Assign oxidation number to the underlined elements in each of the following species:

H2S2O7


Let the oxidation number(O.N.) of S be ‘x’


We know that,


O.N. of H = + 1


O.N. of O = -2


Therefore, we have,


2(+ 1) + 2(x) + 7(-2) = 0


2x = 12


X = + 6


The O.N. of S is + 6.


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