The synthesis of $\ce{CoH[P(OPh)3]4}$ requires the use of an inert atmosphere e.g. nitrogen as the compound is air-sensitive. Is there any way to tell whether it is the oxygen or water vapour in the air specifically? I thought it might be water sensitive seeing as metal hydrides are basic and may therefore react in an acid-base reaction:

$$\ce{CoH[P(OPh)3]4 + H2O ⟶ Co[OH][P(OPh)3]4 + H2}$$

Is rationalisation correct/is there more to it than this?

  • 1
    $\begingroup$ selectively expose it to pure oxygen or pure water and you will know whether only one of those is a problem. $\endgroup$ – matt_black Mar 20 '18 at 21:33

I think that this will be water sensitive for the reasons you present but also oxygen sensitive for two reasons:

  • Due to the relatively low oxidation state of $\ce{Co(I)}$ in the compound which will react with $\ce{O2}$ towards formation of $\ce{Co(II)}$.
  • The $\ce{P^{III}(OPh)3}$ ligand is air sensitive as well and can oxidize to $\ce{OP^V(OPh)3}$ as well as reacting with water: https://www.alfa.com/en/catalog/A18662/.
| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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