# Why do atoms tend towards electrical stability?

If an oxygen atom has six electrons, then it has an unfilled orbital and the oxygen atom may share electrons from two hydrogen atoms (and form water) in order to become more stable.

But why does oxygen, or any other atom for that manner, favor stability over instability? Why can't the oxygen atom just have six electrons and be done with it?

Also, is an atom being stable identical to it being neutral?

• I've deleted an inappropriate comment and several responses to it. There was a possibly-salvageable discussion about whether it's appropriate to assign intent to inanimate objects; that discussion wants to happen in a chat room, rather than in these comments. – rob Feb 10 at 2:46
• Maybe you mean to ask why an oxygen atom can't be stable with 6 electrons. Asking why something favors stability is a different question that can be answered simply by looking at the definition of "stable", which is to be "firmly fixed" or "not likely to give way". Something unstable, by definition, is not fixed and likely to give way. – JoL Feb 10 at 5:45
• Everyone please keep in mind that comments are meant for suggesting improvements or requesting clarifications on the question, and in particular, not for answering. – David Z Feb 10 at 14:06
• There's some confusion of terms here... the title refers to electrical stability, yet the body discusses the stability of completed shells and 8 electrons in the outermost shell (chemistry.stackexchange.com/q/1196). Furthermore, are you referring to stable in terms of "not-going-to-decay", or "will-try-to-gain-them-electrons"? – user191954 Feb 10 at 14:39
• @Chair well the title I originally wrote didn't say "electrical" stability however dotancohen appears to have thought this was a proper amendment to what I was looking for. by "stability" I am NOT referring to radioactive decay. – Korvexius Feb 10 at 17:26

An $$\text{O}^{2+}$$ ion, with 6 electrons, is stable in a vacuum. If you put one in the depths of interstellar space, far from any other matter, then it will sit in that state until the end of time.

Similarly, you can have a high-elevation lake, and it can be absolutely stable if there is no path for water to drain out.

Also, is an atom being stable identical to it being neutral?

So no, as shown by that example, it is not the same thing.

However, of you have a higher-elevation lake and another one whose water level is 10 meters lower, and they're connected by a stream bed, then water will drain out of the higher one until the two surfaces reach an equal level. The stable state is the one that has the minimum energy (minimum total gravitational potential energy of all the water). This is analogous to what happens when oxygen and hydrogen are mixed. The electrons are like the water.

• Ahh, so stability is basically the ability to maintain your properties in light of the environment, right? – Korvexius Feb 9 at 21:59
• > "An ion, with 6 electrons, is stable in a vacuum. If you put one in the depths of interstellar space, far from any other matter, then it will sit in that state until the end of time." That is not so clear cut. In empty vacuum yes, but in real one, there is background EM radiation everywhere that can break and separate molecule's/ion's constituents. This is easy to see from the fact that such isolated system has increasing density of energy levels to infinity when energy approaches 0. – Ján Lalinský Feb 10 at 1:47
• For a system which is not a pure mechanical one, entropy is to be taken in consideration ? For example, at pressure and temperature fixed, the stable state is the minimum of the Gibbs free energy and not the minimum of energy. – Vincent Fraticelli Feb 10 at 19:16

Wanting or favoring are not very good terms in physics. More scientific view on this would be that whenever an oxygen atom comes close to another atom, they interact and provided things are right (such as the number of electrons and their state), the atoms attract and can get closer, while losing part of their initial energy, in the form of radiation or lost electron or transfer some energy to other atom/molecule in a scattering event.

After the binding energy is lost from the pair to the surrounding space, it becomes more probable it will stay together, until required energy for its breaking is supplied from outside the system. This can be EM radiation with the right spectral characteristics, or some other particle that moves nearby and cleaves the bond.

For the air in troposphere, the available mechanisms to supply so much energy (collisions, cosmic particles) can only do so for very little fraction of molecules, so majority of oxygen atoms will exist in pairs, much lower number in triplets and so on. The situation is different in upper layes of atmosphere, where UV light and cosmic particle are more intense, so greater proportion of gas particles may be in exotic form such as unpaired oxygen atom.

