Answer:
In order to have a dipole moment (i.e., to be a polar molecule) a molecule must have polar bonds and must have a molecular geometry which is not symmetrical (i.e., one in which the vector sum of the bond dipoles 0).
In CH2F2 the C-F and C-H bonds are polar and the molecule is not symmetrical; therefore, the molecules is polar and would show a dipole moment.
In CF4 the C-F bonds are polar, but the molecule is symmetrical; therefore, the molecule is non-polar and would not show a dipole moment.
Discuss briefly the relationship between the dipole moment of a molecule and the polar character of the bonds within it. WIth this as a basis, account for the difference between the dipole moments of CH2F2 and CF4.
does anyone have any idea?
the greater the dipole moment of a molecule the more polar it's bonds.
A bond is polar when on atom is more electronegative than the other or the molecule is asymmetrical in geometry.
A bond is polar when an atom is more electronegative than the other. That is true. However, molecules made up of different atoms may have a dipole moment if they have atoms with different electronegativities AND they are not symmetrical. The CF4 molecule is tetrahedral but completely symmetrical; therefore, CF4 has no dipole moment even though C and F have different electronegativities. The individual polar C-F bonds add up to a net zero dipole moment for the molecule because of the symmetry of the tetrahedral arrangement. Replacing 2 of the F atoms with 2 H atoms destroys that symmetry.
what would the dipole moment be for CH2F2
Look on page one for the title "What causes chemical reactions to happen?" on the link below and you will see CH2F2 shows 2.29 Debeys. OR you can go to www.google.com and type in "dipole moment CH2F2" without the parenthese, click on the article and read all about it. I did that and never found the part that dealt with CH2F2 but perhaps I just overlooked it.
http://www.google.com/search?hl=en&q=dipole+moment+CH2F2&btnG=Google+Search
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