Exercises
1. Explain the formation of a chemical bond.
Answer
A chemical bond forms because the bonded state is more stable (has lower energy) than the isolated atoms. When two atoms approach each other, attractive forces operate between the nucleus of one atom and electrons of the other, along with repulsive forces between the two nuclei and between electron clouds. A bond forms when attractive forces dominate at the equilibrium bond length.
Types of chemical bonds:
• Ionic bond: Transfer of electrons from metallic atom to non-metallic atom, forming oppositely charged ions that attract each other (e.g., NaCl).
• Covalent bond: Sharing of electron pairs between atoms, allowing both to attain noble gas configuration (e.g., H₂, Cl₂).
• Coordinate (dative) bond: One atom donates both electrons of the shared pair (e.g., NH₄⁺).
The driving force is attainment of stable electronic configuration (usually the octet).
2. Write Lewis dot symbols for atoms of the following elements: Mg, Na, B, O, N, Br
Answer
• Mg (2 valence electrons): · Mg · (2 dots, one on each side)
• Na (1 valence electron): Na· (1 dot)
• B (3 valence electrons): · B · with dot on top (3 dots total)
• O (6 valence electrons): :Ö: (6 dots — 2 pairs and 2 singles)
• N (5 valence electrons): :N· (5 dots — 1 lone pair + 3 singles)
• Br (7 valence electrons): :Br· (7 dots — 3 lone pairs + 1 single)
Dots are placed up to 2 per side around the element symbol.
3. Write Lewis symbols for: S and S²⁻; Al and Al³⁺; H and H⁻
Answer
• S: 6 valence electrons
⇒ :S̈: (6 dots)
S²⁻: gains 2 electrons
⇒ 8 valence electrons
⇒ [:S̈:]²⁻ (complete octet)
• Al: 3 valence electrons
⇒ · Al · (3 dots)
Al³⁺: loses 3 electrons
⇒ 0 valence electrons
⇒ [Al]³⁺ (no dots)
• H: 1 valence electron → H· (1 dot)
H⁻: gains 1 electron
⇒ 2 electrons
⇒ [H:]⁻ (helium configuration)
4. Draw the Lewis structures for the following molecules and ions: H₂S, SiCl₄, BeF₂, CO₃²⁻, HCOOH
Answer
H₂S: S is the central atom, 2 bond pairs to H, 2 lone pairs on S. Structure: H–S–H (bent, ~92°).
SiCl₄: Si central atom with 4 Cl atoms, each Cl has 3 lone pairs, Si has no lone pairs. Tetrahedral geometry.
BeF₂: Be central atom with 2 F atoms. Be has only 4 electrons (incomplete octet). Structure: F:–Be–:F (linear, 180°).
CO₃²⁻: C central atom, 3 O atoms. Total 24 valence electrons. One C=O double bond and two C–O single bonds (negative charges on O), with 3 equivalent resonance structures. Bond order = 4/3.
HCOOH (Formic acid): Structure H–C(=O)–OH. C has one C=O (carbonyl, sp²), one C–O (hydroxyl), and one C–H. The OH has lone pairs on O.
5. Define octet rule. Write its significance and limitations.
Answer
Octet Rule: Atoms tend to combine so that each has 8 electrons in its valence shell (noble gas configuration), achieving stability.
Significance: Explains ionic bonding (Na⁺, Cl⁻ attain octets), covalent bonding (H₂O, NH₃, CH₄), and predicts valency of elements.
Limitations
1. Incomplete octet: BeCl₂ (4e on Be), BCl₃ (6e on B), AlCl₃ (6e on Al).
2. Expanded octet: PCl₅ (10e on P), SF₆ (12e on S) — due to available d-orbitals in Period 3+.
3. Odd-electron molecules: NO, NO₂, ClO₂ — cannot satisfy octet for all atoms.
4. Does not explain bond lengths, bond energies, or relative stability.
6. Write the favourable factors for the formation of ionic bond.
Answer
1. Low ionization enthalpy of metal: Metal should easily lose electrons (e.g., alkali/alkaline earth metals).
2. High (negative) electron gain enthalpy of non-metal: Non-metal should readily accept electrons (e.g., halogens, chalcogens).
