Why does the $2p$ orbital have lower energy than $2s$?

Asked Jun 28 '21 at 12:42

Active Jun 30 '21 at 08:53

Viewed 73 times

I know there already are questions about this. But I find some error there. So, I asked again....

The answers I have read mostly point out that electrons in 2s orbital have a higher probability to be near to the nucleus than 2p....and so they experience all the charge of nucleus, and shield the 2p electrons from experiencing it.

But, as the radial probability graphs indicate, 2p electrons have a lesser average distance from the nucleus than 2s. Also, the most probable distance of 2p is lesser the most probable distance of 2s. So, mostly, 2p electrons would be nearer to nucleus. And so, shouldn't THEY (2p electrons) shield 2s electrons instead?

I am new to QM....so please give a satisfactory explanation.

edited Jun 30 '21 at 08:53

Qmechanic

201,751

asked Jun 28 '21 at 12:42

user95732

Are we talking about hydrogen and a pure Coulomb field? If so then $2p$ and $2s$ have the same energy: $-13.6/4$eV. – ZeroTheHero Jun 28 '21 at 12:45
No, I am talking about multi-electronic atoms – user95732 Jun 28 '21 at 12:47
2

As you move up the periodic table, the 2s shell fills before the 2p shell: the electron configuration for beryllium is $1s^2 2s^2$, while boron adds a p-shell electron. That suggests the 2s shell is the more tightly bound, lower-energy state, and the premise of the question is wrong. – rob Jun 28 '21 at 13:08
3

Closely related: Why are the higher angular momentum states of a hydrogen atom closer to the nucleus? – Emilio Pisanty Jun 28 '21 at 13:10

Why does the $2p$ orbital have lower energy than $2s$?

0 Answers0