Dipoles can also be induced in polar and non polar compounds, then why don't they dissolve?

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    $\begingroup$ Think of it like two groups of people who speak different languages, it's possible for them to try to mix around and socialise but it's just easier for them to stick to their own groups. Now chemicals are pretty rude and have no idea about social conventions, so they prefer to keep within their own group. $\endgroup$ Commented Oct 2, 2015 at 8:56
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    $\begingroup$ By the way dipoles aren't induced in polar compounds. They're already there. That's why you call them a permanent dipole. Their magnitudes will fluctuate but that's a different matter entirely $\endgroup$ Commented Oct 2, 2015 at 8:59
  • $\begingroup$ Shouldn't non polar dissolve in polar the way non polar dissolve in non polar? $\endgroup$ Commented Oct 2, 2015 at 9:23
  • $\begingroup$ No they should not. The disruption of the solvent is very different. $\endgroup$
    – jimchmst
    Commented Dec 9, 2022 at 4:29
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    $\begingroup$ Does this answer your question? Why does like dissolve like? $\endgroup$ Commented Apr 15 at 4:41

2 Answers 2


Very simply, you explain the reason for this solubility rule by taking in consideration the energy requirements for the breaking of intermolecular forces between the molecules in the solute and the solvent.

Note: this is only a simplified explanation as it also depends on other factors such as change in entropy

Here is some background information on intermolecular forces. In non-polar substances, there are dispersion forces between each molecule. These dispersion forces are relatively weak and hence only require little energy to break them. In polar substances, there are dipole dipole and hydrogen bonding (depending on the substance) between each molecule. These forces are much stronger than dispersion forces and require more energy to break.

Now lets consider the following cases:

Non-polar Solute and Solvent

For the solute to dissolve, the dispersion forces between the molecules in the solute and solvent need to break. This only requires very little energy. However when the solute dissolves into the solvent, they are able to be make dispersion forces with each other. The making of these forces releases very little energy. So simply put, very little energy is required to break the forces and very little energy released when making the forces. Hence overall everything balances out and the process occurs.

Non-polar Solute and Polar Solvent (and vice versa)

For the solute to dissolve into the solvent, both dispersion forces and dipole dipole forces are broken which require large amount of energy. However the molecules in the solute and solvent are only able to make dispersion forces with each other (as they aren't both polar). This only releases very little energy. Therefore overall, more energy is required than released and hence the process won't happen.

Polar Solute and Solvent

For the solute to dissolve into the solvent, dipole dipole forces are broken which require large amount of energy. However when they do dissolve, the molecules in the solute and solvent are able to form dipole dipole forces which releases large amount of energy. Therefore overall, everything balances out and the process occurs.

  • $\begingroup$ I was thinking the same but in my textbook there are some orders given like NaI is less soluble than LiI in Non polar solvent because as per fajan's rule LiI is more covalent but its very vague statement because I think dipole moment of NaI is greater than LiI because Na^+ IS BIGGER than Li^+ and also polarizing power of Na+ Is also less so positive and negative charge density is well separated so its dipole -induced dipole interaction with solvent will be high and also its lattice energy is less than LiI so it should dissolve more than LiI but it doesn't how? $\endgroup$ Commented Jan 5, 2022 at 11:12


It's because the enthalpy changes of a solution generally don't favor dissolution.

A longer version:

To explain this, usually the enthalpy change explanation is given. For the sake of understandability, let's see what happens when two compounds dissolve. Take ethanol dissolving in water as an example. Here's the gist of what happens:

  1. The intermolecular forces (i.e. hydrogen bonds in this case) in water break apart. $\rm \color{green}{(endothermic)}$
  2. The intermolecular forces in ethanol break apart. $\rm \color{green}{(endothermic)}$
  3. A new force and attraction is formed between the ethanol and water molecules. $\rm \color{red}{(exothermic)}$
    $\hspace{22ex}$ Ethanol and water forming hydrogen bonds; Source

This happens for any two species that will dissolve in each other. For ionic solutes, the 'bond cleavage' is actually the lattice breaking apart. So you'd expect an endothermic process with a gaining of energy equal to the enthalpy of formation of lattice.

You have to be careful about two things:

  • Solubility isn't binary. We usually have to indicate it with short and comprehensible (i.e. by a large audience) words, and that's why we use it. Is gypsum as insoluble in water as calcium carbonate or as soluble as propanoic acid? (Propanoic acid is miscible in water in RTP and STP)
  • That "nonpolar doesn't dissolve in polar" isn't accurate. Nonpolar solutes are generally insoluble in polar solvents. We can easily think of exceptions. Bromine water is an example for a start, but certainly not the most remarkable example.

So the question that comes to mind is,

If the process for dissolution is going to be the same for polar or nonpolar molecules and crudely the same for molecules with hydrogen bond and for ionic compounds, why are some solutes insoluble in some solvents?

As we saw, there are two endothermic processes and one exothermic process involved. Very simply put, a rudimentary answer to your question is that "because induced dipoles are known as one of the weakest intermolecular interactions and thus the solvent-solute interactions wouldn't release enough energy while being formed, so $\Delta H > 0$. This would mean that thermodynamically it's more favorable for the solvent-solvent interaction not to be broken and hence, no dissolution".

