A step-by-step tutorial on assigning the transitions in the UV-vis spectrum of a d7 system using a Tanabe-Sugano Diagram.
Пікірлер: 37
@guit4rpl4y3r5 жыл бұрын
Thank you for being the only person on here with a clear English speaking lecture on the topic
@monoclinico4 жыл бұрын
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.
@y.a.59172 ай бұрын
Thanks for clearly pointing out how to use the diagrams
@AllForgottenNow4 жыл бұрын
dude thank u so much u literally saved my life
@fawziaiphon30046 жыл бұрын
Thank you for this tutorial. It was. Wet usefull.
@marilynnaeem61365 жыл бұрын
this is short and super cool!
@hassanfakhri78905 жыл бұрын
the way you write 8 is disturbing, but that was very helpful, thank you :D
@vatrareksa40424 жыл бұрын
thank you for this good explanation
@naserianikambaine49344 жыл бұрын
Very helpful Thank you
@kaushikumarihami19822 жыл бұрын
Very useful video. thank you
@NubMG4 жыл бұрын
Thanks very much bro, you help my lab report
@oliveiraaaa1237 жыл бұрын
Thanks!
@avengersnewbie23485 жыл бұрын
Thank you sir
@Hannah-lr1uc4 жыл бұрын
can you only assume both graphs have a gradient of 1 if they look like they have a similar value?
@JP-ue4jr2 ай бұрын
THANK U
@Pierrot1101946 жыл бұрын
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?
@EricVictor6 жыл бұрын
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.
@suneeshcv3 жыл бұрын
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.
@EricVictor3 жыл бұрын
You're correct, I never caught that the figure I used had the wrong ground state Mulliken term for the low-spin region.
@katiesilverthorne84904 жыл бұрын
I thought bipy was a strong field ligand. Wouldn't that make it low-spin?
@EricVictor3 жыл бұрын
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.)
@katiesilverthorne84903 жыл бұрын
If only I knew this before my inorganic exam last year 😂 thanks for the reply, very informative!
@EricVictor3 жыл бұрын
@@katiesilverthorne8490 sorry, I just now saw it!
@lucasgobel1930 Жыл бұрын
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
@lucasgobel1930 Жыл бұрын
had*
@Whatever142534 жыл бұрын
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?
@EricVictor4 жыл бұрын
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.
@Whatever142534 жыл бұрын
@@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
@jimenat3574 жыл бұрын
Where did you get the 27 from?
@EricVictor4 жыл бұрын
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.
@jimenat3574 жыл бұрын
Eric Victor oh ok, thanks!
@sousou-positive54073 жыл бұрын
@@EricVictor pouvez vs m'expliquer ce diagramme en français svp ??
@EricVictor3 жыл бұрын
@@sousou-positive5407 je ne parle malheureusement pas français
@pelirojopajaro5 жыл бұрын
Yes indeed. You cannot see excited state absorptions with conventional UV vis