For #H_3CNO_2#, we have #3xx1(H)+4(C)+5(N)+2xx6(O)=24# #"valence electrons"# to distribute over 7 centres.
A Lewis representation of #H_3C-stackrel(+)N(=O)(-O^(-))# is standard, and this distributes the 12 electron pairs. Because there are 6 electrons formally associated with nitrogen (4 from the bonds, and 2 inner core), this centre is said to be #"quaternized"#, and thus bears a formal positive charge (7 electrons are associated with the neutral nitrogen atom). The anionic oxygen is associated with 9 electrons (2 inner core) instead of the eight of neutral oxygen, and so bears a formal negative charge. Of course, we can distribute this negative charge to the other oxygen by resonance.
Is this what you want? I have a feeling you asked a question whose answer I have not addressed.
