- Dr. Sanjay Kumar Pawar
Chemical Bonding – Exercises 4.1 to 4.25
Step-by-step answers for beginner students
Chapter: Chemical Bonding and Molecular Structure
📚 Quick Contents
- Formation of Chemical Bond
- Lewis Dot Symbols
- Lewis Symbols of Atoms and Ions
- Lewis Structures
- Octet Rule
- Formation of Ionic Bond
- VSEPR Shapes
- NH₃ vs H₂O Bond Angles
- Bond Strength and Bond Order
- Bond Length
- Resonance in CO₃²⁻
- H₃PO₃ Resonance
- Resonance Structures
- Electron Transfer
- CO₂ and H₂O Dipoles
- Applications of Dipole Moment
- Electronegativity
- Polar Covalent Bond
- Increasing Ionic Character
- Lewis Structure of CH₃COOH
- Why CH₄ Is Not Square Planar
- Dipole Moment of BeH₂
- NH₃ vs NF₃
- Hybridisation
- Hybridisation of Al in AlCl₄⁻
4.1 Explain the formation of a chemical bond.
Answer
A chemical bond is the attractive force that holds atoms or ions together in a molecule or compound.
Atom A + Atom B
↓ ↓
lose/gain/share electrons
↓
Chemical bond
↓
Stable species
The bond forms because the bonded arrangement has a suitable lower energy and there is electrostatic attraction between the particles.
4.2 Write Lewis dot symbols for Mg, Na, B, O, N and Br.
Step 1: Find the number of valence electrons
| Element | Electronic configuration | Valence electrons |
|---|---|---|
| Na | 2, 8, 1 | 1 |
| Mg | 2, 8, 2 | 2 |
| B | 2, 3 | 3 |
| N | 2, 5 | 5 |
| O | 2, 6 | 6 |
| Br | 2, 8, 18, 7 | 7 |
Step 2: Draw the Lewis symbols
•
Na Na•
•
Mg •Mg•
•
• B •
••
• N •
•
••
• O •
••
••
• Br •
••
•
The dots represent the valence electrons.
4.3 Write Lewis symbols for the following atoms and ions.
(a) S and S²⁻ (b) Al and Al³⁺ (c) H and H⁻
(a) Sulfur: S and S²⁻
Sulfur has atomic number 16:
S = 2, 8, 6
Therefore sulfur has 6 valence electrons. When sulfur gains 2 electrons:
S + 2e⁻ → S²⁻
S atom: S²⁻ ion:
•• ••
• S • • S •
•• ••
••
[ S ]²⁻
S²⁻ has 8 valence electrons, so it has an octet.
(b) Aluminium: Al and Al³⁺
Aluminium has 3 valence electrons.
Al → Al³⁺ + 3e⁻
Al atom: Al³⁺ ion: • [Al]³⁺ • Al •
The Al³⁺ ion has lost its three valence electrons.
(c) Hydrogen: H and H⁻
Hydrogen has 1 electron.
H + e⁻ → H⁻
H atom: H⁻ ion: H• [ H•• ]⁻
H⁻ has two electrons and follows the duplet rule.
4.4 Draw the Lewis structures for H₂S, SiCl₄, BeF₂, CO₃²⁻ and HCOOH.
(a) H₂S
..
H — S — H
..
Shape: Bent.
(b) SiCl₄
Cl
|
Cl — Si — Cl
|
Cl
Each Cl has three lone pairs around it.
Shape: Tetrahedral.
(c) BeF₂
F — Be — F
Shape: Linear.
(d) CO₃²⁻ — Carbonate ion
C = 4
3 O = 18
2− charge = 2 extra electrons
Total = 4 + 18 + 2 = 24 electrons
O
||
[ O — C — O ]²⁻
There are three equivalent resonance structures:
O O⁻ O⁻
|| | |
[ O⁻ — C — O⁻ ] ↔ [ O = C — O⁻ ] ↔ [ O⁻ — C = O ]
The actual carbonate ion is a resonance hybrid. The negative charge is delocalised over the oxygen atoms.
(e) HCOOH — Formic acid
The correct connectivity is:
O
||
H — C — O — H
So the Lewis structure is: H–C(=O)–O–H.
4.5 Define octet rule. Write its significance and limitations.
Definition
The octet rule states that atoms tend to gain, lose or share electrons until they have eight electrons in their valence shell, similar to a noble gas.
Less stable More stable
Valence shell → gain / lose / share electrons
↓
8 electrons
↓
Stable
Significance
- Helps us understand why atoms form chemical bonds.
- Helps predict the formulae of many compounds.
- Helps us draw Lewis structures.
- Explains electron transfer in many ionic compounds.
- Explains electron sharing in many covalent compounds.
Limitations
- Incomplete octet: Some atoms have fewer than eight electrons. Example: BeCl₂ and BF₃.
