A) \[+\,6\]
B) \[-\,6\]
C) \[+\,\,4\]
D) \[+\,\,8\]
Correct Answer: A
Solution :
Key Idea: Oxidation state of oxygen m peroxide linkage is -1. \[H-O-\underset{O}{\overset{O}{\mathop{\underset{||}{\overset{||}{\mathop{S}}}\,}}}\,-O-O-\underset{O}{\overset{O}{\mathop{\underset{||}{\overset{||}{\mathop{S}}}\,}}}\,-O-H\] \[\therefore \]In \[{{H}_{2}}{{S}_{2}}{{O}_{8}}\]there is one peroxide linkage \[\therefore \]Oxidation state of two oxygens is -1 and oxidation state of 6 oxygens is -2 Let oxidation state of S in\[{{H}_{2}}{{S}_{2}}{{O}_{8}}=x\] \[\therefore \]\[(+\,1\,\times 2)+2x+(6x-2)+(2x-1)=0\] or \[2x+2-12-2=0\] or \[2x=12\] \[\therefore \] \[x=+\,6\]
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