# Question #139d0

Jul 24, 2017

It's likely because the $6 s$ orbital penetrates more than the $4 f$ orbital (is closer to the nucleus), which somewhat blocks it from relaxing electrons down into a valid destination orbital. I don't think it's impossible to radiate... but it might be hard.

Well... for instance, platinum is an example of an element with a $4 f$ orbital. It is the lanthanides that are the first elements in the periodic table to use $4 f$ orbitals (and platinum is past the lanthanides). Let's think about what it means to radiate... it means an electron that was previously excited into the $4 f$ orbital is anticipated to relax down to a lower energy level.

A lower energy level tends to be of a lower principal quantum number $n$, and for simplicity, must be (per the selection rules) different by exactly one integer step of the angular momentum quantum number $l$ if one electron moves:

$\Delta L = {\sum}_{i} {l}_{f} - {\sum}_{i} {l}_{i} = \pm 1$
$\Delta S = {\sum}_{i} {m}_{s , f} - {\sum}_{i} {m}_{{s}_{i}} = 0$ In simpler terms, only diagonal transitions into adjacent orbital columns are allowed.

So, the $4 f$ would only be expected to be allowed to radiate down to the $\underline{4 d}$ or $\underline{3 d}$ orbitals (no $p$ orbitals and no $s$ orbitals can take this radiating electron).

The problem is, the $3 d$ and $4 d$ are supposed to be nearer the nucleus, and the $4 f$ is outer-core. Consider the radial density distributions here, which show the electron density of these orbitals: The most probable locations of the electron in the $4 d$ and $6 s$ orbitals are farther outside the atom than the $4 f$, so we might expect the $4 f$ electron to relax down into the $3 d$ orbital.

In a heavy element like platinum, there are significant scalar relativistic effects that lead to the $6 s$ orbital being contracted, since its electrons approach the speed of light.

That means the $6 s$ electron(s) penetrate the atom down into the nucleus (seen where the $6 s$ graph reaches down into the core of the atom, near $r = 0$), while the $4 f$ orbital has a hard time...

and I would expect that that makes it difficult to radiate the $4 f$ electron(s) into a valid lower energy level---because the $6 s$ electron(s) repel them away.