- Odd-electron molecules: Molecules such as NO contain an odd number of electrons and cannot satisfy the octet rule for all atoms.
- Expanded octet: Elements from the third period and beyond can have more than eight electrons around the central atom. Examples: PCl₅ and SF₆.
- The octet rule does not explain all aspects of molecular shape and bonding.
4.6 Write the favourable factors for the formation of an ionic bond.
An ionic bond forms when one atom transfers electron(s) to another atom, producing oppositely charged ions.
Metal Non-metal
↓ ↓
loses electron gains electron
↓ ↓
Cation (+) Anion (−)
\ /
\ /
Ionic bond
Important favourable factors
- Low ionisation enthalpy of the metal: The metal should lose electrons easily.
- High electron gain tendency of the non-metal: The non-metal should accept electrons readily.
- High lattice enthalpy: Strong attraction between the ions makes the ionic compound stable.
- Large electronegativity difference: A large difference between the two atoms favours electron transfer.
- Suitable ionic sizes: The sizes of the ions influence the strength of attraction and lattice stability.
4.7 Discuss the shape of the following molecules using the VSEPR model.
Molecules: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S and PH₃
VSEPR means Valence Shell Electron Pair Repulsion.
Electron pairs around a central atom repel each other and arrange themselves as far apart as possible.
| Molecule | Electron regions | Shape | Approx. bond angle |
|---|---|---|---|
| BeCl₂ | 2 | Linear | 180° |
| BCl₃ | 3 | Trigonal planar | 120° |
| SiCl₄ | 4 | Tetrahedral | 109.5° |
| AsF₅ | 5 | Trigonal bipyramidal | 90°, 120° |
| H₂S | 4 electron pairs: 2 bond pairs + 2 lone pairs | Bent | about 92° |
| PH₃ | 4 electron pairs: 3 bond pairs + 1 lone pair | Trigonal pyramidal | about 93.5° |
BeCl₂ BCl₃
Cl — Be — Cl Cl
\
B — Cl
/
Cl
SiCl₄ AsF₅
Cl F
| |
Cl — Si — Cl F — As — F
| / \
Cl F F
H₂S PH₃
H H
\ /
S P
/ / \
H H H
4.8 Why is the bond angle in water less than that in ammonia?
Both NH₃ and H₂O have a tetrahedral arrangement of electron pairs around the central atom, but their molecular bond angles are different.
| Molecule | Lone pairs | Bond pairs | Bond angle |
|---|---|---|---|
| NH₃ | 1 | 3 | 107° |
| H₂O | 2 | 2 | 104.5° |
Step-by-step explanation
NH₃: H₂O:
.. ..
N O
/ | \ / \
H H H H H
one lone pair two lone pairs
107° 104.5°
Therefore: 104.5° < 107°
4.9 How do you express bond strength in terms of bond order?
Bond order is the number of bonds between two atoms.
| Bond | Bond order | Relative strength |
|---|---|---|
| C–C | 1 | Lower |
| C=C | 2 | Higher |
| C≡C | 3 | Highest |
Generally:
Bond order ↑
↓
Bond strength ↑
↓
Bond length ↓
Thus, a triple bond is generally stronger and shorter than a double bond, and a double bond is stronger and shorter than a single bond.
4.10 Define bond length.
Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule.
Atom A Atom B
●-----------------------------●
←── bond length ──→
It is usually measured in pm (picometres) or Å (angstrom).
4.11 Explain important aspects of resonance using CO₃²⁻.
What is resonance?
Sometimes a molecule or ion cannot be represented correctly by only one Lewis structure. Two or more valid Lewis structures can be written. These are called canonical or resonance structures.
Carbonate ion
The carbonate ion is:
CO₃²⁻
O O⁻ O⁻
|| | |
O⁻ — C — O⁻ ↔ O = C — O⁻ ↔ O⁻ — C = O
Important points
- The atoms do not move from their positions.
- Only the arrangement of electrons changes.
- The three structures are equivalent.
- The actual carbonate ion is a resonance hybrid.
- The negative charge is delocalised over the oxygen atoms.
- All three C–O bonds have the same average character.
- The actual C–O bond order is between a single and a double bond.
4.12 Can the two given structures of H₃PO₃ be taken as canonical forms?
Answer: No.
Reason
Canonical forms of resonance must have:
- the same arrangement of atoms,
- the same total number of electrons, and
- the same overall charge.
Only the positions of electrons, such as π electrons or lone pairs, may change.
RESONANCE:
Atoms stay fixed
↓
A—B—C A=B—C
electrons move
↓
A=B—C ↔ A—B=C
NOT RESONANCE:
A—H + B ↔ A + H—B
Here the position of H changes.
That is a change in atomic arrangement.
In the structures shown for H₃PO₃, the hydrogen/atom connectivity is not the same. Therefore they cannot be considered canonical resonance structures.
