Answer

In $\mathrm{PH_3}, \mathrm{P}$ is $s p^{3}$ hybridized. Three orbitals are involved in bonding with three hydrogen atoms and the fourth one contains a lone pair. As lone pair-bond pair repulsion is stronger than bond pair-bond pair repulsion, the tetrahedral shape associated with $s p^{3}$ bonding is changed to pyramidal. $\mathrm{PH_3}$ combines with a proton to form $\mathrm{PH_4}^{+}$in which the lone pair is absent. Due to the absence of lone pair in $\mathrm{PH_4}^{+}$, there is no lone pair-bond pair repulsion. Hence, the bond angle in $\mathrm{PH_4}^{+}$is higher than the bond angle in $\mathrm{PH_3}$.

$\mathrm{PH_4}^{+}$

$\mathrm{PH_3}$

Answer

White phosphorous dissolves in boiling $\mathrm{NaOH}$ solution (in a $\mathrm{CO_2}$ atmosphere) to give phosphine, $\mathrm{PH_3}$.

$$ \mathrm{P_4}+3 \mathrm{NaOH}+3 \mathrm{H_2} \mathrm{O} \longrightarrow \underset{\text{Phophine}}{\mathrm{PH_3}}+ \underset{\text{Sodium hypophosphite}}{3 \mathrm{NaH_2} \mathrm{PO_2}} $$