I was looking at how the strongest acid was made and I found something I did not understand. On the Wikipedia page for antimony trifluoride it shows the production of $\ce{SbF3}$ using $\ce{HF}$: $$\ce{Sb2O3 + 6 HF -> 2 SbF3 + 3 H2O}$$ The page also shows production of $\ce{SbF5}$ using $\ce{F2}$:
$$\ce{SbF3 + F2 -> SbF5}$$ However on the page for fluoroantimonic acid it shows production of:

$$\ce{SbF5 + 2 HF -> SbF6- + H2F+}$$

So my question is why the production of $\ce{SbF5}$ seems to require fluorine insted of hydrogen fluoride? And is it possible to use hydrogen fluoride to make $\ce{SbF5}$?


1 Answer 1


It can be prepared using hydrogen fluoride, in a double displacement reaction:

$\ce{SbCl5 + 5 HF -> SbF5 + 5 HCl}$

However, hydrogen holds onto fluorine more tightly than does antimony in $\ce{SbF5}$.

Consider the reverse of your proposed reaction, which is preferred energetically:

$\ce{SbF5 + H2 -> SbF3 + 2HF}$

  • 2
    $\begingroup$ @EdV, sorry, I don't have that info. And, as you state, it's not something I'd try to measure. Does Derek Lowe have an article on SbF5, similar to this in his series Things I Won't Work With blogs.sciencemag.org/pipeline/archives/2010/02/23/…? $\endgroup$ Jun 19, 2019 at 19:27
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    $\begingroup$ @EdV, your welcome, but I have Wikipedia to thank. $\endgroup$ Jun 19, 2019 at 21:17
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    $\begingroup$ so my takeaway is keep your Antimony pentafluoride away from your hydrogen. $\endgroup$
    – zenot fun
    Jun 20, 2019 at 2:37

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