4.13 Write the resonance structures for SO₃, NO₂ and NO₃⁻.
(a) SO₃
Sulfur trioxide can be represented by equivalent resonance forms. A simplified Lewis representation is:
O O⁻
|| |
O — S — O ↔ O = S — O⁻
↔
other equivalent forms
The actual structure is a resonance hybrid with delocalised bonding.
(b) NO₂
NO₂ has an odd number of valence electrons, so it is an odd-electron species.
O = N — O· ↔ ·O — N = O
The dot represents the unpaired electron.
(c) NO₃⁻
O O⁻ O⁻
|| | |
O⁻ — N — O⁻ ↔ O = N — O⁻ ↔ O⁻ — N = O
There are three equivalent resonance structures. The negative charge is delocalised over the oxygen atoms.
4.14 Show electron transfer to form cations and anions.
(a) K and S (b) Ca and O (c) Al and N
(a) K and S
K has 1 valence electron. Sulfur needs 2 electrons. Therefore two K atoms are required.
K → K⁺ + e⁻ K → K⁺ + e⁻ S + 2e⁻ → S²⁻ Therefore: 2K + S → 2K⁺ + S²⁻ Formula: K₂S
(b) Ca and O
Ca loses two electrons and O gains two electrons.
Ca → Ca²⁺ + 2e⁻ O + 2e⁻ → O²⁻ Therefore: Ca + O → Ca²⁺ + O²⁻ Formula: CaO
(c) Al and N
Al loses 3 electrons and N gains 3 electrons.
Al → Al³⁺ + 3e⁻ N + 3e⁻ → N³⁻ Therefore: Al + N → Al³⁺ + N³⁻ Formula: AlN
4.15 Why is H₂O polar while CO₂ is non-polar although both are triatomic molecules?
The important factor is molecular geometry.
CO₂
O = C = O
← dipole dipole →
cancel
Linear molecule
180°
The two C=O bonds are polar, but they point in exactly opposite directions. Therefore their dipole moments cancel.
Net dipole moment of CO₂ = 0
H₂O
O
/ \
H H
Bent molecule
The two O–H bonds are polar and the molecule is bent. Therefore the bond dipoles do not cancel.
Net dipole moment of H₂O ≠ 0
| Molecule | Shape | Dipole cancellation | Polarity |
|---|---|---|---|
| CO₂ | Linear | Complete | Non-polar |
| H₂O | Bent | Incomplete | Polar |
4.16 Write the significance/applications of dipole moment.
Dipole moment measures the separation of positive and negative charges in a molecule.
μ = q × r
where q is the magnitude of charge and r is the distance between the charges.
Applications
- Determining molecular polarity: A molecule with a non-zero dipole moment is polar.
- Predicting molecular geometry: Dipole moment can help distinguish between possible structures.
- Finding ionic character: The dipole moment of a bond can give information about its polarity.
- Distinguishing isomers: Some geometrical isomers have different dipole moments.
- Understanding solubility: Polar molecules tend to dissolve better in polar solvents.
Charge separation
↓
Dipole moment
↓
Information about
polarity + structure
4.17 Define electronegativity. How does it differ from electron gain enthalpy?
Electronegativity
Electronegativity is the tendency of an atom in a molecule to attract the shared pair of electrons towards itself.
For example, in H–Cl, chlorine attracts the shared electrons more strongly than hydrogen.
H —————→ Cl
shared
electrons
Cl attracts the electrons more strongly.
Electronegativity vs Electron Gain Enthalpy
| Electronegativity | Electron gain enthalpy |
|---|---|
| Ability of an atom in a molecule to attract shared electrons. | Energy change when an isolated gaseous atom gains an electron. |
| It is a relative quantity. | It is an experimentally measurable energy change. |
| Applies to an atom participating in a chemical bond. | Refers to an isolated gaseous atom. |
| Usually expressed using scales such as Pauling scale. | Expressed in kJ mol⁻¹. |
4.18 Explain a polar covalent bond with a suitable example.
A polar covalent bond is formed when two atoms share electrons unequally because they have different electronegativities.
Example: HCl
H — Cl Cl is more electronegative. Hδ+ — Clδ−
The shared electron pair is attracted more towards chlorine. Therefore:
- Hydrogen gets a partial positive charge: δ⁺
- Chlorine gets a partial negative charge: δ⁻
Thus H–Cl is a polar covalent bond.
4.19 Arrange the bonds in increasing order of ionic character.
Compounds: LiF, K₂O, N₂, SO₂ and ClF₃
Ionic character generally increases as the electronegativity difference between the bonded atoms increases.
