[Polleke]

hallo all,
everyone knows the "general" saying: polar molecules "like" polar ones and non polar like non polar ones.
(like = dissolve)

But how do you prove this theoreticly, how to show this?

I know that the interaction between polar molecules themself is different from the interaction between non polar ones.
But there is also a difference in interaction between the non polar ones themself (depening on how they are bounded).

I mean: I can see that when you have 2 polar substances togheter , the interaction to break their bounds(A-A and B-B) will be "the same"  as when they are mixed(A-B + A-B).
But what with the non polar ones: there you have so many sorts of "non polar" .



So how do you clarify this statement?

[Astrokel]

I mean: I can see that when you have 2 polar substances togheter , the interaction to break their bounds(A-A and B-B) will be "the same"  as when they are mixed(A-B + A-B).
But what with the non polar ones: there you have so many sorts of "non polar" .

I don't get this part, can you provide an example?

[macman104]

I don't get this part, can you provide an example?

Seconded, I'm not sure I 100% understand what your asking either...

[Polleke]

well they explained it to me like this:

imagine you have a molecule that exists out of A-A interactions, and another one that exists out of B-B molecules.

and when you put them together they form A-B boundings if the A-B interaction is about the same as a A-A or B-B interaction.
(meaning A and B interactions need to be similar)

Thus you need to have 2 polar ones or 2 non polar ones.
(both A-A and B-B need to be polar or non polar)

I do understand the general view of this, however I do not see it when you have 2 different types of non polar boundings...
I mean: not al non polar boundings are of the same interaction (depending on their geometry etc..)

[Astrokel]

Does A-A and B-B, the dash represents a bond or intermolecular attraction?

[Vidya]

A---A is a nonbonding interaction
B----B is also nonbonding inteaction
 A---B interaction  is also nonbonding
 if A---B interaction is as strong as A---A and B----B then  A and B are soluble in each other.Like dissolves like is based on it only.

[Polleke]

A---A is a nonbonding interaction
B----B is also nonbonding inteaction
 A---B interaction  is also nonbonding
 if A---B interaction is as strong as A---A and B----B then  A and B are soluble in each other.Like dissolves like is based on it only.

this is what I mean, but its not really a proof.
I want to know what is behind this.

[Vidya]

You know even in non polar molecules there is instantaneous dipoles
due electrons and their distribution around nucleus .This can induce dipoles in other molecules .I think that can be the reason for their solubility.

[nj_bartel]

Yes, but if they're becoming charged, and that's what creates their solubility, why would they not be soluble in a polar compound, where there are also charges, and where the charges in the nonpolar molecule could still be formed?

[ARGOS++]

Dear Polleke;

IMHO: The A-A, B-B, A-B, B-A model is wrong, because it is pure static, but we need a dynamic model to describe solubility and find then the major players of this “game”.

If we declare the crystal as A-A, and the Solvent as B-B:
In my model the driving force is only each particular concentration difference which attracts/forces the molecules in both directions: into the crystal, and into the solution/solvent!
The measure would be both Diffusion Coefficients: fd_{A in B–B}(...), and fd_{B in A-A}(...).
Therein fd_x() depends very strong at the polarity difference of A and B, and maybe at some other substance specific properties.
As limiting parameter we must use also the saturated concentrations of each in each.
Maybe we need as a contrary force the gap between both saturated concentrations too.

I believe, that only if we use such a model as above we are able to describe, what will going on during the solution task, and what solubility “finally” means.

I have drawn some pictures, but I haven’t finished the whole model with all equations yet.
But that's maybe only my opinion!
Good Luck!
                    ARGOS⁺⁺

[Vidya]

Yes, but if they're becoming charged, and that's what creates their ...

There are no permanent dipoles as we can see in polar or polar covalent molecules .These are instantanous dipoles which  appear/disappear and can be the reason that they are insoluble in polar compounds .

[nj_bartel]

But what attraction is there between 2 nonpolar molecules if there are no dipole moments?  And if there is no attraction other than the temporary dipole, why don't polar molecules form equally strong or stronger attractions with these temporary dipoles than other temporary dipoles?

[macman104]

But what attraction is there between 2 nonpolar molecules if there are no dipole moments?

There are temporary instantaneous dipoles just like vidya said.  Have you heard about London Dispersion Forces?

[nj_bartel]

Sorry, dipole moment was the wrong word - meant LDF's.  Why do polar molecules not cause LDF's in nonpolar molecules, as nonpolar molecules do to themselves?

[Polleke]

ok, without reading the other responses, I am gonna write an explenation here that I got today about the theory.


when you have 2 substances that are both non polar, they mix because you break 2 non polar bonds (with vanderwaals forces) to form 2 new non polar bonds (with again vanderwaals forces) => energybalans is about the same, so they mix.

When you have a substance with polar bonds and one with non polar bonds and they WOULD mix , then you would break 1 non polar bond (vanderwaals forces) and 1 polar bond where you would break vanderwaals forces and strong polar -polar bonds.... when you put those 2 substances togheter you would only get vanderwaals forces back (non-polar + polar)  and thus the energy balance would not be the same.
You would need to put a lot more energy in the system (to break the polar-polar bonds) and would not get the same energy out the system when forming (if they would) non polar + polar bonds.