Chemistry - Why is tin(II) fluoride more effective in turning apatite into fluorapatite?

Solution 1:

Found an answer to this question in a book called 'Pediatric Dentistry'.

Apparently, when Stannous Fluoride reacts with dental enamel (hydroxyapatite), both the fluoride and the tin react to form Stannous Trifluorophosphate $(\ce{Sn3F3PO4})$ -- This is highly resistant to acid attack.

At high concentrations of stannous fluoride (I don't know what counts as "high", the book doesn't say), apparently the following compound is formed as well: Trifluorostannate -- This is also highly acid resistant.

In addition, Calcium Fluoride is also produced, which then further reacts with hydroxyapatite forming Fluorhydroxyapatite (which is also rather acid resistant, although it seems that the stannous trifluorophosphate is more resistant.)

Sodium fluoride only produces Fluorhydroxyapatite as its reaction.

It appears that Stannous Fluoride is a superior agent because it produces 1 or 2 additional compounds (that become part of the surface enamel) that are more acid resistant than just the one that Sodium Fluoride produces. These other compounds (which become a part of the surface of the tooth) also have an anti-microbial effect, thus inhibiting the growth of the very bacteria which are directly responsible for producing the majority of the acid found in dental decay.

Solution 2:

$\ce{SnF2}$ appears to have a predominantly covalent structure while $\ce{NaF}$ is ionic. So I think $\ce{SnF2}$ might release fluoride ions slower (more reluctantly) than $\ce{NaF}$.

Since toothpaste contains $\ce{Ca}$ containing minerals as well (as mentioned in the Wikipedia page), highly insoluble $\ce{CaF2}$ is formed easily when more fluoride ions are present. This is bad since $\ce{CaF2}$ will be inactive.

So since $\ce{SnF2}$ doesn't release fluoride ions as much (unless it comes into contact with water – it undergoes hydrolysis fast. Again this is good since more fluoride will be available only while you are brushing your teeth.), it has a longer shelf life than $\ce{NaF}$ I guess.

This also means that it doesn't matter what fluoride you use as long as its covalent. However I suspect tin is used because it happens to be one of the covalent fluorides relatively harmless to us (see this).

This is all just a guess so take it with a grain of salt... I don't have many sources to back it up (except Wikipedia of course!). Do point out any mistakes.

EDIT

As Georg has mentioned in the comments, toothpastes have a significant amount of water. Apparently toothpastes containing stannous fluoride contain a strong complexing agent which prevents oxidation of the $\ce{Sn^2+}$ ion (I don't have access to the paper but I read the abstract here).

Also the abstract mentions that $\ce{Sn^2+}$ has a antibacterial effect (most probably due to it being a heavy metal ion) and this is another reason why it is used (in contrast to $\ce{Na+}$ which has almost no antibacterial effects).