
dipole moment is very useful in understanding the nature of chemical bonds and molecular structure.

Main applications:
✅ To determine whether a molecule is
polar or non-polar.
✅ To estimate the
percentage ionic character in a bond.
✅ To determine the
molecular geometry or shape of molecules.
✅ To distinguish between molecules having
similar structures but different polarity.

Explanation (Step by Step)
(i) Determination of polarity
👉 If the dipole moment of a molecule is
zero, the molecule is generally
non-polar.
👉 If the dipole moment is
non-zero, the molecule is
polar.

(ii) Percentage ionic character
👉 Dipole moment helps in estimating the
ionic character of a bond.
👉 A higher dipole moment generally indicates
greater ionic character.

(iii) Determination of molecular geometry
👉 By measuring dipole moment, scientists can predict the
arrangement of atoms in a molecule.
👉 Example:
• CO₂ → μ = 0 →
Linear structure.
• H₂O → μ ≠ 0 →
Bent structure.

(iv) Distinguishing isomers
👉 Dipole moment can differentiate between molecules that have the
same formula but different structures.

Did You Know?
👉 Dipole moment is measured in the unit
Debye (D).

👉 The formula for dipole moment is:
μ = q × d
where q = magnitude of charge and d = distance between charges.

👉 Dipole moment helps chemists understand
bond polarity and molecular structure.

Q.17: Define
electronegativity
. How does it differ from
electron gain enthalpy
?

Answer
✅
Electronegativity is the tendency of an atom in a molecule to attract the shared pair of electrons towards itself.

✅
Electron gain enthalpy is the amount of energy released or absorbed when an isolated gaseous atom gains an electron to form a negative ion.

Main Difference:
👉 Electronegativity refers to atoms
in a bonded state.
👉 Electron gain enthalpy refers to
isolated gaseous atoms.

Explanation (Step by Step)
(i) Electronegativity
👉 It measures the ability of an atom to attract
bonding electrons towards itself.
👉 It is a
relative property and has no unit.
👉 Example: Fluorine has the
highest electronegativity.

(ii) Electron Gain Enthalpy
👉 It is the
energy change that occurs when an electron is added to an isolated gaseous atom.
👉 It is a
measurable thermodynamic quantity.
👉 It is expressed in
kJ mol⁻¹.

(iii) Key differences
• Electronegativity → property of atoms in a molecule.
• Electron gain enthalpy → property of isolated gaseous atoms.
• Electronegativity → dimensionless value.
• Electron gain enthalpy → has energy units.

Did You Know?
👉 Electronegativity generally
increases across a period and
decreases down a group in the periodic table.

👉 Halogens usually have
very high electron gain enthalpy because they need only one electron to complete their octet.

👉 Fluorine has the highest electronegativity value
(≈ 4.0 on Pauling scale).

Q.18: Explain with the help of a suitable example a
polar covalent bond
.

Answer
✅ A
polar covalent bond is a covalent bond in which the
shared pair of electrons is unequally distributed between the two atoms due to a difference in their
electronegativity.

✅ As a result, one atom develops a
partial negative charge (δ−) and the other develops a
partial positive charge (δ+).

Example:
In
HCl, chlorine is more electronegative than hydrogen, so the shared electron pair shifts towards chlorine.
H^δ+ — Cl^δ−

Explanation (Step by Step)
(i) Formation of covalent bond
👉 In a covalent bond, atoms share electrons to achieve a
stable electronic configuration.

(ii) Unequal sharing of electrons
👉 If the two atoms have
different electronegativities, the shared electrons are pulled closer to the more electronegative atom.
👉 This unequal sharing results in
partial charges on the atoms.

(iii) Example: HCl molecule
👉 Chlorine is more electronegative than hydrogen.
👉 Therefore, the bonding electrons shift towards chlorine.
👉 This creates a
dipole in the molecule.

H^δ+ → Cl^δ−

Did You Know?
👉 If the electronegativity difference between two atoms is
very small, the bond is
non-polar covalent (e.g., H₂, Cl₂).

👉 If the difference is
very large, an
ionic bond may form.

👉 Polar covalent bonds are common in molecules such as
HCl, H₂O, and NH₃.

Q.19: Arrange the following molecules in the order of
increasing ionic character of bonds
:
LiF, K₂O, N₂, SO₂, ClF₃
.

Answer
✅ The correct order of
increasing ionic character is:

N₂ < SO₂ < ClF₃ < LiF < K₂O

Explanation (Step by Step)
The ionic character of a bond mainly depends on:

👉
Difference in electronegativity (Δχ)
👉
Nature of atoms involved (metal or non-metal)
👉
Structure of the molecule

(i) N₂
👉 Both atoms are identical, so
Δχ = 0
👉 Therefore it is a
pure covalent bond and has the
least ionic character.

(ii) SO₂
👉 The electronegativity difference between
S and O is moderate.
👉 Hence it is
polar covalent and more ionic than N₂.

(iii) ClF₃
👉 The electronegativity difference between
Cl and F is greater than in SO₂.
👉 Therefore it has
greater ionic character than SO₂.

