8.2 What are the oxidation number of the underlined elements in each of the following and how do you rationalize your results
O.N of potassium ()= +1
O.N of hydrogen()= +1 (In case of metalic hydride, -1)
O.N of Oxygen() = -2 ( In case of peroxide and superoxide it wil be different ON)
1*1 + 3*x = 0
x = (-1/3)
average O. N. Of is . But it is wrong because O.N cannot be fractional. So lets try with structure of
O.N of = -1 (because a coordinate bond is formed between and ion. Hence O. N of three atoms are 0,0 and -1, O.N 0 in moleculeand -1 in ion.)
Assume O.N of S is x
Fractional O.N is not possible so try with structure -
The two S atom present in the middle has zero (0) O.N and other two have (+5) O.N [ -2 from two O atom and -1 from OH]
If you calculate the oxidation number of Fe in it would be 8/3 and however, O.N cannot be in fractional.
Here one iron atom has +2 O.N and the other two are of +3 O.N.
let assume carbon has x oxidation Number
So,[ x + 1(3) +x +1(2) +(-2)+1 = 0]
2x = -4
In this molecule two carbon atoms present in different enviroments. Hence, they cannot have the same O.N.Thus, C exhibits the O. S of -3 and -1.
suppose the oxidation number of Carbon is x.
If we calculate the O.N of x we get x=0
However, 0 is average O.N. of C atoms. In this molecule two carbon atoms present in different enviroments. Hence, they cannot have the same O.N.Thus, C exhibits the O.S of +3 and –3 in CH3COOH This can be more understood by structure-
Here we can see that at right C, +3 O.N (-1 from OH and -2from O atom) and in left C, -3 O.N(contribution from H atom only)