The reaction of white phosphorus with aqueous | vedclass

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(C) The reaction of white phosphorus $(P_4)$ with aqueous $NaOH$ is given by:$P_4(s) + 3NaOH(aq) + 3H_2O(l) \rightarrow PH_3(g) + 3NaH_2PO_2(aq)$In th

Vedclass · Accepted Answer

disproportionation reaction; $-3$ and $+1$

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(C) The reaction of white phosphorus $(P_4)$ with aqueous $NaOH$ is given by:$P_4(s) + 3NaOH(aq) + 3H_2O(l) \rightarrow PH_3(g) + 3NaH_2PO_2(aq)$In this reaction,the oxidation state of phosphorus changes from $0$ in $P_4$ to $-3$ in $PH_3$ and $+1$ in $NaH_2PO_2$. Since the same element is simultaneously oxidized and reduced,it is a disproportionation reaction.However,if we consider the final product formed upon further reaction or heating as $Na_3PO_4$,the oxidation state of $P$ in $Na_3PO_4$ is $+5$. Given the options provided,the reaction is a disproportionation reaction where phosphorus is reduced to $-3$ in $PH_3$ and oxidized to $+1$ in $NaH_2PO_2$ (or $+5$ in $Na_3PO_4$ depending on the context of the question). Based on standard textbook options for this specific question,the correct answer is disproportionation reaction with oxidation states $-3$ and $+1$ (or $+5$ if considering the final phosphate product). Given the choices,option $C$ is the most accurate representation of the initial products.

The reaction of white phosphorus with aqueous $NaOH$ gives phosphine along with another phosphorus-containing compound. The reaction type and the oxidation states of phosphorus in phosphine and the other product are respectively:

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