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In general chemistry we often encounter $\ce{H3PO4}$, which got me wondering if I can replace the $\ce{P}$ in $\ce{H3PO4}$ with $\ce{N}$ giving us $\ce{H3NO4}$ , does such a compound exist?

Both phosphorous and nitrogen are in group 15 in the periodic table implying they have same number of valence electrons making the lewis structures for both $\ce{H3NO4}$ and $\ce{H3PO4}$ the same with octet configuration.

Orthonitrate ion and phosphate ion

I do notice compounds exist which do not follow the octet rule (eg: $\ce{S}$ in $\ce{SO3}$) which makes the octet rule wrong but still why do we not find $\ce{NO4-}$ ions in nature?

Possibly helpful info: As far as I know $\ce{H3NO4}$ has never been observed in real life but $\ce{Na3NO4}$ has been observed.

Formation of Na3NO4

The compound forms white crystals that are very sensitive to atmospheric moisture and $\ce{CO2}$:

Decomposition of Na3NO4

X-ray structural analyses have shown that the $\ce{NO4^3-}$ ion has regular $T_d$ (op note: $T_d$ means tetrahedral I think) symmetry with the unexpectedly small N - O distance of 139pm. This suggests that substantial polar interactions are superimposed on the N - O single bonds since the $d_\pi$ orbitals on N are too high in energy to contribute significantly to multiple covalent bonding. It further implies that $d_\pi$-$p_\pi$ interactions need not necessarily be invoked to explain the observed short interatomic distance in the isoelectronic oxoanions $\ce{PO4^3-}$, $\ce{SO4^3-}$ and $\ce{ClO4^-}$. [source, page 472]

But how did they observe $\ce{Na3NO4}$ was present? In the above book's link a source was mentioned: https://onlinelibrary.wiley.com/doi/epdf/10.1002/anie.197705341. This source's abstract says

The existence of the $\ce{NO4^3-}$ ion has now been proved by Raman spectroscopy. X-ray structure analysis has so far been precluded by the non-availability of single crystals of the extremely moisture- and $\ce{CO2}$-sensitive compound $\ce{Na3NO4}$.

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  • $\begingroup$ Orthonitrate exist but not in nature due to its instability. Nitrogen can form minimalized octets but it rarely forms expanded octets. Sulfur and Phosphorus on the other hand more readily form expanded octets and so sulphate and phosphate are stable $\endgroup$
    – ChemLover68
    Commented Sep 2 at 7:32
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    $\begingroup$ Related: chemistry.stackexchange.com/questions/136084/… $\endgroup$
    – Buck Thorn ♦
    Commented Sep 2 at 8:09
  • $\begingroup$ The parallel formation of NaOH and NaHCO3 instead of Na2CO3 + H2O seems suspicious $\endgroup$
    – Maurice
    Commented Sep 2 at 9:35
  • $\begingroup$ @poutnik maybe it works. Assuming that sodium hydtixide is first formed, it would form a nucleophilic adduct with the carbon dioxide which is the bicarbonate. But whether this reacts with more sodium hydroxide to give sodium carbonate would depend on transferring protons in the solid state. $\endgroup$
    – Oscar Lanzi
    Commented Sep 2 at 10:53
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    $\begingroup$ To the OP: please edit and pick one question. Also, the answer you had posted did not answer your own question and included new questions (reason for merging). $\endgroup$
    – Buck Thorn ♦
    Commented Sep 2 at 11:12

1 Answer 1

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Modern molecular orbital theory posits that in tetrahedral species with terminal oxygen atoms, pi bonding of the central atom to oxygen exists because formal "lone pairs" on the oxygen overlap sigma-antibonding orbitals between the central atom and its other neighbors. But this seems to work better in phosphate ion and in another relatively stable species, trifluoramine oxide ($\ce{ONF3}$), than in orthonitrate ion.

Since trifluoramine oxide has the same central atom as orthonitrate, we can compare pi back-donation in these two species via nitrogen-oxygen bond lengths and illustrate the value of having strongly polar sigma bonds as the pi acceptor. While the reference in the question points out that the $\ce{N – O}$ bond is shortened to 139 pm in orthonitrate, Wikipedia citing [1] reports that in trifluoramine oxide it is only 115.8 pm. The pi back-donation is clearly stronger in the more stable compound.

For pi back-donation in a tetrahedral molecule to work well, the sigma bonds should be highly polarized so that the antibonding orbitals lie mostly on the central atom where the overlap occurs. The $\ce{N – F}$ sigma bonds in trifluoramine oxide appear polar enough to promote pi back-donation from oxygen to the central atom, and (given the lower electronegativity of phosphorus), the same would be true of the $\ce{P – O}$ bonding in phosphate. But the $\ce{N – O}$ sigma bonds would be less polar, and so is a poorer pi acceptor from other $\ce{N – O}$ bonds than either the $\ce{N – F}$ bonds in trifluoramine oxide or the $\ce{P – O}$ bonds in phosphate ion.

Cited reference

  1. Plato, Vernon; Hartford, William D.; Hedberg, Kenneth (November 1970). "Electron-Diffraction Investigation of the Molecular Structure of Trifluoramine Oxide, F3NO". The Journal of Chemical Physics. 53 (9): 3488–3494. doi:10.1063/1.1674522.
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  • $\begingroup$ In "tetrahedral species with terminal oxygen atoms, pi bonding of the central atom to oxygen exists because formal "lone pairs" on the oxygen overlap sigma-antibonding orbitals " why isn't the bolded text 'bonding orbital' because a bonding orbital is lower in energy, stabilising the pi back donated electrons more than antibonding? Or does it turn out that you have to fill antibonding orbitals in this example as they are the last left to fill? Let me try to make the diagram. $\endgroup$
    – Saif
    Commented Sep 3 at 4:19
  • $\begingroup$ I was only familiar with orbital diagrams for diatomic molecules, so I tried learning for polyatomic molecules eg: CO2. As I understand, the bond making process is the same, we can use the same molecular orbitals as we used in diatomic molecules and fill it in with all the electrons in the polyatomic molecule. I also assume the molecular orbital energies for 3rd shell follows the same pattern as 2nd shell, as I needed 3rd shell to fill all the electrons in the MO diagram for NO4-. It went all the way till sigma star 3p, I'll put a picture of this in the question, I think this is wrong because $\endgroup$
    – Saif
    Commented Sep 3 at 4:39
  • $\begingroup$ because 3rd shell is involved here, what went wrong. Though I am getting the last electrons are put in antibonding which fits your statement. $\endgroup$
    – Saif
    Commented Sep 3 at 4:40
  • $\begingroup$ Oh I forgot non bonding orbitals for the lone pairs, let me fix it. But which electrons do I choose to go into the antibonding orbital? $\endgroup$
    – Saif
    Commented Sep 3 at 4:57
  • $\begingroup$ "For pi back-donation in a tetrahedral molecule to work well, the sigma bonds should be highly polarized so that the antibonding orbitals lie mostly on the central atom where the overlap occurs." Is the antibonding orbital's electrons pulled closer to the ligand and away from the central atom, stabilising the pi back bond electrons coming to the central atom as they don't have to be in an antibonding orbital? So you are saying NO4- is less stable because of poor pi back donation while phosphate and trifluoroamide is stable due to good pi back donation? $\endgroup$
    – Saif
    Commented Sep 3 at 5:45

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