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

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.

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

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).

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.

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

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.

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.

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.

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 ↑

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

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.

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.

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.

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⁻.

(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)

(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.

(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.

(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−

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

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.

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.

(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.

(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

(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.

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⁻) .

(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.

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.

(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.

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.

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.

(c) In ethanol, all carbon atoms form only single bonds → sp³.

(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.

(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.

(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.

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.

(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.

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⁻

(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

(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.

(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.

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.