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 \[{{S}_{2}}O_{4}^{2-}<SO_{3}^{2-}<{{S}_{2}}O_{6}^{2-}\]You need to login to perform this action.
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