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