Atomic Radius clarification

[icewave667]

I have a question about atomic radius. Usually, the general rule is that Cations are always smaller than Anions, meaning Anions are always bigger than Cations, do you agree?

If we are comparing two elements, let's say [Ar] and Molybdenum (Mo4+) (Molybdenum here let's hypothetically say that it has lost 4 electrons) so it is now cation.

Which atomic radius is bigger?

I would think that Argon [Ar] is smaller because it's all the way on the right side and [Mo] is closer to the left side (bigger atomic radius)

However, because Mo lost 4 electrons, the atomic radius is smaller than Ar?

This doesn't make sense to me because doesn't Mo have more shells than Ar? [meaning it should be bigger]

Can someone tell me what they think the answer is between a transition metal that has lost its electrons (now cation) and an element on the far right, which is supposed to be smaller, but in this situation, bigger.

I hope my question makes sense. Thanks

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[DrknoSDN]

Molybdenum is 2 periods lower on the table. It's going to be bigger.
http://ptable.com/#Property/Radius/Calculated

Looking at that you might get a relative feel for how radius works but basically when they are talking about cation/anion they are talking about size relative to the uncharged atom. If you lose or gain a shell it's going to drastically change the radius.

The number of shells an atom has is the biggest factor in it's radius. After that you have to do with effective nuclear charge (z) makes atoms on the right of the periodic table smaller than left atoms. Lastly losing an electron forming a cation is always going to make the atom smaller because you have the same nuclear charge (force), with less electron-electron steric repulsion within the outer shell.

Xenon vs Molybdenum... the Noble gas is much smaller within the same period.

QofQuimica explains it well in the QandA gen chem sticky found here:
http://forums.studentdoctor.net/threads/general-chemistry-topics-with-explanations.210973/