[Kate]

Hi. I've been reading about van der Waals forces but I'm a bit confused as to how many van der Waals forces exist and what exactly is the difference between them.

So, under the van der Waals forces there are:
- dipole - dipole interactions (Keesom forces)
- dipole - induced dipole interactions (Debye forces)
- dispersion or London forces

Is this correct ? And could anyone explain me the differences between each force ? Also, I was reading Chang's "Chemistry" book and he mentions ion - induced dipole interactions and I don't know where this fits in the van der Waals forces.

[renge ishyo]

The polarity of the interacting molecules determines the strength of the van der walls forces. A "dipole" means that you have a distinct region of positive charge at one end of a molecule and a distinct region of negative charge at the other end of a molecule. A "nonpolar molecule" is roughly neutral across its entire surface.

For dipole-dipole interactions both molecules are dipoles, and you have the positive end of one dipole molecule attracting the negative end of another dipole molecule. Even if the attraction isn't strong enough to exchange electrons with each other (as in an ionic interaction) both molecules are still attracted to each other to some extent.

Weaker than the dipole-dipole interactions are the dipole-induced dipole interactions. This is an interaction between a dipole molecule and a second, non-polar molecule. The charge separation on the dipole can induce a slight charge separation on the non-polar molecule. For instance, the negative end of the dipole as it approaches the normally non-polar molecule will repel the negative electrons on the non-polar molecule, but not its positive protons. This "induces" a charge separation in the non-polar molecule that allows the dipoles negative end to get closer to the positive proton area that has now been exposed on the formally non-polar molecule. This is weaker than the dipole-dipole interaction because in the prior case both molecules were already polarized and you didn't need to "spend energy" to repel the electrons in the non-polar molecule to induce it to become a dipole like you have to do in the latter case.

Finally, the weakest of them all are the london forces. These are interactions between a non-polar molecule and another non-polar molecule. Normally they are both nuetral, but every once in awhile the electrons on one non-polar molecule might sway to one side of the atom or the another creating a momentary weak dipole. This can then interact with the momentary weak dipole on the second non-polar molecule that just happens to have the same thing happening at that moment. The interactions are far weaker in this case than in the previous two cases above where at least one species was a permanent dipole the whole time.

Hope that made sense 

[Kate]

It did made sense, thank you for your explanation.

Just have one more question though. Which compound, CH₃F or CCl4, has stronger intermolecular forces and why ?

[renge ishyo]

You should be able to handle that question now    The stronger interaction will be for the *molecule* with the largest overall dipole moment. So draw out the Lewis structures for CH₃F and CCl₄ first. Next, consider the difference in electronegativity for each *bond* one at a time (this determines the size of the dipole arrow you will draw for that bond above it...the greater the electronegativity difference the larger the arrow you will draw pointing towards the more electronegative atom). Draw the dipole arrows next to each bond in the moelcule. The molecule with the greatest net overall dipole (most unbalanced arrows) will be the one with the greatest intermolecular attraction.

For the case of CCl₄ be careful. If you have arrows pointed exactly opposite each other, they "cancel" each others effect out. So in CH₃F for example, if you consider the dipoles for the C-H and C-F bonds opposite to each other then you might see:

   
-   ----
H - C - F

This indicates that the left dipole for the CH bond cancels the CF bond dipole out somewhat. The net dipole is still to the right (because the CF bond is more polar than the C-H bond), but it will be smaller than the C-F bond dipole by itself. If the arrows for the dipoles on each of these bonds were equal in magnitude and opposite in direction there would be no net dipole for the molecule. What significance does this have for CCl₄?

If you want to check your work with us before turning it in just post your answer on here and we'll take a look 

[Kate]

I know CCl₄ is non-polar and there's no net dipole for the molecule. So, that should mean that CH₃F has stonger intermolecular forces than CCl₄. But the boiling points for CH₃F and CCl₄ is -78 ºC and -23 ºC, respectively. How is this explained ?

[cliverlong]

Jim Clarke writes that dispersion forces have greater impact on mp & bp than dipole or Hydrogen bonds

http://www.chemguide.co.uk/atoms/bonding/vdw.html

So although the C-F bond is highly polar and CH₃F is assymetric, and one expects hydrogen bonding between the CH₃F molecules - which is not present in CCl₄, the fact the Cl atoms are larger and the resulting molecule is larger, the dispersion forces are much greater in  CCl₄ and this raises the bp of CCl₄.

To be honest, I find that surprising as it contradicts prety much all that I have read and believed about the relative strength of these forces !!

Clive

[renge ishyo]

I know CCl4 is non-polar and there's no net dipole for the molecule. So, that should mean that CH3F has stonger intermolecular forces than CCl4. But the boiling points for CH3F and CCl4 is -78 ºC and -23 ºC, respectively. How is this explained ?

