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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.

danej317
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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 '19 at 17:23
  • Could you elaborate? This doesn’t really explain the “why,” even if I have it backwards. – danej317 Jan 31 '19 at 17:31
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    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 '19 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. – knoftrix Jan 31 '19 at 18:06
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    @SauravSingh The question is certainly on topic on [Chemistry.se], even if there are misconceptions within it. – Martin - マーチン Jan 31 '19 at 18:21
  • Related: Hypervalency and the octet rule – Martin - マーチン Jan 31 '19 at 18:21
  • Possible duplicate of Why are atoms with eight electrons in the outer shell extremely stable? – Mithoron Jan 31 '19 at 18:51
  • This is imho a great question (even if probably a duplicate) and Andrew does a good job of answering. The intuition seems right, you have to consider what the electron was doing before it entered the Cl atoms shell. The chlorine $is$ excited when it accepts the electron, relative to the configuration with the electron at infinite distance, but that is countered by the electron being lower in energy residing closer to the nucleus thanks to the electrostatic interaction. – Buck Thorn Jan 31 '19 at 18:52
  • I suggest reading https://chemistry.stackexchange.com/questions/26833/whats-the-relation-between-the-octet-rule-and-quantum-exclusion-and-how-is-it-r and the answer by Wildcat – Alchimista Feb 02 '19 at 15:08
  • Reading carefully thus rather than too broad should have been closed as a duplicate. – Alchimista Feb 02 '19 at 15:10
  • I was very aware of similar problems being asked, but they all focused on how quantum numbers allow 8 electrons in a shell. I was instead curious about why rudimentary electron orbital diagrams and basic approaches to ionization didn't seem consistent with literature ionization energies and electron affinities. This is why I didn't think my question was a duplicate and why I appreciate Andrew's answer—the electrostatic potential was something I overlooked and clears up the energetics I was confused about. Wildcat's answer on a related post was also helpful in outlining binding energy effects. – danej317 Feb 03 '19 at 17:47
  • I also would like to say I tagged quantum mechanics just because I wanted to apply what I've been learning in QM to a rule of thumb that freshmen are taught to blindly accept. I was curious about how quantum mechanics would go about explaining the drive for ionization in order to satisfy the octet rule. Maybe I'll post this another time. – danej317 Feb 03 '19 at 18:13
  • Maybe you should rather read https://chemistry.stackexchange.com/questions/710/fundamental-forces-behind-covalent-bonding – Mithoron Feb 04 '19 at 00:02

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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 $\frac{q^+q^-}{r}$.

Where molecular orbital theory comes in is when we calculate the charge and the distance r. Simplistically, orbitals that are higher in energy on the same atom are higher in energy because they have a longer average distance from the nucleus (denominator in PE term is bigger) and because there is more electron density from other electrons between them and the nucleus, so the "effective" positive charge of the nucleus is reduced (numerator in PE is smaller). This is called "shielding".

Slater's rules are simple heuristics that provide a fairly good approximation of the effective nuclear charge experienced by an electron based on the atomic number and orbital.

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.

Andrew
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