Why do we put a formal charge on nitrogen in ammonium cation?

Sep 5, 2016

$N {H}_{3} \left(a q\right) + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s N {H}_{4}^{+} + H {O}^{-}$

Are we agreed that the above reaction is balanced with respect to mass and charge?

Explanation:

Ammonia is clearly a neutral molecule and its Lewis structure reflects this. Each hydrogen shares 1 electron from the $N - H$ bond to balance its nuclear charge. The nitrogen atom has 2 inner core electrons, gets 3 electrons from the $N - H$ bonds, and gets 2 electrons from the formal lone pair: $2 + 3 + 2 = 7 {e}^{-}$. 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 $4 \times N - H$ bonds. So thus $4$ electrons, + $2$ inner core electrons, do not balance the $+ 7$ nuclear charge. Nitrogen, in ammonium ion is formally cationic. This is consistent with the stoichiometric equation.

Note that all of the $N - H$ in ammonium are equivalent; the $N {H}_{4}^{+}$ cation is a tetrahedron. What hybridization would we assign the free base and the acid?