Solution:

(a) CH₄ vs. CH₃CH₃

Higher BP: CH₃CH₃ (ethane: -89°C vs. methane: -162°C)

Both are nonpolar molecules with only London dispersion forces. Ethane has more electrons (18 vs. 10) and greater surface area, leading to stronger dispersion forces.

(b) CH₃OH vs. CH₃SH

Higher BP: CH₃OH (methanol: 65°C vs. methanethiol: 6°C)

Both have similar structures, but:

• CH₃OH can hydrogen bond (O-H bonds)
• CH₃SH has dipole-dipole forces only (S-H cannot H-bond effectively because S is not electronegative enough)
• Hydrogen bonding >> dipole-dipole forces

(c) HF vs. HCl

Higher BP: HF (19.5°C vs. HCl: -85°C)

Despite HCl having more electrons (18 vs. 10):

• HF exhibits strong hydrogen bonding (F is highly electronegative)
• HCl has dipole-dipole forces only
• The hydrogen bonding in HF is so strong it overcomes HCl's larger dispersion forces

Key concept: Hydrogen bonding (when present) usually dominates other IMF effects.

Solution:

Given: Three molecules Find: Identify all IMFs present in each

Step 1: Analyze CH₄ (methane)

Lewis structure: C with 4 H atoms, tetrahedral

Molecular geometry: Tetrahedral (SN = 4, 0 lone pairs)

Polarity check:

• C-H bonds have small polarity (ΔEN = 0.4)
• Symmetrical tetrahedral geometry
• All C-H dipoles cancel out
• Nonpolar molecule

IMFs present:

• London Dispersion Forces (LDF) only
• All molecules have LDF
• No permanent dipole → no dipole-dipole
• No O-H, N-H, or F-H → no hydrogen bonding

Answer for (a): London dispersion forces only

Step 2: Analyze HCl (hydrogen chloride)

Lewis structure: H-Cl with 3 lone pairs on Cl

Molecular geometry: Linear (diatomic)

Polarity check:

• H-Cl bond is polar (ΔEN = |3.0 - 2.1| = 0.9)
• Only one bond (diatomic)
• Polar molecule with permanent dipole
• δ+ on H, δ- on Cl

IMFs present:

1. London Dispersion Forces (LDF) ✓

   • All molecules have LDF

2. Dipole-Dipole Forces ✓

   • HCl is polar
   • Permanent dipole moment
   • Molecules align: H(δ+) ↔ Cl(δ-)

3. Hydrogen Bonding? ✗

   • Has H, but H is bonded to Cl (not N, O, or F)
   • Cl is not electronegative enough
   • No hydrogen bonding

Answer for (b): Dipole-dipole forces and London dispersion forces

Step 3: Analyze NH₃ (ammonia)

Lewis structure: N with 3 H atoms and 1 lone pair

Molecular geometry: Trigonal pyramidal (SN = 4, 1 lone pair)

Polarity check:

• N-H bonds are polar (ΔEN = |3.0 - 2.1| = 0.9)
• Asymmetric trigonal pyramidal geometry (lone pair)
• N-H dipoles do NOT cancel
• Polar molecule

IMFs present:

1. London Dispersion Forces (LDF) ✓

   • All molecules have LDF

2. Dipole-Dipole Forces ✓

   • NH₃ is polar
   • Permanent dipole moment

3. Hydrogen Bonding ✓

   • H bonded directly to N (nitrogen) ✓
   • Other NH₃ molecules have lone pair on N to accept H-bond ✓
   • Both requirements met!

Hydrogen bonding in NH₃:

• Each NH₃ can form up to 4 H-bonds
• 3 H atoms as donors
• 1 lone pair as acceptor

$\ce{H-N-H ... N-H}$

Answer for (c): Hydrogen bonding, dipole-dipole forces, and London dispersion forces

Summary:

| Molecule | LDF | Dipole-Dipole | H-bonding | Strongest IMF | |----------|-----|---------------|-----------|---------------| | CH₄ | ✓ | ✗ | ✗ | LDF | | HCl | ✓ | ✓ | ✗ | Dipole-dipole | | NH₃ | ✓ | ✓ | ✓ | Hydrogen bonding |

