Disproportionation of hypophosphite and phosphite

The disproportionation of NaH2PO2, NaH2PO3, Na2HPO3 and their mixtures was studied by TG-DSC, XRD and P-31 NMR. NaH2PO2 reacted in three steps to yield PH3, Na5P3O10 and H-2 by disproportionation and oxidation with water released in condensation reactions. In the first step at 310 degrees C NaH2PO2 reacted to yield PH3, Na2H2P2O5, Na2HPO3, Na4P2O7 and H-2. H-2 rather than H2O was the coproduct of the disproportionation, because H2O oxidized hypophosphite and phosphite at elevated temperatures, in agreement with DFT results that show that the reaction of H3PO3 with H2O is exothermic. The oxidation of phosphite by H2O has a twofold effect on gas formation. First, it diminishes gas formation because water is not released and, second, because H2O oxidizes phosphite, less phosphite is available for disproportionation to phosphate and, thus, less PH3 is formed. As a consequence, the maximum PH3 efficiency in the disproportionation of NaH2PO2 was not 50% but 40%. Phosphate was never observed, only Na4P2O7, which appeared as a primary product of the oxidation of Na2H2P2O5 by H2O. Na2HPO3 did not react below 450 degrees C when heated alone, but in the presence of compounds which release H2O it was oxidized to Na4P2O7 below 400 degrees C. NaH2PO3 first reacted to yield PH3 and Na2H2P2O5 and then to Na5P3O10 and Na3P3O9. Heating a 1 : 1 mixture of NaH2PO3 and Na2HPO3 led to PH3, Na2H2P2O5 and Na2HPO3, then to Na4P2O7 and eventually to Na5P3O10 and Na3P3O9. In all cases some red phosphorus was formed by the decomposition of PH3, especially at higher temperatures.