UnkleRhaukus
  • UnkleRhaukus
What is the range and magnitude of the Van der Waals force
Chemistry
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
If you mean dispersion (London) forces, they fall off like 1/r^6, and the magnitude is directly proporional to the number of electrons in the interacting molecules and the polarizability of the electron orbitals. Does that help?
UnkleRhaukus
  • UnkleRhaukus
yeah the london dispersion forces one of the Van der Waals forces, what i really want to know is how come it is 1/r^6, not 1/r^2. i cant visualize where the 1/r^6 is coming from
anonymous
  • anonymous
It's a dipole-induced dipole force. The field created by the instantaneous dipole on one molecules falls of as 1/r^3, that being the nature of an electric dipole field. That means the strength of the induced dipole on the other molecule goes like 1/r^3. Then the field of the induced dipole falls off like 1/r^3 on its way back to the original molecule, so to speak. So that gives you two factors of 1/r^3, for a total of 1/r^6. As for why the dipole field itself falls off like 1/r^2: because the Coulomb field itself falls off like 1/r^2, and the only way you can tell a dipole field exists is if you are close enough to "see" both charges (otherwise it looks like a net charge of zero, from far enough away). So that gives you an extra factor of d/r, where d = effective charge separation, because as d/r -> 0 you are too far away to "see" the charge separation.

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anonymous
  • anonymous
Woops, the first sentence of the second para should read "As for why the dipole field itself falls off like 1/r^3..."
UnkleRhaukus
  • UnkleRhaukus
ok that is a real good answer i just need some pictures to visualize

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