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 : $$\ce{SiO_2 + 4 HF -> SiF_4 + 2 H_2O} ......... (1)$$
$$\ce{SiCl_4 + 2 H_2O -> SiO_2 + 4 HCl}........ (2)$$ 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 $$\ce{SiO_2 + 6 HF -> H_2SiF_6 + 2 H_2O}$$ $$\ce{H_2SiF_6 <=> SiF_4(g) + 2 HF } $$$\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...