(iv) LiF
👉 Combination of
metal (Li) and
non-metal (F).
👉 The electronegativity difference is very large, so the compound is
highly ionic.

(v) K₂O
👉
Potassium is a very electropositive metal.
👉 When combined with
oxygen, the ionic character becomes extremely high.
👉 Hence among the given compounds it shows the
maximum ionic character.

Conclusion
Therefore, the increasing order of ionic character is:

N₂ < SO₂ < ClF₃ < LiF < K₂O

Did You Know?
👉 When the
electronegativity difference (Δχ)
between atoms increases, the
ionic character generally increases.

👉 Bonds between
identical atoms such as
N₂ or O₂ are
pure covalent.

👉 Compounds formed between
metals and non-metals
usually show
high ionic character.

Q.20: The skeletal structure of
CH₃COOH (acetic acid)
shown above is correct, but some of the bonds are shown incorrectly.
Write the
correct Lewis structure for acetic acid.

Answer
The correct Lewis structure of
acetic acid (CH₃COOH) is:

H₃C — C(=O) — O — H

👉 The second carbon atom forms a
double bond with one oxygen (C=O) and a
single bond with another oxygen.

👉 The second oxygen is bonded to hydrogen forming an
–OH group.

Explanation (Step by Step)
Step 1: Total valence electrons
👉 C = 4 electrons × 2 = 8
👉 O = 6 electrons × 2 = 12
👉 H = 1 electron × 4 = 4

👉 Total =
24 valence electrons.

Step 2: Skeleton structure
👉 Carbon atoms form the backbone of the molecule.
👉 The structure becomes:
CH₃ – C – O – H

Step 3: Complete octets
👉 To satisfy the octet rule, one oxygen forms a
double bond with carbon.
👉 The other oxygen forms
single bonds with carbon and hydrogen.

Step 4: Lone pairs
👉 Each oxygen atom contains
two lone pairs of electrons.
👉 This satisfies the
octet rule for all atoms.

Did You Know?
👉
Acetic acid is the main component of
vinegar.

👉 The functional group present in acetic acid is the
carboxyl group (–COOH).

👉 The
C=O bond is shorter and stronger than the
C–O bond.

Q.21: Apart from
tetrahedral geometry
, another possible geometry for
CH₄
is
square planar
, where four hydrogen atoms are at the corners of a square and carbon is at the centre.
Explain why
CH₄
is not square planar.

Answer
✅ The molecule
CH₄
is not square planar because square planar geometry requires
dsp² hybridisation.

✅ Carbon is a
second period element and does not have
d orbitals in its valence shell.

✅ Therefore, carbon cannot undergo
dsp² hybridisation.

✅ In CH₄, carbon undergoes
sp³ hybridisation,
resulting in a
tetrahedral geometry.

Explanation (Step by Step)
Step 1: Requirement for square planar geometry
👉 In a
square planar arrangement,
four bonds lie in the same plane and the bond angles are
90°.
👉 This geometry generally arises from
dsp² hybridisation.

Step 2: Limitation of carbon
👉 Carbon has atomic number
6 and belongs to the
second period.
👉 In the second period, only
2s and 2p orbitals are available.
👉
2d orbitals do not exist,
so carbon cannot form dsp² hybrid orbitals.

Step 3: Actual hybridisation in CH₄
👉 In methane, carbon mixes
one 2s orbital and
three 2p orbitals to form
sp³ hybrid orbitals.
👉 These four hybrid orbitals arrange themselves to minimize repulsion, forming a
tetrahedral shape.
👉 The H–C–H bond angle is approximately
109.5°, not 90°.

Step 4: Conclusion
👉 Because carbon does not have
d orbitals,
it cannot adopt square planar geometry.
👉 Therefore, the stable geometry of
CH₄
is
tetrahedral.

Did You Know?
👉 Elements of the
second period such as
C, N, O, and F
generally cannot form
expanded octets because they lack d orbitals.

👉 The
tetrahedral shape of CH₄
is also explained by the
VSEPR theory, where four bond pairs arrange themselves to minimize repulsion.

👉 Square planar geometry is commonly observed in some
transition metal complexes, but not in methane.

Q.22: Explain why
BeH₂
molecule has a
zero dipole moment
although the
Be–H bonds are polar
.

Answer
✅ The
Be–H bonds in BeH₂ are slightly
polar because hydrogen is more electronegative than beryllium.

✅ However, the BeH₂ molecule has a
linear geometry with a bond angle of
180°.

✅ Due to this
symmetrical arrangement, the two bond dipole moments act in opposite directions and
cancel each other.

👉 Therefore, the
net dipole moment of BeH₂ is zero.

Explanation (Step by Step)
(i) Nature of Be–H bond
👉 Hydrogen is slightly more
electronegative than beryllium.
👉 Therefore, each Be–H bond has a
bond dipole.

(ii) Molecular geometry
👉 According to
VSEPR theory, BeH₂ has two bonding pairs and no lone pairs around Be.
👉 The molecule therefore adopts a
linear shape.

H — Be — H (180°)

(iii) Cancellation of dipole moments
👉 The dipole moments of the two Be–H bonds are equal and act in
opposite directions.
👉 Hence they
cancel each other.

