What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
a.`Kul(I)_(3)` b. `H_(2)ul(S)_(4)O_(6)` c. `ul(Fe)_(3)O_(4)`
d. `ul(C )H_(3)ul(C )H_(2)OH`
e. `ul(C )H_(3)ul(C )OOH`

Explanation

`KunderlineI_(3)`
In `KI_(3)` , the oxidation number (O.N.) of K is +1 . Hence the average oxidation number of I is `-(1)/(3)` . However , O.N. cannot be fractional . Therefore , we will have to consider the structure of `KI_(3)` to find the oxidation states .
In a `KI_(3)` molecule , an atom of iodine forms a coordinate covalent bond with an iodine molecule .
`overset(+1)(K^(+)) [ overset(0)(I) - overset(0)(I) larr overset(-1)(I)]^(-)`
Hence , in `KI_(3)` molecule , the O.N. of the two I atoms forming the `I_(2)` molecule is 0 , whereas the O.N. of the the I atom forming the coordinate bond is `-1` .
(b) `H_(2)underlineS_(4) O_(6)`
` overset(+1)(H_(2))overset(x)(S) O_(4) overset(-2)(O_(6))`
Now , `2(+1) + 4(x) + 6 (-2) = 0`
`implies 2 + 4x - 12 = 0`
`implies 4x = 10 `
`implies x = +2(1)/(2)`
However , O.N. cannot be fractional . Hence , S must present in different oxidation states in the molecule .
`H-O-overset(O)overset(||)underset(O)underset(||)(S)overset(+5)(-) overset(O)(S)-overset(O)(S)-overset(O)overset(||)underset(O)underset(||)(S)overset(+5)(-)O-H`
The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is O .
`underline(Fe)_(3)O_(4)`
On taking the O.N. of O as `-2` , the O.N. of Fe is found to be `+2(2)/(3)` . However , O.N. cannot be fractional .
Here , one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3 .
`overset(+2)(F)eO, overset(+3)(F) e_(2)O_(3)`
(d) `underlineCH_(3) underlineCH_(2)OH`
`overset(x)(C_(2))overset(+1)(H_(6)) overset(-2)(O)`
`2(x) + 4(+1) + 1(-2) = 0 `
`implies 2x + 6 -2 = 0`
`implies x = -2`
Hence , the O.N. of C is `-2` .
(e) `underlineCH_(3)underlineCOOH`
`overset(x)(C_(2))overset(+1)(H_(4)) overset (-2)(O_(2))`
`2(x) + 4(+1) + 2(-2) = 0`
`implies 2x + 4 -4 = 0`
`implies x = 0`
However , 0 is average O.N. of C . The two carbon atoms present in this molecule are present in different environments . Hence , they cannot have the same oxidation number .
Thus , C exhibits the oxidation states of +2 and `-2` in `CH_(3)COOH`.