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Saturday, July 10, 2010

Group II - Thermal decomposition

Group II metal carbonates, nitrates and hydroxides are capable of decomposition to give the corresponding metal oxide and release CO2, NO2 and O2, and H2O respectively. The following are the equations.

MCO3 (s) -> MO (s) + CO2 (g)
M(NO3)2 (s) -> MO (s) + 2NO2 (g) + ½O2 (g)
M(OH)2 (s) -> MO (s) + H2O (g)

The ease of decomposition decreases as we go down the group. I will use the metal carbonates as an example to illustrate this trend. The reason used will be applicable to describe the ease of decomposition of Group II nitrates and hydroxides.

1. Charge density and Polarising power of cation.

The Group II metals share a common charge of +2, but down the group the ionic radius increases. Hence, the charge density of the cation decreases and its polarising power becomes weaker. Hence, the cation is less able to polarise the anion and more importantly is less able to distort the covalent bond found in the carbonate ion.


This distortion weakens the covalent bond and hence making the decomposition easier.

Interestingly, this reason is why Group I nitrates decompose to give MNO2 and O2 instead of forming MO, NO2 and O2 as products. Only LiNO3 decomposes like a Group II nitrate. This is not surprising as Li and Mg have a diagonal relationship. Their small size and charge will lead to a high charge density which make Li+ more polarising than the other Group I cations.

2. Percentage change in lattice energy.

This reason is not often used as it requires some mathematical conceptualisation.

The decomposition of metal carbonates form metal oxides. The different between the lattice energy (LE) of the two ionic compounds is the sum of ionic radius. Since both compound have the same charges for the cation and that for the anions, we can conclude that the difference in lattice energy will determine whether the decomposition is more favourable or less.


A smaller cation will have a larger percentage difference in the two LE while the bigger cation will have a smaller difference. You can make use of the above picture to conceptualise this statement.

Therefore, down the group the difference in lattice energy becomes smaller and hence decomposition becomes more difficult.

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Article written by Kwok YL 2010.
Disclaimer and remarks:
  • If you would like to use this source, kindly drop me a note by leaving behind a comment with your name and institution. I am all for sharing as the materials on this blog is actually meant for the education purpose of my students.
  • This material is entirely written by the author and my sincere thanks will be given to anyone who is kind, generous and gracious to point out any errors.

8 comments:

RL said...

"The Group II metals share a common charge of +2, but down the group the ionic radius increases. Hence, the charge density of the cation increase and it polarising power becomes stronger. Hence, the cation is able to polarise the anion and more importantly is able to distort the covalent bond found in the carbonate ion."

I'm just learning this and trying to understand. Since the ionic radii increase shouldn't that mean the charge density decreases thus larger cations are less polarizing (as you explained for the Li+ cation)? Thus larger cations would distort the carbonate ion less so more heat energy would be required to break off carbon dioxide.

Mr Kwok said...

Hi RL,

Well, I think you need to be cautious. Charge density is dependent on charge and radius of the cation. When we increase the cation radius, it is likely for charge density to decrease, provided that the charge remains the same.

Now, we are comparing Li+ with Mg2+ (for example). Now, clearly Li+ will be smaller than Mg2+, but the charge is greater in the latter. Hence, it will not be easy for us to conclude (using estimation) whether the charge density of Li+ will be smaller than Mg2+.

However, my sense is this. Since Mg(NO3)2 can be decomposed to give me an oxide, but LiNO3 will give me LiNO2 instead, I am likely to say that Mg2+ will have a larger charge density.

Hope this helps.

RL said...

I should have been clearer in my previous comment (sorry). Right now I'm focusing mainly on the trend for the thermal stability of group II carbonates, but thank you for the information about Lithium too

Mr Kwok said...

Haha, now worries. Btw RL, you are?

Well, with regards to your query on group II carbonates, yes, a larger cation will have a less polarising effect and hence causes a smaller distort in the electron cloud and hence weakens the covalent bonds in the carbonate ions by a smaller extent.

Hence, by doing so, it will actually cause the carbonate to become more stable. Thus more heat is needed to break it down.

Anonymous said...

Mr Kwok,

According to Briggs, the smaller the cation (of say for example Group II), the stronger the charge density. So Mg2+ would have a strong polarising effect than Ba2+ on the carbonate. MgCO3 would be thermally unstable but BaCO3 would be much stable to heat.

Mr Kwok said...

Hi you are?

Yes, you are right and that was exactly what I am trying to explain. However, in my write up, I offered two perspectives and both explain how the trend of decomposition decreases down the group.

The first view is to examine the polarising power of the cation. The second view is to examine the difference between the LE of MCO3 and LE of MO, the difference is smaller down the group and hence resulting in the compound to be more thermally stable aka more heat required to decompose.

Anonymous said...

Umm do your statements on this page not contradict eachother? You day the larger cations have a greater polarising power, (why? their charge is spead over a larger molecule and therefore the charge density is less and should have less polarising power) but if this were true then they would release CO2 easier and decompose easier, which is not the case.

Mr Kwok said...

To the person who made the most recent comment, thank you so much for spotting the error. :) I must have been writing those lines with two thoughts running at the same time, although I should have prove read further. Thanks again! :)