3. High lattice energy: Energy released when ions form the crystal lattice must be high (higher for small, highly charged ions).
4. Large electronegativity difference: Generally >1.7 on Pauling scale favours ionic bonding.
5. Noble gas configuration: The resulting ions should attain stable noble gas configurations.
7. Discuss the shape of the following molecules using the VSEPR model: BeCl₂, BCl₃, SiCl₄, AsF₅, H₂S, PH₃
Answer
BeCl₂: 2 bp, 0 lp
⇒ sp hybridisation
⇒ Linear, 180°.
BCl₃: 3 bp, 0 lp
⇒ sp² hybridisation
⇒ Trigonal planar, 120°.
SiCl₄: 4 bp, 0 lp
⇒ sp³ hybridisation
⇒ Tetrahedral, 109.5°.
AsF₅: 5 bp, 0 lp
⇒ sp³d hybridisation
⇒ Trigonal bipyramidal. Equatorial: 120°; axial-equatorial: 90°.
H₂S: 2 bp, 2 lp
⇒ sp³ (approximately)
⇒ Bent (V-shaped), ~92° (S uses more p-character due to size).
PH₃: 3 bp, 1 lp
⇒ sp³ hybridisation
⇒ Trigonal pyramidal, ~93.5°.
8. Although geometries of NH₃ and H₂O molecules are distorted tetrahedral, bond angle in water is less than that of ammonia. Discuss.
Answer
Both NH₃ (107°) and H₂O (104.5°) have sp³ hybridisation but different bond angles due to lone pairs:
NH₃: N has 3 bonding pairs and 1 lone pair. One lp–bp repulsion compresses bond angle from 109.5° to 107°.
H₂O: O has 2 bonding pairs and 2 lone pairs. Two lp–lp repulsions + two lp–bp repulsions (lp–lp > lp–bp > bp–bp) cause greater compression of the bond angle to 104.5°.
Since H₂O has more lone pairs (2) than NH₃ (1), it experiences greater repulsion and has a smaller bond angle.
9. How do you express the bond strength in terms of bond order?
Answer
Bond order is the number of bonds (shared electron pairs) between two bonded atoms. Higher bond order
⇒ stronger bond
⇒ higher bond enthalpy
⇒ shorter bond length.
• Single bond (order 1): e.g., C–C (154 pm, 347 kJ/mol)
• Double bond (order 2): e.g., C=C (134 pm, 611 kJ/mol) — stronger and shorter
• Triple bond (order 3): e.g., C≡C (120 pm, 837 kJ/mol) — strongest and shortest
In MO theory: Bond Order = ½ × (bonding electrons – antibonding electrons)
10. Define the bond length.
Answer
Bond length is the equilibrium distance between the nuclei of two bonded atoms in a molecule — the distance at which potential energy is minimum (attractive and repulsive forces balance). Measured in picometres (pm).
Factors: (1) Higher bond order
⇒ shorter bond; (2) Larger atomic size
⇒ longer bond; (3) Higher s-character in hybridisation
⇒ shorter bond (sp < sp² < sp³ bond lengths).
11. Explain the important aspects of resonance with reference to the CO₃²⁻ ion.
Answer
Resonance: When a molecule/ion cannot be accurately described by a single Lewis structure but by two or more structures differing only in electron arrangement (not atomic positions), the actual structure is the resonance hybrid of all canonical forms.
CO₃²⁻ (carbonate ion): Total valence electrons = 4 + 18 + 2 = 24. Three equivalent resonance structures can be drawn, each with one C=O and two C–O⁻ bonds, with the double bond shifting among the three oxygens.
Key aspects of resonance in CO₃²⁻:
• All three C–O bonds are equivalent (bond length 129 pm — between C–O 143 pm and C=O 122 pm).
• Bond order = 4/3 (1.33) for each C–O bond.
• The molecule is more stable than any single canonical form (resonance stabilisation energy).
• The charge (–2) is equally distributed among all three oxygen atoms.
12. H₃PO₄ can be represented by structures 1 and 2. Can these be taken as canonical forms of the resonance hybrid? If not, give reasons.