(iv) Result
👉 Net dipole moment (μ) =
0.

Did You Know?
👉 A molecule may contain
polar bonds but still be
non-polar if the bond dipoles cancel due to symmetry.

👉 Similar examples include
CO₂ and CCl₄.

👉 Dipole moment is therefore very useful in determining
molecular geometry.

Q.23: Which out of
NH₃
and
NF₃
has a
higher dipole moment
and why?

Answer
✅
NH₃ has a
higher dipole moment than
NF₃.

✅ In NH₃, the dipole moments of the
N–H bonds and the
lone pair on nitrogen act in the
same direction, increasing the overall dipole moment.

✅ In NF₃, the dipole moments of the
N–F bonds act in the
opposite direction to the lone pair dipole, which reduces the net dipole moment.

Explanation (Step by Step)
(i) Molecular geometry
👉 Both NH₃ and NF₃ have a
trigonal pyramidal geometry due to the presence of one lone pair on nitrogen.

(ii) Case of NH₃
👉 Nitrogen is more electronegative than hydrogen.
👉 Therefore, the bond dipoles of the N–H bonds point
towards nitrogen.
👉 These dipoles act in the
same direction as the lone pair dipole.
👉 This
increases the net dipole moment.

(iii) Case of NF₃
👉 Fluorine is more electronegative than nitrogen.
👉 The bond dipoles of N–F bonds point
away from nitrogen.
👉 These dipoles act
opposite to the lone pair dipole.
👉 Hence they partially
cancel each other, resulting in a smaller dipole moment.

(iv) Result
NH₃ → Dipole moment ≈
1.46 D
NF₃ → Dipole moment ≈
0.24 D

Did You Know?
👉 Even though fluorine is highly electronegative, NF₃ has a
smaller dipole moment due to cancellation of dipoles.

👉 Dipole moment depends on both
bond polarity and
molecular geometry.

👉 Therefore, molecules with similar shapes can still have
different dipole moments.

Q.24: What is meant by
hybridisation of atomic orbitals
? Describe the shapes of
sp, sp², and sp³
hybrid orbitals.

Answer
✅
Hybridisation is the process in which atomic orbitals of nearly equal energy mix together to form
new equivalent hybrid orbitals.

Types and shapes of hybrid orbitals:

•
sp hybridisation →
Linear shape (bond angle ≈ 180°)
•
sp² hybridisation →
Trigonal planar shape (bond angle ≈ 120°)
•
sp³ hybridisation →
Tetrahedral shape (bond angle ≈ 109.5°)

Explanation (Step by Step)
(i) Meaning of hybridisation
👉 Atomic orbitals such as
s and p orbitals combine to form new orbitals called
hybrid orbitals.
👉 These hybrid orbitals have
equal energy and identical shape.

(ii) sp hybridisation
👉 One s orbital mixes with one p orbital.
👉 Number of hybrid orbitals formed =
2.
👉 Arrangement →
Linear geometry.
👉 Bond angle ≈
180°.
Example:
BeCl₂, CO₂.

(iii) sp² hybridisation
👉 One s orbital mixes with two p orbitals.
👉 Number of hybrid orbitals formed =
3.
👉 Arrangement →
Trigonal planar geometry.
👉 Bond angle ≈
120°.
Example:
BF₃, SO₃.

(iv) sp³ hybridisation
👉 One s orbital mixes with three p orbitals.
👉 Number of hybrid orbitals formed =
4.
👉 Arrangement →
Tetrahedral geometry.
👉 Bond angle ≈
109.5°.
Example:
CH₄.

Did You Know?
👉 Hybridisation helps explain
molecular geometry and bond angles in many molecules.

👉 The number of hybrid orbitals formed is always equal to the
number of orbitals mixed.

👉 Hybridisation is widely used to explain bonding in molecules like
CH₄, BF₃, and BeCl₂.

Q.25: Describe the
change in hybridisation
(if any) of the
Al atom
in the following reaction:
AlCl₃ + Cl⁻ → AlCl₄⁻

Answer
✅ In this reaction, the hybridisation of the
Al atom changes.

• In
AlCl₃,
hybridisation of Al =
sp².

• In
AlCl₄⁻,
hybridisation of Al =
sp³.

Therefore, the hybridisation changes as:
sp² → sp³

Explanation (Step by Step)
Step 1: Hybridisation in AlCl₃
👉 In
AlCl₃,
aluminium forms
three bonds with chlorine atoms.
👉 Therefore, there are
three electron density regions around Al.
👉 This leads to
sp² hybridisation and a
trigonal planar geometry.

Step 2: Addition of Cl⁻
👉 The
chloride ion (Cl⁻)
has a lone pair of electrons.
👉 It donates this electron pair to aluminium, forming a
coordinate bond.
👉 Now aluminium becomes surrounded by
four electron density regions.

Step 3: Hybridisation in AlCl₄⁻
👉 With four bonding regions, aluminium undergoes
sp³ hybridisation.
👉 The geometry becomes
tetrahedral.

Step 4: Conclusion
👉 Thus, during the reaction the hybridisation of aluminium changes from
sp² (in AlCl₃) → sp³ (in AlCl₄⁻)
.

