# 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

In simple terms electrons repel each other, so distributing them over a geater number of bonds will reduce the repulsive force between them so resulting in a lower energy state. In the case of phenol the oxygen p orbital lone pair at right angles to the ring can overlap with the delocalised p orbital above and below the ring enabling a degree of delocalisation into the ring system so leaving a slight + charge on the oxygen. This in turn will reduce the electron density in the OH bond so weakening it and accounting for the weak acidity of phenol.

• @user36790 You were probably thinking of a,b-unsaturated carbonyl that is commonly drawn as the oxygen carrying the partial charge. Now think of a,b-saturated carbonyl. It has the same amount of electrons but more space for the electrons to fly in. The oxygen does not draw more electrons than it can handle, the partial charge at the oxygen will not increase to infinity when you add double bonds to the other end. Nov 26 '14 at 18:14

Q1.How does shifting of π electrons let the body lose internal energy? What is the cause?

Delocalization of electrons constitute a current in the molecule which leads to an increase in randomness/ entropy in the molecule that leads to stability. Also, according to molecular orbital diagram, the pi-electrons finally achieve an overall lower energy level on interaction with other pi-electrons

Q2. Why there is partial negative charge on the body? And why does oxygen carry partial positive charge? What is the cause?

The first thing you need to understand is that the resonance structures (canonical structures or contributing structures) are hypothetical and individually do not represent any real molecule. They contribute to the actual structure in proportion to their stability

In phenol, three resonating structures contain '+' charge on O . Thus, in the final structure, there is a slightly positive charge on oxygen and also, partial negative charge on the benzene ring

• "They contribute to the actual structure in proportion to their stability" that is absolutely wrong. You say before that, that they are purely hypothetical and do not represent the molecule. They are not stable in any sense of the word. And please do not call them resonating structures. The concept is called resonance. The structures do not resonate with each other. It is a mathematical trick. Sep 10 '20 at 8:55
• @Martin-マーチン link Head to page 20 of the pdf. Sep 11 '20 at 0:51
• It's very likely that ncert is getting this wrong; they are one of the worst offenders when it comes to outdated, deprecated, sloppy, and incomplete information. Also, it should be fairly obvious, that in any sense of the words, hypothetical and stable are mutually exclusive. Sep 11 '20 at 9:49