Answer
No. For two structures to be resonance (canonical) forms, they must: (1) have identical molecular formula, (2) have the same atomic arrangement (skeleton), and (3) differ only in the arrangement of electrons. If the two H₃PO₄ structures differ in the positions of hydrogen atoms (i.e., different H atoms connected to different atoms), they represent different structural isomers (tautomers), not resonance forms. True resonance requires that only electrons (bonds and lone pairs) are relocated, never atoms. Therefore, such structures of H₃PO₄ where atomic connectivity changes are tautomers, not canonical resonance forms.
13. Write the resonance structures for SO₂, NO₂, and NO₃⁻.
Answer
SO₂ (18 valence electrons): Two resonance structures — one with S=O double bond on left and S–O on right, and vice versa. S also has a lone pair. Both S–O bond lengths are equivalent (143 pm, intermediate).
NO₂ (17 valence electrons — odd electron molecule): Two resonance structures — N double bonded to one O and single bonded to other, with an unpaired electron on N. The two N–O bonds are equivalent. NO₂ is paramagnetic (1 unpaired electron).
NO₃⁻ (24 valence electrons): Three equivalent resonance structures, each with N double bonded to one O and single bonded to two O⁻. All three N–O bonds are equivalent (bond order 4/3). Similar to CO₃²⁻.
14. Use Lewis symbols to show electron transfer: (a) K and S (b) Ca and O (c) Al and N
Answer
(a) K and S: 2K· + :S̈:
⇒ 2K⁺ + [S²⁻]
⇒ K₂S (2 K atoms each lose 1e⁻ to S)
(b) Ca and O: ·Ca· + :O:
⇒ Ca²⁺ + [O²⁻]
⇒ CaO (Ca loses 2e⁻ to O)
(c) Al and N: ·Al· + :N̈
⇒ Al³⁺ + [N³⁻]
⇒ AlN (Al loses 3e⁻ to N)
15. Although both CO₂ and H₂O are triatomic molecules, CO₂ is linear while H₂O is bent. Explain using dipole moment.
Answer
CO₂ (O=C=O): C has sp hybridisation, no lone pairs, linear structure. The two C=O bond dipoles are equal and point in exactly opposite directions (180°) and cancel completely. Net dipole moment = 0 D.
H₂O: O has sp³ hybridisation, 2 lone pairs, bent structure (104.5°). The two O–H bond dipoles do not cancel — they have a resultant net dipole. Lone pairs on O also contribute. Net dipole moment = 1.85 D (non-zero).
The zero dipole moment of CO₂ confirms its linear structure (symmetrical cancellation). The non-zero dipole moment of H₂O confirms its bent, asymmetric structure.
16. Write the significance/applications of dipole moment.
Answer
1. Determine molecular geometry: Zero μ
⇒ symmetric molecule (CO₂ linear, not bent; CCl₄ tetrahedral).
2. Assess bond polarity: Higher μ
⇒ more polar bond
⇒ more ionic character.
3. Distinguish geometrical isomers: cis-isomers have non-zero μ; trans-isomers often have zero μ.
4. Calculate % ionic character: % = (observed μ / theoretical μ for full ion pair) × 100.
5. Predict physical properties: Polar molecules (high μ) have higher boiling points and are soluble in polar solvents.
6. Identify lone pairs: Deviations from expected geometry (detected via μ) indicate the presence of lone pairs.
17. Define electronegativity. How does it differ from electron gain enthalpy?
Answer
Electronegativity: Relative tendency of a bonded atom in a molecule to attract the shared electron pair. It is dimensionless, measured on the Pauling scale (F = 4.0). It is a property of an atom in a bond and varies with hybridisation.
Electron gain enthalpy (ΔegH): Enthalpy change when a neutral gaseous atom gains one electron: X(g) + e⁻
⇒ X⁻(g). Measured in kJ/mol for isolated atoms.
Key differences
• Electron gain enthalpy: isolated atom; Electronegativity: atom in a bond.
• Electron gain enthalpy: measurable thermodynamic quantity; Electronegativity: empirical, relative concept.
