[Big-Daddy]

The data in my data book suggests that, for an ionic compound with a certain theoretically estimated lattice energy (e.g. by the Kapustinskii or Born-Landé equation), the more polarizing the cation and the less polarizing the anion (and thus the more covalent the lattice as a whole), the greater the positive difference between the theoretical energy calculated from the purely ionic model and the experimental lattice energy found from the Born-Haber cycles. This would suggest that lattice enthalpy is increased (made more exothermic) based on the degree of covalent character.

However, this seems to be in a contradiction with the explanations I have at hand for why carbonates and nitrates become more stable (less soluble, higher temperature points of decomposition, etc.) as you descend Groups 1 and 2: the common explanation is that they are made more ionic and thus they remain more stable. Surely stability and lattice enthalpy are linked to each other by the same factors, by definition of lattice enthalpy, so is it increased covalent character or ionic character which brings about increased stability, and how do you explain the opposite trend observed in one of these cases?

Can anyone explain where this discrepancy came from? Feel free to use undergraduate chemistry to explain in detail if that would be good, I only posted in the High School forum because this seems to pertain to the chemistry we meet at high school-level.

[Corribus]

Everything has to do with the size of the ions involved.  But rather than typing out a lengthy explanation, I will direct you to such an explanation that already exists on the web.

http://www.chemguide.co.uk/inorganic/group2/thermstab.html

I think that should answer all your questions.  If not, you know where to post any others you may have.

[Big-Daddy]

Everything has to do with the size of the ions involved.  But rather than typing out a lengthy explanation, I will direct you to such an explanation that already exists on the web.

http://www.chemguide.co.uk/inorganic/group2/thermstab.html

Thanks for the link. I have read the page but I'm still confused about the effect of polarization. If "highly polarized" means "more covalent", then according to my data-book the extra covalency should cause the energy required to dissociate the lattice to increase (when compared with the theoretical estimates) rather than decrease.

Perhaps the confusion arises here: the author writes "If it is highly polarised, you need less heat than if it is only slightly polarised." What I would like to ask is: why?

Maybe then I can answer my own question.

[Corribus]

Just think of the reaction CO₃^2- O^2- + CO₂(g)

The carbonate ion is in a resonance structure, with all oxygens being equivalent.  This makes carbonate in principle a pretty stable structure.  In fact, the primary driver for decomposition of carbonate is really entropy, because CO₂ is a very stable gas.  Otherwise, carbonate is pretty stable from an enthalpic point of view.

The stability of carbonate to heat is going to depend, then, on how easy it is to break the carbonate apart and form carbon dioxide.  This requires breaking one of the C-O bonds to give an isolated oxygen with a -2 formal charge.  In completely isolated carbonate, each bond has an order somewhere between a sigma and pi bond, and each oxygen has a formal charge of 2/3.  (Two negative charges delocalized over three oxygen atoms.)  For the reaction above to proceed then, you have to break a pretty strong "almost pi pond" and concentrate a whole lot of electron density that is spread over a fairly large molecule onto a single oxygen atom, all in one shot.  Breaking a strong bond + large electronic rearrangement = needs lot of energy to happen.   That's a fairly unfavorable process, and the only reason it happens at all is because of the entropic driving force, which gets greater at higher temperatures.   

But let's not confuse things with entropy - we'll assume that doesn't change as a function of the counterion.  That means to determine relative stability we only need to consider the enthalpy, which is related to the actual breaking of bonds and rearrangement of electron density.  With me so far?

Ok, so where does polarisation come in?  Well, as we've said, a neutral carbonate is very enthalpically stable.  Takes a lot of energy to break one of those bonds and rearrange the electron density.  However one thing I'll point out is that polarization of the electron density is just another way of saying "rearrange the electron density".  In fact, polarization of carbonate is required to make it decompose into oxide and carbon dioxide.  You have to put two formal charges on one oxygen, leaving the rest of the molecule neutral.  That's the very definition of polarization.  So, if you have a situation where carbonate starts off partially polarized, that is, with a lot of electron density already on one of the oxygen atoms, then you're already halfway to decomposition.  In the limit of complete polarization, all electron density is localized on one oxygen already, and there is no bond at all between that oxygen and the carbon - the bond order is zero.  The reaction has already happened in that case.  In essence, in a carbonate that's already partially polarized, the bond order of the C-O bond is reduced (and the bond order of the other two C-O bonds is increased) and the amount of electronic rearrangement that must still occur for decomposition is less, which translates into less energy required for the reaction to go - which means the carbonate is enthalpically less stable and the driving force for reaction is greater. 

