What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your result?
(a) `KI_(3)`
(b) `H_(2)S_(4)O_(6)`
(c)`Fe_(3)O_(4)`
(d) `CH_(3)CH_(2)OH`
(e)`CH_(3)COOH`

Explanation

(A) in `KI_(3)` since the oxidation number of K is +1 therefore the average oxidation number of iodine =-`1//3` but the oxidation number cannot be fractional therefore we must consider its structure `K^(+)[I-lrarr]^(-)` here a coordinate bond is formed between `I_(2)` moleceul and `T^(-)` ion the oxidation number of two iodin e atoms forming the `IO_(2)` molecure is zero while that of iodine forming the coordinate bond is -1 thus the O.N of there I atoms atoms in `KI_(3)` are 0,0 and -1 respectively
(b) By conventional method O.N of S in `H_(2)S_(4)O_(6) =overset((+1)H_(2)overset(x)S_(4)overset(-2)O_(6) or 2(+1)`
(b) By conventional method O.N of S in `H_(2)S_(2)O_(6)=H_(2)S_(4)O_(6)`
or 2(+1) +4 x+6 (-2) =0 or x=+2.5 (wrong)
but it is wrong because all the foiu S atoms cannot be int he same oxidation state by chemical bonding method the structure of `H_(2)S_(4)O_(6)` is show below
`H-O overset(+5)-underset(O)underset(||)overset(O)overset(||)S-overset(S)-overset(0)S-underset(O)underset(||)overset(O)overset(||)Soverset((+5)-OH`
The O.N of each of the S atoms linked with each other in the middle is zero while that of each of the remaining two S atoms is +5
(c ) By conventional method O.N of Fe in `Fe_(3) O_(4)^(-2)` therefore of Fe in `Fe_(3)O_(4)^2` and +3
(d) By conventional method O.N of Cin `H_(3)CH_(2)OH =C_(2)H_(6)O`
or 2x+6(+1)+1 (-2) =0 or x=-2
by chemical bonding `C_(2)` is attached to three H atoms (less electronegative than carbon ) and one `CH_(2)OH` group (more electronegative than carbon) therefore
`H overset(2)-underset(H)underset(|)overset(H)overset(|)C-underset(H)underset(|)overset(H)overset(|)C-OH`
O.N of `C_(2)=3(+1)+x1(-1)=0 or x=-2`
`C_(1)` is however attached ot one OH (O.N =-1) and one `CH_(3)`(O.N)=+1group
(e ) By conventional method `CH_(3)COOH=C_(2)H_(4)O_(2) or 2x+4-4=0 or x=0`
by chemical bonding method `C_(2)` is attached to three h atoms and one -COOH group (more electronative than carbon)
`H-underset(H)underset(|)overset(H)overset(|)C-overset(O)overset(||)C-OH` n
`C_(1)` is however attached to one oxygen atom by double bond one OH(O.N =-1) and one `CH_(3)(O.N=+1)` group therefore therefore O.N of `C_(1) =+1+x+1(-2)+1 =0 OH or x=+2`