A) \[+6\]
B) \[-6\]
C) \[+4\]
D) \[+8\]
Correct Answer: A
Solution :
The oxidation of S in \[{{H}_{2}}{{S}_{2}}{{O}_{8}}\] is \[2+2x+(8\times -2)=0\], \[2+2x-16=0\] \[2x=14=7\] This is wrong. Because this is perdi-sulphuric acid. It is called Marshal acid also. \[H-O-\underset{O}{\mathop{\underset{||}{\mathop{\overset{O}{\mathop{\overset{||}{\mathop{S}}\,}}\,}}\,}}\,-\overset{-1}{\mathop{O}}\,-\overset{-1}{\mathop{O}}\,-\underset{O}{\mathop{\underset{||}{\mathop{\overset{O}{\mathop{\overset{||}{\mathop{S}}\,}}\,}}\,}}\,-O-H\] \[2+2x+(2\times -1)+(6\times -2)=0\] \[2+2x-2-12=0\] \[2+2x-14=0\] \[2x=12=6\]
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