Step 1: Compare the bonds
| Bond | Nature | Approximate ionic character |
|---|---|---|
| N–N in N₂ | Non-polar covalent | Lowest |
| S–O in SO₂ | Polar covalent | Higher |
| Cl–F in ClF₃ | Polar covalent | Higher |
| K–O in K₂O | Ionic | Very high |
| Li–F in LiF | Strongly ionic | Very high |
Answer
N₂ < SO₂ < ClF₃ < K₂O < LiF
4.20 Write the correct Lewis structure for acetic acid, CH₃COOH.
The skeletal arrangement of acetic acid is:
CH₃–COOH
Step 1: Arrange the atoms
CH₃ — C — O — H
|
O
Step 2: Give carbon four bonds
The second carbon is bonded to the first carbon, one oxygen by a single bond, and one oxygen by a double bond.
O
||
H₃C — C — O — H
Correct Lewis structure
..
O
||
H₃C — C — O — H
..
..
Each oxygen has two lone pairs.
CH₃–C(=O)–OH
4.21 Why is CH₄ not square planar?
CH₄ has four bond pairs around carbon and no lone pairs. According to VSEPR theory, the four electron pairs repel each other and move as far apart as possible.
Tetrahedral arrangement
H
|
H — C — H
/
H
Bond angle ≈ 109.5°
Why not square planar?
In a square planar arrangement, adjacent H–C–H angles would be 90°. This causes greater electron-pair repulsion.
Square planar:
H — C — H
|
H
|
H
Many 90° interactions
↓
Greater repulsion
↓
Less stable
The tetrahedral arrangement gives a larger average separation between the four bond pairs.
Therefore CH₄ is tetrahedral, not square planar.
4.22 Why does BeH₂ have zero dipole moment although Be–H bonds are polar?
The Be–H bonds are polar because Be and H have different electronegativities.
However, BeH₂ is linear.
Hδ− ← Be → Hδ−
180°
Bond dipoles are equal and opposite.
←────────────→
cancel
The two bond dipoles are equal in magnitude and opposite in direction. Therefore they cancel each other.
Net dipole moment = 0
4.23 Which has a higher dipole moment: NH₃ or NF₃? Why?
NH₃ has a higher dipole moment than NF₃.
NH₃
..
N
/ | \
H H H
Bond dipoles and lone-pair effect
are in the same general direction.
The N–H bond dipoles and the effect of the lone pair combine to give a relatively large resultant dipole moment.
NF₃
..
N
/ | \
F F F
N–F bond dipoles oppose
the lone-pair contribution.
Fluorine is much more electronegative than nitrogen. The N–F bond dipoles are directed towards fluorine and tend to oppose the dipole associated with the lone pair.
| Molecule | Result |
|---|---|
| NH₃ | Higher dipole moment |
| NF₃ | Lower dipole moment |
4.24 What is hybridisation? Describe the shapes of sp, sp² and sp³ hybrid orbitals.
Definition
Hybridisation is the mixing of atomic orbitals of comparable energy to form a new set of equivalent hybrid orbitals.
(a) sp hybridisation
One s orbital + one p orbital → two sp hybrid orbitals.
sp hybridisation
180°
Orbital ←──── A ────→ B ────→
central atom
Shape: Linear
Angle: 180°
Example: BeCl₂
(b) sp² hybridisation
One s orbital + two p orbitals → three sp² hybrid orbitals.
A
/
/
C
/ \
B D
Shape: Trigonal planar
Angle: 120°
Example: BF₃
(c) sp³ hybridisation
One s orbital + three p orbitals → four sp³ hybrid orbitals.
A
|
C
/ \
B D
\
E
Shape: Tetrahedral
Angle: 109.5°
Example: CH₄
| Hybridisation | Orbitals mixed | Number of hybrid orbitals | Shape | Angle |
|---|---|---|---|---|
| sp | 1s + 1p | 2 | Linear | 180° |
| sp² | 1s + 2p | 3 | Trigonal planar | 120° |
| sp³ | 1s + 3p | 4 | Tetrahedral | 109.5° |
sp → 2 orbitals → linear
sp² → 3 orbitals → trigonal planar
sp³ → 4 orbitals → tetrahedral
4.25 Describe the change in hybridisation of Al in AlCl₃ + Cl⁻ → AlCl₄⁻.
Step 1: Consider AlCl₃
In the simple molecular description, aluminium in AlCl₃ has three bonding regions around it.
Cl
\
Al
/ \
Cl Cl
3 electron regions
Hybridisation = sp²
Shape = trigonal planar
Step 2: Chloride ion donates a lone pair
The chloride ion, Cl⁻, has lone pairs. It can donate one lone pair to the electron-deficient aluminium atom.
