In analyzing the solubility among $\ce{CaF2, CaCl2, CaBr2}$, I found that $\ce{CaF2}$ is insoluble due to high ionization enthalpy.

But comparing among $\ce{AgF, AgCl, AgBr }$, it was said that $\ce{AgF}$ is more soluble because of Fajans' rules (consequence of polarization). $\ce{AgCl, AgBr}$ have high polarizing effect and so they have high covalent characteristics.

Now question is why can't I use the Fajans' rules at $\ce{CaF2}$? Where can I use Fajan's Rule and Where should I use Lattice enthalpy rule?

This problem is also present at comparing the melting point of $\ce{NaCl, KCl}$. Fajans' rules cannot be applied here also.


2 Answers 2


The thing you have to understand is that the enthalpy of dissolution is actually the sum of lattice enthalpy and hydration enthalpy. So being a thermodynamic concept, we will have to turn to Gibbs free energy i.e. ΔG= ΔH-TΔS for completely understanding the solubility trends. For the process to be spontaneous, ΔG(dissolution)<0 . Now, entropy will generally increase while dissolving, so we can safely assume ΔS>0 for now. So obviously, the factor of -TΔS will be negative for T>0, so this contributes to the solvation of the said compound.(From now on, ΔH will be referring to enthalpy of dissolution)

Now, it all comes down to the ΔH part : it's magnitude and sign. If ΔH is positive and initially greater than -TΔS, then ΔG will be positive, and as temperature is increased, the -TΔS factor will ultimately overcome the positive ΔH and then, ΔG will become negative. And indeed, this is why you will find that $\ce{PbCl2}$ is insoluble in cold but quite soluble in hot water. Similarly, you can consider the implications of variations in the magnitude and sign of ΔH on solubility (in a polar solvent) yourself.

Now, coming to your specific problem, you see, you first of all have to consider the configurations of $\ce{Ag+}$ and $\ce{Ca^2+}$

$\ce{ Ag^+ - [Kr] 4d10}$ and $\ce{ Ca^2+ - [Ar]}$

So you see, the calcium ion has 8 electrons in it's outermost shell ( 2 in 3s and 6 in 3p) while the silver ion has 18 electrons in it's outermost shell( 4s2 4p6 4d10). Now, due to the ineffective shielding by the d-electrons, effective nuclear charge on the valence shell and hence it's polarising power increases tremendously. Here, the calcium ion possesses an "inert gas configuration" while the silver cation has a kind of "pseudo noble gas configuration" (as it is not exactly a noble gas configuration, but still the cation is quite stable). On moving further down the periodic table,you get cations like $\ce{Tl^+, Bi^3+}$ etc. which have neither 8 nor 18 valence electrons (hence they can be called "non-inert gas configurations" ) and owing to the even poorer shielding by f and d electrons, there polarising power is even higher than the previous two configurations. So, the general order of polarising power of cations can be : Non-inert> Pseudo-inert> Inert gas configuration .( Note: All this can also be derived quantitavely by actually calculating effective nuclear charge on valence electrons by Slater's rule)

So, between $\ce{Ag^+}$ and $\ce{Ca^2+}$, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as compared to $\ce{Ca^2+}$ . As to your question about when to apply which concept: You are basically required to compare the sum of the hydration enthalpy and lattice enthalpy. If the lattice is very strong(example: LiF due to compatibility of ions and less interatomic distance) then the hydration enthalpy can be ignored, while for something like $\ce{LiNO3}$ the lattice is weak and hydration enthalpy is highly negative, so lattice enthalpy can be ignored for the final ΔH. You can use any concept as per your convenience, be it Fajan's rule, or extent of hydration or lattice strength; they will genrally complement each other provided you apply them correctly

For alkali metal chlorides like KCl and NaCl, the cations have only got a unit positive charge (which is not much to cause a lot of polarization) and more importantly, these cations belong to the inert gas configuration type, so here lattice energy will be the dominant factor

  • $\begingroup$ It might be a good idea to brush up your usage of MathJax a bit. I'm quite certain you mean $\ce{Ca^2+}$ ($\ce{Ca^2+}$), not $\ce{Ca2+}$ ($\ce{Ca2+}$). If you want to know more, please have a look here and here. Please don't use any edit statements, if someone really needs to know what has changed and how, the edit history is available. Otherwise these are just superfluous, distracting noise for the (first time) reader. $\endgroup$ Commented Dec 3, 2018 at 12:49

