The equivalent weight of Hypo in the reaction $[\mathrm{M}=$ molecular weight] $2{\mathrm{Na}}_2{ \mathrm{S}}_2{\mathrm{O}}_3+{\mathrm{I}}_2⟶2\mathrm{Nal}+{\mathrm{Na}}_2{ \mathrm{S}}_4{\mathrm{O}}_6$ is

The oxidation state of S  in ${\mathrm{Na}}_2{ \mathrm{S}}_2{\mathrm{O}}_3$ is $+2$

The oxidation state of S  in ${\mathrm{Na}}_2{ \mathrm{S}}_4{\mathrm{O}}_6$ is $+2.5$

The reaction is an oxidation reaction, there is a loss of $0.5$ electrons form reactant to product

There are$2$ moles of reactant and hence $2\times 0.5=1$ is the no. of electrons participate in reaction

$\mathrm{Equivalent} \mathrm{weight} =\mathrm{Mass}/\mathrm{no}. \mathrm{of} \mathrm{electrons} \mathrm{participate} \mathrm{in} \mathrm{reaction}$

Therefore, equivalent weight of $\mathrm{Hypo}=\mathrm{M}/1$