Bounty Ended with 50 reputation awarded by Hema occurred Dec 5 '18 at 14:02 3 added 21 characters in body edited Dec 3 '18 at 14:21 YUSUF HASAN 51711 gold badge55 silver badges1616 bronze badges 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 $$\ce{Ca2+}$$$$\ce{Ca^2+}$$ $$\ce{ Ag+ - [Kr] 4d10}$$$$\ce{ Ag^+ - [Kr] 4d10}$$ and $$\ce{ Ca2+ - [Ar]}$$$$\ce{ Ca^2+ - [Ar]}$$ 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 $$\ce{Tl+, Bi3+}$$$$\ce{Tl^+, Bi^3+}$$ 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 Ag+$$\ce{Ag^+}$$ and Ca2+$$\ce{Ca^2+}$$, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as comapred to Ca2+ . 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 Edit 1: 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 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 $$\ce{Ca2+}$$ $$\ce{ Ag+ - [Kr] 4d10}$$ and $$\ce{ Ca2+ - [Ar]}$$ 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 $$\ce{Tl+, Bi3+}$$ 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 Ag+ and Ca2+, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as comapred to Ca2+ . 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 Edit 1: 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 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 comapred to Ca2+ . 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 2 added 288 characters in body edited Nov 30 '18 at 4:20 YUSUF HASAN 51711 gold badge55 silver badges1616 bronze badges 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 $$\ce{Ca2+}$$ $$\ce{ Ag+ - [Kr] 4d10}$$ and $$\ce{ Ca2+ - [Ar]}$$ 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 $$\ce{Tl+, Bi3+}$$ 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 Ag+ and Ca2+, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as comapred to Ca2+ . 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 Edit 1: 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 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 $$\ce{Ca2+}$$ $$\ce{ Ag+ - [Kr] 4d10}$$ and $$\ce{ Ca2+ - [Ar]}$$ 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 $$\ce{Tl+, Bi3+}$$ 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 Ag+ and Ca2+, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as comapred to Ca2+ . 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 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 $$\ce{Ca2+}$$ $$\ce{ Ag+ - [Kr] 4d10}$$ and $$\ce{ Ca2+ - [Ar]}$$ 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 $$\ce{Tl+, Bi3+}$$ 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 Ag+ and Ca2+, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as comapred to Ca2+ . 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 Edit 1: 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 1 answered Nov 30 '18 at 2:32 YUSUF HASAN 51711 gold badge55 silver badges1616 bronze badges 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 $$\ce{Ca2+}$$ $$\ce{ Ag+ - [Kr] 4d10}$$ and $$\ce{ Ca2+ - [Ar]}$$ 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 $$\ce{Tl+, Bi3+}$$ 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 Ag+ and Ca2+, Fajan's rule will be much more applicable to the silver halides as the silver cation will generate much more polarization as comapred to Ca2+ . 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