In a conclusion, I would say "yes, the polarity of the solute/solvent plays an important rule in determining solubility or insolubility. But that's not the half of it." A small review of the matter can be found below.

$\color{gray}{\textit{I don't even know why I'm doing this.}\\\ \textit{A general, prescriptive rule can be explained mainly by another simplistic view of the matter.}\\\ \textit{Don't read on if you're not interested.}}$

However, just taking enthalpy into account isn't scientifically accurate. At least, that's not what's happening in real life. The meaning of enthalpy is associated with constant temperature and pressure. That's not what is happening in real-life chemistry.

For a start, it's best if we consider the hydrophobic effect (related to entropy), size of the solute species, rate of dissolution, the common ion effect, ionic strength.

The hydrophobic effect:

Just like a system favors the least potential energy, it favors disorder. The hydrophobic effect can better explain why some nonpolar molecules can't dissolve in water:

The hydrophobic effect is the observed tendency of nonpolar substances to aggregate in aqueous solution and exclude water molecules. This occurs because interactions between the hydrophobic molecules allow water molecules to bond more freely, increasing the entropy of the system. The word hydrophobic literally means "water-fearing," and it describes the segregation and apparent repulsion between water and nonpolar substances. - The hydrophobic effect, Wikipedia

Simply put, the reason for this isn't well-understood. A simplified explanation is that the structure of water allows it to have three degrees of freedom and it can form four hydrogen bonds. If it does so, it can't orientate as easily as it could and thus the entropy would decrease. So, to favor entropy, this needs to happen minimally.

If you want to study entropy of mixing, the linked Wikipedia article is very nice.

The size of species:

$\ce{AgCl}$ is less soluble in water than $\ce{AgNO3}$. This could best be described with the fact that silver and chlorine ions are almost the same size, and hence they may be packed tighter together. i.e. they're harder to break apart and 'dissolve'.

Note that "like dissolves like" or any similar rules can't explain this. Actually, the impressive delocalization of the electron in the nitrate ion can explain the massive range of soluble nitrates.

Rate of dissolution:

Would you still call a species soluble if it dissolves in the solvent in a wide enough time span? I'd imagine not, since the radioactive species with long enough half-life periods are commonly attributed the 'stable' medal.

Rate of dissolution is not a thermodynamic property, but a kinetical one.

Dissolution is not always an instantaneous process. It is fast when salt and sugar dissolve in water but much slower for a tablet of aspirin or a large crystal of hydrated copper(II) sulfate. These observations are the consequence of two factors: the rate of solubilization (in kg/s) is related to the solubility product (dependent on temperature) and the surface area of the material. The speed at which a solid dissolves may depend on its crystallinity or lack thereof in the case of amorphous solids and the surface area (crystallite size) and the presence of polymorphism. - Rate of dissolution, Wikipedia

Additives (Dispersants):

There's no obligation that we only consider the existence of solute and the solvent. What would you do if you needed to dissolve a fatty acid (which is hydrophobic) in water?

Getting help from micelles is a way. This is crudely the same way fats are being transported in the blood, and the same mechanism soaps use to cleanse oil from your skin.

Solubilization is distinct from dissolution because the resulting fluid is a colloidal dispersion involving an association colloid. This suspension is distinct from a true solution, and the amount of the solubilizate in the micellar system can be different (often higher) than the regular solubility of the solubilizate in the solvent. - Micellar solubilization, Wikipedia

$\hspace{7ex}$Micellar solubilization of fatty substance in water with the use of a dispersant - Andreas Dries; Source

Additives (The common ion effect):

The common ion effect uses the Le Chatelier's principle to explain the lesser solubility of a certain precipitate because of the existence of a similar ion in the solution.

For instance, a dilute solution of magnesium sulfate is less soluble if some copper(II) sulfate is dissolved.

Ionic strength:

To expand on the concepts related to the common ion effect, ionic strength is defined:

The ionic strength of a solution is a measure of the concentration of ions in that solution. Ionic compounds, when dissolved in water, dissociate into ions. The total electrolyte concentration in solution will affect important properties such as the dissociation or the solubility of different salts. One of the main characteristics of a solution with dissolved ions is the ionic strength.

The ionic strength, $I$, of a solution is a function of the concentration of all ions present in that solution.$$I = \frac{1}{2} \sum\limits^n_{i=1} c_iz_i^2$$

where $c_i$ is the molar concentration of ion $i$ (M, mol/L), $z_i$ is the charge number of that ion, and the sum is taken over all ions in the solution. - Ionic strength, Wikipedia (Emphasis mine)

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    $\begingroup$ I thought I should expands on activity coefficients and how they lead to deviations from ideal behaviors of ideal solutions, esp. from Raoult's law, but I think that's enough for now. :P Still I didn't expand on pressure's effect, since it would require explanations regarding partial pressures. $\endgroup$
    – M.A.R.
    Commented Oct 2, 2015 at 14:10

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