The oxidation states of sulphur in the anions $\mathrm{SO}_{3}^{2-}, \mathrm{S}_{2} \mathrm{O}_{4}^{2-}$ and $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$ follow the order

• A

  $\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$
• B

  $\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{SO}_{3}^{2-}$
• C

  $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}<\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}$
• Correct answer
  D

  $\mathrm{S}_{2} \mathrm{O}_{4}^{2-}<\mathrm{SO}_{3}^{2-}<\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$

Solution

According to chemical bond method:

Hence, the correct answer is option: 'D'.