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With the loss of an electron from $\ce{Li}$ ($\mathrm{1s^2\ 2s^1}$), we get $\ce{Li+}$ ($\mathrm{1s^2}$), then the 2nd shell where the electron had been revolving in $\ce{Li}$ will get disappeared or remain there around the nucleus but empty after the loss of the electron?

I'm actually asking for all the other atoms as well, like generally, are shells constructed already upto the very highest ones containing f orbitals in them and just remain empty when there are no electrons in them?

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    $\begingroup$ Chem+Math formatting reference: MathJax Basics / Expressions/formulas/equations / Upright vs Italics / Math SE Mathjax tutorial // MathJax is preferred not to be used in CH SE Q titles. $\endgroup$
    – Poutnik
    Commented Jan 11 at 13:45
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    $\begingroup$ Well, the shell keeps existing in peoples minds, so nothing changes - it's only an abstraction. $\endgroup$
    – Mithoron
    Commented Jan 11 at 15:18
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    $\begingroup$ It might be easier to think of a ladder of energy levels which can still exist even if they contain zero, one or two electrons $\endgroup$
    – porphyrin
    Commented Jan 11 at 16:09

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In quantum mechanics, an atomic orbital is a function describing the location and wave-like behavior of an electron in an atom. They are pen-and-paper concepts that we use to try to explain the arrangement of electrons in an atom around the nucleus. They don't really exist in real life, per se. So, whether the single electron in an orbital is ionized or not, it never exists. But the energy level represented by that orbital is always accessible to that atom, and if a proper amount of energy is supplied to the atom, the electron in a lower subshell will excite that previously empty subshell. Truly speaking, no orbitals, whether empty or containing electrons don't exist. However, when an orbital is empty, it does not exist anymore and does not have any influence on the behaviour of the atom, until it is filled again. Proof of empty orbitals still existing (or can be made to contain to electrons again): In Peter Atkins' Physical Chemistry, while talking about transitions, he says

The Lewis concept of a ‘lone pair’ of electrons is represented in molecular orbital theory by a pair of electrons in an orbital confined largely to one atom and not appreciably involved in bond formation. One of these electrons may be excited into an empty π* orbital...

This is one of the many statements that give the proof of the existence of orbitals even if they are empty, though they do not truly affect the immediate existence and properties of the atom unless electron addition or excitation is being considered.

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    $\begingroup$ I think you are mixing some ideas in your answer. For example, reading your answer it may be understood that the energy level of $Li^+ (^1S)$ and $Li(^2S)$ are the same. They can't, one system has two electron while the second has three and you have to take that into account. $\endgroup$
    – PAEP
    Commented Sep 22 at 11:50
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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Sep 22 at 11:50
  • $\begingroup$ @PAEP what I meant is that that orbital still exists, and an electron which is in the 1s subshell can be excited to that level with the addition of energy. Hope that clarifies what I actually meant. Could you point out which line gave the impression to you so I can edit it? $\endgroup$
    – LordOfMordor
    Commented Sep 22 at 11:58
  • $\begingroup$ @LordOfMordor I get your point but the orbital does not exist. Empty orbitals are useful abstractions. Anyway, its properties will be different in both systems. In this case you should improve your answer, think that the person that reads it may not the comments or the comments may not help to clarify the answer. $\endgroup$
    – PAEP
    Commented Sep 22 at 12:04
  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented Sep 25 at 14:14

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