White phosphorus is heated with concentrated $\mathrm{NaOH}$ in $\mathrm{CO}_2$ atmosphere to form a gas $A$…

White phosphorus is heated with concentrated $\mathrm{NaOH}$ in $\mathrm{CO}_2$ atmosphere to form a gas $A$ and compound $B$. When $A$ is bubbled into aqueous $\mathrm{CuSO}_4$ solution copper phosphide and $C$ are formed, $B$ and $C$ are respectively
  1. $\mathrm{PH}_3, \mathrm{H}_2 \mathrm{SO}_4$
  2. $\mathrm{NaH}_2 \mathrm{PO}_2, \mathrm{H}_2 \mathrm{SO}_4$
  3. $\mathrm{NaHPO}_2, \mathrm{CuS}$
  4. $\mathrm{NaH}_2 \mathrm{PO}_2, \mathrm{Cu}_2 \mathrm{~S}$

Solution

(i) When white phosphorus is heated with concentrated $\mathrm{NaOH}$ in inert atmosphere of $\mathrm{CO}_2$, it gives $\mathrm{PH}_3(g)$ and $\mathrm{NaH}_2 \mathrm{PO}_2$ as follows : $ \begin{aligned} & \mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \stackrel{\mathrm{CO}_2}{\longrightarrow} \underset{(A)}{\mathrm{PH}_3(g)} \\ & +\quad 3 \mathrm{NaH}_2 \mathrm{PO}_4 \\ & \text { (Sodium hypophosphite) } \\ & \end{aligned} $ (B) ii) When $(A)$, i.e. $\mathrm{PH}_3(g)$ is bubbled into aqueous $\mathrm{CuSO}_4$ solution, copper phosphide $\left(\mathrm{Cu}_3 \mathrm{P}_2\right)$ and $\mathrm{H}_2 \mathrm{SO}_4(\mathrm{C})$ is formed. The reaction occurs as follows : $ 2 \mathrm{PH}_3+3 \mathrm{CuSO}_4 \longrightarrow \underset{\begin{array}{c} (\text { Copper } \\ \text { phosphide) } \end{array}}{\mathrm{Cu}_3 \mathrm{P}_2}+3 \mathrm{H}_2 \mathrm{SO}_4 $ Hence, $(B)=\mathrm{NaH}_2 \mathrm{PO}_2,(C)=\mathrm{H}_2 \mathrm{SO}_4$ Hence, option (b) is the correct answer

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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