If one electron is gotten rid of from an atom of $\ceK$, why does the then have actually 0 valence electrons as it says in my basic piersonforcongress.com textbook? I would certainly think that if this happened, the would have the same specific electron configuration together $\ceAr$, which has actually 8 valence electrons, correct? What"s walking on here?

What you say is correct. The construction we room left v does have 8 valence electrons. However I think that is just semantics. Elemental potassium has an 4s1 electron configuration. One would say it has one valence electron. If us take that one valence electron away, it makes sense come say that it now has actually zero valence electrons due to the fact that "1 - 1 = 0".

You room correct in that it has actually 8 valence electrons, not 0. If we look at the electron construction of potassium (K), we check out that it has one electron: 4s1. Obviously, removing that electron provides us (same configuration as K1+), which is a noble gas and has 8 electrons.

Valence electrons are generally regarded together being "the outermost electrons" for a given atom. Therefore, through neutral potassium, there is one valence electron. If we take far the outermost electron, us now have actually a brand-new set of outermost electrons being the 8 electron in the 3s and also 3p orbitals. Ron is exactly in his evaluate of semantics.

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I think you require to think about that if you take it last electron from the orbital it"s there, but it"s empty. Orbital itself is a bit abstract thing and if you accept that it deserve to be empty then you have the right to have 0 valence electrons - that"s why your book says what that says.

If I"m not best it will certainly be a lesson because that me too :)

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