Expected boiling point order: NH₃ > HCl > CH₄

Actual values:

• CH₄: -161°C (weakest IMF)
• HCl: -85°C (dipole-dipole)
• NH₃: -33°C (H-bonding, strongest IMF)

Key takeaway:

• Nonpolar → LDF only
• Polar (no N-H/O-H/F-H) → DD + LDF
• Has N-H, O-H, or F-H → H-bonding + DD + LDF

Solution:

(a) CH₄ vs. CH₃CH₃

Higher BP: CH₃CH₃ (ethane: -89°C vs. methane: -162°C)

Both are nonpolar molecules with only London dispersion forces. Ethane has more electrons (18 vs. 10) and greater surface area, leading to stronger dispersion forces.

(b) CH₃OH vs. CH₃SH

Higher BP: CH₃OH (methanol: 65°C vs. methanethiol: 6°C)

Both have similar structures, but:

• CH₃OH can hydrogen bond (O-H bonds)
• CH₃SH has dipole-dipole forces only (S-H cannot H-bond effectively because S is not electronegative enough)
• Hydrogen bonding >> dipole-dipole forces

(c) HF vs. HCl

Higher BP: HF (19.5°C vs. HCl: -85°C)

Despite HCl having more electrons (18 vs. 10):

• HF exhibits strong hydrogen bonding (F is highly electronegative)
• HCl has dipole-dipole forces only
• The hydrogen bonding in HF is so strong it overcomes HCl's larger dispersion forces

Key concept: Hydrogen bonding (when present) usually dominates other IMF effects.

Solution:

Key factors:

H₂O - Boiling point: 100°C

• Can form hydrogen bonds (H bonded to highly electronegative O)
• Each water molecule can form up to 4 H-bonds (2 as donor via H atoms, 2 as acceptor via O lone pairs)
• Creates extensive 3D network of H-bonds
• O-H···O hydrogen bonds are very strong (~20 kJ/mol each)

H₂S - Boiling point: -60°C

• Cannot form hydrogen bonds (S is not electronegative enough)
• Only dipole-dipole forces and London dispersion
• Although H₂S has more electrons (34 vs. 18) → stronger dispersion forces
• S-H bonds are much less polar than O-H bonds → weaker dipole-dipole

Electronegativity values:

• O: 3.44 (highly electronegative, strong H-bonding)
• S: 2.58 (not electronegative enough for effective H-bonding)

Conclusion: The extensive hydrogen bonding network in water requires much more energy to break than the dipole-dipole and dispersion forces in H₂S. Hydrogen bonding is so strong it overcomes the mass/size advantage of H₂S.

ΔH_vap values:

• H₂O: 40.7 kJ/mol (much higher)
• H₂S: 18.7 kJ/mol

Solution:

Key factors:

H₂O - Boiling point: 100°C

• Can form hydrogen bonds (H bonded to highly electronegative O)
• Each water molecule can form up to 4 H-bonds (2 as donor via H atoms, 2 as acceptor via O lone pairs)
• Creates extensive 3D network of H-bonds
• O-H···O hydrogen bonds are very strong (~20 kJ/mol each)

H₂S - Boiling point: -60°C

• Cannot form hydrogen bonds (S is not electronegative enough)
• Only dipole-dipole forces and London dispersion
• Although H₂S has more electrons (34 vs. 18) → stronger dispersion forces
• S-H bonds are much less polar than O-H bonds → weaker dipole-dipole

Electronegativity values:

• O: 3.44 (highly electronegative, strong H-bonding)
• S: 2.58 (not electronegative enough for effective H-bonding)

Conclusion: The extensive hydrogen bonding network in water requires much more energy to break than the dipole-dipole and dispersion forces in H₂S. Hydrogen bonding is so strong it overcomes the mass/size advantage of H₂S.

