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Van Der Wal's force doesn't add up

  1. Dec 1, 2006 #1
    I was given the explanation that as you go down a group in the periodic table, the boiling points of the elements increase as the radius of the atom increase. There are more electrons and therefore the greater distance they can move in which the temporary dipoles can cover and so the Van Der wal's force is bigger.

    And here comes the question. I thought that Van der wal's force is a temporary imbalance of electrons in an atom in which one end is slightly negative than the other. Therefore if you have larger atomic radius then wouldn't there be smaller van der wal's force as they electrons are more likely to spread out over a bigger area creating lots of little dipoles which are insignificant to the overall charge of the atom or cancel each other out instead of creating a big temporary dipoles by being concentrated on one side.

    Can anyone help me out or is there something I'm missing?
  2. jcsd
  3. Dec 1, 2006 #2
    More electrons means that there will be more electron imbalances. Therefore, the London dispersion forces will be stronger.
  4. Dec 2, 2006 #3
    But if the nuclear radius increase wouldn't this counteract the van der wal's forces because with bigger radius there would be less chance of all the electrons causing an imbalance as they would be more likely to "spread out".
  5. Dec 2, 2006 #4


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    Bigger atom => greater polarizability => greater dipole moment fluctuations => greater van der waals interactions.

    For a rule of thumb relationship between atomic size and polarizability, see Griffiths' Introduction to Electrodynamics.

    PS: Note that the atomic size is not correlated to the nuclear radius in any way.
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