What are the oxidation number to the underlined elements in each of the following and how do you rationalise your results?
a) `Kunderline(I_(3))` b) `H_(2)underline(S_(4))O_(6)` c) `underline(Fe_(3))O_(4)`

Explanation

a) `KI_(3)toK^(+)+I_(3)^(-)`
`I_(3)^(-)` ion is formed by combining `I^(-1)` with `I_(2)`.
The average oxidation number is `1/3` but `I^(-)` is in -1 oxidation state while `I_(2)` is zero oxidation state.
b) `H_(2)S_(4)O_(6)` has the following structure `HO-underset(O)underset(||)overset(O)overset(||)(S_(1))-S_(2)-S_(3)-underset(O)underset(||)overset(O)overset(||)(S_(4))-OH`
The oxidation states of `S_(2)andS_(3)` are zero but the oxidation states of `S_(1)andS_(4)` are +5.
The average oxidation state is `10/4` = 2.5
c) `Fe_(3)O_(4)` contain FeO and `Fe_(2)O_(3)`
In FeO oxidation state of Fe in FeO = +2
In `Fe_(2)O_(3)` oxidation state of Fe in `Fe_(2)O_(3)` = +3
So average oxidation of Fe
`=((+2)+2xx(+3))/3=8/3=2.67`