This is the impact of polarization.  Counterions which have small, compact charges - like berylium - polarize carbonate to a much greater degree than counterions which have large, diffuse charges - like barium.  Therefore we can say that in calcium carbonate, the carbonate is already highly polarized, the C-O bond that needs to break is significantly reduced in bond order, and the electronic arrangement already partially resembles that of the product.  The amount of energy to drive the reaction is therefore small, and the compound is therefore relatively unstable.  On the other hand, in barium carbonate, the carbonate is less polarized, the C-O bond that needs to break is closer to that of a strong pi bond, and the electronic arrangement is highly delocalized.  This translates into a a large amount of energy needed to drive the reaction, and the compound is therefore relatively stable.

In general, the degree of covalency increases as you go down a group, because orbitals are more diffuse and shielding by core electrons is greater.  Therefore we might say that the bond covalency between the counterion and carbonate is correlated with carbonate stability.  This makes sense: polarization in bonds with high covalent character is less than polarization in bond with high ionic character.  More covalent bonds have less polarization, which renders carbonate more stable.  Which explains the trends observed. 

So: "highly polarized" means "less covalent", which should agree with your data book, right?

Does this clarify things for you?

[Big-Daddy]

Just think of the reaction CO₃^2- O^2- + CO₂(g)

The carbonate ion is in a resonance structure, with all oxygens being equivalent.  This makes carbonate in principle a pretty stable structure.  In fact, the primary driver for decomposition of carbonate is really entropy, because CO₂ is a very stable gas.  Otherwise, carbonate is pretty stable from an enthalpic point of view.

The stability of carbonate to heat is going to depend, then, on how easy it is to break the carbonate apart and form carbon dioxide.  This requires breaking one of the C-O bonds to give an isolated oxygen with a -2 formal charge.  In completely isolated carbonate, each bond has an order somewhere between a sigma and pi bond, and each oxygen has a formal charge of 2/3.  (Two negative charges delocalized over three oxygen atoms.)  For the reaction above to proceed then, you have to break a pretty strong "almost pi pond" and concentrate a whole lot of electron density that is spread over a fairly large molecule onto a single oxygen atom, all in one shot.  Breaking a strong bond + large electronic rearrangement = needs lot of energy to happen.   That's a fairly unfavorable process, and the only reason it happens at all is because of the entropic driving force, which gets greater at higher temperatures.   

But let's not confuse things with entropy - we'll assume that doesn't change as a function of the counterion.  That means to determine relative stability we only need to consider the enthalpy, which is related to the actual breaking of bonds and rearrangement of electron density.  With me so far?

Ok, so where does polarisation come in?  Well, as we've said, a neutral carbonate is very enthalpically stable.  Takes a lot of energy to break one of those bonds and rearrange the electron density.  However one thing I'll point out is that polarization of the electron density is just another way of saying "rearrange the electron density".  In fact, polarization of carbonate is required to make it decompose into oxide and carbon dioxide.  You have to put two formal charges on one oxygen, leaving the rest of the molecule neutral.  That's the very definition of polarization.  So, if you have a situation where carbonate starts off partially polarized, that is, with a lot of electron density already on one of the oxygen atoms, then you're already halfway to decomposition.  In the limit of complete polarization, all electron density is localized on one oxygen already, and there is no bond at all between that oxygen and the carbon - the bond order is zero.  The reaction has already happened in that case.  In essence, in a carbonate that's already partially polarized, the bond order of the C-O bond is reduced (and the bond order of the other two C-O bonds is increased) and the amount of electronic rearrangement that must still occur for decomposition is less, which translates into less energy required for the reaction to go - which means the carbonate is enthalpically less stable and the driving force for reaction is greater. 

