$\ce{BF3}$ 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?

  • 4
    $\begingroup$ 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. $\endgroup$
    – orthocresol
    Jan 20 '16 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 '16 at 21:04
  • $\begingroup$ There was no need for you to remove your newer questions imo, duplicates also have their place here. $\endgroup$
    – Mithoron
    Jan 22 '16 at 1:16
  • $\begingroup$ I lost reputation because of them plus the links which were attached to my posts did not answer my questions and satisfy me @Mithoron $\endgroup$
    – YAZO
    Jan 22 '16 at 17:48

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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