Equivalent weight of ${\mathrm{H}}_3{\mathrm{PO}}_2$ in a reaction is found to be half of its molecular weight. It can be due to its

(a) It is monobasic acid, hence only one ${\mathrm{H}}^{+}$ can be neutralized $\mathrm{E} = \frac{\mathrm{M}}{1}$

${\mathrm{H}}_3\stackrel{+1}{\mathrm{P}}{\mathrm{O}}_2\to \stackrel{-3}{\mathrm{P}}{\mathrm{H}}_3 ; \mathrm{E}=\frac{\mathrm{M}}{4}$

${\mathrm{H}}_3\stackrel{+1}{\mathrm{P}}{\mathrm{O}}_2\to {\mathrm{H}}_3\stackrel{+3}{\mathrm{P}}{\mathrm{O}}_3 ; \mathrm{E}=\frac{\mathrm{M}}{2}$

${\mathrm{H}}_3\stackrel{+1}{\mathrm{P}}{\mathrm{O}}_2\to {\mathrm{H}}_3\stackrel{+5}{\mathrm{P}}{\mathrm{O}}_4 ; \mathrm{E}=\frac{\mathrm{M}}{4}$