Find out the oxidation number of (i) C in `CO_2` (ii) Cr in `Cr_2O_7^(2-)` (iii) Pb in `Pb_3O_4`.

Explanation

(i) Let x be the oxidation number of C in `CO_2`
Oxidation number of each O atom = -2
Sum of oxidation number (O.N) of all atoms in `CO_2 = x + 2 (-2) = x - 4`
As `CO_2` is a neutral molecule, `x - 4 = 0` or `x + 4`
As `CO_2` is a neutral molecule, `x - 4 = 0` or `x = +4`
(ii) Let x be the oxidation number of Cr in `Cr_2O_7^(2-)`.
O.N. of O atom = `-2, Cr_2 o_7^(2-) = 2x + 7(-2) = 2x - 14`
`2x - 14 = -2 or x = +6`
(iii) Let x be the oxidation number of Pb in `Pb_3O_4`
O.N. of O atom = `-2, 3x + 4(-2) = 0` or `3x - 8` or `x = 8//3`