Is the energy of subshell s bigger than p?

1 Answer
Truong-Son N.
Apr 16, 2018

Surely not... And this can be shown quantitatively, as #"Be"# (#[He]2s^2#) has a higher ionization energy than #"B"# (#[He]2s^2 2p^1#):

[

[https://physics.nist.gov/PhysRefData/ASD/ionEnergy.html

It is easier to ionize out of a higher-energy orbital.

The angular momentum of an #s# orbital is #l = 0# (which is why it is a sphere), and that of a #p# orbital is #l = 1# (which is why it is a dumbbell). This increases the number of angular nodes in a #p# orbital of the same #n#.

Any general chemistry textbook must show this as an introduction into electron configurations.

https://www.learner.org/

Thus, #E_(2p) > E_(2s)# and #E_(3p) > E_(3s)#. Is it easier to ionize out of a #3p# orbital or a #3s# orbital?

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