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Why is $\ce{SF6}$ inert towards hydrolysis? Would not $\ce{H - F}$ bonds be created in water which is much stable and has a high negative enthalpy of formation. This hydrogen bond making must disintegrate the whole compound.

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marked as duplicate by Mithoron, Ivan Neretin, Jon Custer, aventurin, Todd Minehardt Jun 13 '18 at 1:11

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In short its kinetically stabilised, attack at S is close to impossible due to steric hindrance and rigidity of FSF angles. Thermodynamically with water it is very unstable with respect to HF$_{aq}$ and H$_2$SO$_4$. Creepy ...

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  • $\begingroup$ Hello I was wondering if the fluorine atoms would make hydrogen bonds with water molecules, since the HF bond is very stable this bonding must release enough energy to break the SF bonds,but I suppose I'm wrong, can you tell me why? $\endgroup$ – Sulphur Jun 17 '18 at 15:56
  • $\begingroup$ The group electronegativity (see Huheey) of -SF$_5$ is about 3.0, that means F in F-SF$_5$ is not very strongly polarized (F is anyway quite hard), so its a bad hydrogen bond acceptor, thus the F-S bond is not really substantially weakend. The only "dangerous" attack would be a nucleophilic one at sulphur by water lone pairs, but that is sterically hindered. $\endgroup$ – Rudi_Birnbaum Feb 11 at 14:42

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