CH₃F does indeed have the stronger intermolecular attractions as you say. The boiling point is dependent on many factors, and one of the main factors is the molecular weight. Compounds with higher molecular weights are harder to "lift" into the air than are compounds with lower molecular weights and these heavy compounds have higher boiling points in general if you ignore intermolecular attractions.

Compare the molecular weights of the two compounds in question:

CH₃F = 34 g/mol

CCl₄ = 152 g/mol

Note that there is a great discrepancy in the weight of the two compounds, but there is only a 50 degree difference in their boiling points. The reason why the difference in boiling points is so small is because CH₃F has all those intermolecular attractions that raise its boiling point. If it lacked such attractions it would boil off at much lower temperatures. For example, look at the non-polar methane:

CH₄ = 16 g/mol

This isn't that much smaller than the the weight for CH₃F, but look at what the lack of intermolecular forces in methane does to its boiling point:

Boiling point: -161.6 °C

[Kate]

I understand now, thank you for your explanations. But shouldn't the boiling point be a measure of the intensity of the intermolecular forces between molecules of the same compound ? So if A has a higher boiling point than B, shouldn't the intermolecular forces be stronger in A than in B ?

By the way...

one expects hydrogen bonding between the CH₃F molecules

... I don't think there is hydrogen bonding between CH₃F molecules. Right ? 

[Arctic-Nation]

As renge ishyo said, boiling points are dependent on many factors, intermolecular forces only comprising a few of them. If you simply look at molecules with very similar intermolecular forces, you will notice that the one with the higher molecular weight will always have the higher boiling point. Take, for example, benzene and toluene, methanol and ethanol, hexane and heptane: the second molecule is the heaviest, and has the highest boiling point.

As for fluoromethane, there indeed is very little hydrogen bonding involved.

[Yggdrasil]

CH₃F does indeed have the stronger intermolecular attractions as you say. The boiling point is dependent on many factors, and one of the main factors is the molecular weight. Compounds with higher molecular weights are harder to "lift" into the air than are compounds with lower molecular weights and these heavy compounds have higher boiling points in general if you ignore intermolecular attractions.

This is absolutely wrong!  CCl₄'s higher boiling point is not due to it being harder to lift into the air.  Here are two pieces of evidence to convince you:

1)  Gravitational forces are extremely weak.
Let's calculate the amount of energy needed to lift one mole of CCl₄ molecules 1m into the air.  E_grav = mgh = 1.50 J/mol.  Now compare this to the amount of thermal energy available at 0^oC:  RT = 2.3 kJ/mol.  The energy required to lift molecules against gravity is orders of magnitude lower than the energy contributed by thermal energy (much less intermolecular forces) and therefore contributes little to the energy required to vaporize a substance.

2)  CCl₄ has a higher melting point than CH₃F
If CCl₄'s boiling point were due to stronger gravitational forces and not due to stronger intermolecular forces, then one would expect CCl₄ to have a lower melting point than CH₃F since "lifting" molecules out of a liquid would not be involved in a changing from a solid to a liquid.  However, the melting points of CCl₄ and CH₃F are -23^oC and -142^oC, respectively (note, the boiling point of CCl₄ is misquoted above.  CCl₄ boils at 77^oC.  Else I would be worried about that bottle of liquid carbon tetrachloride I have in lab).  From these data we can conclude that the intermolecular forces between CCl₄ molecules are indeed stronger than the intermolecular forces between CH₃F molecules.


Now why do CCl₄ molecules have stronger intermolecular forces than CH₃F?  CCl₄ molecules are very polarizable because they are big and have many electrons.  It is very easy for these molecules to develop instantaneous dipoles and even easier for these instantaneous dipoles to induce dipoles in neighboring molecules.  True, CH₃F does have a slight permanent dipole, but people tend to underestimate the strength of dispersion forces in comparison to forces between permanent dipoles especially when the difference in molecular size is large (see the link posted by cliverlong).

Yes, it is true that factors other than the strength of intermolecular forces determine a compounds boiling point.  Gravitational forces, however, are not among them.

[renge ishyo]

Yggdrasil, it is true that the gravitational force is far far weaker than the electrical force. There would be a difference due to gravity from the earth to be sure, but the effect would be small as noted. It is an empirical fact that boiling point increases with molecular weight if you compare molecules with similar intermolecular forces (which is where my "wording" comes from). For instance, the noble gases show the trend of increasing BP's as they go up in weight and it is recognized that london forces are the only electromagnetic interactions between the noble gases so you can also say that these interactions generally increase with molecular weight as well (due to the polarizability of the larger electron clouds increasing).

As for saying that the molecules are "lifted" I guess that is my fault for trying to find a feeling for the idea, and I apologize for it. What I meant is that it is harder to lift a larger molecule away from another larger molecule and was not considering really lifting the molecules relative to the earth itself. At least I can console myself with the fact that I never actually mentioned the gravitational force in my discussion (although I suppose it was strongly implied by the idea of "lifted").

