[Big-Daddy]

How do I identify which isomers are unstable and thus unlikely to form, just from looking at them once I've drawn them?

Take S₄O₆^2- (dithionate ion) for example. The actual shape can be found from a simple Wikipedia search so I won't outline it. What I will show is my first (badly wrong) attempt to draw it (all I drew on was the S₄O₆^2- formula):



This fills all criteria; 4 S atoms, 6 O atoms, 2 negative formal charges; O's formal charges and bonds are as they should be and each S is well within its limit of 6 bonds (in fact each has a lone pair remaining, i.e. pyramidal structure). But this is very very wrong and the only reason I can guess is something is energetically unstable about this molecule. How can I tell by looking at molecules in general (we can start with this one if you like, or provide a more instructive example, I just want to learn) which isomers are likely to be the more stable and thus real/correct ones?

[Corribus]

Were I to give an off the cuff explanation for the specific example mentioned here, I'd say the form you drew is less stable because it has fewer viable resonance structures (= less charge delocalization).  Pi-bonds between sulfur atoms are not very stable because the overlap integral is low.  This is why sulfur tends to form linear sigma-bonded chains but oxygen tends to form pi-bonded diatomic molecules.  Effectively, in your structure the negative charges can each only be delocalized over two oxygens.  In the real structure they can be delocalized each over three.

[Big-Daddy]

Were I to give an off the cuff explanation for the specific example mentioned here, I'd say the form you drew is less stable because it has fewer viable resonance structures (= less charge delocalization).  Pi-bonds between sulfur atoms are not very stable because the overlap integral is low.  This is why sulfur tends to form linear sigma-bonded chains but oxygen tends to form pi-bonded diatomic molecules.  Effectively, in your structure the negative charges can each only be delocalized over two oxygens.  In the real structure they can be delocalized each over three.

Ok ... so would a generalized approach be to look out for the isomer where charges can be delocalized over the most possible atoms (thus this form would have the most resonance structures)? Any other factors I should consider?

[Corribus]

I think this would be one good general guideline.  Minimization of charge localization is usually a controlling factor in stability because it is a fairly unfavorable situation.  Of course, it's not always easy to see what form gives the most charge delocalization.  In your example you would only know this if you know that third row elements don't typically like to pi-bond amongst themselves.

[Big-Daddy]

I think this would be one good general guideline.  Minimization of charge localization is usually a controlling factor in stability because it is a fairly unfavorable situation.  Of course, it's not always easy to see what form gives the most charge delocalization.  In your example you would only know this if you know that third row elements don't typically like to pi-bond amongst themselves.

I suppose there is something to be said for general chemical familiarity. My mind may have leapt straight to S=O parts flying around because I have seen C=O (and indeed S=O) quite a lot whereas I subconsciously avoided S=S because I'd simply never heard of it before! (Obviously if I'd known to discard this isomer I'd move on to alternatives and might have thought of an S=S, but probably not before the real structure which isn't that complicated anyway). Am I right in thinking pi bonds between atoms in period 3 (where both are in period 3) are even more unstable than when one is in period 3 and one is not?

[Corribus]

Yes.  You do see pi-bonds between period 2 and period 3 elements (in the form of phosphates, sulfates, etc.).  However you rarely see them between two period 3 elements.  As I said, this is because the 3p orbitals are more diffuse than the 2p orbitals, and the overlap between adjacent 3p orbitals in the z-axis direction (perpendicular to the bond axis) is especially low, which weakens the bonding interaction.  As a result, I wouldn't tend to give resonance structures which involve these kinds of bonding interactions very much weight.

On the other hand, the sigma bonding interaction between many period 2 elements (with the exception of carbon) is weak compared to the pi-bonding interaction, due to mutual repulsion of lone pairs.  This is why peroxides and nitrogen-nitrogen sigma bonds are fairly unstable - you don't see long chains of oxygen and nitrogen. 

