Why Do Noble Gases Remain So Stable?

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SUMMARY

Noble gases exhibit exceptional stability due to their full valence shell, typically containing eight electrons, which results in a net charge of zero. This octet configuration minimizes energetic disharmony, making these atoms less reactive compared to alkali and halogen elements that seek to achieve a full octet through electron loss or gain. The discussion highlights the role of electron orbitals, specifically s and p orbitals, in achieving this stable state, while also touching on the complexities introduced by transition metals and their d and f orbitals. The inquiry into the physics behind this stability, particularly the conservation of angular momentum, remains a focal point for further exploration.

PREREQUISITES
  • Understanding of atomic structure and electron configuration
  • Familiarity with quantum numbers and orbital types (s, p, d, f)
  • Basic principles of chemical reactivity and stability
  • Knowledge of conservation laws in physics, particularly angular momentum
NEXT STEPS
  • Research the role of angular momentum in atomic stability
  • Explore the concept of electron orbitals and their configurations in detail
  • Study the differences in reactivity between noble gases, alkali metals, and halogens
  • Investigate the implications of quantum mechanics on atomic structure and behavior
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Students and educators in physics and chemistry, particularly those studying atomic theory, electron configurations, and the stability of elements in the periodic table.

jancarlo
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Hi, I am a student currently engaged in AP Physics B and AP Physics C, and I am curious as to why it is that most of the Noble Gases exhibit and exist in such a stable state. In AP Chem last year I learned that having a configuration in which the valence shell of an atom has eight electrons leads to a very electrically neutral and stable atom. I like to apply the knowledge I learn in physics to other fields, so I was just interested as to WHY the octet configuration is extremely stable. I asked my physics teacher, and he said it had something to do with conservation of angular momentum, so if anyone can elaborate on this I would appreciate it.

PS
I have looked this up in various textbooks but I have not gained the information I wish to obtain.
 
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A full octet of electrons is sort of an equilibrium point for atoms. Everything is even, there's a lot less disharmony in their configuration. All atoms on the Periodic table want to take the quickest path to a full octet they can. So, for the Alkali Metals, the quickest path for them is to lose an electron, giving them a positive charge and a full octet (or full s-orbital, in the case of Lithium). Similarly, the Alkali Earth Metals want to lose two electrons. On the other side of the chart, the Halogen family wants to pick up an electron, giving them a negative charge and a full octet.

The Noble Gases are the most stable because they have a full octet *and* a net charge of 0.

So they're basically willing to take on a charge if it means they can have a full octet, because of how it brings them to that sort of equilibrium.

Of course, the transition metals get more complicated because of their d-orbitals and f-orbitals, but that's the basis of it.
 
I understand the notion that atoms typically want to have an octet- metals like to lose electrons and non-metals, like fluorine for example, like to gain electrons to form such an octet. My question is WHAT makes an octet so favorable- in other words- why is there less "disharmony" in such a configuration, and why is it an equilibrium point for atoms? Why eight and not ten, or twelve, etc.?
 
That has to do with the electron orbitals. The smallest are s-orbitals, with each holding two electrons. The next up is the p-orbital, holding six electrons. If I recall correctly, all of the elements on the left and right sides of the periodic table have s or p orbitals as their 'outermost' orbital (there are also d and f orbitals, which are bigger, and are what trans-metals have as their outermost orbitals). Thus, filling their last set of s and p orbitals gives them an octet, which makes them happy.

This gets complicated quickly as you move into the concepts of quantum numbers and electron configuration if you haven't heard about the s, p, d, and f orbital set ups before.
 
I totally understand what you are telling me, with regards to the 4 quantum numbers and anything relating to it. What I am really looking for is a explanation from a physics standpoint. Like I mentioned earlier, what does angular momentum and its conservation have to do with such stable equilibrium?
 
Helium does not have an 'octet'. Of the remaining noble gases, only Neon and Argon have octets in their valence shell. Krypton and Xenon have 18 electrons in their valence shell, Radon has 32... not exactly 'octets' unless you define octets to mean that the outermost (valence) shell contains completely full s and p orbitals.
 
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Thanks for the clarification, but YES what I meant was the "octet" of electrons that are part of valence shell. So, what is so great about having these configurations? I am sure there has to be a reason why "eight" is the lucky number.
 
So your question is why does an 's' orbital contain 2 electrons and a 'p' orbital contain 6?
 
Not really. What I am aiming toward is more of the mechanics behind the configuration of this system. How is it synergistically balanced? Is the underlying concept conservation of angular momentum, or something else?
 
  • #10
I also want to know the answer to this. I think you're asking what are the mechanics of the atom, so that it "wants" to have 8 electrons in its outer shell. Can someone please give a good answer to this? I want to know and I feel like (no offense) no one who answered understood the question.
 

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