White phosphorus,when heated with conc. NaOH | vedclass

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Question

(C) White phosphorus $(P_4)$ undergoes disproportionation reaction when heated with concentrated $NaOH$ solution to form phosphine $(PH_3)$ and sodium

Vedclass · Accepted Answer

$+1$

Vedclass · Answer

(C) White phosphorus $(P_4)$ undergoes disproportionation reaction when heated with concentrated $NaOH$ solution to form phosphine $(PH_3)$ and sodium hypophosphite $(NaH_2PO_2)$.The chemical equation is: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.In the salt $NaH_2PO_2$ (sodium hypophosphite),let the oxidation state of $P$ be $x$.The sum of oxidation states is: $(+1) + 2(+1) + x + 2(-2) = 0$.$1 + 2 + x - 4 = 0$.$x - 1 = 0$.$x = +1$.

White phosphorus,when heated with conc. $NaOH$ solution in an inert atmosphere of $CO_2$,forms phosphine and a sodium salt of oxoacid of phosphorus,$'X'$. The oxidation state of phosphorus in $'X'$ is

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