March 27, 2017

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I am having trouble understanding dipole moments. I can't figure out why that a tetrahedral CF4 molecule, trigonal planar SO3 molecule and a octahedral H2SF4 molecule are non-polar while a H2CF2 is polar? I hope I explained that well and thanks!

For CF4, think of it this way.
There is a dipole between each C and F since the electronegativities are different (C abouat 2.1 and F about 4.0). BUT CF4 is a tetrahedral molecule and the symmetry of the tetrahedral shape cancels the resultant dipole moment and the dipole moment of the molecule is zero. It may be easier to see with the SO3 since that is trigonal planar. Let dots represent oxygens.
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You can see the symmetry of the triangle. With S in the middle of the triangle, bonds from S to O have a dipole moment but the resultant is zero for the overall molecule. Same thing for octahedral. It would help if we could draw structures on the computer but the boards aren't so equipped. Be sure to follow up if you don't get it but please explain exactly what you don't understand.

Well from my very limitted knowledge of organic chemistry... The first three molecules you mentioned are completely symmetric in the distribution of the Hs Os and Fs about the carbon. With the H2CF2 however you have a tetrahedral molecule in which the Fs occupy one side of the molecule and the Hs the other side. Since the Fs attract the electrons more strongly than the Hs, the F side of the molecule is more - creating the dipole moment. To help visualize this google search molecular geometry and find a picture of the tetrahedral configuration. As you will see any location of the Fs and Hs results in this unsymmetric configuration. Hope this helps.

I'm sorry that I wasn't too clear. I understand that if all dipoles cancel then that molecule is non-polar but what I don't understand is how they cancel? Is it all based on symmetry and if it is; does the directions of the electronegativity have to pointing in the same direction when using symmetry?

yes and yes. Thus, all the bonds are equal in CF4. CF3Cl would have a dipole moment, as will CF2Cl2 and CFCl3. But CCl4, CF4, CH4, and CI4 will not. For octahedral, MX6 will be symmetrical but MX5Y will not. MX6 will not have a dipole moment because of the symmetry and all of the M-x dipoles canceling but all of the small individual M-X won't cancel when we have 5 M-x and 1 M-y for the MX5Y molecule. The symmetry thing is how we know H2O is not a linear molecule. If it were linear, as in H-O-H the the individual H-O dipole would cancel the O-H dipole and the result would be zero for H2O. BUT we can measure a dipole moment for H2O which means it must NOT be linear. In fact, the H2O molecule is quite easy to see how the symmetry makes the dipoles cancel if the molecule is linear.

Thanks, to both of you, you helped me out a lot! Now I'll be able to pass the "lovely" Chem. quiz on Tuesday! Have a g'night.

It just dawned on me what you meant by direction and my answer of yes may or may not be on the mark depending upon how you look at it. Look at the H2O molecule again.
The H-O-H linear arrangement has a slight + charge on the left H and a slight negative charge on the O. Mark that on a sheet of paper. Then on the other side, we have a another slight - on O and the H is +. Now draw an arrow on the left side going from - to + and on the other side another arrow going from - to +. See the arrows go in opposite directions so their effect cancels.

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