Wolfram
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Nitrogen N4
How do I fuse 4 nitrogens in a tetraeder?
What poperties would such a compound have? Any guesses?
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Marvin
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You wouldnt, and it doesnt.
If you forced 4 nitrogen atoms into that position it would form 2 nitrogen molecules which would drift apart, not a single tetrahedron.
You have to ask yourself the question not how stable a molecule is,, but how stable the products would be, and how easily they'd form.
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Wolfram
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I have read about this ..
I have read about this and it was said it already has been done . The compound was said to be 20 times stronger explosive than TNT if I remember
right.
I guess the methods to do this would be rather complicated becouse I have not heard of any comercial or military application for this.
[Edited on 12-5-2004 by Wolfram]
[Edited on 12-5-2004 by Wolfram]
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Blind Angel
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Maybe you can but it would be very hard and almost impossible in an amateur environnement.
It's like the pentazonium anion (N<sub>5</sub><sup>-</sup>
/}/_//|//) /-\\/|//¬/=/_
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chemoleo
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Marvin, although the products of N4 decomposition are surely more stable than N4 itself, it doesn't necessarily mean that it doesn't exist -
there is such a thing called activation energy.
The question you have to ask is how high this activiation energy is (admittedly it may even be so low that it decomposes at -200 deg C- but it
doesn't mean it doesn't exist ).
Of course you know that so don't take offense please - just thought it worth pointing out.
PS Also, isnt nitrogen naturally a tetrahedron? 3 p's and one s orbital? wouldn't that make this structure possible? I.e. the empty orbitals
all pointing away from the center of the tetrahedron?
[Edited on 12-5-2004 by chemoleo]
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Wolfram
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There also seem to exist..
There also seem to exist N8 cubane , but I dont know for how long though..
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Wolfram
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Here is a paper..
Here is a paper about tetra N4 and N8 cubane which doesn´t exist accoring to Marvin the national hazard.
http://aros.ca.sandia.gov/~mlleinin/Docs/jp970258k.pdf
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Polverone
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did you read the paper you linked to?
Those were computational studies of the molecules. They didn't actually make N4 or N8. It's easy to simulate compounds that you
couldn't even make a millimole of in the lab.
PGP Key and corresponding e-mail address
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vulture
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...which doesn´t exist accoring to Marvin the national hazard.
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There's no use in getting personal. Unless you like getting sanctioned ofcourse...
One shouldn't accept or resort to the mutilation of science to appease the mentally impaired.
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Wolfram
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What is the problem
What is the problem? Marvin clearly wrote that they dont exist..
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Blind Angel
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we call this "flame" and it's not good, you insult someone, or try to diminish it
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frogfot
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Huh, here one would make use of things one learned at quantum chemistry..
I tryed to assemle tetrahedral N4 with p's of each N in center of the tetrahedron, and then there appear some strange overlaps between rest of
the p's.. This even becomes more complicated when the energy levels are drawn.. can somebody throw some light on this.. 
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Marvin
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Chemleo,
Avoiding eximers, which dont really apply to the thread,
Activation energy results because a high energy state seperates the products from the reactants. This results entirly because the atoms are not in
the right positions to form the products from the reactants, that is to say the change is not purely electronic.
For example, in the case of hydrogen azide gas, the atoms would be much happier as ammonia and nitrogen. The azide part of the molecule is stable
however, because releasing N2 would leave a species behind, H-N which is much higher in energy than the amount gained. That is to say it is
electronically unfavourable for the molecule to break up. If it wasnt, the molecule wouldnt even form.
Looking at 4 nitrogen atoms in a tetrahedral arangement there is nothing the stop the atoms at the first sign of asymmetry choosing which of the atoms
its next to it wants to form a triple bond with, and the 2 nitrogen molecules drifting apart.
Thus the conclusion, the molecule is not stable and there is no activation energy for its decomposition to nitrogen.
frogfot,
The first stage would be to hybridise to get the tetrahedral arangement. For each of the 4 nitrogens, you have 5 electrons 3 of which go into bonds,
one with each other nitrogen, and the two that remain go into the non bonding orbital (lone pairs). In essence its very similar to ammonia as an
electronic problem.
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chemoleo
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Hmm...
although I agree with all of what you say, I don't follow this to your conclusion (i.e. that it doesn't exist). For one thing, not for no
reason have people been trying to make it for a long time (which they wouldn't if it theoretically can't exist).
The point I was making is simply that the formation of 2N2 from N4 is favoured, and that N4 is unstable, as you say. That doesn't mean though it
doesnt exist at all - it could be stable enough at 50 K to exist for a few microseconds, or less, or more (or at some other exotic conditions).
Besides... there is bond breakage required for the formation of N2 from N4. Bond breakage does require some activation energy, regardless whether the
resulting formed bonds are energetically favoured?
THis bond breakage would correspond to the A.E., would it not?
PS have a look at that paper from Wolfram, there is a lot of debate about N4, and others. According to energy landscapes it should exist.
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vulture
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Besides... there is bond breakage required for the formation of N2 from N4. Bond breakage does require some activation energy, regardless whether the
resulting formed bonds are energetically favoured?
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If there is bond breakage, the N4 molecule must first be formed. What Marvin is pointing out is that it wouldn't be formed because 4N atoms
together, even in excited state will fall back to N2 immediatly, there is no incentive for the atoms to form N4. Add to that the steric hindrance....
I think many people here are making the fault of comparing carbon cage molecules with nitrogen cage molecules. Carbon cages are much more likely, not
only because it being tetravalent and having no electron pair that stresses the bonds, but because there is no such thing as a stable carbon dimer
with a insanely high bond enthalpy like N2.
One shouldn't accept or resort to the mutilation of science to appease the mentally impaired.
