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.