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Intermolecular forces

  1. Jan 28, 2006 #1
    I have been trying to get my head around this question, but do not understand it.

    Consider the two isoelectronic substances, bromine (Br2) and Iodine monochloride (ICl). Based upon your knowledge of intermolecular forces, explain the difference in their boilings pts. (Bromine: 59 degrees C; iodine monochloride: 97 degrees C)

    Well, I know that both bromine and iodine monochloride have 70 electrons per molecule, therefore, the strength of the London dispersion forces between the molecules of each should be the same. Now, what I'm not sure of is how to assess their polarities.

    VSPER Theory doesn't really help me out because they are both of the same shape, so one would assume that they would both be either polar or non-polar.

    However, if I assess their respective electronegativities...

    BR-BR (must be the same; therefore has to be non-polar?)
    I-CL (is different; therefore polar?)

    Therefore, ICL should have stronger dipole-diple-force, and a higher boiling point.

    Are these conclusions justified?

    Thanks for your prompt response.
  2. jcsd
  3. Jan 28, 2006 #2
    Looks good to me! You may want to wait for a more qualified person to confirm, but that's probably how I would've gone about doing it.
  4. Jan 29, 2006 #3


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    Yeah, I think that's a good assessment. Although the london dispersion interactions relate to the polarity of the molecule, in that you'll need to consider the individual atoms and their quantum numbers. The farther away an electron is from the nucleus, the effective nuclear charge is relatively less, the electron is not being held as strongly. The orbit is farther away from the nucleus (quantum number) and there's relatively more electron shielding.
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