Question about Friday's lecture

[205743684]

Hi! I remember from my notes on Friday's lecture that Dr. Lavelle mentioned that after the atomic number (Z) = 20, the d is lower in energy than s. I am a bit confused what this means and how we have to take into account this information when writing the electron configuration. For example, in 1E.17 For each of the following ground-state atoms, predict the type of orbital (1s, 2p, 3d, 4f, etc.) from which an electron will be removed to form the +1 ion: (a) Ge; (b) Mn; (c) Ba; (d) Au.

For b) I put the answer as 3d, yet the correct answer is 4s. Why is this the case and is it an exception or if we have both d and s, we must write d before s always (thus removing an electron from s). How does this relate to the Z=20 information above?

I hope I am being clear!
Thank you,
Nicolena

[Carina H - 2G]

Hi Nicolena!

I think if I understand correctly, the reason that we put d before s even though technically the s shell fills before d does (if we think about it chronologically) is because 3d is technically a lower energy level than 4s. So, when we write electron configurations, we go by order of energy rather than the order the elements on the periodic table express themselves. In terms of Z=20, that just refers to the atomic number. Z=20 is where there is a 4s orbital, and after that number, we must incorporate the d orbital into our writings of electron configuration.

Hope this helps!

[ashna kumar 3k]

After Z=20, we enter the 3d block where the 3d orbital becomes lower energy than 4s. When we fill electron orbitals, we follow the periodic table in which 4s comes first. However, once we start filling 3d, it becomes closer to the nucleus and thus lower in energy than the 4s orbital. So, for Mn, the valence electrons are actually the 4s shell even though we filled it first as they are higher in energy.

Hope this makes sense!