Did You Know?
👉
AlCl₃ behaves as a
Lewis acid because aluminium is
electron deficient.

👉 The
Cl⁻ ion acts as a
Lewis base by donating an electron pair.

👉 When hybridisation changes, the
geometry of the molecule or ion
also changes.

Q.26: Is there any change in the
hybridisation of B and N atoms
as a result of the following reaction?
BF₃ + NH₃ → F₃B·NH₃

Answer
✅ In the molecule
BF₃, boron has
sp² hybridisation with a
trigonal planar structure.

✅ When
NH₃ donates its lone pair to boron, a
coordinate bond is formed.

👉 As a result, the hybridisation of
boron changes from sp² to sp³.

👉 The hybridisation of
nitrogen remains sp³.

Explanation (Step by Step)
(i) Hybridisation in BF₃
👉 Boron has three valence electrons and forms three σ-bonds with fluorine atoms.
👉 Therefore, boron undergoes
sp² hybridisation.
👉 Geometry →
Trigonal planar.

(ii) Role of NH₃
👉 Nitrogen in NH₃ has a
lone pair of electrons.
👉 This lone pair is donated to boron forming a
coordinate (dative) bond.

(iii) Change in hybridisation
👉 After bond formation, boron becomes surrounded by
four electron pairs.
👉 Therefore, boron changes its hybridisation to
sp³ and the geometry becomes
tetrahedral.

👉 Nitrogen already has
sp³ hybridisation, so its hybridisation
does not change.

Did You Know?
👉
BF₃ acts as a
Lewis acid because boron has an incomplete octet.

👉
NH₃ acts as a
Lewis base because it donates a lone pair.

👉 The compound
F₃B·NH₃ is called a
Lewis acid–base adduct.

Q.27: Draw diagrams showing the formation of a
double bond and a triple bond
between carbon atoms in
C₂H₄ and
C₂H₂ molecules.

Answer
1. Formation of double bond in C₂H₄ (Ethene)

👉 Each carbon atom undergoes
sp² hybridisation.
👉 One
σ (sigma) bond is formed by
sp²–sp² overlap between carbon atoms.
👉 One
π (pi) bond is formed by
sidewise overlap of unhybridised p-orbitals.

H      H
  /
C = C
/
H      H

Double bond = 1 σ bond + 1 π bond

2. Formation of triple bond in C₂H₂ (Ethyne)

👉 Each carbon atom undergoes
sp hybridisation.
👉 One
σ bond is formed by
sp–sp overlap between carbon atoms.
👉 Two
π bonds are formed by overlap of
two unhybridised p-orbitals.

H — C ≡ C — H

Triple bond = 1 σ bond + 2 π bonds

Explanation (Step by Step)
Step 1: Hybridisation in C₂H₄
👉 Each carbon atom forms three σ bonds, so it undergoes
sp² hybridisation.
👉 The remaining unhybridised p-orbital forms a
π bond.

Step 2: Hybridisation in C₂H₂
👉 Each carbon atom forms two σ bonds, so it undergoes
sp hybridisation.
👉 Two unhybridised p-orbitals remain.
👉 These form
two π bonds.

Step 3: Bond composition
👉
Double bond =
1 σ + 1 π.
👉
Triple bond =
1 σ + 2 π.

Did You Know?
👉
σ bonds are formed by
head-on overlap of orbitals.

👉
π bonds are formed by
sidewise overlap of p-orbitals.

👉
Triple bonds are shorter and stronger than
double bonds.

Q.28: What is the
total number of σ (sigma) and π (pi) bonds
in the following molecules?
(a) C₂H₂ (b) C₂H₄

Answer
(a) C₂H₂ (Acetylene)
Structure:
H – C ≡ C – H

• Number of
σ bonds =
3
• Number of
π bonds =
2

(b) C₂H₄ (Ethene)
Structure:
H₂C = CH₂

• Number of
σ bonds =
5
• Number of
π bonds =
1

Explanation (Step by Step)
(i) Nature of σ and π bonds
👉 A
sigma (σ) bond is formed by
head-on overlap of orbitals.
👉 A
pi (π) bond is formed by
sidewise overlap of orbitals.

(ii) Bonds in C₂H₂
👉 One C≡C triple bond contains:
•
1 σ bond
•
2 π bonds
👉 Two C–H bonds contribute
2 σ bonds.

👉 Total =
3 σ and
2 π bonds.

(iii) Bonds in C₂H₄
👉 One C=C double bond contains:
•
1 σ bond
•
1 π bond
👉 Four C–H bonds contribute
4 σ bonds.

👉 Total =
5 σ and
1 π bond.

Did You Know?
👉 A
single bond contains
1 σ bond.

👉 A
double bond contains
1 σ + 1 π bond.

👉 A
triple bond contains
1 σ + 2 π bonds.

Q.29: Considering the
x-axis as the internuclear axis
, which of the following orbitals will
not form a σ bond
and why?
(a) 1s and 1s
(b) 1s and 2p_x
(c) 2p_y and 2p_y
(d) 1s and 2s

Answer
✅ The orbitals that
will not form a σ bond are:

because these orbitals are oriented
perpendicular to the internuclear (x) axis
and cannot overlap in a head-on manner.

