6.5 Electronic Configurations

[@thatbossguy123]

Hey Chad, just really want to thank you for all of your help this year! I found you while taking Orgo this year and you saved my ass a couple times on things like substitution and elimination reactions haha. I have taken gen chem three times now in my life: Honors chem in Highschool, AP Chem in highschool and college gen chem last year. Throughout the 3 times I have learned about quantum numbers I never truly understood the concept but this video cleared so many things up for me! Not only this video but all your other videos always clear up my confusion and I am able to completely understand each concept so well now! You are really such an awesome teacher and I will be continuing to watch your entire gen chem series as I study for the MCAT. Take care! Best, Greg

[@ChadsPrep]

Awesome Greg and thanks for taking the time to leave such a thoughtful response! And best wishes on the MCAT!!!

[@thatbossguy123]

@@ChadsPrep Thanks Chad!

[@aryanerry7004]

Hey Chad, first of all a huge thanks for your videos, they are really informative and the way you teach is phenomenal! However, I had one small doubt that I was hoping you could answer. Just like the 4s electron has been transferred to the 3d orbital in case of Chromium and Copper, why is the 2s electron not transferred to the 2p orbital in Fluorine and other halogens? Shouldn't Fluorine also follow the same pattern as Copper and Chromium? I did a little reading and saw that it has to do with the exchange energy, but in case we fill up the 2p orbital for Fluorine using 1 electron from the 2s orbital, it turns out that the new configuration has more exchange energy than the one in which the electron is not transferred. So why is this the case?

[@ChadsPrep]

Hello Aryan! It really comes down to the fact that the difference in energy between the 4s and 3d orbitals is relatively small whereas that between the 2s and 2p is larger. We generally consider the 4s orbital to be lower energy than the 3d orbital as we fill the 4s first before filling the 3d as in the examples of potassium and calcium. But once you begin adding electrons into the 3d orbitals this actually switches. Due to the different shapes of the s and d orbitals electrons experience repulsion from the other electrons in the atom/ion differently. Due to this difference in the experienced repulsion (we often call this shielding) an electron now in a 3d orbital is lower energy than an electron in a 4s orbital. This is why when we ionize a transition metal we remove the 4s electrons before the 3d electrons...the 4s are higher energy and easier to remove. And this difference in the energies that takes into account the difference in shielding is what we actually attribute the electrons filling differently for Cr and Cu (and others). So the energies of electrons in particular for transition metals are more complicated than we typically present in a undergraduate freshman class, and we often get closer to the truth in an advance inorganic course often taught to senior chemistry majors. So back to the period 2 elements. The 2s and 2p orbitals are not as close together in energy as the 4s and 3d orbitals and doesn't switch for different elements like the 4s and 3d do for the transition elements. As a result any difference in shielding between an electron in a 2s orbital as compared to a 2p orbital isn't significant enough to make a difference with the larger gap between the 2s and 2p orbitals which is why we don't see any sort of 'exceptions' in the period 2 elements like we do in the transition metals. So the truth is complicated and even more complicated than this. Note that we often only require students to know 5 exceptions for ground state electron configurations (Cr, Mo, Cu, Ag, Au), but there are numerous other exceptions and somewhere along the way somebody decided that it wouldn't be helpful to require freshman to know all these exceptions or to rigorously understand why there are exceptions in the first place. I think they did us all a favor! Happy studying!

[@aryanerry7004]

@@ChadsPrep Thanks a lot for clearing my doubt!

[@Azzarrati]

Hey Chad, thanks for the info. But what about Copper? How many valence electrons? Is it 1 ? As 3d is 10 because of the exception, left with 4s1. So that means is it 1 or is it 11 4s2 3d9 (9+2). Thanks

[@parwanalutfi8044]

Hi Chad! What happens in the scenario that you have to eliminate electrons from the higher shell number and stabilize the d shell. Do you stabilize d shell first and then eliminate from the s shell?

[@ChadsPrep]

Nope. Don't try to apply the filled or half-filled d subshell exceptions to cations when you're removing electrons. Simply remove the higher shell electrons first no matter what (like removing 4s before 3d). Hope this helps!