While I was preparing for my upcoming exams, I stumbled upon this sentence which is bothering me quite a bit:

The contraction of the lanthanoids is due to the imperfect shielding of one electron by another in the same subshell.

Though I am well aware of lanthanoids contraction and the poor shielding effect of the f- and d-orbitals, but still according to what I have studied and understood so far is the shielding effect is the shielding of valence electrons by the electrons between the nucleus and the valence electrons from the nuclear charge.

Then how can an electron shield another electron of the same subshell?

  • $\begingroup$ @Zhe So does this actually effect the atomic radus of the lanthanides $\endgroup$ – ayush paul Feb 9 at 20:15
  • $\begingroup$ Normally, as you add electrons, after adding several, you'd need to expand to a higher energy -- and larger -- orbital. But with lanthanides, you're basically sticking all of the electrons in the same energy level where shielding is imperfect. So the shielding does not grow commensurate with the increase in nuclear charge. $\endgroup$ – Zhe Feb 9 at 22:43

Shielding is simply the cancellation of some of the effective charge of the nucleus.

It's a lot easier to envision an electron to shield another electron that, by time average, is likely to be farther from nucleus, e.g., $1s$ shielding $3p$.

But for 2 electrons in the same subshell, on average, you can expect one to be closer to the nucleus than the other roughly 50% of the time. While it's true that a good deal of the time that other electron may be on the other side of the nucleus, for d- and f- orbitals, we're looking at more than 1 electron, so there's some partial shielding. Partial, because the shielding electrons don't actually shield a good portion of the time due to relative distances from the nucleus.


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