Ammonia is clearly a neutral molecule and its Lewis structure reflects this. Each hydrogen shares 1 electron from the N-HN−H bond to balance its nuclear charge. The nitrogen atom has 2 inner core electrons, gets 3 electrons from the N-HN−H bonds, and gets 2 electrons from the formal lone pair: 2+3+2=7e^-2+3+2=7e−. These 7 electrons are precisely balanced by the 7 protons in the nitrogen nucleus (which nuclear charge must be there if it is a nitrogen atom).
Now when ammonia is quaternized, the nitrogen is conceived to have a half share only of the the electrons in the 4xxN-H4×N−H bonds. So thus 44 electrons, + 22 inner core electrons, do not balance the +7+7 nuclear charge. Nitrogen, in ammonium ion is formally cationic. This is consistent with the stoichiometric equation.
Note that all of the N-HN−H in ammonium are equivalent; the NH_4^+NH+4 cation is a tetrahedron. What hybridization would we assign the free base and the acid?