• Electronegativity varies with hybridisation (sp > sp² > sp³); electron gain enthalpy is fixed per atom.
18. Explain with the help of a suitable example polar covalent bond.
Answer
A polar covalent bond has unequal sharing of the bonding electron pair due to electronegativity difference between the bonded atoms. The more electronegative atom acquires partial negative charge (δ–), the other gets partial positive charge (δ+).
Example — HCl: Electronegativity: Cl (3.0) > H (2.1). Cl pulls electrons toward itself.
δ+H–Clδ–: dipole moment = 1.03 D.
Other examples: H₂O (O–H bonds), NH₃ (N–H bonds), HF (very polar, μ = 1.91 D).
Degree of polarity increases with electronegativity difference.
19. Arrange the bonds in order of increasing ionic character: LiF, K₂O, N₂, SO₂, ClF.
Answer
Ionic character ∝ electronegativity difference (Δχ):
• N₂: N–N, Δχ = 0 (purely covalent)
• ClF: Cl(3.0)–F(4.0), Δχ = 1.0
• SO₂: S(2.5)–O(3.5), Δχ = 1.0
• K₂O: K(0.8)–O(3.5), Δχ = 2.7
• LiF: Li(1.0)–F(4.0), Δχ = 3.0 (most ionic)
Increasing ionic character: N₂ < SO₂ ≈ ClF < K₂O < LiF
20. The skeletal structure of CH₃COOH is given. Write the correct Lewis structure for acetic acid.
Answer
Acetic acid structure: CH₃–C(=O)–OH
• Methyl C (sp³): bonded to 3 H (single bonds) and carboxyl C (single bond)
• Carboxyl C (sp²): double bonded to O (carbonyl, :O:), single bonded to OH oxygen
• Carbonyl O: 2 lone pairs, double bond to C
• Hydroxyl O: 2 lone pairs, single bond to C and single bond to H
Total: 3 C–H + 1 C–C + 1 C=O + 1 C–O + 1 O–H = 7 bonds, with lone pairs on all O atoms.
21. Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar. Explain why CH₄ is not square planar?
Answer
1. No d-orbitals available: C is in Period 2 with only 2s and 2p orbitals. Square planar requires dsp² hybridisation (needs a d-orbital). Carbon has no d-orbitals in its valence shell.
2. sp³ hybridisation is more stable: The four sp³ orbitals point toward tetrahedral corners (109.5°), maximising distance between bonding pairs. Square planar geometry (90° angles) would have greater electron-pair repulsion.
3. VSEPR: 4 bond pairs with no lone pairs adopt tetrahedral geometry (minimum repulsion).
Therefore, CH₄ is tetrahedral (sp³), not square planar.
22. Explain why BeH₂ molecule has a zero dipole moment although the Be–H bonds are polar.
Answer
In BeH₂, Be is sp hybridised
⇒ linear structure (180° H–Be–H). Each Be–H bond is polar (χBe=1.5, χH=2.1), but the two bond dipoles point in exactly opposite directions and are equal in magnitude. The vector sum = 0. Net dipole moment of BeH₂ = zero, despite polar individual bonds. Symmetrical molecules with polar bonds but geometry that causes cancellation are non-polar overall (e.g., CO₂, CCl₄, BF₃).
23. Which out of NH₃ and NF₃ has higher dipole moment and why?
Answer
NH₃ (μ = 1.46 D) > NF₃ (μ = 0.24 D).
NH₃: N (χ=3.0) more electronegative than H (χ=2.1). Bond dipoles point from H
⇒ N (toward N). Lone pair on N also points in the same direction. Bond dipoles and lone pair add constructively
⇒ large net dipole.
NF₃: F (χ=4.0) more electronegative than N (χ=3.0). Bond dipoles point from N
⇒ F (away from N). Lone pair on N points opposite to bond dipoles. Bond dipoles and lone pair partially cancel
⇒ small net dipole.