[Big-Daddy]

Yes.  You do see pi-bonds between period 2 and period 3 elements (in the form of phosphates, sulfates, etc.).  However you rarely see them between two period 3 elements.  As I said, this is because the 3p orbitals are more diffuse than the 2p orbitals, and the overlap between adjacent 3p orbitals in the z-axis direction (perpendicular to the bond axis) is especially low, which weakens the bonding interaction.  As a result, I wouldn't tend to give resonance structures which involve these kinds of bonding interactions very much weight.

On the other hand, the sigma bonding interaction between many period 2 elements (with the exception of carbon) is weak compared to the pi-bonding interaction, due to mutual repulsion of lone pairs.  This is why peroxides and nitrogen-nitrogen sigma bonds are fairly unstable - you don't see long chains of oxygen and nitrogen.

Thanks. What other factors are there I should consider?

How would I go about this sort of problem: Write the Lewis structures for the 3 most energetically stable resonance forms of N₅⁺, and the 5 most energetically stable resonance forms of cyclic N₅^-. Draw the molecular geometries of both.

What should be my strategy, with these problems? (So that I could then try and draw on some guidelines to relate to N's general behaviour in polyatomic ions, for instance.)

[Corribus]

How would I go about this sort of problem: Write the Lewis structures for the 3 most energetically stable resonance forms of N₅⁺, and the 5 most energetically stable resonance forms of cyclic N₅^-. Draw the molecular geometries of both.

Is this a problem you made up or did you find it somewhere?  You could probably use some kind of modified Huckel theory do estimate it but as far as I know, there are no stable nitrogen allotropes like this.  Most of them (e.g., hexazine) are purely theoretical and have never been synthesized.

[Big-Daddy]

How would I go about this sort of problem: Write the Lewis structures for the 3 most energetically stable resonance forms of N₅⁺, and the 5 most energetically stable resonance forms of cyclic N₅^-. Draw the molecular geometries of both.

Is this a problem you made up or did you find it somewhere?  You could probably use some kind of modified Huckel theory do estimate it but as far as I know, there are no stable nitrogen allotropes like this.  Most of them (e.g., hexazine) are purely theoretical and have never been synthesized.

This is definitely not asking you to estimate how energetically stable the forms are - just to come up with them. I got it from an IChO paper, pretty reliable source (as it's scrutinized by hundreds of Team Leaders, usually with PhDs, trying to get the very best marks they can out of their teams' papers). What are the general principles?

I don't mind using some other example instead. I just want to understand how to draw the most energetically stable forms for an ion when given the formula.

[Corribus]

Huh, who knew?  Guess they're not technically allotropes since they're ions, but apparently they exist for some finite period of time:

http://en.wikipedia.org/wiki/Pentazenium
http://en.wikipedia.org/wiki/Pentazole

The first link even has an answer to your question, or at least part of one (they draw six structures, your problem asked for three).  Obviously you know that like charges aren't going to want to be next to each other, so that'd be a place to start.  I'm not sure which of those six are the most stable three.  Maybe you could figure it out by considering the provided bond length and angle data.

[Big-Daddy]

Huh, who knew?  Guess they're not technically allotropes since they're ions, but apparently they exist for some finite period of time:

http://en.wikipedia.org/wiki/Pentazenium
http://en.wikipedia.org/wiki/Pentazole

The first link even has an answer to your question, or at least part of one (they draw six structures, your problem asked for three).  Obviously you know that like charges aren't going to want to be next to each other, so that'd be a place to start.  I'm not sure which of those six are the most stable three.  Maybe you could figure it out by considering the provided bond length and angle data.

I should rephrase the question as I stated it wrong. I asked for the "most energetically stable" forms of the species. These aren't needed, as the question actually said: "Write the Lewis structures for 3 energetically stable resonance forms of N₅⁺, and 5 energetically stable resonance forms of cyclic N₅^-. Draw the molecular geometries of both."

Thanks for the Wikipedia page, but whilst it answers the question completely with respect to N₅⁺ it doesn't help much with the overall approach. My original attempts at drawing the isomers included 2 N-N double bonds and no triple bonds. However, simply saying that the triple bond is very stable doesn't quite explain this situation, as there are still double bonds in the resonance structures given.