The following combinations
can form σ bonds:
• 1s – 1s
• 1s – 2p_x
• 1s – 2s

Explanation (Step by Step)
Step 1: Condition for σ bond formation
👉 A
σ (sigma) bond forms when orbitals overlap
head-on along the internuclear axis.
👉 In this question, the
internuclear axis is the x-axis.

Step 2: Analysis of each pair

(a) 1s and 1s
👉 s orbitals are
spherically symmetrical.
👉 They can overlap directly along the internuclear axis.
✅ Therefore, they form a
σ bond.

(b) 1s and 2p_x
👉 The
2p_x orbital lies along the
x-axis.
👉 Hence it can overlap head-on with the 1s orbital.
✅ Therefore, a
σ bond is formed.

(c) 2p_y and 2p_y
👉 The
2p_y orbitals are oriented along the
y-axis.
👉 Since the internuclear axis is
x-axis, head-on overlap is not possible.
❌ Therefore, they
cannot form a σ bond.
👉 Instead, they may overlap
sidewise to form a
π bond.

(d) 1s and 2s
👉 Both are
s orbitals and are spherical.
👉 They can overlap along the internuclear axis.
✅ Therefore, a
σ bond can form.

Did You Know?
👉 Which p orbital forms a
σ bond depends on the
internuclear axis.

👉 If the axis is
x-axis, then
p_x forms σ overlap, while
p_y and
p_z usually form
π bonds.

👉 Because
s orbitals are spherical, they can form σ bonds in almost any direction.

Q.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

Answer
(a) In
CH₃–CH₃, both carbon atoms use
sp³ hybrid orbitals.

(b) In
CH₃–CH=CH₂, the first carbon uses
sp³ hybridisation while the second and third carbons use
sp² hybridisation.

(d) In
CH₃–CHO, the methyl carbon has
sp³ hybridisation while the carbonyl carbon has
sp² hybridisation.

(e) In
CH₃COOH, the methyl carbon has
sp³ hybridisation while the carboxyl carbon has
sp² hybridisation.

Explanation (Step by Step)
Hybridisation depends on the number of σ-bonds formed by a carbon atom.

👉 If carbon forms
four σ-bonds, hybridisation is
sp³ and geometry becomes
tetrahedral.

👉 If carbon forms
three σ-bonds and one π-bond, hybridisation is
sp² with
trigonal planar geometry.

Application to given molecules

(a) In ethane, each carbon forms four σ-bonds →
sp³.

(b) In propene, the carbon atoms involved in
C=C double bond are
sp² hybridised.

(d) In aldehyde group
(–CHO), the carbon of
C=O double bond is
sp² hybridised.

(e) In carboxylic acid group
(–COOH), the carbonyl carbon is
sp² hybridised due to the
C=O double bond.

Did You Know?
👉
sp³ hybridised carbon forms
tetrahedral geometry (109.5°).

👉
sp² hybridised carbon forms
trigonal planar geometry (120°).

👉
Double bonds always involve
one σ-bond and one π-bond.

Q.31: What do you understand by
bond pairs and lone pairs of electrons
? Illustrate by giving
one example of each.

Answer
🔹 A
bond pair is a pair of electrons
shared between two atoms to form a
covalent bond.

Example: In
H₂, the two hydrogen atoms share one pair of electrons.
This shared pair is called a
bond pair.

🔹 A
lone pair is a pair of electrons
present on one atom but not shared with another atom.

Example: In
NH₃, nitrogen has
one lone pair of electrons which is not involved in bonding.

Explanation (Step by Step)
(i) Bond Pair
👉 When two atoms form a
covalent bond, they share a pair of electrons.
👉 This shared pair is called a
bond pair.

👉 Example: In
H₂, the bond between the two hydrogen atoms is formed by
one shared electron pair.

(ii) Lone Pair
👉 Some electron pairs remain
unshared on an atom.
👉 These non-bonding electrons are called
lone pairs.

👉 Example: In
NH₃, nitrogen forms three bonds with hydrogen and keeps
one lone pair.

👉 Lone pairs strongly influence
molecular shape according to
VSEPR theory.

Did You Know?
👉
Lone pairs occupy more space than bond pairs because they are attracted by only one nucleus.

👉 Due to this, the repulsion order in molecules is:
Lone pair – Lone pair > Lone pair – Bond pair > Bond pair – Bond pair
.

👉 This is the reason why molecules like
NH₃ and H₂O have distorted shapes.

Q.32:
Distinguish between
a
sigma (σ) bond
and a
pi (π) bond
.

Answer
✅ Both
σ (sigma)
and
π (pi)
bonds are
covalent bonds, but they differ in the way they are formed, their strength, rotation ability, and stability.

✅ Generally, a
σ bond is stronger,
while a
π bond is comparatively weaker.

