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` a fraction?

Options

A. `CO`
B. `CO_(2)`
C. Carbon suboxide
D. All

Answer

C

Explanation

Highest oxidation state of `C` is in (a)
`C_(3)O_(2)` (carbon suboxide): `3x-4= 0 implies x=(4)/(3)`