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chemoleo
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Hmm... so you are implying that the formation of N4, by some tricky unknown synthesis route, is theoretically and therefore practically impossible?
And therefore there's no need to discuss the decomposition in the first place, because it can't exist?
I am not following this.
I thought theoretically N4 is possible (See W's paper). Yet no synth. route has been discovered. I don't see where you get the conviction
that 'because 4N atoms together, even in excited state will fall back to N2 immediatly' and that it therefore can't exist.
Neither do I understand the conviction for 'Looking at 4 nitrogen atoms in a tetrahedral arangement there is nothing the stop the atoms at the
first sign of asymmetry choosing which of the atoms its next to it wants to form a triple bond with, and the 2 nitrogen molecules drifting
apart'.
If it 'chooses' to form a triple bond, first other bonds have to be broken (and THIS is what may stop it from forming 2N2 right away!). This
requires energy, regardless by the energy generated with the formation of the triple bond. Hence there is a possibility that it is stable at certain
conditions.
What's wrong with this?
PS and no, I am not talking about carbon cage analogues.
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vulture
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If it 'chooses' to form a triple bond, first other bonds have to be broken
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That's the fault in your logic. You automatically imply that N4 is formed first.
A synthesis is likely going to require extreme conditions, leaving N in a loneley excited state which automatically leads to it having no bonds at
that moment. It then has the choice to go to N4 or 2N2... which would it choose?
So, any synthesis is likely going to need cold temperatures and a non-excited transition state.
[Edited on 16-5-2004 by vulture]
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Marvin
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I think it doesnt help to start by using rather inaccurate organic chemistry terms for a problem that shouldnt even be in the organic section. Atoms
arnt fused together, if the atoms want to stick, the bond forms because this is a lower energy state. The actual errors seem to stem from confusing
molecular terms, with electronic ones. This isnt a molecular problem, its an electronic one.
vulture is essentially right in his earlier post, but knowing one solution is more likley doesnt prevent the other one forming, you have to show its
the only solution. If you put 4 nitrogen atoms in a tetrahedral arangement the resulting molecular orbital must follow the same symmetry. This means
we have a potential molecule to start our thought experiment with. The question is if you let go, will it fall apart straight away. In this case its
pretty easy to see what will happen, when the symmetry breaks the nitrogen atoms will pair up. The nitrogen atom will be attracted to one of other
atoms more than the others and the distance will shrink. The bond between them will get stronger as the distance shrinks and the other two bonds will
get weaker as those distances increase, then the highly distorted species will seperate and 2 nitrogen molecules will go off in seperate directions.
In electronic terms you never need to break a bond before making another one. Activation energies are entirly concerned with atoms being in the wrong
locations.
Acrobat is currently incompatable with my system, but Ive now read the paper and the species being discussed is an eximer.
I stand by my earlier conclusion, tetrahedral N4 is not stable, and there is no activation energy for its decomposition.
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unionised
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Just a thought, could N4 (tetrahederal, square, linear ot whatever) exist as a Van der Walls dimer?
Dreadfully unstable, essentially impossible except as a low pressure gas (so b***er all use) but possible I think
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madscientist
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The electrons in a N<sub>4</sub> structure would immediately fall toward N<sub>2</sub> structures due to the immediate gain in
stability it would afford for the system. Marvin's right, N<sub>4</sub> can't exist.
I weep at the sight of flaming acetic anhydride.
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frogfot
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Marv, I had a second thoat on the NH3 problem. As I was tought, all orbitals have permanent positions and hybridisation is only an overlap between two
or more orbitals (the more like orbitals, the stronger is the hybridisation.. orbitals comes kinda in a resonance..). Where I was trying to come to,
in a tetrahedral N4 there would be 60* between each bond and that would be vey unstable (since hybridisation is weak).
I was thinking that if each N had 3 p orbitals, one side of p from each N would point in the center of tetrahedron, and the interactions between the
rest of p orbitals is easy to assemble... This sounds to be more stable, but I still can't draw the energy levels for this thing.. uh, I gonna
draw the p orbitals now in mspaint..
EDIT:
Ok, something like this:
http://img37.photobucket.com/albums/v113/frogfot/n4orbitals....
I've simplified the p orbitals to thin lines, just so it would be possible to draw..
See how many overlaps there are between the neighboring lobes? Specially the ones that comes in the centrum of tetrahedron (eng?) However, only upper
N's orbitals have a bit higher energy positions (meaning this tetrahedron(eng?) wouldn't have equal sides).. but if you draw that pic from
above there is a way to point them that will minimise energy..
Now, this is just a bigass simplification, but I wanned to point out that lobes of p orbitals can interact in many different ways..
[Edited on 17-5-2004 by frogfot]
[Edited on 17-5-2004 by frogfot]
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vulture
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What's even more is that the free electron pair will reduce the angle between the bonds, further straining the molecule.
White phosphorus is a P4 monomer with "banana" like bonds because they're being forced to bend at the extreme bond angle of 60C. We all
know how stable white phosphorus in air is.
Red phosphorus on the contrary is a polymer without the extreme bond angles and is pretty stable.
Note that phosphorus is a bigger atom than N4 and that it can make use of d orbitals.
[Edited on 17-5-2004 by vulture]
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unionised
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Madscientist; why do you think a V der W dimer is impossible? http://content.aip.org/JCPSA6/v88/i2/642_1.html
Looks perfectly plausible to me.
[Edited on 17-5-2004 by unionised]
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Marvin
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A van der Waals dimer is certainly possible, but the bonds would not be in perpendicular planes from their center (as LDF is based primarily on
dipoles) and the structure could not be tetrahedral anyway from the differences in bond lengths.
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