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Question

(B) The reaction of white phosphorus $(P_4)$ with concentrated $NaOH$ solution is a disproportionation reaction: $P_4 + 3NaOH + 3H_2O \rightarrow 3NaH

Vedclass · Accepted Answer

$+1, -3$

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(B) The reaction of white phosphorus $(P_4)$ with concentrated $NaOH$ solution is a disproportionation reaction: $P_4 + 3NaOH + 3H_2O \rightarrow 3NaH_2PO_2 + PH_3$ Here,the salt '$X$' is sodium hypophosphite $(NaH_2PO_2)$ and the gas '$Y$' is phosphine $(PH_3)$. In $NaH_2PO_2$,the oxidation state of $P$ is calculated as: $1 + 2 + x + 2(-2) = 0 \Rightarrow x = +1$. In $PH_3$,the oxidation state of $P$ is calculated as: $x + 3(1) = 0 \Rightarrow x = -3$. Thus,the oxidation states are $+1$ and $-3$ respectively.

White phosphorus on heating with concentrated $NaOH$ solution in an inert atmosphere of $CO_2$ gives a salt '$X$' and gas '$Y$'. The oxidation state of central atom in $X$ and $Y$ is respectively

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