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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?

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    $\begingroup$ 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
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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/.
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