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In the MO diagram of $\ce{N2^2-}$, does s-p mixing happen to a significant degree to change the ordering of orbitals - $\pi_\mathrm{2p}$ orbital below $\sigma_\mathrm{2p}$? As it is isoelectronic with oxygen, I think that the s-p mixing shouldn't change the ordering.

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In order to test whether there may be $\ce{s}$-$\ce{p}$ mixing and also to check whether there may be actually a bound state, I optimised this molecule (at PBE0-D3/def2-TZVPD) using NWChem 6.6 and analysed it using natural bonding orbitals (NBO 5.9). I found following minimum geometry:

N2-2

Figure 1: Minimum geometry found for $\ce{N2^{-2}}$ (distance in angstrom)

So there seems to be a bound state.

Natural charge on each $\ce{N}$ is -1.0 as would be expected (natural electron configuration was found to be $\ce{1s^{2.00} 2s^{1.65} 2p^{4.17} 3s^{0.06} 3p^{0.05} 3d^{0.07} 4d^{0.01}}$). Wiberg bond order in the NAO basis for the $\ce{N-N}$ bond was found to be 2.2511.

Two bonding NBOs were found:

    (Occupancy)   Bond orbital/ Coefficients/ Hybrids
-------------------------------------------------------------------------------
  1. (2.00000) BD ( 1) N  1- N  2
              ( 50.00%)   0.7071* N  1 s(  0.00%)p 1.00( 99.74%)d 0.00(  0.24%)
                                                 f 0.00(  0.02%)
              ( 50.00%)   0.7071* N  2 s(  0.00%)p 1.00( 99.74%)d 0.00(  0.24%)
                                                 f 0.00(  0.02%)
  2. (2.00000) BD ( 2) N  1- N  2
              ( 50.00%)   0.7071* N  1 s( 36.78%)p 1.70( 62.66%)d 0.01(  0.51%)
                                                 f 0.00(  0.06%)
              ( 50.00%)   0.7071* N  2 s( 36.78%)p 1.70( 62.66%)d 0.01(  0.51%)
                                                 f 0.00(  0.06%)

That is,

$$\pi_{\ce{N1-N2}} = 0.7071 \ce{p}_{\ce{N1}} + 0.7071 \ce{p}_{\ce{N2}}$$

pi bond

Figure 2: $\pi_{\ce{N1-N2}}$ bond of $\ce{N2^{-2}}$

$$\sigma_{\ce{N1-N2}} = 0.7071 \ce{sp^{1.70}}_{\ce{N1}} + 0.7071 \ce{sp^{1.70}}_{\ce{N2}}$$

sigma bond

Figure 3: $\sigma_{\ce{N1-N2}}$ bond of $\ce{N2^{-2}}$

Furthermore,

NATURAL BOND ORBITALS (Summary):

                                                    Principal Delocalizations
          NBO                 Occupancy    Energy   (geminal,vicinal,remote)
===============================================================================
Molecular unit  1  (N2)
  1. BD ( 1) N  1- N  2        2.00000     0.18488
  2. BD ( 2) N  1- N  2        2.00000    -0.43475  10(g),42(g)

