Determining the term symbol of Sc

In summary, the term symbols of the ground states of the following atoms are: a) H, b) F, c) Cu, d) F- , e) P, f) Na, g) Sc
  • #1
physicisttobe
56
13
Homework Statement
electron configuration of SC
Relevant Equations
...
I've solved further Term Symbol problems. The task is: derive the term symbols of the ground states of the following atoms: a) H, b) F, c) Cu, d) F- , e) P, f) Na, g) Sc

Why is the d-orbital considered here in g)? Why isn't the s orbital taken into account here? The 4s orbital is energetically higher than the 3d orbital, so why don't you look at the 4s orbital. I also post the solution and my approach to this problem. I can't explain why my approach is incorrect. Can someone help me?

my approach:
C5D1F174-C00B-4242-A072-B400DE853580.jpeg


solution:
ECB91335-C5CF-470D-94EE-3DA783A57DF4.jpeg
 
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  • #2
The s orbital is full. The net contribution of all filled subshells to L and S is zero, so you only need to consider partially filled subshells. (And I assume 3d14 is a typo for 3d1.)

Oh, and if you meant Sc you should have said so. SC is something very different!
 
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  • #3
mjc123 said:
The s orbital is full. The net contribution of all filled subshells to L and S is zero, so you only need to consider partially filled subshells.
All right, but what about F-? We have the configuration [He]2s^2 2p^6, both orbitals are full.
I thought it would be 1P0 but it's wrong. The solution is 1S0, but I can't explain why 1S0.
I solved this problem like this:

EE4B9EA0-09D4-4259-8706-EEB25FFD77E8.jpeg

mjc123 said:
(And I assume 3d14 is a typo for 3d1.)
Yes, it's a typo, sorry for that.
mjc123 said:

Oh, and if you meant Sc you should have said so. SC is something very different!
Oh, sorry, I mean Sc, of course.
mjc123 said:
The s orbital is full. The net contribution of all filled subshells to L and S is zero, so you only need to consider partially filled subshells. (And I assume 3d14 is a typo for 3d1.)

Oh, and if you meant Sc you should have said so. SC is something very different!
 

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  • #4
physicisttobe said:
All right, but what about F-? We have the configuration [He]2s^2 2p^6, both orbitals are full.
I thought it would be 1P0 but it's wrong. The solution is 1S0, but I can't explain why 1S0.

What is your reasoning for selecting ##P## in your solution ##^1P_0##?
A full orbital subshell has zero net orbital angular momentum.
 
Last edited:
  • #5
physicisttobe said:
Oh, sorry, I mean Sc, of course.
{Fixed in thread title now)
 
  • #6
TSny said:
What is your reasoning for selecting ##P## in your solution ##^1P_0##?
A full orbital subshell has zero net orbital angular momentum.
All right, I get it. Thank you so much for your help!
If any questions arise, I'll get in touch here
 
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1. What is the term symbol of Sc?

The term symbol of Sc is ^2D3/2, which represents the ground state of scandium.

2. How is the term symbol of Sc determined?

The term symbol of Sc is determined by considering the electronic configuration of the atom, specifically the number of electrons in each subshell and their spin orientations. This information is then used to construct the appropriate term symbol using the Russell-Saunders coupling scheme.

3. What is the significance of the term symbol of Sc?

The term symbol of Sc provides important information about the electronic state of the atom, including the total angular momentum, orbital angular momentum, and spin angular momentum. It also helps in predicting the spectroscopic properties of the atom.

4. How does the term symbol of Sc differ from other elements?

The term symbol of Sc is unique to scandium and is determined by its specific electronic configuration. While other elements may have similar term symbols, they will differ in their quantum numbers and energy levels.

5. Can the term symbol of Sc change?

Yes, the term symbol of Sc can change if the electronic configuration of the atom changes. This can occur through the absorption or emission of photons, which can cause electrons to jump to different energy levels and alter the term symbol. However, the ground state term symbol of Sc will remain ^2D3/2.

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