AlCl₃ + Cl⁻ → AlCl₄⁻
lone pair
Cl⁻ ─────────────→ Al
coordinate bond
Step 3: Four bonding regions are now present
Cl
|
Cl — Al — Cl
|
Cl
Four electron regions
Hybridisation = sp³
Shape = tetrahedral
Final answer
| Species | Hybridisation of Al | Shape |
|---|---|---|
| AlCl₃ | sp² | Trigonal planar |
| AlCl₄⁻ | sp³ | Tetrahedral |
sp² → sp³
🧠 One-Page Revision: Important Ideas
| Topic | Remember |
|---|---|
| Lewis symbol | Shows valence electrons as dots. |
| Octet rule | Atoms often become stable with 8 valence electrons. |
| Ionic bond | Formed mainly by electron transfer and electrostatic attraction. |
| Covalent bond | Formed by sharing electron pairs. |
| Polar covalent bond | Electrons are shared unequally. |
| VSEPR | Electron pairs repel and arrange themselves as far apart as possible. |
| Resonance | Only electrons move; atoms remain in the same positions. |
| Bond order | Higher bond order → stronger and generally shorter bond. |
| Dipole moment | Measures charge separation in a molecule. |
| sp | Linear, 180° |
| sp² | Trigonal planar, 120° |
| sp³ | Tetrahedral, 109.5° |
CHEMICAL BONDING MAP
Valence electrons
↓
┌─────────┴─────────┐
↓ ↓
Transfer Sharing
↓ ↓
Ionic Covalent
│ │
Cation + Anion Single/Double/Triple
│
↓
Bond order
│
┌────────────┴────────────┐
↓ ↓
Bond strength ↑ Bond length ↓
MOLECULAR SHAPE
VSEPR theory
↓
Electron-pair repulsion
↓
Molecular shape
↓
Molecular dipole moment
↓
Polarity
🧪 Chemical Bonding & Molecular Structure
Exercises 4.26 – 4.40
Beginner-friendly questions, answers, explanations and visuals
📚 Contents
- 4.26 Hybridisation of B and N
- 4.27 Formation of Double and Triple Bonds
- 4.28 Number of Sigma and Pi Bonds
- 4.29 Which Orbitals Cannot Form a Sigma Bond?
- 4.30 Hybrid Orbitals of Carbon
- 4.31 Bond Pairs and Lone Pairs
- 4.32 Sigma Bond vs Pi Bond
- 4.33 Formation of H₂ by Valence Bond Theory
- 4.34 Conditions for Molecular Orbital Formation
- 4.35 Why Be₂ Does Not Exist
- 4.36 Stability and Magnetic Properties of Oxygen Species
- 4.37 Meaning of + and − Signs in Orbitals
- 4.38 Hybridisation and Bond Length in PCl₅
- 4.39 Hydrogen Bond vs van der Waals Force
- 4.40 Bond Order of N₂ and Oxygen Species
4.26 Is there any change in the hybridisation of B and N in the reaction?
BF₃ + NH₃ → F₃B·NH₃
Step 1: Look at BF₃
Boron in BF₃ forms three bonds with fluorine. Therefore, there are three electron regions around B.
F
/
B
/ \
F F
BF₃
3 electron regions
↓
sp² hybridisation
↓
Trigonal planar
Step 2: Look at NH₃
Nitrogen in NH₃ has three N–H bonds and one lone pair. That means four electron regions.
..
N
/ | \
H H H
NH₃
3 bond pairs + 1 lone pair
↓
sp³ hybridisation
↓
Trigonal pyramidal
Step 3: Formation of the coordinate bond
NH₃ has a lone pair on nitrogen. Boron in BF₃ is electron deficient, so nitrogen donates its lone pair to boron.
lone pair
NH₃ ─────────────→ BF₃
N → B
coordinate bond
Step 4: After reaction
After accepting the lone pair, boron has four bonds around it. Therefore, its hybridisation changes.
| Atom | Before reaction | After reaction |
|---|---|---|
| B | sp² | sp³ |
| N | sp³ | sp³ |
Answer: B changes from sp² → sp³, while N remains sp³.
4.27 Draw diagrams showing the formation of a double bond and a triple bond between carbon atoms in C₂H₄ and C₂H₂.
(A) Double bond in C₂H₄
Ethene is:
C₂H₄ = CH₂=CH₂
Step 1: Hybridisation
Each carbon atom is sp² hybridised. One unhybridised p orbital remains on each carbon.
H H
| |
C = C
| |
H H
C = C
σ bond + π bond
Step 2: How is the double bond formed?
The C=C double bond contains:
- One sigma (σ) bond
- One pi (π) bond
p orbital p orbital
↑ ↑
| |
C = C
| |
↓ ↓
Sideways overlap of p orbitals
↓
π bond
Head-on overlap of sp² orbitals
↓
σ bond
(B) Triple bond in C₂H₂
Ethyn(e) / acetylene is:
C₂H₂ = H–C≡C–H
Step 1: Hybridisation
Each carbon is sp hybridised. Two unhybridised p orbitals remain on each carbon.