This is the impact of polarization.  Counterions which have small, compact charges - like berylium - polarize carbonate to a much greater degree than counterions which have large, diffuse charges - like barium.  Therefore we can say that in calcium carbonate, the carbonate is already highly polarized, the C-O bond that needs to break is significantly reduced in bond order, and the electronic arrangement already partially resembles that of the product.  The amount of energy to drive the reaction is therefore small, and the compound is therefore relatively unstable.  On the other hand, in barium carbonate, the carbonate is less polarized, the C-O bond that needs to break is closer to that of a strong pi bond, and the electronic arrangement is highly delocalized.  This translates into a a large amount of energy needed to drive the reaction, and the compound is therefore relatively stable.

In general, the degree of covalency increases as you go down a group, because orbitals are more diffuse and shielding by core electrons is greater.   Therefore we might say that the bond covalency between the counterion and carbonate is correlated with carbonate stability.  This makes sense: polarization in bonds with high covalent character is less than polarization in bond with high ionic character.  More covalent bonds have less polarization, which renders carbonate more stable.  Which explains the trends observed. 

So: "highly polarized" means "less covalent", which should agree with your data book, right?

Does this clarify things for you?

Thanks for the explanation. However, I was under the impression that "highly polarized" means "more covalent", as they are both dependent on the same factors: for something to be highly polarized, the cation must be small and of high charge, and the anion large and of high charge, and the same is true for covalency (the smaller and more highly charged the cation, the more it will attract the electrons from the anion towards itself, so the more polarized/less ionic/more covalent the bond will be). For example, Cs⁺ compounds are less covalent than Li⁺ compounds (same charge but smaller).

I think three realms are at work here and I'm not sure how they interplay:

If you look at the Kapustinskii or Born-Landé theoretical lattice energy estimates, to show how stable a compound is, they show that the larger the charges and the smaller the radii of the two ions, the greater the lattice energy of the compound. Thus something like Al₂O₃ or to a lesser extent AlF₃ has about the highest lattice energy you can get (something like CsI, with small charges and large radii, has about the lowest lattice energy you can get, working purely ionically).

To add to this is the effect observed in my data book that "the greater the degree of covalency, compared to a purely ionic form" (i.e. the more polarizing the cation and the more polarizable the anion, i.e. the smaller and more highly charged the cation), "the greater the difference between the theoretical and experimental lattice energies" - and, as I observe from the data in my book, this difference is always positive (experimental lattice energies always have a larger magnitude than theoretical ones). Thus, the lattices with smaller and highly charged cations have two advantages: first, a higher theoretical lattice energy; and second, a greater boost from the degree of covalency.

But this indicates that smaller and more highly charged cations, which have more polarising power, lead to greater covalency and thus to greater stability, whereas this seems to conflict with what you've said?

[Corribus]

BigDaddy,

I've read your post like ten times and I guess I would have to read your source because I'm having trouble following the way you've transcribed it here. 

That said:

Lattice energy as a concept typically applies to ionic crystals - that is, solids consisting of ions held together by electrostatic interactions of discrete charges - and is an estimation of how much energy it would take to break apart (or form, depending on convention) a single crystal into its individual ioniccomponents.  The Kapustinskii equation predicts that the lattice energy is proportional to the magnitude of the charges of the ions in the crystal, and generally inversely proporitional to the radii of the ions.  Large radii would give rise to small lattice energies, and vice-versa.  The Born-Lande equation predicts something similar.  The latter equation also incorporates a factor related to the crystal structure (geometry), so any direct comparisons between lattice energies of two crystals must take this into consideration.

What I'm getting at is that these methods to estimate lattice energy assume that all bonds in the crystal are strictly ionic.  However I would expect deviations between the predictions of these equations and real values (such as they can be determined) in cases where there is bond covalency, and those deviations would become larger as the degree of covalency increases.  Remember, Born-Lande and Kapustinskii are predictive models, and pretty simple ones at that.  In any predictive model, you have to know what assumptions they make in order to understand where they break down.

