Question

White phosphorus is heated with concentrated $NaOH$ in $CO _{2}$ atmosphere to form a gas $A$ and compound $B$. When $A$ is bubbled into aqueous $CuSO _{4}$ solution copper phosphide and $C$ are formed, $B$ and $C$ are respectively

• A. $\ce{PH_3 , H_2SO_4}$
• B. $\ce{NaH_2PO_2, H_2SO_4 }$
• C. $\ce{NaHPO_2, CuS }$
• D. $\ce{ NaH_2 PO_2, Cu_2S}$

Answer 1

Answer: (B)

$\ce{NaH_2PO_2, H_2SO_4 }$

Answer 2

(i) When white phosphorus is heated with concentrated $NaOH$ in inert atmosphere of $CO _{2}$, it gives $PH _{3}(g)$ and $NaH _{2} PO _{2}$ as follows :

$P_4 + 3NaOH + 3H_2O \ce{->[CO_2][\Delta]} \underset{(A)}{PH_3(g)} + \underset{\overset{\text{(Sodium hypophosphite)}}{(B)}}{3NaH_2PO_4}$

(ii) When $(A)$, i.e. $PH _{3}(g)$ is bubbled into aqueous $CuSO _{4}$ solution, copper phosphide $\left( Cu _{3} P _{2}\right)$ and $H _{2} SO _{4}(C)$ is formed. The reaction occurs as follows:

$2 PH _{3}+3 CuSO _{4} \longrightarrow \underset{\text { (Copper phosphide)}}{ Cu _{3} P _{2}}+\underset{( C )}{3 H _{2} SO _{4}}$

Hence, $(B)= NaH _{2} PO _{2},( C )= H _{2} SO _{4}$