1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 ...

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '19 at 14:42

Not the answer you're looking for? Browse other questions tagged or ask your own question.