# Ions Sizes Question

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#1
Hi,

I'm having trouble ordering the following ions in order of size of radius:

La^(3+), Eu^(3+), Gd^(3+), Yb^(3+).

If we were to disregard the ionization and consider the atoms, we'd get
La > Eu > Gd > Yb, since La is farthest to the left in the periodic table etc...
But when the atoms are ionized, it seems to me that only La would loose all the electrons in its outer shell when we consider La^(3+) and reach noble gas configuration [Xe]. Thus, according to my thinking, La^(3+) may be smaller in size than Eu^(3+). However, the answer to the question should be:
La^(3+)> Eu^(3+)> Gd^(3+)> Yb^(3+).

Also, I'm not sure if my notation is correct/appropriate for this forum, so comments on that too would be highly appreciated.

Last edited by OrangeJuiceBlues; 1 year ago
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1 year ago
#2
I think it’s to do with the poor shielding and penetration of the 4f orbitals. The 4f orbitals shield the nuclear charge poorly and don’t penetrate very far towards the core of the ion (look up the radial distribution function to see this). This means that when a electron is added into the 4f shell (which is where each additional electron goes as you go across the lanthanide series), and a proton is also added as you move to the element, the effective nuclear charge on the electrons (=nuclear charge - shielding constant) increases rapidly. This causes a marked decrease across the period, overriding any potential ligand field effects.

Also, regarding your comment about La^3+ being smaller than Eu^3+, this is very unlikely given that Eu is towards the centre of the period so will have a much higher nuclear charge, causing the ion’s orbitals to be more contracted giving a smaller radius. Even with La^3+ compared to Ce^3+ there is a decrease, probably do to the poor shielding of the 4f orbitals causing the effective nuclear charge to increase rapidly which reduces the ion size.

If anyone thinks this is wrong/has missed points please correct me!
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