Question

`P_4` undergoes disproportionation in basic medium to give `PH_3` (phosphine) and `H_2PO^(ɵ)` (dihydrogen hypophoshite ion). Atomic weight of P is 31.

Answer 1

`P_4to4H_2PO_2^(ɵ)+4e^(-)(n=4)` (oxidation)
`12e^(-)+P_4to4PH_3(n=12)` (reduction)
`Ew(P_4)=(Mw(P_4))/(4)+(Mw(P_4))/(12)`
`=(31xx4)/(4)+(31xx4)/(12)`
`=31+(31)/(3)=31+10.33=41.33g`
Alternatively,
`Ew(P_4)=Mw(P_4)((1)/(4)+(1)/(12))=Mw(P_4)xx(1)/(3)`
`:.` n-factor of `P_4=3`
second method :
`P_(4)to4H_(2)PO_(2)^(ɵ)+cancel(4e^(-))]xx3`
`underline(12^(-)+P_(4)to4PH_(3))`
`underline(4P_(4)to4H_(2)PO_(2)^(ɵ)+4PH_(3))`
Total change in number of electrons `=12`
Effective molecular weight of `P_(4)=Mw(P_(4))+3Mw(P_(4))`
`=Mw(P_(4))`
`Ew(P_(4))=(4Mw(P_(4)))/(12)=(4xx4xx31)/(12)=41.33`g
`therfore` n-factor of `P_(4)=(12)/(4)=3`