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

(a) O.N of iodine in `KI_(3)`: By conventional method, the oxidaiton number of iodine in the `I_(3)^(-)` ion may be expresed as:
`overset(+1)Koverset(x)I_(3)`
`1+3x=0 ` or `x= -1//3`
Explanation: The oxidation number of iodine (I) comes out to be fractional which doies not seem tto be possible.Let us consider the structure `I_(3)^(-)` ion).In it , two atoms of iodine are linked to each other by covalent bond (I-I).The iodide ion `(I^(-))` is linked to the molecule by co-ordinate bond `[I-larrI]^(-)`. The molecule may be represented as `K^(+)[I-larrI]^(-)` Now, inthe anion, the oxidation number of the two I atoms is zero While `I^(-)` ion has -1 oxidation number.The over all oxidation number of `I_(3)^(-)` ion is :
`1//3[overset(0)I-overset(0)Il rarr overset(-1)I]=-1//3`
`(b) O.N of S in H_(2)S_(4)O_(6)`: By conventinal method, the oxidation number of sulphur may be calculated as:
`overset(+1)H_(2)overset(x)S_(4)overset(-2)O_(6)`
`2+4x+6(-2)=0 or 4x=12-2=10 or x=5//2`
Explanation: In order to acount for the fractional value,let us assing oxidation numbers to different sulphut atoms in the structural formula of the acid. The oxidation number of the two middle suplhur atoms is zero while the two atoms at the terminal positions have + 5 oxidtion number

`therefore` Average O.N opf `S= (1)/(4)[5+0+0+5]=5//2`
(c) O.N of Fe in `Fe_(3)O_(4)`: By conventional method, oxidation number of Fe may be claclulated as:
`overset(x)Fe_(3)overset(-2)O_(4)`
3x+4(-2)=0 or x= `8//3`
Explanation :`Fe_(3)O_(4)` is la mixed oxide and is an equimolar mixture of `overset(+2)Feoverset(-2)O and overset(+2)Feoverset(-2)O and overset(+3)Fe_(2)overset(-2)O_(3)`
`therefore Average O.N of Fe=1/3(2+2x3)=8//3`
(d) O.N Of C in `CH_(3)CH_(2)OH`: by conventinal method, the oxidation nuimber of carbon in ethanol molecule may be calculated as:
`CH_(3)CH_(2)or overset(x)C_(2)overset(+1)H_(6)overset(-2)O`
`2x+6(+1)+(-2)=0 or 2x+6-2=0 or 2x=-4 or x=-2`