1. Module 2 Intro

1.22. Page 4

Lesson 4

Module 2—Chemical Compounds

Read


Read “Using Dipole-Dipole and London Forces to Predict Boiling Points” on pages 107 and 108 in your textbook. After reading this section, answer the questions below. Send your answers to your teacher.

  1. Does the data shown in "Table 1" on page 107 support the hypothesis relating number of electrons and boiling point? Explain your reasoning.

  2. State the name of the type of intermolecular bonding force that is influenced by number of electrons.

  3. Was the comparison between hydrocarbons in the lab, Boiling Points of Hydrocarbons, a valid set of compounds to examine to study this type of relationship? Support your answer by making specific reference to the structure and polarity of the compounds involved.

Self-Check

 

SC 1. Complete "Practice" problem 1 on page 109 of your textbook.

 

Check your work.
Self-Check Answer

 

SC 1.    a.   London, dipole-dipole (Water is a polar substance.)
            b.   London
            c.   London
            d.   London, dipole-dipole (Ethanol is a polar substance.)
            e.   London, dipole-dipole (Ammonia is a polar substance.)
            f.   London

 

SC 2. Complete "Practice" problem 2 on page 109 of your textbook.

 

Check your work.
Self-Check Answer

 

SC 2.

  1. hydrogen fluoride. The difference in electronegativity between H and F is greater than the difference between H and Cl, resulting in stronger bond dipoles.

  2. CH3Cl. The bond dipole between C and Cl is greater than C and I.

  3. Ammonia, NH3. The bond dipole between N and Hl is greater than N and Br.

  4. Water. The bond dipole between H and O is greater than H and S.

 

SC 3. Complete "Practice" problem 3 on page 109 of your textbook.

 

Check your work.
Self-Check Answer

 

SC 3.


  1. Compound

    Number of Electrons

    methane

    10

    ethane

    18

     

    Ethane will have stronger London forces since it has the greater number of electrons.


  2. Compound

    Number of Electrons

    oxygen

    12

    nitrogen

    14

     

    Nitrogen will have stronger London forces since it has the greater number of electrons.


  3. Compound

    Number of Electrons

    sulfur dioxide

    32

    nitrogen dioxide

    23

     

    Sulfur dioxide will have stronger London forces since it has the greater number of electrons.


  4. Compound

    Number of Electrons

    methane

    10

    ammonia

    10

     

    The two compounds being compared are isoelectronic; therefore, if the London force is the only intermolecular bonding force, then they should have the same boiling point.

 

SC 4. Complete "Practice" problem 4 on page 109 of your textbook.

 

Check your work.
Self-Check Answer

 

SC 4.


  1. Compound

    Number of Electrons

    Stereochemical Shape

    Polar

    boron trifluoride

    32

    trigonal planar

    no

    nitrogen trifluoride

    34

    trigonal pyramidal

    yes

     

    Nitrogen trifluoride would have the higher boiling point since it has the larger number of electrons; therefore, has stronger London forces, is polar, and will have additional dipole-dipole forces.

     

    Boron trifluoride is nonpolar, has only London forces, and has fewer electrons; therefore, it has relatively weaker attractive forces.


  2. Compound

    Number of Electrons

    Stereochemical Shape

    Polar

    chloromethane, CH3Cl

    26

    tetrahedral

    yes

    ethane, C2H6

    18

    tetrahedral

    nonpolar

     

    Chloromethane would have the higher boiling point of the two compounds but stronger London forces and dipole-dipole forces due to its polarity.

 

Module 2: Lesson 4 Assignment

 

Retrieve the copy of the Module 2: Lesson 4 Assignment that you saved to your course folder earlier.

 

Complete questions 2.a. and b. of the assignment.