# How are octet-rule models supported by thermodynamics and quantum theory? [closed]

I understand that the four quantum numbers dictate how many electrons can "fit in a shell," but I don't understand how gaining a full octet is particularly energetically favorable.

The example of the ionization of sodium seems to make sense: the atom loses a single 3s electron, and the energy of the system is lowered significantly down to the 2p energy level.

However, the ionization of chlorine isn't as obvious to me: it is indeed granted its "full octet," but filling the last available space in the 3p suborbital would not lower the energy of the system. Rather, it would raise it slightly, as the newly added electron would experience some electron/electron repulsion.

Is the main drive for this ionization "reaction" the drop in energy in the formation of the cation? Do I have to consider molecular orbital theory to answer this question for covalent interactions?

I'm mainly wondering why there's a drive to form a full octet, particularly for the halogens, chalcogens, and pnictogens.

## closed as too broad by Mithoron, A.K., airhuff, Soumik Das, TyberiusFeb 1 at 15:41

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• You have it backwards. No element gives up an electron willingly. Those which achieve a full octet by doing so (like Na) are just the least unwilling. On the contrary, Cl and many others would take an extra electron quite eagerly. – Ivan Neretin Jan 31 at 17:23
• Could you elaborate? This doesn’t really explain the “why,” even if I have it backwards. – danej317 Jan 31 at 17:31
• Well, not much can be said besides the obvious. One atom wants an electron really badly, and another is like "alright, dude, I'd give you mine, but I sorta need it myself". Then comes a huge energy of crystal lattice formation, and that's the main drive. – Ivan Neretin Jan 31 at 18:06
• This question is off topic or not explained fully if you are asking 'why does it quantum theory or thermodyanamics affect stability of electronic configuration' then you must know a lot about quantum mechanics or thermodynamics or studying it but your explanation does not reflect that. You are comparing your opinions based on 'old theories' with quantum mechanics. You have to consider quantum mechanics if you want answer based on quantum mechanics same for thermodyanamics. Also no explanation is given in MOT or any 'old theories' it is just an observation. – Saurabh Singh Jan 31 at 18:06
• @SauravSingh The question is certainly on topic on Chemistry, even if there are misconceptions within it. – Martin - マーチン Jan 31 at 18:21

You can't really compare the energy of $$\ce{Cl-}$$ to $$\ce{Cl.}$$, since the two states have different composition. Instead, you're interested in the difference in energy between $$\ce{Cl-}$$ and $$\ce{Cl. + e-}$$, where the electron is completely free and resting in a vacuum with no interactions with anything else. We give that electron a reference potential energy of 0. Then the question becomes, how much potential energy is lost by interacting with various atoms, and the thermodynamically favorable state is the one with the most negative potential energy.
If we consider nuclei to be point positive charges and electrons to be negative charge distributed in a much wider space around the nucleus, we find that as that free electron gets closer to the nucleus, it experiences an electrostatic interaction with the positively charged nucleus even if the atom overall is neutral. We can simplistically estimate the magnitude of the potential energy loss based on basic principles of electrostatics, where potential energy is a function of $$\frac{q^+q^-}{r}$$.
Conceptually, then, the answer to your question is that adding an electron to a neutral atom like $$\ce{Cl.}$$ is most favorable when the electron experiences the best combination of effective nuclear charge and short average distance to the nucleus. This happens when adding the last electron to an octet because that is the element of that period with the greatest nuclear charge. Adding an electron to the next neutral element Argon is much less favorable because it has to go in the next shell (n=4) meaning that it has a larger r (larger denominator) and has much more electron density between it and the nucleus, so the effective nuclear charge is less (smaller numerator) even though the atomic number of Ar is one more than Cl.
For $$\ce{Na+}$$, it isn't correct to say that the system has lower energy after an electron is removed, but rather that it takes a lot less energy to remove that electron than an electron from neutral neon, which hopefully makes sense based on the discussion above.