Question

Iodine oxidises ${S_2}{O_3}^{ - 2}$ ion into :
$S{O_3}^{ - 2}$
$S{O_4}^{ - 2}$
${S_4}{O_6}^{ - 2}$
${S^{ - 2}}$

Answer 1

Hint: Iodine belongs to group 17 , the halogen family . All the halogens have a strong tendency to accept an electron and therefore act as a strong oxidising agent . Their oxidising power decreases from ${F_2}$ to ${I_2}$ .

Complete step by step answer:
An oxidizing agent is referred to as a chemical compound that readily transfers oxygen atoms or a substance that gains electrons in a redox chemical reaction . It is a substance which oxidises the other compound and itself gets reduced .
Iodine is the weakest oxidising agent in the halogen family as oxidising power of halogens decreases from ${F_2}$ to ${I_2}$ .
When iodine oxidises ${S_2}{O_3}^{ - 2}$ the following reaction takes place .
${I_2} + 2{S_2}{O_3}^{ - 2} \to {S_4}{O_6}^{ - 2} + 2{I^ - }$
As you can see in the above reaction iodine oxidises ${S_2}{O_3}^{ - 2}$ to ${S_4}{O_6}^{ - 2}$ and itself gets reduced to iodide ion .
The valency of sulphur changes from +2 to +2.5 ( this is the average oxidation state since in this ion all the sulphurs have different oxidation states ) , whereas the oxidation of iodine changes from 0 to +1 .

So, the correct answer is Option C .

Note:
In ${S_4}{O_6}^{ - 2}$ the valency of two sulphur atoms which form a bridge that is they are attached to each other have a valency of -1 whereas the other two sulphur atoms have a valency of +6 as they are attached to three oxygen atoms each . Therefore the average valency of all the four sulphur atoms is +4 .