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.