# What is the explanation of the changes in stability going down a group for carbonates, bicarbonates, fluorides, and chlorides?

For carbonates and bicarbonates, I know that stability increases down the group, and for chlorides and fluorides, stability decreases down the group. Why does this happen? Can someone explain this in detail?

(I am talking about S block alkali metals)

• What do you mean by "stability"? – hBy2Py Mar 31 '15 at 15:10
• Well how should i explain :-P!I mean less reactive :-)! – user14857 Mar 31 '15 at 15:13
• Reactive with what? I'm not trying to be difficult; the terms 'stable' and 'reactive' encompass a lot of different areas & answering your question well depends on exactly what you're referring to. Can you provide more context to your question? – hBy2Py Mar 31 '15 at 15:15
• Ok this is a doubt that occured due to my textbook and here is the direct quotation : "As the electropositive character increases down the group, the stability of the carbonates and hydorgencarbonates increases." – user14857 Mar 31 '15 at 15:19
• If there is relevant information in the pdf, please include it in the question. Responders shouldn't have to search for it. – jerepierre Mar 31 '15 at 16:18

## Carbonates

Carbonates of alkaline earth metals are insoluble in water and can be precipitated by addition of a sodium or ammonium carbonate solution to a solution of a soluble salt of these metals. The solubility of carbonates in water decreases as the atomic number of the metal ion increases. All the carbonates decompose on heating to give carbon dioxide and the oxide. Beryllium carbonate is unstable and can be kept only in the atmosphere of CO2. The thermal stability increases with increasing cationic size.

So the stability that you are referring to is thermal stability. This is an important detail. So what is thermal stability? It's how resistant a molecule is to decomposition at higher temperatures.

## What's happening to cause thermal instability?

So, lets look at the carbonate ion here:

This is just an illustration, and in reality the negative charge we see on the two $\ce{O}$ atoms is localized due to resonance.

Below the illustration shows where the negative charge is likely to be concentrated (colored in red).

So, when we create a carbonate complex like the example below, the negative charge will be attracted to the positive ion.

Because of this polarization, the carbon dioxide will become more stable and energetically favorable.

## How does going down a group play into this?

Well as you go down the group, the charged ion becomes larger. The larger the ion, we see a lower charge density. Charge density is basically the amount of charge in a given volume. So, if a small ion has the same charge as a larger ion, the charge density will be greater for that small ion. Greater charge density, means a greater pull on that carbonate ion, and a greater pull causes the delocalized ions, and a more stable $\ce{CO2}$ molecule. So, the larger the ion, the lower the charge density, the less polarizing of an effect, and reduced stability of a $\ce{CO2}$ molecule, favoring the $\ce{CO3}$.

## Chloride and fluoride stability

Stability of fluorides, chlorides, and other halogens, are likewise related to thier size. The halogens, specifically fluouride, is known for their electronegativity. Electronegativity, is the tendency to attract electrons to itself. As you move up the group, you see an increase in electronegtivity. This results in the creation of polar bonds.

Illustrated below, you see that as charge of the positive ions increase, polarizability increases (left), and as the halogen ion increases, polarizability and electronegativity decrease (right). When the ions electron cloud, is less polarized, the bond is less strong, leading to a less stable molecule.

Information and illustrations on carbonate ions were sourced from here.

By Fajan's Rule you should be getting the answer and then more electropositive metal will have more ionic character and then that will increase stability.

• Welcome to chemistry.SE! Could you please be a little more elaborate? Your answer might sound comment-like to some people, and I don't think it will solve the OP's problem, really. – M.A.R. Apr 1 '15 at 12:28