Suggest a list of the substances where carbon can exhibit oxidation states from–4 to + 4 and nitrogen from –3 to + 5.
The substances where carbon can exhibit oxidation states from -4 to + 4 are following:-
Let the O.N. of C be ‘x’
Substances | O.N. of carbon | Calculation |
CH4 | -4 | x + 4(+ 1) = 0 x = -4 |
H3C-CH3 | -3 | 3(1) + x + x + 3(1) = 0 X = -3 |
CH3Cl | -2 | x + 3(+ 1) + 1(-1) = 0 x = -2 |
HC = CH | -1 | 1 + x + x + 1 = 0 X = -1 |
CH2Cl2 | 0 | x + 2(1) + 2(-1) = 0 x = 0 |
ClC = CCl | + 1 | 1(-1) + x + x + 1(-1) = 0 X = + 1 |
CO | + 2 | x + 1(-2) = 0 x = + 2 |
Cl3C-CCl3 | + 3 | 3(-1) + x + x + 3(-1) = 0 X = + 3 |
CCl4 | + 4 | x + 4(-1) = 0 x = + 4 |
The substances where Nitrogen can exhibit oxidation states from -3 to + 5 are following:-
Let the O.N. of N be ‘y’
Substances | O.N. of N | Calculations |
NH3 | -3 | y + 3(1) = 0 y = -3 |
N2H4 | -2 | 2y + 4(1) = 0 y = -2 |
N2H2 | -1 | 2y + 2(1) = 0 y = -1 |
N2 | 0 | 2y = 0 y = 0 |
N2O | + 1 | 2y + 1(-2) = 0 y = + 1 |
NO | + 2 | y + 1(-2) = 0 y = + 2 |
N2O3 | + 3 | 2y + 3(-2) = 0 y = + 3 |
NO2 | + 4 | y + 2(-2) = 0 y = + 4 |
N2O5 | + 5 | 2y + 5(-2) = 0 y = + 5 |