In Allenes, the $\ce{C=C=C}$ structure makes sure that the $p$-orbital of central carbon overlaps with 2 other carbons and the $p$-orbital of central carbon spreads out over the 3 carbons.

In Alkynes, the $\ce{C#C-C}$ structure makes sure that the energy released during formation of alkynes.
I also read that more the number of non-delocalized $\pi$-electrons on an atoms, the lesser its stability, though I do not know why.

When, it comes to comparing Allenes and their isomeric Alkynes,it has always been stated that "Allenes are more stable than Alkynes".

One reason I found was that $\pi$-electrons in Allenes are less strongly bound than those in Alkynes.

  1. Why are $\pi$-electrons less strongly bound? If the reason in incorrect, what is the actual reason?

  2. Why is it that as the number of non-delocalized $\pi$-electrons increases on a carbon, the smaller its stability?

  1. The shape of the two bonds (see the first page of this .PDF) shows why pi bond electrons are less tightly bound. Sigma bonds are formed pretty much along the line connecting the two atomic nuclei, whereas the pi bonds are formed above and below that same line. Naturally, then, the distribution of the latter less concentrated (having more space to cover) than that of the former. Loosely stated, the shape of the pi bond distribution is more roundabout whereas the sigma bond directly touches the two nuclei which makes the electrons in it less strongly bound.

  2. When it can, delocalization of electrons provides greater stability for the molecule because the electron density is distributed well over the delocalized regions, rather than, say, confining it in a small high-energy region surrounded by low-energy regions.

Think about resonant structures. Benzene (C6H6) is a good example. So the inverse of that is: less delocalization, less stability. Hope this helps.


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