Iodine oxidises ${S_2}O_3^{ - 2}$ ion into:
A. $SO_3^{ - 2}$
B. $SO_4^{ - 2}$
C. ${S_4}O_6^{ - 2}$
D. ${S^{ - 2}}$

Hint: Iodine considers the ability to oxidize the ions of another and changes down the group. When iodine reacts with the ions of another from a salt solution the iodine ion causes the reactions. Iodine appears either as a red solution if chlorine in little amounts is used or as a dark grey precipitate if the chlorine is in excess. Iodine won’t oxidize any of the other halide ions, except possibly the extremely radioactive and rare astatide ions.

Complete step by step answer:
The thiosulphate ion $({S_2}O_3^{ - 2})$ is oxidized by iodine.
The reaction of oxidises ${S_2}O_3^{ - 2}$ by using iodine is:
 $2{S_2}O_3^{ - 2} + {I_2} \to {S_4}{O_6}^{ - 2} + 2NaI$
    (+2) (+2.5)
As we see that the thiosulphate ion contains +2 oxidation number and ${S_4}O_6^{ - 2}$ contains +2.5 oxidation number. If the oxidation number increases, then reaction occurs. Here, the oxidation number increases in this reaction by the react with iodine.

Therefore, option (C) is the correct answer.

Additional information:
Iodine clock reaction is used with sodium, potassium or ammonium persulfate to oxidize iodine ions to iodine. Sodium thiosulphate is used as a medication to treat cyanide poisoning, pityriasis versicolor. It is used in gold mining, water treatment, analytical chemistry, and medicines.
Thiosulphate also known as an oxyanion of Sulphur where thio indicates the sulphate ion with one oxygen replaced by Sulphur.

Note:
A technique to use the fact that going from iodine, on the reactants side to the iodine anion, on the products side. Here thiosulfate anion is acting as a reducing agent because it is reducing iodine to iodide anions and iodine is acting as oxidizing agent because it oxidized the thiosulphate anion to the tetrathionate.