${\mathrm{P}}_4\stackrel{{\mathrm{OH}}^{-}}{\to } {\mathrm{PH}}_3+{\mathrm{H}}_2{\mathrm{PO}}_2^{-}.$ Equivalent Weight of Phosphorous in the reaction is (M= molar mass of Phosphorous)

 ${\stackrel{0}{\mathrm{P}}}_4\to {\mathrm{H}}_2\stackrel{+1}{\mathrm{P}}{\mathrm{O}}_2^{-}$                                                                      ${\stackrel{0}{\mathrm{P}}}_4\to \stackrel{-3}{\mathrm{P}}{\mathrm{H}}_3$                                                                 

increase in ox.state/atom of 'P' = 1                   decrease in ox.state/atom of 'P' = 3

increase in ox.state/mole of 'P' =4(1)                 decrease in ox.state/mole of 'P'=4(3)

                                              $\mathrm{E}.\mathrm{W} \mathrm{of} \mathrm{Phosphorous}=\frac{\mathrm{M}}{4}+\frac{\mathrm{M}}{12}$