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\text { Find the oxidation number of } \mathrm{S} \text { in } \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8}

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It would have been plus seven if possible, however the maximum is the number of valence electron of sulfur to be donated, so that is clearly +6 (six). i.e. 2 +2x -16=0 =>x =7, but for sulphur x cannot be greater than 6... so this means , there is a peroxy linkage in the given compound... and we know that ,  peroxy oxygen have oxdn. no. -1 whenever there is peroxy linkage , sulphur is always in its highest oxdn. state i.e. +6 so here, oxdn. state of sulphur is +6 now u can also find the no. of peroxy linkages.. i.e.  2+2(6) +n(-1) -(n-8)(-2) = 0 so , here n = 2.  so we have two peroxy oxygens and one peroxy linkage.
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