Calculate the oxidation number of C in
`C C l_(4)`

Explanation

(i) Let the oxidaton number of C be x.
`{:(x,,+1,,-2+1),(C_(2),,H_(5),,OH):}`
`2x + 5 - 2 + 1 = 0, 2x = -4`
`therefore x = -2`
(ii) Let the oxidation number of C be x.
`C Cl_(4)`
`x + (-1) xx 4 = 0" "x - 4 = 0`