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Q.
The oxidation state of sulphur in $ {{H}_{2}}S{{O}_{5}} $ and of chromium in $ {{K}_{2}}C{{r}_{2}}{{O}_{7}} $ respectively is
Rajasthan PMTRajasthan PMT 2011
Solution:
(i) The structure of $ {{H}_{2}}S{{O}_{5}} $ is $ H-O-\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\overset{\begin{smallmatrix} O \\ || \end{smallmatrix}}{\mathop{S}}}\,-O-O-H $ As it is clear by the structure that two oxygen atom has peroxide linkage and their oxidation state is -1. Thus, we can calculate oxidation number (state) of sulphur in $ {{H}_{2}}S{{O}_{5}} $ as $ \underset{(for\,\,H)}{\mathop{2\times (+1)}}\,+\underset{(for\,\,S)}{\mathop{x}}\,+\underset{(for\,\,O-O)}{\mathop{2\times (-1)}}\, $ $ \underset{\begin{smallmatrix} (for\,\,other \\ three\,\,O\,\,atoms) \end{smallmatrix}}{\mathop{+3\times (-2)=0}}\, $ or $ x=+6 $ (ii) Oxidation state (number) of $ Cr $ in $ {{K}_{2}}C{{r}_{2}}{{O}_{7}} $ $ \underset{(for\,\,K)}{\mathop{2\times (+1)}}\,\underset{(for\,\,Cr)}{\mathop{+2\times x}}\,\underset{(for\,\,O\,\,atoms)}{\mathop{+7\times (-2)=0}}\, $ or $ x=+6 $