Reaction of Silicon Dioxide with Hydrofluoric Acid
It is surprising that Silicon halides behave differently with water according to the choice of the halogen fluor or chlorine. The following reactions are known to happen : $$\ce{SiO_2 + 4 HF -&gt; SiF_4 + 2 H_2O} ......... (1)$$
$$\ce{SiCl_4 + 2 H_2O -&gt; SiO_2 + 4 HCl}........ (2)$$ These equations are exactly the opposite from one another. Why does reaction ($1$) not proceed in the other way like ($2$) ? I don't know, but I would like present a personal interpretation. Maybe it is due to the fact that reaction ($1$) proceeds in two steps, with the intermediate formation of $\ce{H_2SiF_6}$. Maybe ($1$) proceeds like that $$\ce{SiO_2 + 6 HF -&gt; H_2SiF_6 + 2 H_2O}$$ $$\ce{H_2SiF_6 &lt;=&gt; SiF_4(g) + 2 HF } $$$\ce{H_2SiF_6}$ is well known to be formed if $\ce{HF}$ is in excess. And of course HCl cannot form $\ce{H_2SiCl_6}$, because the chlorine atom is too big to act as a ligand this octahedral structure. $\ce{H_2SiCl_6}$ does not exist.
This is a personal idea. I would be pleased to hear about any comment.
Sorry for presenting these lines as an Answer. Alas the text is much too long for a comment...