About oxidation states

We only have oxidation number for an element in a compound, not for the compound itself.It is WRONG to say that the oxidation numbers of NO and NO2 molecules are 0. We can only say that the CHARGE on NO and NO2 molecules are 0.
When we calculate the oxidation number of an element in a compound, we IMAGINE/ASSUME that ionic bonds were formed between atoms. The sum of oxidation numbers must be equal to the overall charge on the compound and the oxidation number of each element will be numerically equal to the charge that would exist on their atoms if ionic bonds were formed.
In ionic compounds, the oxidation numbers of elements forming mono-atomic ions are the same as the charges they carry.
For covalent compounds and polyatomic ions, we shall ASSUME that ionic bonds were formed between atoms and that shared electrons in covalent bonds are assigned to the more electronegative element.
In NO and NO2, O is more electronegative than N and ASSUMES that ionic bonds were formed between O and N, O will get all the shared electrons to become O2- (remember the rule : oxidation number of O is always -2 except in peroxides and when combined with fluorine).
So:
NO :
- overall charge on NO molecule = 0
- oxidation number of O = hypothetical charge on O atom = -2
- oxidation number of N = hypothetical charge on N atom
= overall charge - hypothetical charge on O atom
= 0 - (-2) = +2
NO2:
- overall charge on NO2 molecule = 0
- oxidation number of O = hypothetical charge on O atom = -2
- oxidation number of N = hypothetical charge on N atom
= overall charge - hypothetical charges on 2 O atoms
= 0 - (-2 x 2) = +4

first of all, all molecules like NO2 and NO have an oxidation number 0
if we want to know about the O.N. of N in NO2
we can calculate in this way
O.N. of N + (O.N. of O)*2 = 0
O.N. of N + (-2)*2 = 0
O.N. of N = +4
this is because O.N. of O is always -2
similarly, if we want to know about the O.N. of N in NO
we can calculate in this way
O.N. of N + O.N. of O = 0
O.N. of N + (-2) = 0
O.N. of N = +2