The “strong field” vs “weak field” nature of a ligand isn’t the only factor when determining high spin vs low spin. The metal and charge have an influence as well, as is the case here. Typically 2+ charged first row transition metals have smaller octahedral splitting values compared to 3+ charges or second and third row metals. The only way that you can ultimately determine the spin state of a metal is by performing the experiments (UV-Vis, magnetometry, etc.)
I have this doubt in a recent test. In this article they do experimental and theorical calculations and give the following result "For cobalt (II) tris-2,2’-bipyridine our combined experimental and computational study reveals ~40% low-spin and ~60% high-spin state components". It is a "tris" compound, but still... bib-pubdb1.desy.de/record/428846/files/CoBpy_02Mar2019.pdf
First of all thank you very much for this tutorial, ist was indeed very usefull. However I still have a question. Why didn't you take the 4T1g to 4A2g excitation in consideration but instead only focusd on the 4T1g and 4T2g ecxitations?
nicolas dobler The slope of the 4A2g is around 2 which is for a two-electron transition. They are less likely and therefore not typically observed in an electronic absorption spectrum, whereas the 4T2g has a slope of one, indicating a single electron transition, which is much more likely to be the correct assignment in this case.
Eric, Thank you for the good presentation. The ground Mullicken term for d7 low spin complex should be 2Eg. Right? Here you have shown it as 2T2g. Please clarify.
I'm kind of cunfused as to why the calculated octahedral splitting is larger than the lowest excitation wavenumber. How can a transition occur if the energy that is necessary to promote the electron is greater then the energy I put into the system?
Short answer: The energy of observed absorption isn't strictly due to the promotion of an electron from a t2g to an eg orbital as would be expected if the energy was only related to the octahedral splitting energy, but rather the change in state of the system from the 4T1g microstate (ground state) to the 4T2g microstate (excited state). I will try to post a video in a day or two, and go into more detail on why the lowest energy transition doesn't directly equal the octahedral splitting energy for any system that isn't d1.
![Orgel Diagram - How to use them [Easiest Explanation]](http://i.ytimg.com/vi/XfO6x4_wCfU/mqdefault.jpg)
Thank you for being the only person on here with a clear English speaking lecture on the topic
i've been reading the housecroft and kettle books but none of these were as clear as this video!! keep it up, congrats!!
and thank you very much.
dude thank u so much u literally saved my life
Thanks for clearly pointing out how to use the diagrams
the way you write 8 is disturbing, but that was very helpful, thank you :D
I thought bipy was a strong field ligand. Wouldn't that make it low-spin?
The “strong field” vs “weak field” nature of a ligand isn’t the only factor when determining high spin vs low spin. The metal and charge have an influence as well, as is the case here. Typically 2+ charged first row transition metals have smaller octahedral splitting values compared to 3+ charges or second and third row metals. The only way that you can ultimately determine the spin state of a metal is by performing the experiments (UV-Vis, magnetometry, etc.)
If only I knew this before my inorganic exam last year 😂 thanks for the reply, very informative!
@@katiesilverthorne8490 sorry, I just now saw it!
I have this doubt in a recent test. In this article they do experimental and theorical calculations and give the following result "For cobalt (II) tris-2,2’-bipyridine our combined experimental and computational study reveals ~40% low-spin and ~60% high-spin state components". It is a "tris" compound, but still...
bib-pubdb1.desy.de/record/428846/files/CoBpy_02Mar2019.pdf
had*
this is short and super cool!
First of all thank you very much for this tutorial, ist was indeed very usefull. However I still have a question. Why didn't you take the 4T1g to 4A2g excitation in consideration but instead only focusd on the 4T1g and 4T2g ecxitations?
nicolas dobler The slope of the 4A2g is around 2 which is for a two-electron transition. They are less likely and therefore not typically observed in an electronic absorption spectrum, whereas the 4T2g has a slope of one, indicating a single electron transition, which is much more likely to be the correct assignment in this case.
@@EricVictor Oh Ok I didn't know that the slope of the plot contributes to the amount of excited electrons.
Thank you nontheless for the explanation
Eric, Thank you for the good presentation. The ground Mullicken term for d7 low spin complex should be 2Eg. Right? Here you have shown it as 2T2g. Please clarify.
You're correct, I never caught that the figure I used had the wrong ground state Mulliken term for the low-spin region.
thank you for this good explanation
Very useful video. thank you
Thank you for this tutorial.
It was. Wet usefull.
I'm kind of cunfused as to why the calculated octahedral splitting is larger than the lowest excitation wavenumber. How can a transition occur if the energy that is necessary to promote the electron is greater then the energy I put into the system?
Short answer: The energy of observed absorption isn't strictly due to the promotion of an electron from a t2g to an eg orbital as would be expected if the energy was only related to the octahedral splitting energy, but rather the change in state of the system from the 4T1g microstate (ground state) to the 4T2g microstate (excited state).
I will try to post a video in a day or two, and go into more detail on why the lowest energy transition doesn't directly equal the octahedral splitting energy for any system that isn't d1.
can you only assume both graphs have a gradient of 1 if they look like they have a similar value?
Thanks very much bro, you help my lab report
Very helpful
Thank you
Where did you get the 27 from?
Jimena T From the y-axis for the higher energy state. It isn’t a perfect 1.95 ratio between E1 and E2, but it is pretty close.
Eric Victor oh ok, thanks!
@@EricVictor pouvez vs m'expliquer ce diagramme en français svp ??
@@sousou-positive5407 je ne parle malheureusement pas français
THANK U
Yes indeed. You cannot see excited state absorptions with conventional UV vis
Those eight's tho
Thank you sir
Thanks!