question_answer

The oxidation states of sulphur in the anions \[SO_{3}^{2-},{{S}_{2}}O_{4}^{2-}\] and \[{{S}_{2}}O_{6}^{2-}\] follow the order

A)  \[SO_{3}^{2-}<{{S}_{2}}O_{4}^{2-}<{{S}_{2}}O_{6}^{2-}\]

B)  \[{{S}_{2}}O_{4}^{2-}<{{S}_{2}}O_{6}^{2-}<SO_{3}^{2-}\]

C)  \[{{S}_{2}}O_{6}^{2-}<{{S}_{2}}O_{4}^{2-}<SO_{3}^{2-}\]

D)  \[{{S}_{2}}O_{4}^{2-}<SO_{3}^{2-}<{{S}_{2}}O_{6}^{2-}\]

Correct Answer: D

Solution :

 The chemical bond method gives the O.N.