H — C ≡ C — H C ≡ C 1 σ bond + 2 π bonds
Step 2: Formation of triple bond
C ≡ C
σ
│
↓
head-on overlap
+
π₁
+
π₂
↓
two sideways p-orbital overlaps
Therefore:
Triple bond = 1σ + 2π
| Molecule | Bond | Composition |
|---|---|---|
| C₂H₄ | C=C | 1σ + 1π |
| C₂H₂ | C≡C | 1σ + 2π |
4.28 What is the total number of sigma and pi bonds in the following molecules?
(a) C₂H₂ (b) C₂H₄
(a) C₂H₂
Structure:
H — C ≡ C — H
The C≡C triple bond contains:
- 1 sigma bond
- 2 pi bonds
There are also two C–H sigma bonds.
| Bond | Number | σ / π |
|---|---|---|
| C–H | 2 | 2σ |
| C≡C | 1 | 1σ + 2π |
Total = 3σ + 2π
(b) C₂H₄
Structure:
H H
| |
C = C
| |
H H
There are four C–H bonds and one C=C double bond.
C=C = 1σ + 1π
Total = 5σ + 1π
| Molecule | Sigma bonds | Pi bonds |
|---|---|---|
| C₂H₂ | 3 | 2 |
| C₂H₄ | 5 | 1 |
4.29 Considering the z-axis as the internuclear axis, which orbital combination will NOT form a sigma bond and why?
Options:
- 1s and 1s
- 1s and 2pz
- 2py and 2py
- 1s and 2s
What is a sigma bond?
A sigma bond is formed by head-on overlap along the internuclear axis.
Internuclear axis = z-axis
Atom A Atom B
●─────────────────●
z-axis
Head-on overlap
↓
σ bond
Check each option
| Combination | Overlap | σ bond? |
|---|---|---|
| 1s + 1s | Head-on | Yes |
| 1s + 2pz | Head-on along z-axis | Yes |
| 2py + 2py | Sideways with respect to z-axis | No |
| 1s + 2s | Head-on | Yes |
z-axis → 2pᵧ 2pᵧ ) ( ) ( ) ( Their lobes are perpendicular to the z-axis. Therefore they cannot form a sigma bond along z-axis.
Answer: (c) 2pᵧ and 2pᵧ
They can participate in sideways overlap to form a π bond, not a sigma bond along the z-axis.
4.30 What 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
Easy rule
| Number of electron regions around carbon | Hybridisation |
|---|---|
| 4 | sp³ |
| 3 | sp² |
| 2 | sp |
(a) CH₃–CH₃
CH₃ — CH₃ C₁: 4 single bonds → sp³ C₂: 4 single bonds → sp³
Answer: sp³, sp³
(b) CH₃–CH=CH₂
CH₃ — CH = CH₂ ↑ ↑ ↑ C₁ C₂ C₃ C₁ → sp³ C₂ → sp² C₃ → sp²
Answer: sp³, sp², sp²
(c) CH₃–CH₂–OH
CH₃ — CH₂ — OH C₁ → sp³ C₂ → sp³
Answer: sp³, sp³
(d) CH₃–CHO
The structure is:
O
||
CH₃ — C — H
C of CH₃ → sp³
Carbonyl C → sp²
Answer: sp³, sp²
(e) CH₃COOH
O
||
CH₃ — C — O — H
CH₃ carbon → sp³
COOH carbon → sp²
Answer: sp³, sp²
4.31 What do you understand by bond pairs and lone pairs of electrons?
1. Bond pair
A bond pair is a pair of electrons shared between two atoms to form a covalent bond.
H : H ↑ shared pair This pair forms the H—H bond.
2. Lone pair
A lone pair is a pair of valence electrons that is not involved in bonding.
..
H — O — H
..
The two pairs shown as dots
are lone pairs on oxygen.
| Type | Meaning | Example |
|---|---|---|
| Bond pair | Shared between atoms | H–H |
| Lone pair | Not shared in a bond | H₂O oxygen |
4.32 Distinguish between a sigma bond and a pi bond.
| Property | Sigma (σ) bond | Pi (π) bond |
|---|---|---|
| Type of overlap | Head-on overlap | Sideways overlap |
| Electron density | Along internuclear axis | Above and below internuclear axis |
| Strength | Generally stronger | Generally weaker |
| Rotation | Usually allows rotation around bond axis | Restricts rotation |
| First bond between two atoms | Always σ | Never the first bond |
Visual comparison
SIGMA BOND
Atom A Atom B
( ) ─────────── ( )
head-on
overlap
PI BOND
↑ electron cloud ↑
\ /
C===========C
/ \
↓ electron cloud ↓
Sideways overlap of p orbitals
σ = Straight/head-on overlap
π = Parallel/sideways overlap
4.33 Explain the formation of H₂ molecule on the basis of Valence Bond Theory.
Step 1: Electronic configuration of hydrogen
Each hydrogen atom has one electron:
H = 1s¹
Step 2: Approach of two hydrogen atoms
H(1s¹) H(1s¹) ● → ← ● As atoms approach, their 1s orbitals overlap.
