Atomic Radius: Determining Size through Electron Configuration

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    Atomic Radius
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The discussion centers on the factors influencing atomic radius, specifically comparing potassium ions and argon. The key points highlight that the atomic radius is determined by the atomic number, electron shielding, and relative orbital ranks. In the case of potassium cation and argon, both have the same electron configuration, but potassium has one more proton. This additional proton increases the nuclear charge, resulting in a stronger attraction between the nucleus and the electrons. Consequently, the valence electrons in potassium are held closer to the nucleus compared to those in argon, leading to a smaller atomic radius for potassium cation than for argon. The discussion emphasizes the relationship between nuclear charge and atomic size.
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This is pretty simple, just give me a quick concept should be enough, no need to go over the details.

What determines the radius size again? Was it just the electrons?

Say for example which is bigger, potassium ion or argon. Same electron configuration but different proton numbers. Would the extra proton pull the electrons to make it smaller for potassium or would the extra proton make the radius bigger. This isn't homework or anything, use another example if you want.
 
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Here are some of the factors:

1) Atomic number
2) Electron shielding
3) Relative orbital ranks

Back to your question,
Potassium cation has the same electron configuration as argon. (i.e., 1s^2 2s^2 2p^6 3s^2 3p^6).
The extra proton found within the potassium nucleus will, in effect, produce a stronger nuclear charge (positive) to attract each electron. Relatively speaking, argon will have a weaker nuclear charge due to a smaller proton quantity (since 18 < 19).

Therefore, the valence electrons of potassium cation will be held closer to the nucleus than the valence electrons of argon.

Therefore, potassium cation will have a smaller atomic radius than argon.

(In your question, you simply end up comparing atomic numbers.)
 
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thanks, that was very insightful.
 
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