(a) P4 reacts with NaOH. On heating, it gives NaH2PO2 and PH3. P4+3‌NaOH+3H2O
∆
→
3NaH2PO2+PH3 This reaction is a disproportionation reaction as oxidation state of P changes from 0 to +1 and -3 . Which implies that, it undergoes oxidation and reduction simultaneously. S8 reacts with NaOH on heating to given Na2S and Na2S2O3. S8+12‌NaOH→4Na2S+2Na2S2O3+6H2O This reaction is a disproportionation reaction as oxidation state of S changes from 0 to -2 and +2 . Which implies that, it undergoes oxidation and reduction simultaneously. But, N2 is an inert gas. So, it does not react with NaOH. Thus, only P4 and S8 undergo disproportionation when heated with NaOH solution.
