In the test for iodine, when ${\mathrm{I}}_2$ is treated with sodium thiosulphate, ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$ ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3+{\mathrm{I}}_2⟶\mathrm{NaI}+$......

When Iodine is treated with sodium thiosulfate ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$, an iodine solution is decolourized by sodium thiosulfate as sodium tetrathionate ${\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6$ and sodium iodide is formed. Both are colourless and soluble.

Hence, the correct option is A.