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What do people mean when they say that when an atom has a full outer shell, it is in its "lowest energy state" and that this is the most "stable" configuration (thus it is not likely to react with other atoms)? I am a GCSE student and when attempting to research an answer, I have come across equations and physical concepts that I am not familiar with. What exactly does one mean by "energy state"?

I am familiar with the concept of entropy to a basic degree, if this has something to do with it.

I understand that the closer an electron is to the nucleus, it has lower energy, but I am not sure how this relates to my question.

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    $\begingroup$ For the ignorantati such as myself, what is GCSE? $\endgroup$
    – garyp
    Jun 25, 2020 at 13:09
  • $\begingroup$ en.wikipedia.org/wiki/… (junior high school, roughly speaking). $\endgroup$
    – richardb
    Jun 25, 2020 at 22:05

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In QM, energy levels of atoms are quantized. This means that rather than the system being able to have any energy, there are instead only discrete energy levels the system can be observed to be in. The "lowest energy state" is just the state with the smallest amount of energy. This energy can just consider electrostatic interactions between electrons and the nucleus (as you have noted), but one can also bring in other interactions, include relativistic effects, consider atoms in external electric or magnetic fields, etc. as well that give rise to other possible energy levels.

Energy levels are not caused by or related to entropy.

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  • $\begingroup$ These interactions and effects don't give rise to other levels: they simply split the ones already existing that were degenerate before the effects were taken into account. $\endgroup$
    – Ruslan
    Jun 26, 2020 at 6:22
  • $\begingroup$ @Ruslan To me those sound like the same thing. Unless you actually mean to say the energy levels split from an original level are still all the same energy level? $\endgroup$ Jun 26, 2020 at 10:49
  • $\begingroup$ I mean to say that actually there are multiple states that have the same energy (by definition of degeneracy of spectrum). And, if we analyze using perturbation theory, it's these same states that just get shifted (and reshaped a bit). $\endgroup$
    – Ruslan
    Jun 26, 2020 at 11:23
  • $\begingroup$ @Ruslan Right. I think we are saying the same thing. Just semantics at this point. $\endgroup$ Jun 26, 2020 at 11:56
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What exactly does one mean by "energy state"?

It is state at which system has definite value of energy.

To bring the idea closer to home consider some planet (with mass $m$) revolving around very heavy star (with mass $M\gg m$) due to gravitational attraction of the star. For simplicity, assume the planet can move only in circular motion, thus its velocity is given uniquely by its distance $R$ from the star by requirement of gravitational force being centripetal one: $$G\frac{Mm}{R^2}=m\frac{v^2}{R}.$$ The state of the system is then specified uniquely once the distance from the star is specified and the energy of the system is given as a sum of kinetic and potential energy of the planet: $$E=m\frac{v^2}{2}-G\frac{Mm}{R}=-G\frac{Mm}{2R}$$ So the energy state with energy "E" is the state of the system at which planet is at the distance $R$ given by the above formula.

In atoms, the idea remains but there are complications. One of them is that parameter analogical to $R$ is not continuous but discrete. So electrons can jump only between certain allowed values - orbitals. Another is pauli exclusion principle which states, that two electrons cannot occupy the same state. And yet another is the fact that there are several states with the same energy. So energy state with energy $E$ can correspond to more than one actual state (this already happens for the planetary motion as planet can move on an ellipse instead of circle and have the same energy).

Edit: I forgot one more important complication due to quantum mechanics. Because of wavy nature of QM systems, the system does not need to be in any definite energy state, but can be in a state that can be seen as composed of more different energy states. This does not happen in classical physics, as one can always compute definite value of energy. In QM this is not so.

You perhaps heard that in QM electrons don't have definite position in space, but rather they have only some probability of finding them in certain position. Sometimes, the electron can have 100% probability of being found at on single position and then we say the electron is in the "position state" with that position. If the probability is not 100% then it is not in position state. The case with energy states is analogical.

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What do people mean when they say that when an atom has a full outer shell, it is in its "lowest energy state" and that this is the most "stable" configuration (thus it is not likely to react with other atoms)?

In atoms, electrons 'live' in so-called orbitals, as described by Quantum Physics. The outer 'shell', at least in the case of the light elements, is made up of one $\mathrm{s}$ orbital and three $\mathrm{p}$ orbitals ( $\mathrm{p_x}$, $\mathrm{p_y}$ and $\mathrm{p_z}$) Each orbital can accommodate $2$ electrons (one spin 'up', one spin 'down') making a total of $8$, aka an octet. This electron configuration is sometimes symbolically represented as:

Octet structure

This 'full' outer shell structure is of very low energy and thus very stable, as is evidenced by the Noble Gases which all (with exception of $\text{He}$) have a full octet as their outer (aka 'valence') shell. These elements are chemically very inert (unreactive) towards other elements.

The high chemical stability of the alkali metal cations ($\mathrm{M^+}$) and halide anions ($\mathrm{X^-}$) is also explained by it (they all have full octets as outer shells), as well as the electron configuration of many ionic species and organic compounds.

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  • $\begingroup$ Thanks. Is there a reason why the "full" outer shell structure is of very low energy, though? $\endgroup$ Jun 28, 2020 at 19:40
  • $\begingroup$ The energy calculations for electron configurations are done by Quantum Physics. QP shows that an octet as outer shell is much lower in energy than e.g. a singlet or doublet or even a $7$ filled shell. As they say: 'shut up and compute!' $\endgroup$
    – Gert
    Jun 28, 2020 at 20:08
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    $\begingroup$ Thank you! That was really helpful. $\endgroup$ Jun 29, 2020 at 13:43

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