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. `HCHO`
B. `CH_(2)Cl_(2)`
C. `CrO_(5)`
D. `C_(12)H_(22)O_(11)`

Answer

A, B, C, D

Explanation

`CH_(2)O:x+2-2=0impliesx=0`, Oxidation state of `C=0`.
b. `CH_(2)Cl_(2):x+2-2=0impliesx=0`.
Oxidation state of `C=0`
c. `C_(6)H_(12)O_(6):6x+12-12=0impliesx=0`
Oxidation state of `C=0`
`C_(12)H_(22)O_(11):12x+22-22=0impliesx=0`
Oxidation state of `C=0`