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.


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