# Why does delocalization lead to stability? [duplicate]

If the electrons are relatively more stable when they are delocalized than when they are localized between two atoms, then why do we have bonds formed in the first place? Forming of bonds between two atoms means that electrons localized between the two atoms, if electrons tend to delocalize then why not all electrons (including the sigma bond electrons) are just delocalized? What is the driving force for delocalisation ? How does delocalisation the energy of the system?

• The tendency to delocalize is offset by electrostatic attraction (to the nuclei). – Buck Thorn Feb 3 '19 at 12:38
• An electron system delocalises automatically under certain circumstances, namely if it's a conjugated pi system. It's not a property of the electron, but of this specific bond system! – Karl Feb 3 '19 at 13:00
• You have good answers - but I'd like to add: In general in chemistry, if you allow a "thing" to have more options on how to configure itself, it will have a lower energy state and that will present as stability. – Stian Yttervik Feb 3 '19 at 16:23

The electron distribution in a molecule is a compromise of the two "wants" while not trying anything "forbidden". In the $$\ce{H2}$$ molecule, electrons get to enjoy the attraction of two nuclei while having more space available than when bound to an isolated hydrogen atom. Because of the Pauli exclusion principle, only two electrons get to enjoy that low energy.
If you think of an electron as a de Broglie wave, and confine it to a box, you can show that its energy $$E$$ is inversely proportional to the squared inverse of the length $$l$$ of one side of the box, that is, $$E \propto \frac 1{l^2}$$ Therefore, if you make the box larger (increase $$l$$) you lower $$E$$. This is a purely quantum-mechanical result due to the wave property of particles and it applies equally well to electrons confined by the attractive electrostatic potential generated by the positively charged nuclei in a molecule.