Thus, according to the simple Lewis picture provided by NBO,

  1. $\ce{s}$-$\ce{p}$ mixing ($\ce{sp^{1.70}-sp^{1.70}}$) happens for the formation of $\sigma_{\ce{N1-N2}}$, as expected;
  2. the formation of $\pi_{\ce{N1-N2}}$ happens through pure $p$ hybrids; and
  3. $\pi_{\ce{N1-N2}}$ lies in energy above $\sigma_{\ce{N1-N2}}$ by 0.61963 a.u..
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    $\begingroup$ According to "Experimental Search for the Smallest Stable Multiply Charged Anions in the Gas Phase" Physical Review Letters vol. 83, page 3402: "It has also been concluded that no diatomic or triatomic dianions are stable" casey.brown.edu/chemistry/research/LSWang/publications/84.pdf $\endgroup$ – DavePhD Mar 7 '17 at 12:09
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    $\begingroup$ And Physical Review A vol. 60, page 3515 says "Theoretical attempts to understand the stability of the dianions already discovered and directed towards finding ever smaller molecules capable of supporting two extra electrons soon followed [18–30]. A summary of the status of these calculations is given in Scheller, Compton, and Cederbaum [31]. To date, the smallest molecular dianions calculated to be stable are alkali-metal halides of the form MX3 -2 (M=Li, Na, or K; X=F or Cl)" journals.aps.org/pra/pdf/10.1103/PhysRevA.60.3515 $\endgroup$ – DavePhD Mar 7 '17 at 12:33
  • $\begingroup$ You are completely right, this is highly unstable, as thoroughly supported by the articles provided and further citations therein. In fact it has been stated, "a molecular dianion (such as a diatomic) must be stable not only with respect to the ejection of an extra electron but also to fragmentation" (ref. 31 of your second citation). $\endgroup$ – Felipe S. S. Schneider Mar 7 '17 at 19:51
  • $\begingroup$ Thus, either $\ce{[N2]^{-2} -> [N2]^{-1} + e^-}$ (and maybe even $\ce{[N2]^{-1} -> N2 + e^-}$, I'm not sure) or $\ce{[N2]^{-2} -> 2 N^{-1}}$ (or $\ce{[N2]^{-2} -> N^{-2} + N}$?) is a favourable, spontaneous process. This, in turn, is due to very low transition states (fast processes) and/or very stable products. But I have not claimed stability, but the existence of a bound state. At the level of theory applied, I found one bound state and analysed its wavefunction. No more, no less. $\endgroup$ – Felipe S. S. Schneider Mar 7 '17 at 19:59
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    $\begingroup$ The singly-negative charged $\ce{N2-}$ has been calculated as bound between lengths 1.4 and 2.5 Angstroms aip.scitation.org/doi/pdf/10.1063/1.478408 , so I'm surprised that adding another negative charge would make the bond length shorter and still be bound. $\endgroup$ – DavePhD Mar 7 '17 at 21:09
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Well, s-p mixing only occurs when the s and p atomic orbitals are close in energy (it's a necessary but not sufficient condition). Even though $\rm N_2^{2-}$ is isoelectronic with $\rm O_2$, the lower effective nuclear charge on nitrogen should make its s orbitals a little closer to the energies of the p orbitals than they would be in oxygen. I think you still would have s-p mixing.

Clarification: I mean to cast doubt on the conjecture that there's no s-p mixing here. As you point out in the comments, electron repulsion is a consideration too. We can't definitively say which effect is more important without looking at the actual orbital energies.

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  • $\begingroup$ But the two added electrons would cause big repulsion forces, which would increase the difference in energy between s and p, right? $\endgroup$ – RBW Feb 5 '15 at 15:35
  • $\begingroup$ There would be some additional repulsion which would increase the difference, but those two added electrons go in different orbitals, so it wouldn't be that big a difference. Really we'd have to look at the actual orbital energies to answer your question definitively. $\endgroup$ – Fred Senese Feb 5 '15 at 15:38
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    $\begingroup$ I would be delighted if I could get an accurate answer. $\endgroup$ – RBW Feb 5 '15 at 15:43
  • $\begingroup$ @FredSenese Fred wrote, "Well, s-p mixing only occurs when the s and p atomic orbitals are close in energy." It doesn't always occur even then. The lone pairs in water are an example. One lone pair is basically in a p orbital and the other in an sp orbital. That's very different from what was (still is) taught. S-P mixing is difficult (impossible?) to predict without experimental data such as molecular shape, ESCA (for lone pairs), etc. In addition to the water example, s-p mixing in NH3 and PH3 is different, how would you predict this difference without experimental data? $\endgroup$ – ron Feb 5 '15 at 16:18
  • $\begingroup$ @ron, in your NH3/PH3 example, what role would the different effective nuclear charges play on the difference in energy between s and p orbitals? Wouldn't that predict the difference? $\endgroup$ – Fred Senese Feb 5 '15 at 16:39
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It don't think that $\ce{N2^{2-}}$ can exist as a free species.

$\ce{N-}$ is unstable. In other words, the electron affinity of nitrogen is negative.

To create $\ce{N2^{2-}}$ you would need to bring together 2 negatively charged $\ce{N-}$ which are each already unstable due to their negative charge.

According to Experimental Search for the Smallest Stable Multiply Charged Anions in the Gas Phase Physical Review Letters vol. 83, page 3402:

It has also been concluded that no diatomic or triatomic dianions are stable

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    $\begingroup$ And how is this relevant to the question? $\endgroup$ – DHMO Sep 5 '16 at 16:56
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    $\begingroup$ @user34388 If there is no bound state of the ion, how can it be modeled using linear combinations of bound atomic orbitals such as s and p? $\endgroup$ – DavePhD Sep 5 '16 at 23:00
  • $\begingroup$ Fair enough.... $\endgroup$ – DHMO Sep 5 '16 at 23:11

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