Quantum Chemistry 9.15 - Term Symbols Example 1

Short lecture on how to compute term symbols for an atom.
A ground state carbon atom has term symbols 1D2, 3P2, 3P1, 3P0, and 1S0.
Notes Slide: i.imgur.com/RGI6SH8.png
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[@Papuca3000]

Thank you SO MUCH for these videos. This was very poorly explained in the class and I coudnt find a good source on the internet, until I found your channel.

[@PunmasterSTP]

This stuff still makes my head spin, but I'm hopefully it'll all make sense someday. Thank you for making this video and sharing it, as I feel as though it's helping me on my way!

[@arsalaraichand581]

You are a life saver and make chem so much easier! Thank you so much!!!!

[@marcoponts8942]

Why do you call them "determinant"?

[@Emi-tc5ju]

I don´t understand how you get J...

[@jerryli5555]

dual standards were used in this video: when address 3D, only mL=+2 is considered, while address 1D, mL covers all +2, +1, 0, -1, -2. weird?

[@Wymiataczxp]

The 10th combination has L=1, S=0 and J=1. Shouldn't 1P1 be its term symbol?

[@aliciastaley-best7763]

I don't understand how you find the value for J right at the end there. If I follow the formula for J as |L+S| >/= J >/= |L-S| Then I get J = 1, 2 or 3 :( Please help.

[@TMPChem]

Hi Alicia. Each term symbol has its own possible set of J values. for singlet D, L=2, S=0, therefore the only value of J is 2. Likewise for singlet S, the only allowed value is J=0. For triplet P, L=1, S=1, therefore L+S = 2 and L-S=0, therefore J can be 2, 1, or 0.

[@zccau2316]

Can you explain what is the entire point of this? Is it to find peaks on spectroscopy?

[@njomzaisufaj8204]

Hi, I tried your technique for d^2 transition metals. Unfortunatly it does not eliminate the microstates after the second substraction (F-Term). Maybe i do something wrong. I would be very thankful for your help

[@musicislife665]

I tried but I think that when wee see the first tabel where ml =2 we have to think like this: along ms = 0 i have all the values for ml. then I take ml max = 2 and I see that when ml=2 ms = 0. this gives me L=2 and S=0. Multiplicity will be 1, the term symbol for L= 2 will be D and J will be just 2. Tell me if I'm wrong. Thanks.

[@123Handbuch]

So at the end of the video, you once "choose" a whole row and once a 3x3 matrix - you determine the shape of those by using (2S+1) x (2L+1) right?

[@123Handbuch]

What N do you get for L=1 and S=1/2?

[@Ranegolian]

Is it necessary to demonstrate the ones that are energetically degenerate?

[@TMPChem]

Not sure exactly what you mean. As far as the energy ranking of the different term symbols, that's done in the video following these examples on "Hund's Rules" where we can rank the term symbols based on their values of S, L, and J.

[@Ranegolian]

it was a mistake I made early on in my method, thanks for the reply!

[@andresimoes3756]

This was extremely useful, but what if we have something like Be*=1s2 2s1 2p1 ? How do we tackle the situation?

[@TMPChem]

Hi Andre. You can find out what happens in such high-spin systems in Term Symbols Example 3.

[@andresimoes3756]

I didn't notice that there were more examples, duh :P. Thanks so much!

[@raisulawal9673]

So, what will be the transition of electrons? I see the books, show the transition only from singlet to singlet or doublet to doublet or triplet to triplet and so on. I clearly understand the Term Symbol from your videos but how I show allowed transition into one diagram. I think you should make a video of transition diagram between allowed energy states. Thank you very much.

[@TMPChem]

Selection rules for atomic transitions are discussed in video 9.19 on Atomic Spectroscopy.

[@raisulawal9673]

For Na atom, we can get Term Symbols same as H atom. But in Quantum Chemistry book by Donald McQuarrie (1st edition Chapter 8, Figure 8-5, page 325), I see many allowed transitions which is far beyond my imagination. As far as I understand transition will be happened only between energy levels of Term Symbols. So,The diagram of Na atom should be focused on those Term symbols. But why the diagram is complex? I hope I make you understand.

[@TMPChem]

The issue is that H has 1 electron, whereas Na has many (11) electrons. For H, the 2s and 2p orbitals have equivalent energy (i.e. are degenerate, because all other orbitals are unoccupied). For Na, the interaction of electrons splits atomic orbital sub-shells into energetically distinct levels, creating many more transition energies that appear on the spectrum. Additionally there is splitting by J value, leading to even greater fine structure of the spectrum. None of this happens with H, where all orbitals with the same principal quantum number (n) have the same energy, greatly reducing the possible transition energies.

[@MartinBeseda]

Thank you very much for the video! I'm just not sure, which microstates belong to which term symbol. For example, which specific microstates belong to 1D? I can see, that 7th and 9th will, but the remaining three are not distinguishable (e.g. there are 3 states with L=0 and S=0). Thank you for your explanation.

[@TMPChem]

Hi Starec. I don't believe we can say for sure which determinants belong to which term symbol. All we can say is that the singlet D term symbol has 5 states, each of which is a linear combination of states which have +2, 0; +1, 0; 0, 0; -1, 0; and -2, 0 values for mL and mS, respectively. I believe it's arbitrary which 0, 0 state you select to be the 0, 0 state in 1D, and which one is in 1S.

[@MartinBeseda]

@@TMPChem Ok, thank you! If I understand it well, every term/level can be written as a linear combination of several microstates. Do you have any videos about it? Simple use of Clebsch-Gordan series would be great!

[@eriknelson2559]

@@TMPChem Those linear combination "super-positions" must also be anti-symmetric, to enforce Pauli's Exclusion Principle? And the fact that the Atomic State is an (anti-symmetrized) joint state, and not a simple product state (not a simple product of the individual microstates), implies that the valence electrons have become quantum entangled (through intimate contact) ? The properties of one depending crucially upon the other?

[@eriknelson2559]

​@@TMPChem Could you use Slater Determinants to combine the possible "determinant micro-states" and compute the expectation values of L^2, L_z^2, S^2, S_z^2 for said "Slater states" ?

[@123Handbuch]

You never have 1/2 values for L and S, can you make an example with let's say Hydrogen? Or Lithium?

[@123Handbuch]

I think I got it but I'd still recommend adding it as another example. :) I know you have a video about it but using that technique would be anice one too.

[@tbow2912]

Excellent explanation !

[@santiagomachadovelez4131]

i love you with all soul

[@loumnietwixe7438]

Thank you !!! You saved my exams ! haha

[@TMPChem]

Thanks Loumnie. Good luck on the exam.

[@PunmasterSTP]

Hey I know it's been half a decade, but I just came across your comment and was curious. How'd your exams, and the rest of your academic career, go?