What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

H₂S₄O₆

Let the O.N. of S be ‘x’.

Now,2(+ 1) + 4(x) + 6(-2) = 0

2 + 4x-12 = 0

X = + 2

However, the O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

The O.N. of two of the four S atoms is + 5 and the O.N. of the other two S atoms is 0.