White phosphorus is boiled with concentrated aqueous NaOH in an inert atmosphere. It gives a colourless, foul-smelling gas X and a sodium salt Y.

The reaction is the disproportionation $\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}$. Thus $\mathbf{X}=\mathrm{PH_3}$ (phosphine) and $\mathbf{Y}=\mathrm{NaH_2PO_2}$ (sodium hypophosphite). Phosphorus goes from $0$ to $-3$ in $\mathrm{PH_3}$ and to $+1$ in the hypophosphite. In $\mathrm{H_2PO_2^-}$ the P atom carries two $\mathrm{H}$ atoms directly bonded and two $\mathrm{O}$ atoms, so the oxidation number is $2(+1)+2(-2)+x=-1$, giving $x=+1$. The stoichiometry shows 1 mol $\mathrm{PH_3}$ and 3 mol of $\mathbf{Y}$ per mole of $\mathrm{P_4}$, so option D is correct and option C reverses the ratio. Option A is false because the $\mathrm{H{-}P{-}H}$ angle is about $93.5^\circ$, much smaller than $107^\circ$ in $\mathrm{NH_3}$ owing to the lone pair having high s-character and little hybridisation. Option B gives $\mathrm{H_2PO_3^-}$, which is phosphite-type with $+3$.