Oxidation state of sulphur

Question:

Oxidation state of sulphur in anions $\mathrm{SO}_{3}^{2-} \mathrm{S}_{2} \mathrm{O}_{4}^{2-}$ and $\mathrm{S}_{2} \mathrm{O}_{6}^{2-}$ increases in the orders :

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

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

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

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


Correct Option: , 3

Solution:

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