Lanthanum has electronic configuration [$\ce{Xe}$] $4\mathrm{f}^0$ $5\mathrm{d}^1$ $6\mathrm{s}^2$. Why is the $4\mathrm f$ sub-shell not filled even though we know that $4\mathrm f$ is filled before $5\mathrm d$ sub-shell?

I know that for some elements energy level of $4\mathrm f$>$5\mathrm d$. Why is this so?

  • 1
    $\begingroup$ If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected the author elaborates the topic in the question by at least basic own topic review and writes what he/she understands and what is the eventual stumble stone. The "quick questions" without explicitly expressed particular effort are not very welcome, and may be closed. $\endgroup$ – Poutnik Jul 26 '20 at 8:54
  • 1
    $\begingroup$ Question titles should be like book titles. They distinguish the book from other books, give a hint about the content, but do not tell the full story. The content then should elaborate the topic to full depth the author is able to do. By other words, the title too long, the body too short. $\endgroup$ – Poutnik Jul 26 '20 at 8:55

Remember that energy of electrons in orbitals do not have fixed values, but depends on occupation state of other orbitals.

So if we in a thought experiment fill orbitals to get a neutral atom, 1 electron in time, some orbitals switch the order of their energies.

It has the effect the 4s orbital is filled first, having lower energy, but after some 3d orbitals are filled, energy of filled 4s Is higher then energy of 3d.

Similar case happens between 5f and 6d orbitals for first actinoids.

  • $\begingroup$ Sir can you elaborate it, I didn't understand. $\endgroup$ – Amandip Dutta Jul 26 '20 at 12:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.