White phosphorus on reaction with concentrate | vedclass

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Question

(B) The reaction of white phosphorus $(P_{4})$ with concentrated $NaOH$ is a disproportionation reaction:$P_{4} + 3 NaOH + 3 H_{2}O \longrightarrow PH

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The reaction of white phosphorus $(P_{4})$ with concentrated $NaOH$ is a disproportionation reaction:$P_{4} + 3 NaOH + 3 H_{2}O \longrightarrow PH_{3} + 3 NaH_{2}PO_{2} (X)$Here,$(X)$ is sodium hypophosphite $(NaH_{2}PO_{2})$.Upon acidification with $HCl$,$(X)$ forms hypophosphorous acid $(H_{3}PO_{2})$:$NaH_{2}PO_{2} + HCl \longrightarrow NaCl + H_{3}PO_{2} (Y)$In $H_{3}PO_{2}$,there is only one $P-OH$ bond,while two $P-H$ bonds are directly attached to the phosphorus atom. Since only the hydrogen atom attached to the oxygen atom is ionizable,the basicity of $H_{3}PO_{2}$ is $1$.

White phosphorus on reaction with concentrated $NaOH$ solution in an inert atmosphere of $CO_{2}$ gives phosphine and compound $(X)$. $(X)$ on acidification with $HCl$ gives compound $(Y)$. The basicity of compound $(Y)$ is

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