May 13, 2024, 05:12:25 AM
Forum Rules: Read This Before Posting


Topic: Lower energy and higher energy obitals  (Read 1899 times)

0 Members and 1 Guest are viewing this topic.

Offline Praveen27

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
Lower energy and higher energy obitals
« on: June 11, 2015, 11:25:18 AM »
We know that as n (principal quantum number) increases, the energy of nth shell increases.
But when we deal with sub-orbitals, 4f has more energy than 6s.
So, 6th shell has more energy than 4th shell, but the sub-orbital 6s ( which is obviously in 6th shell) has less energy compared to 4f ( 4th shell obviously). I am pretty confused.
This confusion arised when I was reading about lanthanide contraction and shielding effect.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3491
  • Mole Snacks: +532/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: Lower energy and higher energy obitals
« Reply #1 on: June 11, 2015, 01:44:53 PM »
This is only true for multielectronic systems. In hydrogenic atoms, the energy only depends on the principle quantum number - and the nuclear core charge. In multielectronic atoms, electrons in orbitals with different angular momentum values shield other electrons from the core charge to different extents, which changes the orbital filling order.

Chemguide has a useful explanation of filling order, especially when you start dealing with d-orbitals:

http://www.chemguide.co.uk/atoms/properties/3d4sproblem.html
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline Praveen27

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
Re: Lower energy and higher energy obitals
« Reply #2 on: June 11, 2015, 11:32:02 PM »
Thank you Corribus. Thanks a lot. So, in reality, 3d has less energy than 4s. And so, it applies to higher shells too. Which means, 4f has slightly less energy than 6s. (And 6s are also the outermost electrons). Fine. Thanks for the reference too!

Sponsored Links