A) \[+2\]
B) \[+6\]
C) \[+4\]
D) \[+8\]
Correct Answer: B
Solution :
Let the oxidation number of S be x \[{{H}_{2}}{{S}_{2}}{{O}_{7}}\] \[\therefore \] \[2\times 1+2x+7\times -2=0\] \[x+=6\] Oxidation number of \[S=+6\] \[{{H}_{2}}{{S}_{2}}{{O}_{7}}\] is oleum. \[\underset{\begin{smallmatrix} Suphuric \\ acid \end{smallmatrix}}{\mathop{{{H}_{2}}S{{O}_{4}}}}\,+S{{O}_{3}}\to \underset{Oleum}{\mathop{{{H}_{2}}{{S}_{2}}{{O}_{7}}}}\,\xrightarrow{HOH}\underset{\begin{smallmatrix} Sulphuric \\ acid \end{smallmatrix}}{\mathop{2{{H}_{2}}S{{O}_{4}}}}\,\]
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