Bond strength depends on electronegativity of bonded atoms. Math is simple. Let's find difference in electronegativity for each type of bonds. If we have a number larger than two, the bond is ionic.

\begin{array}{lr} \text{Compound} & \Delta(\text{EN})\\\hline \ce{AgF} & 2.03\\ \ce{AgCl} & 1.23\\ \ce{AgBr} & 1.03\\ \ce{CaF2} & 2.96\\ \ce{CaCl2} & 2.16\\ \ce{CaBr2} & 1.96\\ \ce{KCl} & 2.34\\ \ce{NaCl} & 2.18\\ \end{array}

For $\ce{Ag}$-stuff and $\ce{KCl}$ and $\ce{NaCl}$ it's obvious: most ionic is least stable. But why do we have $\ce{Ca}$-stuff, ignoring this rule? The problem is ... of rivalry kind.

The weakness of ionic bond is it localization of electrons - they are comparatively well-localized. For $\ce{Ag}$, $\ce{K}$, $\ce{Na}$ (monovalent atoms) it's true. For bivalent atoms, such as $\ce{Ca}$, delocalization comes into play: those two little anions, fighting for electron pair.

Now I'm going to speak metaphorically, because my poor skills in calculus won't allow me to show anything mathematically gorgeous. Imagine two guys, fighting for a piece of elastic rubber. If one guy is way stronger, he'll easily get rubber (without stretching it, and for us stretching means delocalization and, thus, stability, insolubility, high melting point, etc.) - ionic bond of two monovalent atoms. If guys are equally strong, they will stretch it together - perfectly covalent bond. But what if there will be 3 fighting guys? Let's imagine that one is weak ($\ce{Ca}$) to fight for rubber (electron pair), but other two are fighting. It's also important that he stays, defeated, in the midst of fight, thus rubber is stretched way more far - he won't move from his place, and other fighters too. Strong fighters (such as $\ce{F}$) would stretch the rubber harder than not very strong (such as $\ce{Br}$), delocalizing electrons and making $\ce{CaF2}$ more stable.

I hope my weird explanation helps. I would also appreciate mathematical model of this process, because my "fighting guys model" is way too poorly made.
enter image description here

[edited]Picture source: http://www.chemhume.co.uk/ASCHEM/Unit%201/Ch3Chemstr/Images%203/electronegativity_values.jpg

  • $\begingroup$ 'Bond strength depends on electronegativity of bonded atoms.' This is an assessment so fuzzy it says everything and nothing at all. The difference of electronegativities is also a quite poor indicator of whether a bond is ionic or not. It is also not a sharp line, no matter what arbitrary number you are assigning. AgBr is a salt with almost exclusively ionic bonds and the difference is only about 1.0. HF on the other hand is a covalent compound and the difference is about 1.8. $\endgroup$ Commented Dec 3, 2018 at 13:03
  • 1
    $\begingroup$ 'The weakness of ionic bond is it localization of electrons - they are comparatively well-localized.' That is just plain wrong. Ionic bonds are not localised at all, they have no direction as covalent bonds would have. The dominant force it electrostatic attraction here. And after that how well the ions fit into a crystal lattice, which defines the lattice energy and ultimately how well the crystal can be dissolved. The whole fighting story is very much misleading, because there is nothing like a rubber involved. $\endgroup$ Commented Dec 3, 2018 at 13:12
  • $\begingroup$ Please also add the sources and citations to images you have not compiled yourself. $\endgroup$ Commented Dec 3, 2018 at 13:12
  • $\begingroup$ "Ionic bonds are not localised at all", but electron pairs, which are "bonds", are well-localized, and I've written about electrons (as you can see from citation you've used). The whole story is metaphoric, and I've mentioned that. You can't fight for probabilistic wave with your fellow, at least, not as for physical object. Rubber can be stretched, and it's easy to "physically" imagine delocalization via stretching (the easiest way I know, at least). Thank you for the tip about lattice energy - I will review this topic, +1. $\endgroup$ Commented Dec 3, 2018 at 13:31

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