The valancy of carbons is generally `4`, but its oxidation state may be `-4, -2, 0, +2, -1`, etc. In the compounds containing `C, H`, and `O`, the oxidation number of `C` is calculated as
Oxidation number of `C= (2n_(O)-n_(H))/(n_(C ))`
Where `n_(O), n_(H)` and `n_(C )` are the numbers of oxygen, hydrogen, and carbons, atoms, respectively.
In which of the following compounds is the oxidation state of `C` highest?

Options

A. `HCOOH`
B. `HCHO`
C. `CH_(3)OH`
D. `CH_(4)`

Answer

A

Explanation

a. `overset(+2)(HCOOH)`
b. `overset(0)CH_(2)O:x+2-2=0 impliesx=0`
c. `overset(-2)(CH)_(3)OH:x+4-2=0impliesx= -2`
d. `overset(-4+1xx4)(CH_(4))`: oxidation state of `C= -4`.