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

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

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

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

Correct Answer: C

Solution :

                Oxidation state of ?S? in                 \[SO_{3}^{2-},\,\,x+\,(-2\times 3)=-2,\,\,x=+6-2=+4\]                 Oxidation state of ?S? in                 \[{{S}_{2}}O_{4}^{2-}\,2x\,+(-2\times 4)=-2\]                 \[2x=+8-2=+\,6\]                 \[x=\frac{+6}{2}=+3\]                 Oxidation state of ?S? in                 \[{{S}_{2}}O_{6}^{2-}\,2x+(-2\times 6)=-2\]                                 \[2x=12-2=10\]                                 \[x=\frac{10}{2}=+5\]                 On the basis of structures                                 Hence, increasing order of oxidation state of is                 \[{{S}_{2}}O_{4}^{2-}<SO_{3}^{2-}<{{S}_{2}}O_{6}^{2-}\]