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Forces 2

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Uploaded by on Nov 5, 2009

Determining the intermolecular forces (Van der Waals) of a molecule and the resulting physical properties trends. Van der Waals forces include London dispersion, hydrogen bonding, and dipoles.

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Uploader Comments (EnderlePhD)

  • we are saying it will have high viscocity /boiling point etc ( which has higher force ) means they are all in liquid state but dipole-d interaction may occur in solid also so overall if we need to consider the forces(attractive) then will not we consider all the 3 states our is it just that we are considering liquid state.

  • @markbondking007 Typical questions are for liquid state, but you can consider other states if you want. Additional factors may come in with other states.

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  • @EnderlePhD I don't quite have enough space to explain here, but I'll try a simple explanation. First, Cl is not always included in the list of atoms that qualify for the H-bond. However, that point is not relevant in this case. If CCl4 is only interacting with itself, then there are no H atoms in solution to cause an H-bond. Not only is an H needed, but that H should be bonded to an electronegative atom.

  • @EnderlePhD How is that CH4 CCl4 are both london dispersion forces?? Shouldn't CCl4 be a hydrogen bond? Hydrogen bonds occur in molecules where H is connected to to a highly electronegative atom such as O, F, N and Cl, so why isn't CCl4 written as a hydrogen bond?

  • my force can make you gay

  • @seaviee1 bonded*

  • @EnderlePhD Can you please explain why NH3 has intERmolecular H-bond when the H bond to N does not interact with (N,O,F) from another molecule. Thanks

  • Thank you, this is very helpful!

  • thanks again for the explanation :)

  • I'd say dipole. H-bonding can be a complex issue but most will simply summarize it as this: an H on one molecule that is directly bonded to an electronegative element (N, O, or F) forms an H-bond when it interacts with an electronegative element of another molecule (N, O, or F). Intramolecular H-bonds are also possible if both possibilities are present on a single, large molecule. For CHF3, there is no H atom that is directly bonded to the F atom (the electronegative element).

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