Step 3: Overlap
The two half-filled 1s orbitals overlap head-on. This forms a sigma bond.
1s 1s
( ● ) ( ● )
\ /
\ /
\ /
H — H
σ
Step 4: Stable H₂ molecule
The two electrons pair with opposite spins in the overlapping region. The electrostatic attraction between the nuclei and shared electrons helps hold the atoms together.
H• + •H → H:H H—H One σ bond
H₂ contains one sigma bond.
4.34 Write the important conditions required for linear combination of atomic orbitals to form molecular orbitals.
According to Molecular Orbital Theory, atomic orbitals combine to form molecular orbitals when suitable conditions are satisfied.
Condition 1: Comparable energy
The atomic orbitals should have comparable energies.
Condition 2: Proper symmetry
The orbitals must have suitable symmetry with respect to the internuclear axis.
Condition 3: Effective overlap
The orbitals should overlap appreciably. Greater overlap generally produces stronger interaction.
Atomic orbital A Atomic orbital B
↓ ↓
\ /
\ overlap /
\ ↓ /
\ / \ /
\ / \ /
↓
Molecular orbitals
| Condition | Easy meaning |
|---|---|
| Comparable energy | Orbitals should have similar energies. |
| Proper symmetry | Orbitals must face each other correctly. |
| Good overlap | Orbitals must overlap effectively. |
4.35 Use Molecular Orbital Theory to explain why Be₂ molecule does not exist.
Step 1: Electronic configuration of Be
Beryllium has atomic number 4:
Be = 1s² 2s²
For bonding, focus on the two 2s electrons of each Be atom.
Step 2: Molecular orbitals from 2s orbitals
2s + 2s
↓
σ2s bonding molecular orbital
σ*2s antibonding molecular orbital
Step 3: Fill the electrons
Two Be atoms provide four 2s electrons.
σ*2s ↑↓
─────────
σ2s ↑↓
─────────
Step 4: Calculate bond order
Formula:
Bond order = ½(Nb − Na)
Here:
- Nb = 2 bonding electrons
- Na = 2 antibonding electrons
Bond order = ½(2 − 2) = 0
Bonding electrons = 2
Antibonding electrons = 2
They cancel each other.
Bond order = 0
↓
No net bond
↓
Be₂ is not stable
Answer: Be₂ does not exist as a stable molecule because its bond order is zero.
4.36 Compare the relative stability and magnetic properties of O₂, O₂⁺, O₂⁻ and O₂²⁻.
Important idea
For these oxygen species, electrons are added to or removed from antibonding π* orbitals.
Bond order formula
Bond order = ½(Bonding electrons − Antibonding electrons)
| Species | Bond order | Unpaired electrons | Magnetic nature |
|---|---|---|---|
| O₂⁺ | 2.5 | 1 | Paramagnetic |
| O₂ | 2.0 | 2 | Paramagnetic |
| O₂⁻ | 1.5 | 1 | Paramagnetic |
| O₂²⁻ | 1.0 | 0 | Diamagnetic |
Why does O₂ have unpaired electrons?
π*2p orbitals
↑ ↑
───── ─────
Two unpaired electrons
↓
Paramagnetic
Relative stability
Greater bond order means stronger and generally more stable bonding.
Bond order: O₂⁺ = 2.5 O₂ = 2.0 O₂⁻ = 1.5 O₂²⁻ = 1.0 Therefore: O₂⁺ > O₂ > O₂⁻ > O₂²⁻ in relative bond strength/stability
4.37 Write the significance of a plus (+) and minus (−) sign shown in representing orbitals.
The + and − signs on orbital diagrams do not mean positive and negative electrical charges.
They represent the phase or sign of the wave function.
Same phase → constructive overlap
+ +
( ) ( )
\ /
\_______________/
overlap
Same phase
↓
Constructive overlap
↓
Bonding interaction
Opposite phase → destructive overlap
+ −
( ) ( )
\ /
\_______________/
overlap
Opposite phase
↓
Destructive overlap
↓
Antibonding interaction
4.38 Describe the hybridisation in PCl₅. Why are axial bonds longer than equatorial bonds?
Step 1: Hybridisation
Phosphorus in PCl₅ forms five P–Cl bonds. Therefore, five hybrid orbitals are needed.
One s orbital + three p orbitals + one d orbital combine:
s + p + p + p + d → sp³d
Thus phosphorus is sp³d hybridised.
Step 2: Shape
The five orbitals arrange themselves in a trigonal bipyramidal geometry.
Cl
|
|
P
/ | \
Cl | Cl
|
|
Cl
There are:
2 axial positions
3 equatorial positions
Bond angles
- Axial–equatorial = 90°
- Equatorial–equatorial = 120°
- Axial–axial = 180°
Why are axial bonds longer?
An axial bond has three 90° interactions with equatorial bond pairs. An equatorial bond has only two 90° interactions with axial bond pairs.
