A) +2
B) +4
C) +6
D) +7
Correct Answer: C
Solution :
\[H-O-\underset{\begin{smallmatrix} | \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{S}}}\,-O-O-\underset{\begin{smallmatrix} | \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{S}}}\,-O-H\] \[\because \]\[2O-\]atoms are in peroxide linkage \[\therefore \] Their oxidation state\[=-1\] and, oxidation state of remaining \[6O-\]atoms\[=-2\] Oxidation state of H-atoms\[=+1\] Let oxidation state of S atom\[=x\] \[{{S}_{2}}\] \[{{H}_{2}}\] \[{{O}_{6}}\] \[{{O}_{2}}\] \[\therefore \] \[2x+2(+1)+6(-2)+2(-1)=0\] \[2x+2-12-2=0\] \[x=x+6\]
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