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 carbon is zero?

Options

A. `CH_(4)`
B. `CH_(3)OH`
C. `HCOOH`
D. `C_(6)H_(12)O_(6)`

Answer

D

Explanation

a. `overset(-4+1xx4)(CH_(4))` oxidation state of `C=4-4`
b. `CH_(3)OH: x+4-2=0 impliesx= -2`
c. `HCOOH:x+2-4 =0implies x=2`
d. `C_(6)H_(12)O_(6): 6x+12-12=0 implies x=0`.