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.
You will be redirected in
3 sec