Explanation (Step by Step)

S.No.	Basis	σ Bond (Sigma Bond)	π Bond (Pi Bond)
1	Mode of formation	Formed by head-on overlap of orbitals.	Formed by sidewise overlap of orbitals.
2	Orbitals involved	Can be formed by s–s, s–p, p–p or hybrid orbitals.	Usually formed by unhybridised p–p orbitals (sometimes d orbitals).
3	Rotation	Free rotation around the bond axis is possible.	Free rotation is not possible because sidewise overlap would break.
4	Strength	Stronger and more stable.	Weaker and less stable.
5	Existence	Can exist independently (single bond = 1σ).	Cannot exist alone; it forms along with a σ bond.
6	Role in molecule	Determines the basic framework and geometry of the molecule.	Influences multiple bonding and reactivity.
7	Examples	Single bond = 1σ	Double bond = 1σ + 1π Triple bond = 1σ + 2π

Did You Know?
👉 Because of the
π bond in alkenes,
free rotation around C=C is not possible, which leads to
cis–trans isomerism.

👉 In a
σ bond, electron density lies along the
internuclear axis, making it stronger.

👉
π electrons are more reactive and therefore play an important role in many
organic reactions.

Q.33: Explain the formation of
H₂ molecule
on the basis of
Valence Bond Theory (VBT)
.

Answer
✅ According to the
Valence Bond Theory, the
H₂ molecule is formed by the
overlap of the 1s orbitals of two hydrogen atoms.

✅ When the atoms approach each other,
attractive and repulsive forces act simultaneously.
✅ At a certain distance, the net attraction becomes maximum and the
energy becomes minimum, forming a stable H₂ molecule.

✅ The
bond length of H₂ is about
74 pm and the
bond dissociation energy is about
433 kJ mol⁻¹.

Explanation (Step by Step)
(i) Background of the theory
👉 The
Valence Bond Theory was proposed by
Heitler and London (1927) and later developed by Pauling and others.
👉 This theory explains bond formation on the basis of
orbital overlap and energy changes.

(ii) Initial state
Consider two hydrogen atoms:
👉 Atom A (nucleus N_A, electron e_A)
👉 Atom B (nucleus N_B, electron e_B)
When the atoms are far apart, there is
no significant interaction between them.

(iii) Forces when atoms approach
Attractive forces:
N_A–e_A, N_B–e_B (internal attraction)
N_A–e_B, N_B–e_A (mutual attraction)

Repulsive forces:
e_A–e_B (electron–electron repulsion)
N_A–N_B (nucleus–nucleus repulsion)

(iv) When does stability occur?
👉 As the atoms approach each other, the
total energy decreases because attraction becomes stronger than repulsion.
👉 At a certain distance,
attractive and repulsive forces balance each other.
👉 At this equilibrium distance the energy becomes
minimum and a stable H₂ molecule is formed.
👉 This distance is the
bond length = 74 pm.

(v) Energy change
H(g) + H(g) → H₂(g) + 433 kJ mol⁻¹
👉 Energy is released during bond formation, making H₂
more stable.

Reverse reaction:
H₂(g) → H(g) + H(g)
(ΔH = +433 kJ mol⁻¹)
👉 This energy is called the
bond dissociation energy.

(vi) Conclusion
✅ The H₂ molecule is formed due to effective
1s–1s orbital overlap forming a
σ bond.
✅ The
minimum energy state corresponds to the
maximum stability.

Did You Know?
👉
H₂ is the
simplest diatomic molecule, so it is the classical example used to explain VBT.

👉 The release of energy during H–H bond formation shows that
bond formation is energetically favourable.

👉 The bond formed by
1s–1s axial overlap is called a
σ bond.

Q.34: Write the
important conditions
required for the
linear combination of atomic orbitals (LCAO)
to form
molecular orbitals
.

Answer
The main conditions for the formation of
molecular orbitals by LCAO are:

✅ The
energies of the combining atomic orbitals must be equal or nearly equal.
✅ The
symmetry of the orbitals with respect to the molecular axis must be the same.
✅ There should be
maximum overlap between the atomic orbitals.

Explanation (Step by Step)
1) Energy Compatibility
👉 Atomic orbitals that combine to form molecular orbitals should have
equal or nearly equal energies.
👉 If the energy difference between orbitals is very large,
effective combination does not occur.
Example:
👉 In the same atom,
1s–1s overlap is possible.
👉 The energy difference between
1s and 2s is very large, so effective combination is unlikely.
👉 In different atoms, the orbitals should have
approximately similar energies.

2) Similar Symmetry
👉 The combining orbitals must have the
same symmetry with respect to the molecular axis (usually the z-axis).
👉 If symmetry does not match,
overlap cannot occur, even if the energies are similar.
Example:
👉
2p_z can combine with 2p_z.
👉
2p_z cannot combine with 2p_x or 2p_y because their symmetry is different.

3) Maximum Overlap
👉 The
greater the overlap between atomic orbitals, the stronger the interaction.
👉 Greater overlap increases the
electron density between nuclei, leading to a more stable bonding molecular orbital.

Conclusion
For effective LCAO formation, all three conditions must be satisfied:

Similar Energy + Same Symmetry + Maximum Overlap

Only then a stable
bonding molecular orbital is formed.

Did You Know?
👉 Having similar energy alone is not enough; if
symmetry does not match, orbitals cannot combine.