24. What is meant by hybridisation of atomic orbitals? Describe the shapes of sp, sp², sp³ hybrid orbitals.
Answer
Hybridisation is the mixing of atomic orbitals of slightly different energies to form new hybrid orbitals of equal energy and shape. Number of hybrid orbitals = number of atomic orbitals mixed.
sp hybridisation: 1s + 1p
⇒ 2 equivalent sp orbitals. Each is 50% s, 50% p character. Oriented at 180° (linear). Remaining 2 p-orbitals unhybridised (for π bonding). Example: BeCl₂, C₂H₂.
sp² hybridisation: 1s + 2p
⇒ 3 equivalent sp² orbitals. Each is 33% s character. Oriented at 120° (trigonal planar). One p-orbital unhybridised (for π bond). Example: BCl₃, C₂H₄, SO₃.
sp³ hybridisation: 1s + 3p
⇒ 4 equivalent sp³ orbitals. Each is 25% s character. Oriented at 109.5° (tetrahedral). No unhybridised p-orbitals. Example: CH₄, NH₃, H₂O.
25. Describe the change in hybridisation of Al atom in: AlCl₃ + Cl⁻ → AlCl₄⁻
Answer
AlCl₃ (before): Al has 3 bp, 0 lp
⇒ sp² hybridisation (trigonal planar, 120°, incomplete octet — 6 electrons on Al).
AlCl₄⁻ (after): Al has 4 bp, 0 lp (Cl⁻ donates lone pair to Al's empty p-orbital)
⇒ sp³ hybridisation (tetrahedral, 109.5°, complete octet).
Change: sp²
⇒ sp³. Al changes from electron-deficient trigonal planar to electron-complete tetrahedral.
26. Is there any change in hybridisation of B and N atoms in: BF₃ + NH₃ → F₃B–NH₃?
Answer
B in BF₃: sp² (trigonal planar)
⇒ sp³ in F₃B–NH₃ (tetrahedral). B accepts N's lone pair into its empty p-orbital. Change: sp²
⇒ sp³.
N in NH₃: sp³ with 3 bp + 1 lp (pyramidal)
⇒ sp³ with 4 bp + 0 lp (tetrahedral) in product. Hybridisation type unchanged (sp³
⇒ sp³), but geometry changes from pyramidal to near-tetrahedral as lone pair is used in coordinate bond.
27. Draw diagrams showing the formation of double and triple bonds in C₂H₄ and C₂H₂.
Answer
C₂H₄ (double bond): Each C is sp² hybridised. Three sp² orbitals form: 1 C–C σ bond (head-on overlap) + 2 C–H σ bonds per C. The remaining unhybridised p-orbital on each C (perpendicular to sp² plane) forms 1 C=C π bond (lateral overlap).
C=C = 1σ + 1π. Total: 5σ + 1π bonds.
C₂H₂ (triple bond): Each C is sp hybridised. Two sp orbitals form: 1 C–C σ bond + 1 C–H σ bond per C. The two remaining p-orbitals on each C (px and py) form 2 π bonds (lateral overlap, px–px and py–py).
C≡C = 1σ + 2π. Total: 3σ + 2π bonds.
28. What is the total number of sigma and pi bonds in: (a) C₂H₂ (b) C₂H₄?
Answer
(a) C₂H₂ (H–C≡C–H): 2 C–H σ bonds + 1 C≡C σ bond + 2 C≡C π bonds = 3σ + 2π
(b) C₂H₄ (H₂C=CH₂): 4 C–H σ bonds + 1 C=C σ bond + 1 C=C π bond = 5σ + 1π
29. Considering x-axis as the internuclear axis, which will not form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px (c) 2py and 2py (d) 1s and 2s
Answer
(a) 1s + 1s: Spherically symmetric, overlap head-on along x-axis
⇒ σ bond ✓
(b) 1s + 2px: 2px is along x-axis, head-on overlap with 1s
⇒ σ bond ✓
(c) 2py + 2py: Both perpendicular to x-axis (internuclear axis). They can only overlap laterally
⇒ forms a π bond, NOT a sigma bond. ✗
(d) 1s + 2s: Both spherically symmetric, overlap along x-axis
⇒ σ bond ✓
Answer: (c) 2py and 2py will not form a sigma bond — they form a π bond.
