What is the oxidation number of N in `HNO_4` ?

Explanation

By conventinal method O.N of N is : +1+4(-2)+x=0 or x =+7
this is impossible because N cannot have an O.N of more than + since it has only 5 electrons in valence shell Therefore O.N of N in `HNO_(4)` must be calculated by chemical bonding method the structure of `HNO_(4)` is since it contins a peroxide linkage therefore each of the two oxygen atmos of the peroxide bonds has an O.N of -1
ON of -2 thus if x is the o.n of N then +1-1-1+x-2-2=0 or x=+5