[Corribus]

Well in these problems I think the only trick is to draw structures that don't exceed conventional bonding tendencies (e.g., nitrogen with five bonds) and to know where to put charges (e.g., a nitrogen with four bonds is positively charged).  Then make sure the charges add up to the appropriate number and that like-charged nuclei aren't adjacent to each other.

For instance, ^-N=N⁺=N⁺=N⁺=N^- technically works but it wouldn't be stable (insofar as any of these are stable) because it has three positive charges next to each other.

[Big-Daddy]

Well in these problems I think the only trick is to draw structures that don't exceed conventional bonding tendencies (e.g., nitrogen with five bonds) and to know where to put charges (e.g., a nitrogen with four bonds is positively charged).  Then make sure the charges add up to the appropriate number and that like-charged nuclei aren't adjacent to each other.

For instance, ^-N=N⁺=N⁺=N⁺=N^- technically works but it wouldn't be stable (insofar as any of these are stable) because it has three positive charges next to each other.

My original attempts were ⁺N=N-N=N-N, N-N=N⁺=N-N, and N=N-N⁺-N=N. How was I to deduce these were wrong?

From that I'll try and extrapolate some more general points or questions for you

[Corribus]

Let's take your attempts one at a time:

⁺N=N-N=N-N

First, your charges are wrong.  Nitrogen with only two bonds is formally -1, not +1.  (Do your formal charge calculation when in doubt, but you should learn to recognize it quickly.  Three-bonded nitrogen is neutral, two-bonded nitrogen is -1 and four-bonded nitrogen is +1.  Easy way to remember is that dinitrogen, a neutral compound, is triple bonded and the cations in ammonium salts are NH₄⁺.)  The singly bonded nitrogen on the right hand side would formally be -2, not 0 (so the overall charge of this molecule would be -1, not +1 as required by the problem), which brings me to the second problem: in some cases you'll see a nitrogen with a -1 formal charge (doubly bonded) but almost never -2 (singly bonded).  So this is what I meant by: learn the conventional bonding patterns of common elements.  Nitrogen can commonly accommodate 2, 3 or 4 total bonds.  Your resonance structures should not include nitrogens that are bonded only once or five times, shouldn't include oxygens that are bonded four times and carbons pretty much should always be bonded 4 times (other versions exist but they're for the most part unstable so you shouldn't worry about them). 

N-N=N⁺=N-N

Nitrogens on the ends would each be formally -2 charge (for an overall molecular charge of -3) and as pointed out above, nitrogen shouldn't be bonded only once.

N=N-N⁺-N=N

Nitrogens on the ends would each be formally -1 charge and the nitrogen you've identified as having +1 charge also has a -1 charge (for an overall molecular charge of -3).

[Corribus]

By the way, you'll notice that all the "correct" isomers have the same total number of bonds - 8.  Because the relationship between number of bonds to a nitrogen and it's formal charge is always constant, I got to thinking that for a given number of nitrogens connected in a linear fashion, there should be a constant relationship between the total number of bonds in the molecule, the number of nitrogens in the chain, and the overall formal charge.  I played around with the math and I came up with a formula.

The total number of bonds, D(n) in a linear chain of n nitrogens with overall formal charge C is

[tex]D(n) = \frac{C + 3n}{2}[/tex]

I tested it out on a bunch of test cases and it seems to work.  (In the case under consideration, n = 5 and C = +1, so the total number of bonds is 8.)  Since D(n) has to be a whole number, any charge/n combination that gives a fraction is impossible.  For instance, a chain of five nitrogens with a +2 charge doesn't work (D = 8.5), so there's no way to make a chain of nitrogens bond in such a way that the formal charge is +2, and playing around with the structures confirmed it as far as I can tell.  I could probably modify the formula for branched chains and also ring structures and such, I suppose, but didn't get that far and the math would get more complicated in a hurry. 

Anyway.... yeah, I'm a nerd but I thought it was kind of interesting.