👉 Greater
orbital overlap usually means a
stronger chemical bond.

👉 In Molecular Orbital Theory, properties such as
bond order and magnetic behaviour are explained using these LCAO conditions.

Q.35: Use
Molecular Orbital Theory
to explain why the
Be₂ molecule does not exist.

Answer
According to
Molecular Orbital Theory, the
bond order of Be₂ is
zero.

Since the bond order is zero, there is
no net bond formation between the two beryllium atoms.

Therefore, the
Be₂ molecule does not exist as a stable molecule.

Explanation (Step by Step)
(i) Electronic configuration of Be atom
👉 Beryllium has atomic number
4.
👉 Electronic configuration:
1s² 2s².

(ii) Molecular orbital formation
👉 When two Be atoms combine, their
2s atomic orbitals overlap to form:
• σ2s (bonding orbital)
• σ*2s (antibonding orbital)

(iii) Electron filling
👉 Total valence electrons in Be₂ =
4 electrons.

👉 Distribution of electrons:
σ2s → 2 electrons
σ*2s → 2 electrons

(iv) Bond order calculation
👉 Bond order = ½ (Number of bonding electrons − Number of antibonding electrons)

👉 Bond order = ½ (2 − 2) =
0.

👉 Since the bond order is zero,
no stable bond is formed.

Did You Know?
👉 In
Molecular Orbital Theory, bond stability depends on the
bond order.

👉 If
bond order = 0, the molecule is
unstable and does not exist.

👉 Molecules like
H₂, O₂, and N₂ exist because their
bond order is greater than zero.

Q.36: Compare the
relative stability
of the following species and indicate their
magnetic properties:

O₂, O₂⁺, O₂^– (superoxide),
O₂²⁻ (peroxide)

Answer
The stability of molecules depends on
bond order.
Higher bond order means
greater stability.

Bond order values:
O₂⁺ →
2.5
O₂ →
2
O₂^– →
1.5
O₂²⁻ →
1

Relative stability:
O₂⁺ > O₂ > O₂^– > O₂²⁻

Magnetic properties:
O₂ →
Paramagnetic
O₂⁺ →
Paramagnetic
O₂^– →
Paramagnetic
O₂²⁻ →
Diamagnetic

Explanation (Step by Step)
Step 1: Molecular Orbital configuration of O₂
👉 Oxygen molecule has
16 electrons.
👉 In MO theory, the last two electrons enter the
π* antibonding orbitals.

Step 2: Bond order formula
👉 Bond Order =
(Bonding electrons − Antibonding electrons) / 2

Step 3: Bond order of O₂
👉 Bond order =
2.
👉 Two unpaired electrons →
Paramagnetic.

Step 4: O₂⁺
👉 One electron removed from antibonding orbital.
👉 Antibonding electrons decrease.
👉 Bond order becomes
2.5.
👉 One unpaired electron →
Paramagnetic.

Step 5: O₂^– (Superoxide)
👉 One electron added to antibonding orbital.
👉 Bond order decreases to
1.5.
👉 One unpaired electron →
Paramagnetic.

Step 6: O₂²⁻ (Peroxide)
👉 Two electrons added to antibonding orbitals.
👉 Bond order becomes
1.
👉 All electrons paired →
Diamagnetic.

Step 7: Stability comparison
👉 Higher bond order means
stronger bond and greater stability.
👉 Therefore stability order is:
O₂⁺ > O₂ > O₂^– > O₂²⁻

Did You Know?
👉 The
paramagnetic nature of O₂
was a major experimental proof of
Molecular Orbital Theory.

👉 According to simple bonding theories, oxygen should be
diamagnetic, but experiments show it is
paramagnetic.

👉 This difference was explained correctly only by
MO theory.

Q.37: Write the
significance of the plus (+) and minus (–) signs
shown while representing
atomic orbitals
.

Answer
✅ The signs
(+) and
(–) shown on orbitals do
not represent electric charge of electrons.

✅ These signs indicate the
phase (sign) of the wave function of the orbital.

✅ When orbitals with the
same phase overlap, a
bonding molecular orbital is formed.
✅ When orbitals with
opposite phases overlap, an
antibonding molecular orbital is formed.

Explanation (Step by Step)
(i) Meaning of + and – signs
👉 The
+ and – signs on the lobes of atomic orbitals are mathematical symbols representing the
phase of the wave function.
👉 They do
not indicate positive or negative electric charge.

(ii) Same sign overlap (in-phase overlap)
👉 When two orbitals overlap with the
same phase
(+ with + or – with –),
constructive overlap occurs.
👉 Electron density between the nuclei
increases.
✅ Result: Formation of a
bonding molecular orbital.

(iii) Opposite sign overlap (out-of-phase overlap)
👉 When orbitals overlap with
opposite phases
(+ with –),
destructive overlap occurs.
👉 Electron density between nuclei
decreases and a node may form.
✅ Result: Formation of an
antibonding molecular orbital.

(iv) Conclusion
👉 The + and – signs determine whether orbital overlap will produce a
bonding or
antibonding molecular orbital.
Same phase → Bonding MO
Opposite phase → Antibonding MO

Did You Know?
👉 In molecular orbital diagrams,
antibonding orbitals are usually represented by a star (*) such as
σ* or
π*.