ΔH_vap values:

• H₂O: 40.7 kJ/mol (much higher)
• H₂S: 18.7 kJ/mol

Solution:

Given:

• H₂O: MW = 18 g/mol, BP = 100°C
• H₂S: MW = 34 g/mol, BP = -60°C

Find: Explain why H₂O has higher BP despite lower molar mass

Expected trend violation:

• Normally, higher molar mass → higher BP (stronger LDF)
• But here: H₂S has higher mass but LOWER BP
• Must consider type of IMF, not just molecular size

Step 1: Analyze H₂O structure and IMFs

Lewis structure: H-O-H with 2 lone pairs on O

Molecular geometry: Bent (SN = 4, 2 lone pairs)

Polarity: Polar (asymmetric, dipoles don't cancel)

IMFs present:

1. Hydrogen bonding: ✓

   • O-H bonds present (H bonded to oxygen)
   • Lone pairs on O can accept H-bonds
   • Each H₂O can form 4 hydrogen bonds:
     • 2 H atoms as donors
     • 2 lone pairs as acceptors

2. Dipole-dipole: ✓ (but overshadowed by H-bonding)

3. London dispersion: ✓ (weak, small molecule)

Strongest IMF: Hydrogen bonding (10-40 kJ/mol)

Step 2: Analyze H₂S structure and IMFs

Lewis structure: H-S-H with 2 lone pairs on S

Molecular geometry: Bent (SN = 4, 2 lone pairs)

Polarity: Polar (asymmetric, like H₂O)

IMFs present:

1. Hydrogen bonding: ✗

   • Has H-S bonds, BUT
   • S is in Period 3 (not N, O, or F)
   • S is not electronegative enough for H-bonding
   • Electronegativity: S (2.5) vs. O (3.5)
   • No hydrogen bonding

2. Dipole-dipole: ✓

   • H₂S is polar
   • Weaker than H₂O's dipole (S less electronegative than O)

3. London dispersion: ✓

   • Stronger than H₂O (more electrons: 18 vs. 10)

Strongest IMF: Dipole-dipole forces (5-25 kJ/mol)

Step 3: Compare IMF strengths

H₂O:

• Strongest IMF: Hydrogen bonding (~20 kJ/mol)
• Very strong intermolecular attractions
• Requires high temperature to break IMFs and boil

H₂S:

• Strongest IMF: Dipole-dipole (~5-10 kJ/mol)
• Moderate intermolecular attractions
• Even though it has stronger LDF (more electrons), dipole-dipole is still weaker than H-bonding

Step 4: Explain boiling point difference

Boiling point depends on IMF strength:

$\text{Stronger IMF} \rightarrow \text{Higher energy needed} \rightarrow \text{Higher BP}$

H₂O (BP = 100°C):

• Must break hydrogen bonds to boil
• Requires significant energy
• Each molecule bonded to ~4 neighbors via H-bonds
• Extended H-bonding network

H₂S (BP = -60°C):

• Must break dipole-dipole forces to boil
• Requires much less energy
• Weaker intermolecular attractions

Energy comparison:

• H-bonding in H₂O: ~20 kJ/mol
• Dipole-dipole in H₂S: ~10 kJ/mol
• H₂O requires roughly 2× more energy to boil

Answer:

Water has a much higher boiling point than hydrogen sulfide because water exhibits hydrogen bonding while H₂S does not.

Key points:

1. Hydrogen bonding requires H bonded to N, O, or F

   • H₂O qualifies (O-H bonds)
   • H₂S does not (S is Period 3, not electronegative enough)

2. Hydrogen bonding is much stronger than dipole-dipole

   • H-bonding: 10-40 kJ/mol
   • Dipole-dipole: 5-25 kJ/mol
   • H-bonding in H₂O >> dipole-dipole in H₂S

3. Type of IMF matters more than molar mass

   • H₂S has more electrons (stronger LDF)
   • But H₂O's H-bonding dominates
   • Demonstrates: H-bonding > dipole-dipole + LDF

General principle: For hydrides of Group 15-17 elements, Period 2 hydrides (NH₃, H₂O, HF) have anomalously high boiling points due to hydrogen bonding. Period 3+ hydrides (PH₃, H₂S, HCl) follow normal trend based on molar mass.

Comparison:

• Without H-bonding, H₂O would have BP around -80°C (similar to H₂S)
• H-bonding raises it by ~180°C!

