What is the oxidation number for nitrogen in each of the following?
Answer:
In a free element (unreacted with anything, or diatomic) the oxidation number is always 0.
Oxygen in a compound always has an oxidation number of -2. The oxidation number for nitrogen combined with oxygen will then be [0 - (-2)*number of O atoms].
In a charged particle like NO2- (I'm guessing that this is nitrite - hard to tell without sub or superscripts) it would be [0 - (-2)*number of O atoms] + (ionic charge).
For example; in SO2(2-); the oxidation number for S is going to be;
0 - 2 (number of O) *(-2(oxidation number for O in compound) + -2 (ionic charge) = +2
N2 is covalent, they are sharing
NO, O=-2, N=+2
NO2, O=2 x2=-4, N=+4
N2O, O=-2, N=+1 each
NO2-, O=-2 x 2=-4, N=+3
In N2, the oxidation number of nitrogen is 0 since mitrogen gas is uncharged.
In NO the oxidation number os nitrogen is +2. This is because the overall charge on NO is 0. Thus to balance the -2 oxidation number of oxygen, the oxidation number of nitrogen must be +2.
Using the same principles,
the oxidation number of N
in NO2 is +4,
in N2O is +1,
in NO2- is 0.
Compounds Oxidation number of nitrogen in that
N2 0
N2O +1
NO +2
NO2 +4
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