${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3+{\mathrm{Cl}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2\mathrm{HCl}+\mathrm{S}.$ Incorrect statement about the reaction is

Equivalent weight of ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ is $\frac{\mathrm{MW}}{2}$

Oxidation state of sulphur atoms in ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ are + 6 and -2

${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ undergoes disproportionation

Chlorine acts like oxidizing agent