Why is 4s before 3d for electron configurations?

1s 2s 2p 3s 3p 4s 3d 4p
Why do electron configurations have 3d orbitals out of order from the rest?
Here we break down how radial and angular nodes create a situation where 3d orbitals experience a greater repulsion from other electrons than 4s electrons do. This is in spite of the fact that 4s have a further average distance from the nucleus.
The end result is that 4s and 3d orbital states have similar energies and this creates scenarios where the lowest energy configuration depends on a myriad of factors (nuclear charge, quantum spin pairing, etc.)
This also can help explain why some of the exceptions to Aufbau are seen such as Cr being a 4s1 3d5 instead of 4s2 3d 4.
If you are interested in this video, you might like my new book Teaching Introductory Chemistry www.amazon.com/dp/B0B6XQBC6T

[@hassanhamad1280]

This was a really well done video, thanks.

[@ceejay1684]

Thank you. This video was extremely helpful

[@Life-zu2tv]

Great... Keep up the good works

[@wristocrat]

What an incredible teacher. That high tech board is really cool tool. Definitely added to my studying routine.

[@thomasadams1011]

Student here, please can you confirm my understanding of the logic?
- an e- further away from the nucleus experiences more repulsion than if it were closer (due to the analogy you give)
- we would expect the electron to be found in the lower repelling region, and therefore you'd EXPECT the electron to 'exist' closer to the nucleus (lower value of r, radius)
- because the 4s orbital has radial nodes, it also has peaks of probability (I'm refering to what you are saying at ~6:51) that are closer to the nucleus
- this means that an e- would require less energy, and therefore more likely to be found in the 4s orbital (more specifically in the earlier peaks shown), rather than the 3d orbital
btw, this is the first video I've seen of yours and I really enjoyed it :)

[@blandon93]

Schrodinger's electron should be the answer. Since the peaks are really close, the electron could be found in 3d and 4s. Keep in mind electron is also a wave.

[@AMGAMER90]

Extreme helpful sir

[@justdoit83388]

Excellent.

[@marie-laureblais-crepeau8932]

Great explanation! I’m trying to figure out how to explain this to my students and you video really helped clear some things up for me.

[@aaronpicking4935]

This video made me certain that chemistry was the right choice to study ! Thank you sir

[@lucie4185]

Very Nice, I am very jealous of that whiteboard system too.

[@CarlosEduardo-by7lq]

Amazing 🇵🇹

[@tari7555]

oh i see , that helped a bit thanks

[@ibchemvids]

Scott, great video and explained clearly. I am an IB chemistry teacher who realises that you know a heck of a lot more "real" chemistry than me! I have a question since you are the orbital guru - 1st IE along period 3. Sulphur has a lower IE than expected because it has paired valence 3p electrons, so they repel , making the e- "easier" to remove. Why is this not true of the P and Ar as well? Or is it true? Cheers,
Rich

[@APphyzicks]

Hey Rich, I assure you I know who you are and am quite the fan of your work. I'm assuming you meant Cl and Ar. The impact of pairing does cause the 1st IE of both of those to lower due to the increased electron repulsion, but the additional protons also increase the IE.
This article made a lot of sense to me but someone else I know who knows a lot more than me was skeptical about the explanations given. pubs.acs.org/doi/pdf/10.1021/ed077p1056

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[@spacespace2698]

But dear teacher what's the important of these in our daily life?

[@silentjoe4745]

Depends on how high your aspirations are.

[@gamagarcia3037]

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[@anasibneahsan3493]

nice

[@anasibneahsan3493]

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[@EMPERENIKA]

LET HIM COOK!

[@xxblackjamxx4368]

Well
I’ll add an hour of this to my 24 hours of studying then ;-;

[@anishnallani5918]

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[@ishankumar4471]

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