👉 Changing the
phase relationship between orbitals can completely change the result of orbital overlap.

👉 Phase relationships are very important in understanding the
nature and strength of chemical bonds.

Q.38: Describe the
hybridisation in PCl₅
. Why are the
axial bonds longer as compared to
equatorial bonds?

Answer
In the molecule
PCl₅, phosphorus undergoes
sp³d hybridisation.

This hybridisation produces
five hybrid orbitals arranged in a
trigonal bipyramidal geometry.

👉 Three bonds lie in the same plane and are called
equatorial bonds.
👉 Two bonds lie perpendicular to this plane and are called
axial bonds.

The
axial bonds are longer and weaker than equatorial bonds because they experience
greater repulsion.

Explanation (Step by Step)
(i) Hybridisation of phosphorus
👉 Atomic number of phosphorus =
15.
👉 Valence shell configuration =
3s² 3p³.

👉 One electron from the
3s orbital is promoted to the
3d orbital.
👉 Now phosphorus has
five unpaired electrons which form
five P–Cl σ-bonds.

(ii) Geometry of PCl₅
👉 The molecule has
trigonal bipyramidal geometry.
👉 Bond angles:
Equatorial–Equatorial = 120°
Axial–Equatorial = 90°
Axial–Axial = 180°

(iii) Reason for longer axial bonds
👉 Each
axial bond experiences repulsion from
three equatorial bond pairs.

👉 Due to this greater
bond pair–bond pair repulsion, axial bonds become
longer and weaker than equatorial bonds.

Did You Know?
👉 In a
trigonal bipyramidal structure, atoms prefer the
equatorial position because it experiences
less electron repulsion.

👉 That is why
larger atoms or lone pairs generally occupy
equatorial positions.

👉 The molecule
PCl₅ exists as a
gas or liquid, but in solid state it forms
[PCl₄]⁺ and [PCl₆]⁻ ions.

Q.39: Define the
hydrogen bond
. Is it
weaker or stronger
than the
van der Waals forces
?

Answer
✅ A
hydrogen bond is the attractive force between a hydrogen atom bonded to a highly electronegative atom such as
F, O, or N and another electronegative atom of the same or a different molecule.

✅ A hydrogen bond is generally
stronger than van der Waals forces
but
weaker than a covalent bond.

Explanation (Step by Step)
(i) When does hydrogen bonding occur?
👉 When hydrogen is covalently bonded to a highly
electronegative atom such as
F, O, or N, the hydrogen atom acquires a
partial positive charge (δ⁺).
👉 This partially positive hydrogen is attracted to a
lone pair of electrons on another electronegative atom.
👉 This special intermolecular attraction is called a
hydrogen bond.

(ii) Examples
👉 In water:
O–H···O
👉 In hydrogen fluoride:
H–F···F
👉 In ammonia:
N–H···N

(iii) Comparison with van der Waals forces
👉
Hydrogen bonding is stronger than ordinary
van der Waals forces because it involves stronger dipole–dipole attraction.
👉 However, it is still
weaker than a covalent bond.

Did You Know?
👉 Hydrogen bonding is responsible for the
unusually high boiling point of water.

👉 The
open crystal structure of ice is also due to hydrogen bonding.

👉 Hydrogen bonding plays an important role in stabilizing
biological molecules such as DNA and proteins.

Q.40: What is meant by the term
bond order
? Calculate the bond order of:

N₂, O₂, O₂⁺, O₂⁻

Answer
Bond order is defined as half of the difference between the number of electrons present in
bonding molecular orbitals and
antibonding molecular orbitals.

Formula:
Bond Order = (Bonding electrons − Antibonding electrons) / 2

Bond order values:
N₂ →
3
O₂ →
2
O₂⁺ →
2.5
O₂⁻ →
1.5

Explanation (Step by Step)
Step 1: Bond order formula
👉 Bond order is calculated using the formula:
Bond Order = (Nb − Na) / 2

where:
Nb =
number of bonding electrons
Na =
number of antibonding electrons

Step 2: Bond order of N₂
👉 Total electrons =
14.
👉 Bonding electrons =
10.
👉 Antibonding electrons =
4.
👉 Bond order = (10 − 4) / 2 =
3.

Step 3: Bond order of O₂
👉 Total electrons =
16.
👉 Bonding electrons =
10.
👉 Antibonding electrons =
6.
👉 Bond order = (10 − 6) / 2 =
2.

Step 4: Bond order of O₂⁺
👉 One electron is removed from the
antibonding orbital.
👉 Antibonding electrons become
5.
👉 Bond order = (10 − 5) / 2 =
2.5.

Step 5: Bond order of O₂⁻
👉 One electron is added to an
antibonding orbital.
👉 Antibonding electrons become
7.
👉 Bond order = (10 − 7) / 2 =
1.5.

Did You Know?
👉
Bond order indicates bond strength.
Higher bond order means
stronger and shorter bond.

👉 For example,
N₂ has bond order
3, so it has a very
strong triple bond.

👉 Species with
higher bond order are more stable.