• Good or not, “favoring” is a common term in physics, at least in English, and, particularly in the context of energy equilibria. We say that a reaction favors the lower-energy state. A search on this very site will find many examples. Here are just two: 1 2. A search of abstracts on arXiv finds many more examples. – MJD Feb 10 at 1:31
• @MJD common or not, I think it is not good to use it before explaining the science behind the phenomenon, especially to beginners, because it is likely to confuse them or induce bad thinking habits. For some other uses such as popularization or informal talk between scientists it may be fine. – Ján Lalinský Feb 10 at 1:58

Also, is an atom being stable identical to it being neutral?

No, not at all: many atoms are more stable after they've either absorbed (an) electron(s) or shed (an) electron(s).

A good example is the formation of table salt, $$\mathrm{NaCl}$$, aka sodium chloride.

This compound forms when sodium atoms lose an electron (the valence electron), as in:

$$\mathrm{Na}\to \mathrm{Na^+}+ \mathrm{e^-}$$

Similarly the choride atoms can absorb an electron:

$$\mathrm{Cl_2}+ 2\mathrm{e^-}\to 2\mathrm{Cl^-}+ 2\mathrm{e^-}$$

When those ions combine we get:

$$2\mathrm{Na}+\mathrm{Cl_2}\to 2\mathrm{NaCl}+\Delta H$$

$$\Delta H$$ is the energy released in the process. The arrangement of these elements in the ionic lattice $$\mathrm{NaCl}$$ is more stable than the combination of the (unreacted) elements.

By losing its 'lone' $$\mathrm{3s^1}$$ valence electron, sodium takes on the very stable electron configuration of neon. Similarly, by absorbing an electron into its $$\mathrm{3s^23p^5}$$ valence electrons, it assumes the very stable electron configuration of argon.

• “By losing its 'lone' $\mathrm{3s^1}$ valence electron, sodium takes on the very stable electron configuration of neon.” – This sounds potentially misleading. Removing the electron from a neutral sodium atom requires an ionization energy of 5.14 eV; i.e. the process is endothermal and $\mathrm{Na^+}+ \mathrm{e^-}$ is less stable than $\mathrm{Na}$. – Loong Feb 10 at 15:53
• @Loong: I made it quite clear that stability of NaCl comes from its ionic lattice. But $\mathrm{Na^+}$ cations are very stable in other contexts to, like solvated ions. – Gert Feb 10 at 17:06
• You can't get there from sodium alone. You need something sufficiently "happy" to take the electron to pay the energy cost of taking it away from the sodium. It's just that the cost is rather low, so a lot of substances are willing to do that, including some substances that aren't known for being strong oxidizers. But with nothing to take the electron, it would rather stay with the sodium. – Ian Feb 10 at 22:49
• @Ian: correct but why single out sodium (and other metals), in most cases where changes in orbital structure occurr either a donor or a receptor is required. – Gert Feb 10 at 23:41
• @Gert The point is that it's not just that $\mathrm{Na}^+$ is particularly stable, it's that it consumes less energy to produce it than is released by converting $\mathrm{Cl}$ to $\mathrm{Cl}^-$. Thus your statements about the nice electron structures of $\mathrm{Na}^+$ and $\mathrm{Cl}^-$ in isolation don't really mean anything; it's all relative. – Ian Feb 10 at 23:42

The oxygen atoms when get close to each other, would certainly interact. When they interact, their is a higher probability that the electrons are shared between the atoms. When they share, they will lose energy. They will not break apart unless energy comes again from somewhere and satisfies them to stay separated.

And no being stable is not equal to being neutral. Take oxygen atom for example, it is more stable in the ion form than the neutral oxygen atom.

I think the bottom line here is that reasoning about atoms, ionization and chemical bonds is often done in a rather sloppy way, and the reason for that may be that people have not been taught that entropy often plays a role in these questions.

