Why does ammonia accept a hydrogen to form ammonium when its octet is already complete as $\ce{NH3}$ and it should be, seemingly, stable as $\ce{NH3}$?
$\begingroup$
$\endgroup$
2
-
2$\begingroup$ It is the same reason why $\ce{H3O+}$ exists (formed by $\ce{H2O}$ accepting a $\ce{H+}$ from stronger acid). $\endgroup$– Mathew MahindaratneCommented Apr 26, 2018 at 5:52
-
8$\begingroup$ It is stable and happy as NH3. The problem is that H+ is very unhappy. $\endgroup$– Ivan NeretinCommented Apr 26, 2018 at 6:00
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
"Hydrogen" means different things at different times. It can mean $\ce{H+}$, $\ce{H.}$, or $\ce{H2}$. (Not to mention deuterium and tritium!)
A bare proton ($\ce{H+}$) does not bring an electron to the filled octet on nitrogen, but attaches to the lone pair. The positive charge is now shared over 5 atoms, and the energy of formation of a new bond stabilizes the ammonium cation. The octet on nitrogen is still only an octet.
Interestingly, a hydrogen atom ($\ce{H.}$) can also attach to an ammonia molecule to form ammonium radical. Electrolysis of ammonium salts with a mercury cathode yields ammonium amalgam, which rapidly decomposes to $\ce{NH3}$ plus $\ce{H2}$. In this case, the extra electron in $\ce{NH4.}$ is also shared among the 5 atoms, and is more stable than when it is attached to just one proton. But not that much more stable, and the bond between two hydrogens is even stronger.
answered Apr 27, 2018 at 4:11