Solution:

Given: Three C₅H₁₂ isomers (same molecular formula, different structures)

• n-pentane: CH₃-CH₂-CH₂-CH₂-CH₃ (linear)
• isopentane: CH₃-CH(CH₃)-CH₂-CH₃ (one branch)
• neopentane: C(CH₃)₄ (four CH₃ groups around central C, highly branched)

Find: Predict BP order and explain

Step 1: Identify molecular properties

All three isomers:

• Same molecular formula: C₅H₁₂
• Same molar mass: 72 g/mol
• Same number of electrons: 42 electrons
• All nonpolar: Symmetrical C-H bonds, no overall dipole
• Same IMFs: London Dispersion Forces (LDF) only

Since mass and polarity are identical, what differs?

Answer: Molecular shape / Surface area contact

Step 2: Analyze molecular shapes

n-Pentane (linear chain):

• Extended, elongated shape
• Can align closely with neighboring molecules
• Maximum surface area contact
• Like two logs lying parallel

$\ce{CH3-CH2-CH2-CH2-CH3}$

Isopentane (one branch):

• Somewhat compact
• Less surface contact than linear
• Intermediate surface area contact

$\ce{CH3-CH(CH3)-CH2-CH3}$

Neopentane (highly branched):

• Nearly spherical shape
• Very compact
• Minimum surface area contact
• Like two soccer balls touching (only contact at one point)

$\ce{ CH3 }$ $\ce{ | }$ $\ce{CH3-C-CH3}$ $\ce{ | }$ $\ce{ CH3 }$

Step 3: Relate shape to LDF strength

London Dispersion Forces depend on:

1. Number of electrons (same for all three = 42 electrons) ✓
2. Surface area contact between molecules
   • More contact → more temporary dipoles can interact
   • More contact → stronger total LDF

Shape effect:

• Linear molecules: Maximum surface contact → Strongest LDF
• Branched molecules: Less surface contact → Weaker LDF
• Compact/spherical: Minimum contact → Weakest LDF

Analogy:

• Two pencils lying parallel (n-pentane): Large contact area
• Two partly-branched objects (isopentane): Medium contact
• Two balls touching (neopentane): Small contact area (point contact)

Step 4: Predict boiling points

Boiling point trend:

$\text{Stronger IMF} \rightarrow \text{Higher BP}$

LDF strength order:

$\text{n-pentane} > \text{isopentane} > \text{neopentane}$

Therefore, BP order:

$\text{n-pentane} > \text{isopentane} > \text{neopentane}$

Step 5: Verify with actual data

Actual boiling points:

• n-Pentane: 36°C (highest)
• Isopentane: 28°C (middle)
• Neopentane: 10°C (lowest)

Difference: 26°C range for same molecular formula!

Step 6: Explain the trend

Why does shape matter so much?

n-Pentane (BP = 36°C):

• Linear chain allows molecules to pack closely
• Large surface area in contact
• Many weak LDF interactions add up
• Collective effect creates stronger total attraction
• Requires more energy to separate molecules

Isopentane (BP = 28°C):

• One branch reduces contact area
• Fewer LDF interaction points
• Intermediate attraction

Neopentane (BP = 10°C):

• Compact, nearly spherical
• Minimal contact (molecules touch at small area)
• Fewest LDF interactions
• Weakest total attraction
• Easiest to separate (boils at lowest temperature)

Answer:

Boiling point order: n-pentane > isopentane > neopentane

Reasoning:

All three isomers have identical molecular formulas (C₅H₁₂), molar masses (72 g/mol), and are nonpolar, so they experience only London Dispersion Forces.

The difference in boiling points arises from molecular shape, which affects the surface area available for intermolecular contact:

1. n-Pentane (linear): Maximum surface contact between molecules → strongest LDF → highest BP (36°C)

2. Isopentane (one branch): Intermediate surface contact → intermediate LDF → middle BP (28°C)

3. Neopentane (highly branched, compact): Minimum surface contact → weakest LDF → lowest BP (10°C)

General principle: For isomers with same molecular formula:

• More branching → More compact → Less surface contact → Lower BP
• Linear structure → Extended shape → More surface contact → Higher BP

Key insight: Even though all molecules have the same number of electrons (same inherent LDF capability), the geometric arrangement determines how effectively molecules can interact. This demonstrates that molecular shape is crucial for determining physical properties, even when chemical composition is identical.