I would still answer that the dipole interaction is stronger than the london forces if I was asked to compare the strength between the two. If we make a bad comparison between two molecules that are quite dissimilar (as we did originally when the two molecules differ in their molecular weight by over 100g) then it might seem that either interaction could be stronger depending on which two molecules we are talking about, but if we compare two molecules of similar size and weight (maybe I should say mass?) so that the characteristics of their electron shells aren't too dissimilar, the molecules with the dipole interaction will beat out the molecules with only london forces in a test of strength hands down. After all, even molecules with permanent dipoles have the london force interactions with them as well so they get that interaction plus the dipole interaction whereas the non-polar molecule only gets the london forces. The CH₄ vs. CH₃F comparison demonstrates the difference in the strengths between the two interactions nicely and they both have similar mass, which is why I tossed this in there.

[Yggdrasil]

Yggdrasil, it is true that the gravitational force is far far weaker than the electrical force. There would be a difference due to gravity from the earth to be sure, but the effect would be small as noted. It is an empirical fact that boiling point increases with molecular weight if you compare molecules with similar intermolecular forces (which is where my "wording" comes from). For instance, the noble gases show the trend of increasing BP's as they go up in weight and it is recognized that london forces are the only electromagnetic interactions between the noble gases so you can also say that these interactions generally increase with molecular weight as well (due to the polarizability of the larger electron clouds increasing).

As for saying that the molecules are "lifted" I guess that is my fault for trying to find a feeling for the idea, and I apologize for it. What I meant is that it is harder to lift a larger molecule away from another larger molecule and was not considering really lifting the molecules relative to the earth itself. At least I can console myself with the fact that I never actually mentioned the gravitational force in my discussion (although I suppose it was strongly implied by the idea of "lifted").

Sorry if my criticism came across as a bit harsh, but I just wanted to dispel any notion of gravitational forces playing any effect.  It is a common misconception that I've heard many times on the forum, so I just wanted to make sure that the issue got explained clearly (especially because in our real life experience gravity seems like the strongest force, so the "lifting" explanation makes a lot of intuitive sense).

I would still answer that the dipole interaction is stronger than the london forces if I was asked to compare the strength between the two. If we make a bad comparison between two molecules that are quite dissimilar (as we did originally when the two molecules differ in their molecular weight by over 100g) then it might seem that either interaction could be stronger depending on which two molecules we are talking about, but if we compare two molecules of similar size and weight (maybe I should say mass?) so that the characteristics of their electron shells aren't too dissimilar, the molecules with the dipole interaction will beat out the molecules with only london forces in a test of strength hands down. After all, even molecules with permanent dipoles have the london force interactions with them as well so they get that interaction plus the dipole interaction whereas the non-polar molecule only gets the london forces. The CH₄ vs. CH₃F comparison demonstrates the difference in the strengths between the two interactions nicely and they both have similar mass, which is why I tossed this in there.

The comparison between CH₄ and CH₃F still isn't quite so fair because CH₃F has more than twice the mass and twice the number of electrons as CH₄.  Perhaps a more fair comparison is between ethane (C₂H₆) and CH₃F.  Both have the same number of electrons and similar sizes, but CH₃F has a permanent dipole whereas ethane does not.  Now, take a look at the comparison between boiling points:

C₂H₆,  bp = -88.6 °C
CH₃F,  bp = -78.2 °C

[renge ishyo]

Even that comparison doesn't seem good Yggdrasil, because the volume of ethane occupies nearly twice the volume of fluoromethane and hence it increases the number of possible individual London interactions for the former to a large degree. This can cloud the comparison of individual strengths between separate intermolecular interactions. Now if we try to compensate by substituting the polar ethanol in so that the volumes and shapes of the two molecules are similar we end up right back where we started:

ethane: bp = -88.6 °C
ethanol bp: 78.4 °C

What we are doing I think is demonstrating the problem of using boiling point data in general to compare the relative strength of intermolecular interactions. The shape of the molecule, it's size (I found a new word that is not weight!), the overall polarity of the molecule, the external pressure on the system, the "softness" of its valence electrons, the nature of whatever solvent, inpurities, etc. happen to be present all have roles to play in the boiling point.  To compare the strength of the interactions directly I think requires a less crude method.

[UG]

Your 'debate' has reminded me of this question given by the RSC, it went something like this:
"Speculate as to why D₂O has a slightly higher boiling point than H₂O"
Now I'm probably looking pretty dumb right now  cause I had no idea. So how would you explain this?

[renge ishyo]

Speculate as to why D2O has a slightly higher boiling point than H2O

Oh, that's easy, it is because D20 has a slightly higher molecular weig...ha! Almost caught me