Example



Example

EXAMPLE Predicting Molecular Polarity: 

 Decide whether the molecules represented by the following formulas are polar or nonpolar. (You may need to draw Lewis structures and geometric sketches to do so.)

a. CO    b. OF2     c. CCl4     d. CH2Cl2     e. HCN

Solution:

a. The Lewis structure for CO2 is

The electronegativities of carbon and oxygen are 2.55 and 3.44. The 0.99 difference in electronegativity indicates that the C-O bonds are polar, but the symmetrical arrangement of these bonds makes the molecule nonpolar.

If we put arrows into the geometric sketch for CO2, we see that they exactly balance each other, in both direction and magnitude. This shows the symmetry of the bonds.

b. The Lewis structure for OF2 is

The electronegativities of oxygen and fluorine, 3.44 and 3.98, respectively, produce a 0.54 difference that leads us to predict that the O-F bonds are polar. The molecular geometry of OF2 is bent. Such an asymmetrical distribution of polar bonds would produce a polar molecule.

c. The molecular geometry of CCl4 is tetrahedral. Even though the C-Cl bonds are polar, their symmetrical arrangement makes the molecule nonpolar.

                

d.  The Lewis structure for CH2Cl2 is

The electronegativities of hydrogen, carbon, and chlorine are 2.20, 2.55, and 3.16. The 0.35 difference in electronegativity for the H-C bonds tells us that they are essentially nonpolar. The 0.61 difference in electronegativity for the C-Cl bonds shows that they are polar. The following geometric sketches show that the polar bonds are asymmetrically arranged, so the molecule is polar. (Notice that the Lewis structure above incorrectly suggests that the bonds are symmetrically arranged. Keep in mind that Lewis structures often give a false impression of the geometry of the molecules they represent.)

              

e.  The Lewis structure and geometric sketch for HCN are the same:

The electronegativities of hydrogen, carbon, and nitrogen are 2.20, 2.55, and 3.04. The 0.35 difference in electronegativity for the H-C bond shows that it is essentially nonpolar. The 0.49 difference in electronegativity for the C-N bond tells us that it is polar. Molecules with one polar bond are always polar.

Click here to see a Study Sheet for this task.

Click here to see an exercise that will allow you to try this task yourself. 

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