# Why does delocalization of π electrons bring stability?

My book mentions that

Delocalization of $\pi$ electrons makes a compound stable as the process of delocalization involves loss in internal energy. This is the synopsis of resonance.

But how does shifting of $\pi$ electrons let the body lose internal energy? What is the cause?

While studying resonance, I observed that one structure possesses charge while in other no charge. But in the final structure, there is, according to the book, partial charge present.

Eg. In phenol, three resonating structures contain '+' charge on $\ce{O}$ . Thus, in the final structure, there is a slightly positive charge on oxygen and also, small negative charge on the main body.

Now my question is, why there is partial negative charge on the body? And why does oxygen carry partial positive charge? What is the cause?

• @edkinsella The solution to the "particle in a box" problem shows that energy levels are a function of 1/L^2, where L is the length of the box. So even with just 1 electron in a molecular orbital, where electron repulsion doesn't come into play, the longer we make the box (e.g. the longer the delocalization path) the more stable (lower in energy) the electron. – ron Nov 26 '14 at 21:29