What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
CH3CH2OH
C2H6O
Let the oxidation number(O.N.) of C be ‘x’
We know that,
O.N. of H = + 1
O.N. of O = -2
Therefore , we have,
2(x) + 4(+ 1) + 1(-2) = 0
2x + 6-2 = 0
X = -2
The O.N. of C is -2.