Why is N₂ stable but HCN and C₂H₂ unstable?

Question

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.)

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.