Question

When thiosulphate ion is oxidised by iodine, which one of the following ions is produced?
A. $SO_{3}^{2-}$
B. $SO_{4}^{2-}$
C. ${{S}_{4}}O_{6}^{2-}$
D. ${{S}_{2}}O_{6}^{2-}$

Answer 1

Hint: Oxidation means increase in the oxidation state or charge of an atom. Here iodine will help to oxidise thiosulphate ion and reduce itself.

Complete Step by Step Answer:
The chemical formula of thiosulphate ion is ${{S}_{2}}O_{3}^{2-}$ . It generally exists as sodium thiosulphate. Sodium thiosulphate has the chemical formula of $N{{a}_{2}}{{S}_{2}}{{O}_{3}}$ . When sodium thiosulphate reacts with iodine gas, sodium tetrathionate is formed as a product along with sodium iodide.

The reaction is written as follows:
$2N{{a}_{2}}{{S}_{2}}{{O}_{3}}+{{I}_{2}}\to 2NaI+N{{a}_{2}}{{S}_{4}}{{O}_{6}}$
Thus in this reaction two moles of sodium thiosulphate react with one mole of iodine gas to produce one mole of sodium tetrathionate as a major product along with two moles of sodium iodide.

This sodium tetrathionate undergoes decomposition to give sodium ions along with tetrathionate ions. The decomposition can be written as follows:
$N{{a}_{2}}{{S}_{4}}{{O}_{6}}\to 2N{{a}^{+}}+{{S}_{4}}O_{6}^{2-}$
Thus when thiosulphate ion is oxidised by iodine, tetrathionate or ${{S}_{4}}O_{6}^{2-}$ ion is produced.
Thus the correct option is C.

Note: In this reaction the iodine gas helps to oxidise the sodium thiosulphate and itself get reduced. The thiosulphate ion is oxidised to tetrathionate ion and iodine is reduced to iodide ion. Thus this is an oxidation reduction type reaction.