What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results? H_(2)ul(S)_(4)O_(6)

What are the oxidation number of the underlined elements in each of th

1.	What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results? H_(2)ul(S)_(4)O_(6)
Answer» Solution :The AVERAGE OXIDATION number of S in `H_(2)S_(4)O_(6)` can be calculated as follows. `overset(+1)(H_(2))overset(x)(S_(4))overset(-2)(O_(6))` `(+1)xx2+[(x)xx4]+[(-2)xx6]=0` or `x=+(12-2)/(4)=(5)/(2)=+2.5` Since, the average oxidation number is FRACTIONAL, let us consider the structure of `H_(2)S_(4)O_(6)` as given below. `H-Ounderset((1))(-)underset(O)underset(\|\|)overset(O)overset(\|\|)Soverset(+5)(-)underset((2)) overset(0)S-underset((3))overset(0)S-underset(O)underset(\|\|)overset(O)overset(\|\|)Sunderset((4))-O-H` The oxidation number of S atoms LABELLED as (2) and (3) is zero because they are linked with similar atoms on either sides. Hence, calculating as above, the oxidation numbers of S atoms labelled as (1) and (4) will be +5 each.