2
$\begingroup$

Why ammonium ion has very similar properties to the heavier alkali metals and is often considered a close relative?

Ammonium is expected to behave as a metal ($\ce{NH4+}$ ions in a sea of electrons) at very high pressures. Can anyone explain?

$\endgroup$
  • $\begingroup$ I'm confused about the charge here. Wouldn't the ammonium cation behave like a group I cation, and the ammonium radical ($\ce{NH4}$, neutral molecule) like elemental group 1 metals? Reference $\endgroup$ – Karsten Theis Mar 9 at 14:51
2
$\begingroup$

Who needs high pressure? Ammonium can be amalgamated into mercury like an alkali metal. As with metals generally, there are not really any ammonium molecules in this phase; rather, the ammonium cations are intermingled with mercury cations and delocalized valence electrons.

Nor is all that high a pressure needed for unamalgamated ammonium, relative to metallizing other materials. Bernal and Massey (1954) reported that a pressure of 250 kbar would be sufficient to metallize an ammonia-hydrogen mixture, which may be compared with the rocky material deep in Earth's mantle remaining nonmetallic even at megabar pressures.

Ammonium is easy to metallize because it has, indeed, an alkali-metal electronic structure. The tetrahedra structure common to ammonium and methane, $\ce{CH4}$, is most stable with eight valence electrons -- there are four bonding orbitals, corresponding to the four bonds in methane, and four antibonding orbitals. In methane the neutral molecule matches this optimal electron count, but in ammonium the neutral molecule would have to include a ninth valence electron in its antibonding orbitals, like an alkali metal atom having one to many electrons for a stable noble gas configuration. Either this extra electron breaks down the proposed ammonium molecule (which is what we ordinarily see) or, under moderately high pressure or in a metallically bonded matrix like the amalgam, it goes into the metallic bonding.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.