The direction of physical processes is such that entropy increases overall. This is why a given system will tend to move towards a state of lower internal energy if one is available---it is because the energy thus liberated can be passed on to something else, such as emitted light or vibrations, with the result of a net increase in entropy of the environment of the system, with little change in the entropy of the system itself. Chemical bonds form mostly because the bonded configuration has a lower energy, and this lower energy is adopted mostly because the liberated energy goes to the surroundings, or to vibrations of the material, or something like that, which carry more entropy. Whenever the concept of free energy is being used, then these entropy arguments are in play.

People often assume that systems will adopt whichever state has the least energy. This is ok as a general informal guide, but if you want to know why or how it happens, then you need to notice that such an assumption requires a way for the system to get rid of any extra energy it may have, and that usually means emission of energy into the system's surroundings.

It is important to use careful language to understand what is happening

"What does an oxygen atom favour stability over instability" is a bad way to ask the question and not just because it anthropomorphises the factors causing chemical reactions. The right way to think about the problem is to ask which configurations of atoms and molecules in a dynamic system will have the lowest energy.

Except in a high vacuum atoms and molecules are constantly interacting with other atoms and molecules. When they interact things can happen. Sometimes the atoms and molecules just bounce off each other; sometimes a chemical reaction occurs; sometimes energy is exchanged but no net reaction results. When reactions occur some of them are reversible and some are not.

Since there are many possible reactions occurring in any mixture how do we know what the net result is? The factors that matter are the the energy levels of the possible products in the mixture (I'm simplifying a bit by ignoring entropy) and the degree of reversibility of all the possible reactions occurring.

In the case of a mixture of isolated hydrogen and oxygen atoms a lot of different things can happen (by the way it is very very hard to create such a mixture). One is that isolated oxygens can meet and combine to give oxygen molecules (this releases energy). Or oxygen can meet hydrogen and combine to give an OH radical which can further combine to give a water molecule (releasing a lot of energy). Lots of other reactions can occur. But when the mixture has a lot of water or oxygen in it it requires a large amount of energy to go back to reverse the reaction and generate an oxygen radical. If the mixture doesn't have enough thermal energy to break up a water molecule in some collisions, this reaction is irreversible. It doesn't require any energy input for most of the isolated atoms to react to these products irreversibly.

In a slightly different case, at low enough temperatures a mixture of hydrogen molecules and oxygen molecules will be stable. The molecules banging into each other at room temperature don't usually have enough energy to react or to release the oxygen radical that propagates the reaction leading to water). But it doesn't require a lot of energy input to cause them to react explosively to give water (just enough to break apart a few oxygens or hydrogens to kick start a reaction that will self-sustain because it releases a lot of energy when water is generated.)

In a system as dynamic as a mixture of gases, the possible collisions between the components will explore many possible outcomes. Some of the molecules that result are resistant to further reaction because they sit in an energy well. We call them stable. Isolated atoms don't seek stability, they just happen to react with other things very easily and many of those reaction paths lead to stable molecules. The molecules don't react easily because there isn't enough energy in the system to cause them to break apart (assuming the temperature is not high enough: hydrogen oxygen mixtures are perfectly stable at room temperature unless you are stupid enough to light a cigarette near the mixture, a mistake you will not make twice).

The situation when things are charged is different. The electromagnetic force is strong and causes large forces to exist between oppositely charged ions. These will actively attract each other and so reactions will happen faster than by the normal process of just bumping into each other. Otherwise the same rules apply. But what matter here is that ions might be individually stable as ions but there are strong forces pulling them together with oppositely charged ions into neutral assemblies. What matters is that you can't have, for example, a large clump of chloride ions by themselves. It isn't that the ions are individually unstable, it is just that strong charges produce strong forces attracting opposite charges.

In summary, when we say that something like an oxygen atom isn't stable, we mean that in the normal course of events oxygen atoms can combine irreversibly and very easily into molecules that are stable (which in turn means they have lower energy and sit in a potential well that is hard to escape from). No molecule or atom in a gas mixture seeks anything, but the statistics of molecular collisions will explore every possible potential well in the space of possible products and the deep wells will be full because the isolated atoms fall into them very easily.