What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
a) `H_(2)underline(S)_(4)O_(6)" "b)underline(F)e_(3)O_(4)`
c) `underline(C)H_(3)underline(C)H_(2)OH" "d)underline(C)H_(3)underline(C)OOH`

Explanation

`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`
b) `Fe_(3)O_(4):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`
c) `underline(C)H_(3)-underline(C)H_(2)-OH`
`H-underset(H)underset(|)overset(H)overset(|)(C)-underset(H)underset(|)overset(H)overset(|)(C)-O-H`
When atoms of the same element combine their oxidation states are taken as zero.
The carbon in `CH_(3)` group is in -3 oxidation state.
The carbon in `CH_(2)OH` group is in zero oxidation state.
d) `underline(C)H_(3)underline(C)OOH`
`H-underset(H)underset(|)overset(H)overset(|)(C)-C-overset(O)overset(||)O-H`
The carbon in `CH_(3)` is in -3 oxidation state while the carbon in COOH group is in `+3 oxidation state.