30. Which hybrid orbitals are used by carbon atoms in the following molecules?
Answer
(a) CH₃–CH₃ (b) CH₃–CH=CH₂ (c) CH₃–CH₂–OH (d) CH₃–CHO (e) CH₃COOH
Ans.
(a) CH₃–CH₃ (Ethane): Both C atoms have only single bonds
⇒ sp³ (tetrahedral)
(b) CH₃–CH=CH₂ (Propene): CH₃ carbon
⇒ sp³; the two carbons of C=C
⇒ sp² each
(c) CH₃–CH₂–OH (Ethanol): Both C atoms have only single bonds
⇒ sp³
(d) CH₃–CHO (Ethanal): CH₃ carbon
⇒ sp³; CHO carbon (C=O)
⇒ sp²
(e) CH₃COOH (Acetic acid): CH₃ carbon
⇒ sp³; COOH carbon (C=O)
⇒ sp²
31. What do you understand by bond pairs and lone pairs of electrons? Illustrate with examples.
Answer
Bond pairs: Electron pairs shared between two bonded atoms, forming the covalent bond. Located between nuclei. Example: In H₂O, the 2 pairs of electrons between O and each H are bond pairs (H:O).
Lone pairs: Electron pairs on an atom NOT involved in bonding. Belong exclusively to one atom. Occupy more space than bond pairs, causing greater VSEPR repulsion. Example: In H₂O, oxygen has 2 lone pairs (pairs of electrons not connected to H). In NH₃, N has 1 lone pair.
Lone pairs act as Lewis bases and influence molecular geometry and bond angles.
32. Distinguish between a sigma and a pi bond.
Answer
Property | Sigma (σ) Bond | Pi (π) Bond |
Type of overlap | Head-on (axial) along internuclear axis | Lateral (sideways) of parallel p-orbitals |
Electron density | Maximum between nuclei | Above and below internuclear axis |
Strength | Stronger (greater overlap) | Weaker (less effective overlap) |
Free rotation | Allowed | Not allowed (breaks π bond) |
Occurrence | In all bonds (single, double, triple) | Only in multiple bonds |
Example | H–H, C–C in C₂H₆ | C=C (second bond), C≡C (2nd and 3rd bonds) |
33. Explain the formation of H₂ molecule on the basis of valence bond theory.
Answer
Each H atom: 1s¹ (one electron in 1s orbital).
As two H atoms approach:
• Electronic energy decreases (each nucleus attracts the other's electron — attractive forces dominate).
• Electron density builds up between the nuclei.
• At equilibrium bond length (74 pm), potential energy reaches minimum (–436 kJ/mol stabilisation = bond dissociation energy).
• Further approach causes nuclear repulsion to dominate, raising energy sharply.
• The two 1s orbitals overlap to form a sigma (σ) bond with the shared electron pair between nuclei (electrons with opposite spins per Pauli's principle).
Result: H–H bond (σ bond), bond length 74 pm, bond enthalpy 436 kJ/mol.
34. Write the important conditions required for LCAO to form molecular orbitals.
Answer
1. Similar energy: Combining AOs must have comparable energies (e.g., 1s+1s; 2s+2s; but not 1s+2p of very different elements).
2. Same symmetry: AOs must have the same symmetry with respect to the molecular axis (e.g., s and px along x-axis can combine; py and s cannot form σ-type MO).
3. Effective spatial overlap: Orbitals must have significant spatial overlap (good directional compatibility and proximity).
4. Conservation: Number of MOs formed = number of AOs combined (one bonding + one antibonding per pair).
35. Use molecular orbital theory to explain why the Be₂ molecule does not exist.
Answer
Be: 1s²2s². In Be₂ (8 electrons):
MO configuration: (σ1s)²(σ*1s)²(σ2s)²(σ*2s)²
Bond Order = ½ × (4 bonding – 4 antibonding) = ½ × 0 = 0
Bond order = 0 means no net bonding. The antibonding electrons cancel out all bonding. Be₂ would have no stability advantage over two isolated Be atoms.
Therefore, Be₂ does not exist as a stable molecule.
