What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
(a) `underlineKI_(3) " " (b) H_(2)underlineS_(4)O_(6) " "(c) underline(Fe)_(3)O_(4) " " (d) underlineCH_(3)underlineCH_(2)OH " " (e) underlineCH_(3)underlineCOOH`

Explanation

In `KI_(3)`, since the oxidation numbers 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-I larrI]^(Θ)`. Here a coordinate bond is formed between `I_(2)` molecule and `I^(Θ)` ion. The oxidation number of two iodine atoms forming the `I_(2)` molecule is zero while that of iodine forming the coordinate bond is `-1`.
b. By conventional method:
`overset(+1)(H_(2))overset(x)(S_(4))overset(-2)(O_(6)) or 2(+1)+4x+6(-2)=0`
`x=+2.5` (wrong).
But it is wrong beacuse all the four `S` atoms cannot be in the same oxidation state. By chemical bonding method, the structure of `H_(2)S_(4)O_(6)` is shown below:
`H-Ooverset(+5)(-)underset(O)underset(||)overset(O)overset(||)(S)-overset(0)(S)-overset(0)(S)overset(+5)(-)underset(O)underset(||)overset(O)overset(||)(S)-OH`
The oxidation number 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.
`overset(x)(Fe_(3))overset(-2)(O_(4))` or `3x+4(-2)=0`
or `x=8//3`
By stoichiometry `Fe_(3)O_(4) -= overset(+2)(Fe)overset(-2)(O).overset(+3)(Fe_(2))overset(-2)(O_(3))`
Fe has oxidation number of `+2` and `+3`.
d. By conventional method.
`CH_(3)CH_(2)OH= overset(x)(C_(2))overset(+1)(H_(4))overset(-2)(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)
`Hoverset(2)(-)underset(H)underset(|)overset(H)overset(|)(C )overset(1)(-)underset(H)underset(|)overset(H)overset(|)(C )-OH`
and one `CH_(2)OH` group (more electronegative than carbon).
Therefore, oxidation number of `C_(3)=3(+1)-x+1(-1)=0`
or `x=-2`
`C_(1)` is, however, attached to one `OH` (oxidation number `-1)` and one `CH_(3)` (oxidation number `=+1)`
of `C_(1)=+1+2(+1)+x+1(-1)=0` group,
`:.` Oxidation number of `C_(1)=+1+2(+1)+x+1(-1)=0`
`x=-2`
e. By conventional method:
`CH_(3)COOH=overset(x)(C_(2))overset(+1)(H_(4))overset(-2)(O_(2)) or 2x+4-4=0`
or `x=0`
By chemical bonding method:
`C_(2)` is attached to three `H` atoms (less electronegative then carbon)
`Hoverset(2)(-)underset(H)underset(|)overset(H)overset(|)(C )overset(1)(-)overset(O)overset(||)(C )-OH`
and one `-COOH` group (more electronegative tan carbon.
Therefore, oxidation number of `C_(2)=3(+1)+x+1(-1)=0`
or `x=-2`
`C_(1)` is, however, attached to one oxygen atom by a double bond, one `OH` (oxidation number `=-1)` and one `CH_(3)` (oxidation number `=+1)` group.
Therefore, oxidation number of `C_(1)=+1+x+1(-2)+1(-1)=0`
or `x=+2`