$\ce{BF3}$ can react with $\ce{F-}$ to give $\ce{BF4-}$ where we have for $\ce{B}$ in this molecule the 2s orbital and the three 2p orbitals filled totally (hybrid orbitals).

In case of $\ce{NH3}$ we have the 2s orbitals and the three 2p orbitals of $\ce{N}$ filled;

So i am asking this: can the hydride ion $\ce{H-}$ make a dative bond with the 3s orbital of $\ce{N}$ (which is vacant) to form $\ce{NH4-}$? If not, why?

  • 5
    $\begingroup$ The simple answer is that the 3s orbitals of nitrogen are too high up in energy. You could theoretically have such a species exactly as how you described. But the chances are that it would have so much energy that it would simply fly apart. $\endgroup$ Jan 20, 2016 at 15:24
  • $\begingroup$ You mean the dative bond will not occur since the overlapping of the 3s orbital of NN with the 1s orbital of HH needs high energy?@orthocresol $\endgroup$
    – YAZO
    Jan 20, 2016 at 21:04
  • $\begingroup$ Why the comparison with boron and nitrogen? or ammonia and boranes? $\endgroup$
    – jimchmst
    Jan 10 at 20:00

2 Answers 2


The 3s orbitals are far too high in energy to participate in chemical bonding interactions. Lower in energy are the empty σ* antibonding orbitals of the N–H bonds, which can still be energetically relevant in further reaction chemistry. As a thought experiment, it might be possible for a hydride anion to donate electron density into an empty N–H σ* orbital in a sort of Lewis acid-Lewis base interaction. In fact, this is a hallmark of so-called `hypervalent' bonding of heavier main group elements. Rather than using d-orbitals (as is often incorrectly taught to intro chem students), these acid-base interactions are how the phosphorus(V) center of PF$_5$ bonds to the fifth fluoride ion, forming what is known as a three-center four-electron bond.

Why isn't this type of bonding seen with NH$_3$ and H$^–$ to form NH${_4}^-$? These `hypervalent'-type interactions involving 3c-4e bonds depend on the outer atoms having a high-enough electronegativity to stabilize a positive charge at the central atom. In the case of the hypothetical NH${_4}^-$ anion, this just isn't the case. Hydrogen is in fact far less electronegative than nitrogen.

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    $\begingroup$ So we might expect an interaction through the hydrogen end of the ammonia dipole. As illustrated in another answer, that interaction is known in a more limited form. $\endgroup$ Jan 9 at 0:58

Among isolable species, arguably the closest thing there is to an ammonium anion can be seen in ammine magnesium borohydride, $\ce{Mg(BH4)2•2NH3}$[1]. In this compound hydridic (negatively charged) hydrogen is dihydrogen-bonded to the protic hydrogen in the ammonia. In the molecular-orbital rendering of hydrogen bonding, the electrons from the boron-hydrogen bond overlap with the formally vacant antibonding orbitals from the ammonia (both of these being orbitals polarized towards hydrogen) to delocalize the bonding into the "intermolecular" region. No contribution from outer orbitals on the nitrogen are involved as this is a hydrogen-hydrogen interaction.

As with other cases of hydrogen bonding, the electron source (borohydride hydrogen) remains bonded primarily bonded within its own "molecule" (meaning to boron); the dihydrogen bond is only a secondary interaction. Were the boron-hydrogen bond to be broken, for instance by by thermal decomposition, the hydridic and protic hydrogens would simply combine to form $\ce{H2}$ (hydrogen evolution on heating is the intended result in the referenced study).

Transient ammonia-hydride species have been identified using photoelectron spectroscopy. Snodgrass et al[2] identify $\ce{H^-(NH3)}$ and $\ce{H^-(NH3)2}$, which are considered hydride ions solvated by ammonia molecules and thus would involve protic-hydridic interactions similar to that in the complex borohydride salt described above. They also found a peak attributed to a tetrahedral $\ce{NH4^-}$ isomer in which all the hydrogens are positive; an ordinary ammonium ion accompanied by two Rydberg electrons. Diaz-Tonoco and Ortixz[3] give ab initio calculations supporting the Rydberg configuration and extend the concept to higher $\ce{N_mH_{3m+1}}^-$ species.


  1. Grigorii Soloveichik, Jae-Hyuk Her, Peter W. Stephens, Yan Gao, Job Rijssenbeek, Matt Andrus, and J.-C. Zhao (2008). "Ammine Magnesium Borohydride Complex as a New Material for Hydrogen Storage: Structure and Properties of Mg(BH4)2·2NH3". Inorg. Chem. 47, 10, 4290–4298. https://doi.org/10.1021/ic7023633

  2. Joseph T. Snodgrass, James V. Coe, Carl B. Freidhoff, Kevin M. McHugh and Kit H. Bowen (1988). "Photodetachment spectroscopy of cluster anions. Photoelectron spectroscopy of H–(NH3)1, H–(NH3)2 and the tetrahedral isomer of NH4-". Faraday Discuss. Chem. Soc., 86, 241-256, https://doi.org/10.1039/DC9888600241.

  3. Manuel Díaz-Tinoco, J. V. Ortiz (3019). "Double Rydberg anions with solvated ammonium kernels: Electron binding energies and Dyson orbitals ". _J. Chem. Phys. 151, 054301. https://doi.org/10.1063/1.5113614.


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