In page number 167 of 1st …

According to the formula oxidation state of  N in  HNO₃ is  +5 ,as calculated below.

Let  the oxidation state of N be  x , then applying the rules to calculate O.N. , we get

+1  +(x)   + (- 2 X 3 )  =  0

{tex}\therefore{/tex}   x   =  (  +6  - 1 )

=  +5 

We know that , N is restricted to a maximum covalency of  4 , since only  four (1s  and 3p  )  orbitals are available for bonding , and also due to the absence of a d-orbital in the atom. However , there are cases where   oxidation state of N in compounds like  viz .  N₂ O₅ ,  HNO₃    is  +5 . This is due to resonating structures  of  these compounds .  Their resonating structures well support and justify the oxidation number of N as  +5  in the compound  as follows - 

Resonating structure of HNO₃ results into formation of a coordinate bond , where N donates both the electrons to oxygen atom. 

and the double bond is formed with one of the oxygen atoms  giving  a charge as  +2 to N

A single bond with the other oxygen atom gaining a negative charge over it  gives  a charge of  +2  to  N

&  OH^- group  gives  a charge to N as  =  +1

Adding up all the charges provided to N in HNO₃  ,

Oxidation  No.  of  N  in  HNO₃   =   2+2+1  =  5 

similarly the oxidation state of N  in  N₂ O₅ can  also be  made out through its structure as given above.