Furthermore, while these are great questions you raise, I think you are confusing the concept of lattice energy here with what the original question is after.  I think the Kapustinskii equation is probably an excellent starting point to approach matters of "stability" for simple ionic crystals lie sodium chloride, where you have cations and anions which are single atoms/ions.  It becomes more complicated when the ions involved are complex, as in the case of, say, sodium sulfate, where the application of heat goes into breaking ionic bonds AND covalent bonds (i.e., the ionic bonds between sodium and sulfate and the covalent bonds that make up sulfate).  It's hard to dissociate these processes and simple models like Kapustinskii may not be useful in these circumstances.  (Actually, I don't know this to be a fact - just my suspicion).

Even more complicated is a case like carbonate, which actually decomposes as it is heated due to the entropic driving force of carbon dioxide liberation.  The stability issue you raised in the first post doesn't necessarily have a lot to do with lattice energy of the overall crystal, because there's an actual separate chemical reaction going on that involves just the anion.  The question you raised about carbonate isn't probing the concept of lattice energy so much as it is probing the concept of electronic induction and polarizability of bonds.  That is, the relative size and polarizing power of the metal cation isn't affecting the lattice energy or stability of the ionic crystal (energy required to break metal-carbonate ionic bonds) so much as it is affecting the stability of the carbonate anion itself.  Of course, they affect both, but the latter stability issue is what the question is trying to get you to think about.  If you want to learn about lattice energy and its impact on the stability of ionic crystals, there are probably better examples to use than carbonates, where a competing chemnical decomposition is taking place.

[Needaask]

Everything has to do with the size of the ions involved.  But rather than typing out a lengthy explanation, I will direct you to such an explanation that already exists on the web.

http://www.chemguide.co.uk/inorganic/group2/thermstab.html

I think that should answer all your questions.  If not, you know where to post any others you may have.

hi sorry to hijack this post but when we polarize the carbonate ion, how does that help to break the bond between the C-O and form CO₂? Or is the reason why the C-O bond become weakened is that all the electrons are okay with being around the oxygen that is closest to the +2 charged ion which ultimately breaks the bond with enough heat?

But for OP's question since the ionic radius of the ions decrease down the group shouldn't the lattice energy decrease as well giving rise to a lower melting point as we go down the group. Since for melting point there shouldn't be the other factor that you explained that's the anion having it's own reaction. So by lattice energy, shouldn't the melting point decrease down the group rather than increase?

Thanks for the help

[Corribus]

"Bare" carbonate, CO₃^2-, has three C-O bonds.  From a Lewis structure, it would appear that two of them are single bonds and one is a double bond.  The oxygens attached with single bonds each have a formal charge of -1.  The oxygen attached with a double bond has a formal charge of 0. 

In reality, though, there is resonance in this molecule which delocalizes the 2 negative charges over all three oxygens (each oxygen would have a formal charge of -2/3) and also delocalizes the pi-bond over all three C-O bonds, meaning each C-O bond has a total bond order of 4/3, putting it somewhere in between a sigma bond and a pi bond.


For this reaction:

CO₃^2-  CO₂ + O^2-

One of the C-O bonds needs to break and both negative charges need to appear on the oxygen attached to this broken bond.  A C-O bond with a bond order 4/3 is pretty stable, and localizing two charges on one oxygen is unfavorable.  (Even from purely a kinetic point of view, we might say the statistics don't favor it.  Even if the position of the electrons were completely random - that is, even if there was no energy of interaction - they'd only be localized on one oxygen molecule a small minority of the time.)

Anyway, now say you have a +2 charge locaed near the carbonate.  This makes it much more favorable that both negative charges will be localized in one place (on one oxygen).  This polarizes the carbonate, because there's now an unequal charge distribution.  When that happens, the bond order of the associated C-O will decrease.  Essesntially, what you have is a starting point that looks a lot more like the products in the above reaction than the free carbonate.  You're already halfway there - so we conclude that the more polarized carbonate is, the less stable it is, because it takes less energy to go from the starting material to the product.  As an analogy - if you're trying to drive to San Francisco, it is a lot easier if you're starting in Colorado than if you're starting in New York. 

Regarding your second question - as I explained to the OP in a follow-up post, you have to be careful when you involve the concept of lattice energy in this specific case, because it will lead to confusion.  Stability of the lattice doesn't have anything directly to do with stability of the carbonate.  These are two different chemical processes.  Granted, breaking apart carbonate probably will impact the lattice structure, but these two equilibria are still dependent on different factors.