36. Compare the relative stability of O₂, O₂⁺, O₂⁻, O₂²⁻ and indicate their magnetic properties.
Answer
Using MO theory for O₂ (bond order 2 as base):
O₂⁺ (15e): Remove 1e from π*2p
⇒ Bonding=10, Antibonding=5
⇒ BO = ½(10-5) = 2.5. Most stable. Paramagnetic (1 unpaired electron).
O₂ (16e): BO = ½(10-6) = 2. Paramagnetic (2 unpaired electrons in degenerate π* orbitals).
O₂⁻ (17e): Add 1e to π*2p
⇒ Bonding=10, Antibonding=7
⇒ BO = ½(10-7) = 1.5. Paramagnetic (1 unpaired electron).
O₂²⁻ (18e): Add 2e to π*2p
⇒ Bonding=10, Antibonding=8
⇒ BO = ½(10-8) = 1. Least stable. Diamagnetic (all electrons paired).
Stability order: O₂⁺ > O₂ > O₂⁻ > O₂²⁻
37. Write the significance of a plus and a minus sign shown in representing the orbitals.
Answer
The + and – signs represent the phase (sign) of the wave function (ψ), not electric charge.
• Constructive overlap (+/+ or –/–): Wave functions add
⇒ increased electron density between nuclei
⇒ bonding MO (lower energy, stable).
• Destructive overlap (+/– or –/+): Wave functions cancel
⇒ a node (zero electron density) between nuclei
⇒ antibonding MO (higher energy, destabilising).
The phase determines whether orbital combination leads to a bonding or antibonding molecular orbital.
38. Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds?
Answer
P: [Ne]3s²3p³. Promotes 1 electron to 3d orbital
⇒ uses 3s + 3p + 3p + 3p + 3d = sp³d hybridisation
⇒ 5 sp³d hybrid orbitals.
Geometry: Trigonal bipyramidal — 3 equatorial bonds (120° in horizontal plane) + 2 axial bonds (perpendicular to equatorial plane, at 90° to equatorial).
Why axial bonds are longer: The 3 equatorial bond pairs are at 120° from each other. The 2 axial bond pairs are at 90° from each of the 3 equatorial bond pairs — experiencing greater repulsion (3 interactions at 90° vs fewer for equatorial). This greater bp-bp repulsion at 90° pushes axial Cl atoms farther away, making axial P–Cl bonds longer (219 pm) and weaker than equatorial P–Cl bonds (204 pm).
39. Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?
Answer
Hydrogen bond is the electrostatic attractive force between a hydrogen atom covalently bonded to a highly electronegative atom (F, O, or N) and another electronegative atom. Represented as X–H···Y (dotted line = H-bond; solid = covalent bond). The δ+ on H (due to electron withdrawal by X) is attracted to the lone pair of Y.
Types: (1) Intermolecular H-bonds (between molecules, e.g., H₂O, HF); (2) Intramolecular H-bonds (within molecule, e.g., o-nitrophenol).
Hydrogen bonds are stronger than van der Waals forces. Van der Waals forces: ~0.1–2 kJ/mol; Hydrogen bonds: ~10–40 kJ/mol (O–H···O) to ~160 kJ/mol (F–H···F). However, H-bonds are much weaker than covalent bonds (~350–400 kJ/mol for C–H).
40. What is meant by bond order? Calculate bond order of: N₂, O₂, O₂⁺ and O₂⁻.
Answer
Bond order = ½ × (bonding electrons – antibonding electrons). Higher bond order = stronger, shorter bond.
N₂ (14e): KK(σ2s)²(σ*2s)²(π2p)⁴(σ2pz)²
Bonding = 10, Antibonding = 4
⇒ BO = ½(10-4) = 3 (triple bond). Diamagnetic. Very stable.
O₂ (16e): KK(σ2s)²(σ*2s)²(σ2pz)²(π2p)⁴(π*2p)²
Bonding = 10, Antibonding = 6
⇒ BO = ½(10-6) = 2 (double bond). Paramagnetic.
O₂⁺ (15e): Remove 1e from π*2p of O₂
⇒ Antibonding = 5
⇒ BO = ½(10-5) = 2.5. Paramagnetic.
O₂⁻ (17e): Add 1e to π*2p of O₂
⇒ Antibonding = 7
⇒ BO = ½(10-7) = 1.5. Paramagnetic.
