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Chemistry LibreTexts

Chapter 2 Solutions

 

P2.1:  Both bonds in O3 have a bond order of 1.5, and are the same length – the two resonance contributors shown below are of equal importance in illustrating the bonding in the molecule.

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P2.2:

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P2.3:

 

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P2.4:

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P2.5:

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P2.6:

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The He2 molecule has four electrons to account for in the MO diagram: two are placed in the σ bonding orbital and two in the σ* antibonding orbital.  Thus there is no net attractive force to hold the He2 molecule together (in fact, there is a net repulsive force due to the relative energy level of the σ* molecular orbital).

P2.7:

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P2.8:

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P2.9:

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P2.10:

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P2.11:

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P2.12:

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P2.13:

1= lowest bp

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Reasoning:

a)  2 and 3 have two fluorines and are more polar than 1, so they  have stronger intermolecular interactions.  3 has one more carbon than 2, and therefore stronger van der Waals interactions. 4 is capable of hydrogen bonding, so it has the strongest intermolecular interactions and the highest boiling point.

b) 1 and 2 have only van der Waals interactions, but 2 has more carbons so these interactions are slightly stronger.  3 has a polar carbonyl group, and 4 is capable of hydrogen bonding.

c) 1 is not capable of hydrogen bonding.  2 and 3 both have hydrogen bonding groups, but 3 has one more carbon and therefore stronger overal van der Waals interactions.

d) 1 has only van der Waals interactions. 2 has a polar thiol group, but 3 has a hydroxyl group which is capable of hydrogen bonding.  4 is a salt: the charge-charge interactions are very strong and lead to a very high boiling point.

P2.14:

a) The one on the right is more soluble (fewer hydrophobic carbons)

b) The one on the left is more soluble (ionic phosphate group)

c) The one on the left is more soluble (fewer hydrophobic carbons)

d) The one on the left is more soluble (capable of hydrogen bonding)

e) The one on the right is more soluble (fewer hydrophobic carbons)

P2.15:  Toluene has one more carbon than benzene, and therefore has stronger overal van der Waals interactions and a higher boiling point.  The melting point trend, however, is a different story.  Remember that melting involves a transition from an ordered solid state, where molecules are stacked together, to a disordered liquid state, where molecules move freely  relative to one another.   Benzene is a flat molecule and is able to stack very effeciently (like plates) in the solid phase.  The methyl group of toluene prevents this efficient packing, and thus the overal intermolecular van der Waals interactions in the toluene solid phase are lower than those for benzene, and it requires less energy (heat) to break the tolune molecules apart.

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P2.16:

 

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P2.17: For each structure, the atoms are indicated (with a black dot) which could bear the corresponding negative formal charge, and one example of a resonance contributor is provided.

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P2.18: For each structure, the atoms are indicated (with a black dot) which could bear the corresponding positive formal charge, and one example of a resonance contributor is provided.

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P2.19: For each structure, the atoms to which the positive charge can be delocalized are indicated with a black dot.

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Contributors

  • Organic Chemistry With a Biological Emphasis, Tim Soderberg (University of Minnesota, Morris)