Solutions to IMF Exercises

Answers/Solutions to Exercises

Exercises

Exercise 1. Indicate the intermolecular forces experienced by each of the following molecules.

a) PCl₃
This is a polar molecule.

It experiences both London dispersion forces and dipole-dipole forces of attraction.

b) PCl₅

This is a non polar molecule.

It only experiences London dispersion forces.

c) C₄H₁₀

This is a nonpolar molecule.

It only experiences London dispersion forces.

d) CH₂F₂

This molecule is polar.

It experiences both London dispersion forces and dipole-dipole forces of attraction.

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Exercise 2. Indicate the intermolecular forces that are experienced by each of the following.

a) CH₃OH

This molecule is polar.

It experiences both London dispersion and dipole-dipole forces of attraction. In addition, hydrogen is bonded to oxygen, and it experiences hydrogen bonding.

b) O₂

O₂ is nonpolar. It only experiences London dispersion forces

c) CHCl₃

This molecule is polar. It experiences London dispersion and dipole-dipole forces of attraction.

d) H₂Se

This molecule is polar. It has a bent shape. It experiences both London dispersion and dipole-dipole attractive forces.

e) C₁₀H₂₂

This molecule is nonpolar. It only experiences London dispersion forces.

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Exercise 3. Below are the boiling points of some substances.

methane, CH₄, (-164°C)
methanol, CH₃OH, (64.7°C)

decane, C₁₀H₂₂, (174°C)
1-decanol, C₁₀H₂₁OH, (231°C)

Explain why methanol has a boiling point that is 229 degrees higher than that of methane while 1-decanol boils only 57 degrees higher than decane.

Both methane and methanol have small dispersion forces of attraction. Methanol can hydrogen bond which explains the much higher boiling point.

Both 1-decanol and decane have large similar dispersion forces. The 1-decanol can also hydrogen bond, which is why it has a 57 degree higher boiling point than decane. The London dispersion forces are much stronger in decane and decanol than in methane and methanol.

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Exercise 4. Draw a sketch of the hydrogen bonding in several molecules of methanol, CH₃OH.

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Exercise 5. Which substance in each pair has the highest boiling point?

a) S₂F₂ or S₂Cl₂

S₂Cl₂ has the highest boiling point. Both molecules have a bent molecular geometry about the sulfur atoms, but Cl is larger than F, therefore, the dispersion forces are greater in S₂Cl₂ than in S₂F₂.

b) NH₃ or PH₃

Both molecules are polar, but NH₃ also has intermolecular hydrogen bonding. NH₃ has the highest boiling point.

c) H₂S or H₂Te

H₂Te has the higher boiling point. Both molecules are polar, have a bent molecular geometry, and experience dipole-dipole forces. They both experience London dispersion forces. H₂Te has the higher boiling point because Te is heavier than S and therefore H₂Te experiences stronger dispersion forces.

d) NOF or NOCl

Both molecules have London dispersion forces and dipole-dipole attractive forces. Their molecular geometry is trigonal pyramidal. The NOCl has a higher boiling point because Cl is heavier than F, and therefore the dispersion forces of attraction are stronger in NOCl.

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Exercise 6. Identify the intermolecular forces for each of the following substances.

a) HO(CH₂)₂OH, ethylene glycol

This molecule experiences London dispersion, dipole-dipole, and hydrogen bonding forces of attraction. The hydrogen bonding forces are especially strong since there are two hydroxyl groups.