Important Extra Questions
What is VSEPR theory? How is it used to predict molecular geometry? Give examples.
Answer
VSEPR (Valence Shell Electron Pair Repulsion) theory: Electron pairs in the valence shell of the central atom repel each other and arrange as far apart as possible to minimise repulsion. The molecular geometry is determined by the total number of electron pairs (bonding + lone pairs).
Order of repulsion: lp–lp > lp–bp > bp–bp. Lone pairs compress bond angles.
Examples:
• 2 bp, 0 lp: BeCl₂ → Linear (180°)
• 3 bp, 0 lp: BCl₃
⇒ Trigonal planar (120°)
• 4 bp, 0 lp: CH₄
⇒ Tetrahedral (109.5°)
• 4 bp, 1 lp: NH₃
⇒ Trigonal pyramidal (107°)
• 4 bp, 2 lp: H₂O → Bent (104.5°)
• 5 bp, 0 lp: PCl₅
⇒ Trigonal bipyramidal
• 6 bp, 0 lp: SF₆ → Octahedral (90°)
What are the differences between ionic and covalent bonds? Give examples.
Answer
Property | Ionic Bond | Covalent Bond |
Formation | Electron transfer | Electron sharing |
Electronegativity difference | Large (>1.7) | Small (<1.7) |
Between | Metal + Non-metal | Non-metal + Non-metal |
Physical state (RT) | Solid crystalline | Gas, liquid, or molecular solid |
Melting/boiling point | High | Generally lower |
Solubility | Polar solvents (water) | Non-polar solvents |
Electrical conductivity | Conducts when molten/dissolved | Non-conductor generally |
Examples | NaCl, MgO, CaCl₂ | H₂O, CH₄, CO₂ |
Explain resonance energy with reference to benzene.
Answer
Resonance energy is the difference between the actual (measured) energy of a molecule and the energy calculated for the most stable single canonical structure. It represents extra stability from electron delocalisation.
Benzene (C₆H₆): Two equivalent Kekulé structures (alternating C=C and C–C). If benzene were purely one Kekulé structure, heat of hydrogenation = 3 × 120 = 360 kJ/mol. Actual measured value = 208 kJ/mol.
Resonance energy = 360 – 208 = 152 kJ/mol
Actual benzene has all 6 C–C bonds equivalent (139 pm — between single 154 pm and double 134 pm), with π electrons delocalised over the ring. This makes benzene exceptionally stable (less reactive than expected for a "triene").
Explain the paramagnetic nature of O₂ using MO theory.
Answer
O₂ has 16 electrons. MO configuration:
(σ1s)²(σ*1s)²(σ2s)²(σ*2s)²(σ2pz)²(π2px)²(π2py)²(π*2px)¹(π*2py)¹
The last two electrons go into degenerate π*2px and π*2py orbitals — one each, following Hund's rule.
O₂ has two unpaired electrons. A substance with unpaired electrons is paramagnetic (attracted to a magnetic field).
Lewis structure of O₂ (O=O) incorrectly predicts all electrons paired (diamagnetic). MO theory's correct prediction of paramagnetism was a major triumph — liquid O₂ is indeed attracted between the poles of a magnet.
What is the role of electronegativity in determining the nature of a chemical bond? Explain with examples.
Answer
Electronegativity determines bond character:
• Δχ = 0: Pure covalent bond (equal sharing). Example: H–H, Cl–Cl, N≡N.
• Δχ = 0.1–1.6: Polar covalent bond (unequal sharing). Example: H–Cl (Δχ=0.9), H–O (Δχ=1.4), C–O (Δχ=1.0).
• Δχ > 1.7: Ionic bond (complete transfer). Example: Na–F (Δχ=3.0), K–Cl (Δχ=2.2).
High electronegativity of F, O, N explains: (1) Why these atoms attract electrons in polar bonds; (2) Why they can form hydrogen bonds (lone pairs on electronegative atoms attract δ+ H); (3) Why ionic compounds of F, O with metals have high lattice energies; (4) The polarity and dipole moments of molecules (H₂O, HF, NH₃).