`P_4` undergoes disproportionation in basic medium to give `PH_3` (phosphine) and `H_2PO_2^(ɵ)` (dihydrogen hypophoshite ion). Atomic weight of P is 31.

Explanation

`P_(4) to 4H_(2)PO_(2)^(-)+4e^(-) (n=4) ("oxidation")`
`12e^(-)+P_(4) to 4PH_(3) (n=12) ("reduction")`
`Ew(P_(4))=(Mw(P_(4)))/(4)+(Mw(P_(4)))/(12)`
`=(31 xx 4)/(4)+(31 xx 4)/(12)=31+(31)/(3)=31+10.33=401.33g`
Alternatively,
`Ew (P_(4))=Mw(P_(4)) (1/4+1/12)=Mw (P_(4)) xx 1/3`
n-factor of `P_(4)=3`
(d) `H_(2)O_(2)` undergoes disporportionation in acid medium to give `O_(2) & H_(2)O`
`H_(2)O_(2) to O_(2)+2e^(-)+2H^(+) (n=2) ("oxidation")........(i)`
`2e^(-)+2H^(+)+H_(2)O_(2) to 2H_(2)O "(n=2) (reduction) ...........(ii)"`
`Ew(H_(2)O_(2))=(Mw(H_(2)O_(2))/(2)+(Mw(H_(2)O_(2))/(2)`
`=Mw(H_(2)O_(2)) (1/2+1/2)=Mw(H_(2)O_(3))=34g`
Hence, n-factor of `H_(2)O_(2)` in disproportionation reaction is 1.
(e) Ew of an element or a compound in a Non redox change.