AXIAL Cl
|
|
P
/ | \
/ | \
E E E
Axial bond:
more 90° repulsions
↓
more repulsion
↓
axial bond becomes longer/weaker
Axial P–Cl bonds are longer than equatorial P–Cl bonds.
4.39 Define hydrogen bond. Is it weaker or stronger than van der Waals forces?
Definition
A hydrogen bond is an attractive interaction between a hydrogen atom bonded to a highly electronegative atom such as F, O or N and another electronegative atom having a lone pair.
Example: Water
H
|
Oδ−
/ \
H ······· H
↑
hydrogen bond
|
O
A simple representation is:
O–H ··· O
The dotted line represents the hydrogen bond.
Hydrogen bond vs van der Waals forces
Hydrogen bonds are generally stronger than ordinary van der Waals forces, although the exact strength depends on the type of interaction being compared.
| Interaction | General strength |
|---|---|
| Ordinary van der Waals forces | Weaker |
| Hydrogen bond | Stronger |
4.40 What is meant by bond order? Calculate the bond order of N₂, O₂, O₂⁺ and O₂⁻.
Definition of bond order
Bond order is the number of bonds between two atoms. In Molecular Orbital Theory, it is calculated using:
Bond order = ½ × (Number of bonding electrons − Number of antibonding electrons)
(a) N₂
For N₂, the molecular orbital electron arrangement gives:
- Bonding electrons = 10
- Antibonding electrons = 4
Bond order = ½(10 − 4)
= ½ × 6 = 3
N ≡ N Bond order = 3 Therefore N₂ has a triple bond.
(b) O₂
For O₂:
- Bonding electrons = 10
- Antibonding electrons = 6
Bond order = ½(10 − 6) = 2
O = O Bond order = 2 Therefore O₂ has a double-bond character.
(c) O₂⁺
O₂⁺ has one electron less than O₂. The electron removed comes from an antibonding orbital.
- Bonding electrons = 10
- Antibonding electrons = 5
Bond order = ½(10 − 5)
= 2.5
(d) O₂⁻
O₂⁻ has one extra electron compared with O₂. That extra electron enters an antibonding orbital.
- Bonding electrons = 10
- Antibonding electrons = 7
Bond order = ½(10 − 7)
= 1.5
Final answer
| Species | Bond order | General bond strength |
|---|---|---|
| N₂ | 3 | Highest |
| O₂⁺ | 2.5 | Very high |
| O₂ | 2 | High |
| O₂⁻ | 1.5 | Lower |
BOND ORDER SCALE N₂ O₂⁺ O₂ O₂⁻ | | | | 3 2.5 2 1.5 ↓ ↓ ↓ ↓ stronger weaker
🧠 Final Revision Sheet: 4.26–4.40
| Question | Most important answer |
|---|---|
| 4.26 | B: sp² → sp³; N remains sp³. |
| 4.27 | Double bond = 1σ + 1π; triple bond = 1σ + 2π. |
| 4.28 | C₂H₂ = 3σ + 2π; C₂H₄ = 5σ + 1π. |
| 4.29 | 2pᵧ + 2pᵧ does not form σ along z-axis. |
| 4.30 | CH₃–CH₃: sp³/sp³; alkene carbons: sp²; carbonyl C: sp². |
| 4.31 | Bond pair = shared electrons; lone pair = non-bonding electrons. |
| 4.32 | σ = head-on overlap; π = sideways overlap. |
| 4.33 | H₂ forms by head-on overlap of two 1s orbitals. |
| 4.34 | Comparable energy + proper symmetry + effective overlap. |
| 4.35 | Be₂ bond order = 0, so it is not stable. |
| 4.36 | Stability: O₂⁺ > O₂ > O₂⁻ > O₂²⁻. |
| 4.37 | + and − show orbital phase, not electric charge. |
| 4.38 | PCl₅: sp³d, trigonal bipyramidal; axial bonds are longer. |
| 4.39 | Hydrogen bonds are generally stronger than ordinary van der Waals forces. |
| 4.40 | N₂ = 3; O₂ = 2; O₂⁺ = 2.5; O₂⁻ = 1.5. |
⭐ Three Rules to Memorise
RULE 1 Single bond = 1σ Double bond = 1σ + 1π Triple bond = 1σ + 2π
RULE 2
Higher bond order
↓
Stronger bond
↓
Shorter bond
RULE 3 sp → Linear → 180° sp² → Trigonal planar → 120° sp³ → Tetrahedral → 109.5° sp³d → Trigonal bipyramidal
🔑 Easy Memory Tricks
- σ = head-on
- π = sideways
- Lone pairs repel more strongly than bond pairs.
- Bond order ↑ → bond strength ↑
- Bond order ↑ → bond length ↓
- Unpaired electrons → paramagnetic
- All electrons paired → diamagnetic
- Resonance moves electrons, not atoms.
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