[Needaask]

Oh that's right, in the separation of just the cation and the carbonate anion, the energy required to break the ionic lattice decreases as we go down the group. The last part in the link associated them very well

Thanks Corribus for the awesome help and also to BigDaddy for the enlightening question

[foreverotaku]

Just think of the reaction CO₃^2- O^2- + CO₂(g)

The carbonate ion is in a resonance structure, with all oxygens being equivalent.  This makes carbonate in principle a pretty stable structure.  In fact, the primary driver for decomposition of carbonate is really entropy, because CO₂ is a very stable gas.  Otherwise, carbonate is pretty stable from an enthalpic point of view.

The stability of carbonate to heat is going to depend, then, on how easy it is to break the carbonate apart and form carbon dioxide.  This requires breaking one of the C-O bonds to give an isolated oxygen with a -2 formal charge.  In completely isolated carbonate, each bond has an order somewhere between a sigma and pi bond, and each oxygen has a formal charge of 2/3.  (Two negative charges delocalized over three oxygen atoms.)  For the reaction above to proceed then, you have to break a pretty strong "almost pi pond" and concentrate a whole lot of electron density that is spread over a fairly large molecule onto a single oxygen atom, all in one shot.  Breaking a strong bond + large electronic rearrangement = needs lot of energy to happen.   That's a fairly unfavorable process, and the only reason it happens at all is because of the entropic driving force, which gets greater at higher temperatures.   

But let's not confuse things with entropy - we'll assume that doesn't change as a function of the counterion.  That means to determine relative stability we only need to consider the enthalpy, which is related to the actual breaking of bonds and rearrangement of electron density.  With me so far?

Ok, so where does polarisation come in?  Well, as we've said, a neutral carbonate is very enthalpically stable.  Takes a lot of energy to break one of those bonds and rearrange the electron density.  However one thing I'll point out is that polarization of the electron density is just another way of saying "rearrange the electron density".  In fact, polarization of carbonate is required to make it decompose into oxide and carbon dioxide.  You have to put two formal charges on one oxygen, leaving the rest of the molecule neutral.  That's the very definition of polarization.  So, if you have a situation where carbonate starts off partially polarized, that is, with a lot of electron density already on one of the oxygen atoms, then you're already halfway to decomposition.  In the limit of complete polarization, all electron density is localized on one oxygen already, and there is no bond at all between that oxygen and the carbon - the bond order is zero.  The reaction has already happened in that case.  In essence, in a carbonate that's already partially polarized, the bond order of the C-O bond is reduced (and the bond order of the other two C-O bonds is increased) and the amount of electronic rearrangement that must still occur for decomposition is less, which translates into less energy required for the reaction to go - which means the carbonate is enthalpically less stable and the driving force for reaction is greater. 

This is the impact of polarization.  Counterions which have small, compact charges - like berylium - polarize carbonate to a much greater degree than counterions which have large, diffuse charges - like barium.  Therefore we can say that in calcium carbonate, the carbonate is already highly polarized, the C-O bond that needs to break is significantly reduced in bond order, and the electronic arrangement already partially resembles that of the product.  The amount of energy to drive the reaction is therefore small, and the compound is therefore relatively unstable.  On the other hand, in barium carbonate, the carbonate is less polarized, the C-O bond that needs to break is closer to that of a strong pi bond, and the electronic arrangement is highly delocalized.  This translates into a a large amount of energy needed to drive the reaction, and the compound is therefore relatively stable.

In general, the degree of covalency increases as you go down a group, because orbitals are more diffuse and shielding by core electrons is greater.  Therefore we might say that the bond covalency between the counterion and carbonate is correlated with carbonate stability.  This makes sense: polarization in bonds with high covalent character is less than polarization in bond with high ionic character.  More covalent bonds have less polarization, which renders carbonate more stable.  Which explains the trends observed. 

So: "highly polarized" means "less covalent", which should agree with your data book, right?

Does this clarify things for you?

Thanks a lot for clarifying it.. So basically the more polarized the carbonate, the more easily it is to decompose it, am I right?

[Corribus]

Yes.