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If an electron is removed from an atom of $\ce{K}$, why does it then have 0 valence electrons as it states in my General Chemistry textbook? I would think that if this happened, it would have the same exact electron configuration as $\ce{Ar}$, which has 8 valence electrons, correct? What's going on here?

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What you say is correct. The [Ar] configuration we are left with does have 8 valence electrons. But I think it is just semantics. Elemental potassium has an [Ar] 4s1 electron configuration. One would say it has one valence electron. If we take that one valence electron away, it makes sense to say that it now has zero valence electrons because "1 - 1 = 0".

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You are correct in that it has 8 valence electrons, not 0. If we look at the electron configuration of potassium (K), we see that it has one electron: [Ar]4s1. Obviously, removing that electron gives us [Ar] (same configuration as K1+), which is a noble gas and has 8 electrons.

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    $\begingroup$ The definition of a valence electron is an electron involved in bonding. As the electrons in the 3rd shell of K are not involved in bonding, it does not make sense to call them valence electrons. K forms metallic bonds with its one 4s electron, and ionic bonds by giving it up to form K+. On the other hand, argon does use the electrons in the 3rd shell to form (covalent/dative) bonds, in the very limited number of compounds it forms. This (3) is the longest list I could find endmemo.com/chem/common/argon.php $\endgroup$ – Level River St Sep 29 '14 at 14:00
  • $\begingroup$ So you're saying that even though it has a full octet, the electrons in the outer shell in this case are not considered to be valence electrons? What about an anion with a full octet? $\endgroup$ – Ian Sep 29 '14 at 20:02
  • $\begingroup$ Yes, in an anion with a full octet, the electrons are involved in bonding, so they are valence electrons. Cl- has a full 8 valence electrons (one of which is perhaps captured from a K atom.) BTW, they're still available to be donated to O atoms to form anything up to ClO4-. On the other hand, a K atom had a valence electron but lost it to become K+. That valence electron (if it could be identified) is now somewhere else, orbiting the nucleus of an anion. $\endgroup$ – Level River St Sep 29 '14 at 20:19
  • $\begingroup$ Ar is isoelectronic with Cl-, but due to its lack of charge, it is very difficult to get it to share its valence electrons to form species like ClO4-. K+ is isoelectronic with Cl- and Ar, but due to the positive charge it is literally impossible to get K+ to share or give up its 3s electrons (except in the plasma phase), so they are not "bonding" or "valence" electrons. $\endgroup$ – Level River St Sep 29 '14 at 20:23
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Valence electrons are generally regarded as being 'the outermost electrons' for a given atom. Therefore, with neutral potassium, there is one valence electron. If we take away the outermost electron, we now have a new set of outermost electrons being the 8 electrons in the 3s and 3p orbitals. Ron is correct in his assessment of semantics.

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I think you need to consider that if you take 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 can be empty then you can have 0 valence electrons - that's why your book says what it says.

If I'm not right it will be a lesson for me too :)

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