Compounds with triple bonds generally seem to be unstable. $\ce{HCN}$ and $\ce{C2H2}$ are high-energy, relatively short-lived molecules that will readily polymerise or react with other organic molecules. My naïve mental picture is that they don't really want to have a triple bond because the geometry makes it awkward, so they'll do anything to transfer an electron somewhere else and turn it into a double or a single bond instead.

But $\ce{N2}$ seems to be an exception. By my naïve reasoning, when an $\ce{N2}$ meets an organic molecule it should be eager to shrug off its triple bond and join the party. But in fact this doesn't happen, and nitrogen fixing organisms have to do quite a bit of work to get nitrogen to participate in organic molecules.

So my question is, what is it about $\ce{N2}$ that makes its triple bond so energetically favourable while $\ce{C#C}$ and $\ce{C#N}$ bonds are so unfavourable?

Note that this question isn't about reactivity with $\ce{O2}$ but rather with organic molecules. I'm interested in why it's so difficult for $\ce{N2}$ to react with organic molecules in general, given that other molecules with triple bonds seem to react with them very easily.

(Information to guide the scope of answers: I'm coming at this from a physics background, so thermodynamic concepts can be taken as understood, but I quickly get lost when it comes to electrons and orbitals and so on, hence the rather basic question.)

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    $\begingroup$ I guess the simplest explanation is, the bond energy difference between N$\equiv$N and N$=$N are about 500 kJ mol$^{-1}$, while C$\equiv$C and C$=$C are about 200 kJ mol$^{-1}$.chemwiki.ucdavis.edu/Theoretical_Chemistry/Chemical_Bonding/… For HCN, there are other channel, such as HCN$\rightarrow$ H$^+$ + CN$^-$. The next question is, why the bond energy difference of N$_2$ is much larger than HCCH. The Kohn-Sham HOMO (I know it is problematic, but N$_2^+$ is a failure of Hartree-Fock) of N$_2$ and HCCH are $\sigma$ and $\pi$ orbitals, former is more stable $\endgroup$
    – user26143
    Commented Jun 12, 2014 at 11:00
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    $\begingroup$ Part of the answer must be to do with the possible mechanisms available for a reaction with something to happen. There are no easy reactions starting with dinitrogen, but lots of easy ones starting with acetylene or hydrogen cyanide (but they don't all start with the triple bond). $\endgroup$
    – matt_black
    Commented Sep 5, 2017 at 14:27

1 Answer 1


I think this question can be answered reasonably well with thermodynamic considerations. When we say something is stable, it is (or should be) assumed that the compound does not engage in a chemical reaction. It is important when we discuss the stability of compounds to focus not only on the compound itself but what the products of the reaction are.

The three triple bonds you mention ($\ce{N#N}$, $\ce{C#N}$ and $\ce{C#C}$) all have very high bond dissociation energies (from various sources such as this one they are $\approx950, 940\text{ and }960\:\mathrm{kJ/mol}$) so all of these triple bonds require a large amount of energy to break.

Now, when we think about the stability of acetylene, I suspect most people are thinking about its stability towards ignition (which has been studied, by the way). That gives us a reaction we can explore, namely the combustion of acetylene in the presence of oxygen to produce carbon dioxide and water. WolframAlpha can calculate the thermodynamics of this reaction, giving a result of $-2600\:\mathrm{kJ/mol}$; clearly a very exothermic reaction.

Now let's look at the combustion of nitrogen in the presence of oxygen, which could form nitrogen dioxide. Again, WolframAlpha (Note, if you happen to play around with WA, make sure you put your reaction in quotation marks or it won't be interpreted properly) gives us a result of +66 kJ/mol or an endothermic reaction.

What do we make of these results? Focusing on the products we see that for the acetylene combustion, there is a large amount of energy released when creating the products. The same is not true for the products of nitrogen combustion.

In summary, when considering stability, we must think about we mean by the word, since that meaning will lead to a chemical reaction, and the tools we use to describe chemical reactions help us quantify that stability.

  • $\begingroup$ Thanks. I'm sure that what you say is right, but as I hoped would be clear from the question, I'm not so concerned with the reactions with O2 as with other organic molecules. I'm not asking "why doesn't N2 burn" but "why doesn't it generally react with biomolecules?" $\endgroup$
    – N. Virgo
    Commented Jun 12, 2014 at 14:04
  • $\begingroup$ @Nathaniel you might want to edit the question to make this point clear (especially since you refer to organic molecules but not biomolecules). The answer will be the same (you need to consider the formation energy of the products) only in the case of biomolecules (of which there are many) the reactions will be more complicated. If you have a particular reaction in mind, that might help lead folks to a usable answer. $\endgroup$ Commented Jun 12, 2014 at 14:47
  • $\begingroup$ I specifically don't have a particular reaction in mind. I'm asking why, in general, is it difficult to react N2 with organic molecules, which is why I asked for help from chemists rather than just looking up the delta-G. I will edit the question to clarify that, although it does actually say it already. $\endgroup$
    – N. Virgo
    Commented Jun 13, 2014 at 0:48
  • $\begingroup$ I disagree on your use of stable, the meaning you are implying is the one of unreactive. Stable should be used in the sense of "it is an isolatable chemical species." $\endgroup$ Commented Jun 13, 2014 at 6:12

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