# Why do atoms “want” to have a full outer shell?

Okay, so I know that this is about filling the orbitals of the atom, and I understand that. What I don't understand is why? For example, an Oxygen atom has 8 protons and 8 electrons spinning around it. Oxygen wants to have 8 valence electrons so it takes 2 electrons from other elements when bonding. But, since Oxygen has 8 protons and now has 10 electrons around it, it has a charge of -2. I don't understand how having a charge is more stable than having no charge at all, shouldn't the most stable version of an element be its neutral state? Why do the atoms want to look like the next noble gas?

• Have a look here. – Philipp Sep 30 '14 at 18:23
• Thanks, I looked into it. But it explains why the number of electrons in a specific shell's full state is what it is, not why it's more stable than the neutral state of the atom, which is what I'm asking – Mertcan Ekiz Sep 30 '14 at 18:42
• I've wondered about this issue also. The answers that I've found generally use the rules of quantum chemistry as an explanation. But while these rules are consistent they don't provide the kind of explanation I and others would find more satisfying. As a starting point, it seems to me that there has to be some relationship between a full outer shell and the ability to construct a low potential energy arrangement. Using the case of a diatomic like H2 for simplicity, a low potential energy arrangement exists a the particular distance at which there is neither attracting nor repelling force. – David Rosen Nov 12 '16 at 13:59
• So the question becomes: Why does the creation of a full outer shell for a pair of hydrogen atoms produce a low potential energy arrangement... "bonding distance"... at which there is zero force? – David Rosen Nov 12 '16 at 13:59
• – Gaurang Tandon Jul 7 '18 at 3:00

You are attaching too much importance to Lewis structures. The 8-electron rule and Lewis structures which are derived from it are only rough guidelines for working out the electronic structure of a compound in very broad strokes. Often these broad strokes are accurate enough to make some meaningful statements about molecular properties but it does not accurately describe the true electron or charge distribution in a compound. Take water for example. As you say, the 8-electron rule would predict that the hydrogen atoms each transfer one electron to the oxygen molecule which would then carry a charge of -2. But experiments show a different result: here you can find a paper that determined

that a charge of approximately $0.5 e$ is transferred from each hydrogen atom to the oxygen

which would amount to a partial charge of $-1 e$ on oxygen and here you can find a similar claim:

The charge distribution depends significantly on the atomic geometry and the method for its calculation but is likely to be about $-0.7e$ on the O-atom (with the equal but opposite positive charge equally divided between the H-atoms) for the isolated molecule.

And there not being a $-2e$ charge on oxygen is quite understandable if you look at the electron affinities of oxygen (e.g. here): You see that the first electron affinity is $-142 \, \mathrm{kJ}/\mathrm{mol}^{-1}$ and thus energy is gained by the process of transferring a free electron to a neutral oxygen atom. But the second electron affinity is $+844 \, \mathrm{kJ}/\mathrm{mol}^{-1}$, so another electron transfer is very costly. The quite extreme situation of a -2-charged oxygen described by the 8-electron rule is very rarely found in real compounds. Even in ionic compounds like $\ce{NaCl}$ the electron is not 100% completely transfered from the electropositive to the electronegative element.

• Essentially, its a result of energetics. Energy generally goes down as we approach the 8 e- valence space. – LordStryker Sep 30 '14 at 20:49

Its not a problem about being charged, its about the symmetry. The quantum states of those valence orbitals are stable, in that there is little interaction with the other electrons. Since there are already filled orbitals, electrons will want to bind to complete the shell because as it happens extra binding energy is available. With a full shell maximum binding occurs.