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In fluorite ($\ce{CaF2}$), the coordination number of $\ce{Ca}$ is $8$, and that of $\ce{F}$ is $4$. Is there any logic behind the ratio of coordination numbers being the same as the stoichiometry of the ions?

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  • $\begingroup$ I think it isn't a coincidence that the ratio of cations to anions is 1:2, and the coordination is 8:4. You can have other $\ce{AB2}$ structures, like rutile where the coordination is 6:3. Also, please improve your grammar. $\endgroup$ Commented Oct 28, 2015 at 22:47
  • $\begingroup$ You didn't explained why that happens? $\endgroup$ Commented Oct 28, 2015 at 22:49
  • $\begingroup$ You can, to some extent, rationalise it by looking at the unit cell. For a highly regular structure like that of fluorite, if you reduce the coordination number of Ca to 7, for example, then the ratio of Ca:F will no longer be 1:2. However, I do admit that that is not quite the best "proof". $\endgroup$ Commented Oct 28, 2015 at 23:00
  • $\begingroup$ I just want to know why the ratio of coordination numbers is same as that of ratio of stoichiometric coefficients $\endgroup$ Commented Oct 28, 2015 at 23:02
  • $\begingroup$ Is explanation related to the formula of compound $\endgroup$ Commented Oct 28, 2015 at 23:03

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Surely there is a logic behind that, and a pretty simple one. Let's take a crystal and count all $\ce{Ca}$ atoms in it. We'll get some huge number N (possibly about as huge as $N_A$). Now how many fluorine atoms are there? Each $\ce F$ is a neighbor of some $\ce{Ca}$, so if we just count all neighbors of all $\ce{Ca}$, we won't miss anything... and will end up with 8N. But wait, each $\ce F$ is a neighbor of four $\ce{Ca}$, so it was counted 4 times! We divide by 4 to compensate for that and happily arrive at 2N, which is consistent with the formula $\ce{CaF2}$.

So yes, it is no coincidence that the ratio of coordination numbers is also the ratio of stoichiometric coefficients.

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