Submitted by Lary / Wright State University on Mon, 03/08/2021 - 15:21
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Tanabe – Sugano question for ∆o and B.

Miessler 5th edition, 11.14 (e) has a solution I don’t understand (d) is [VF6]3–d2 given two transitions, I get the book’s answer.  Also Example 11.8  [V(H2O)6]3+ given two transitions, I get the book’s answer.  However, (e) VCl3(CH3CN)3 d2 given two transitions.  I don’t understand the books answer –or mine (I am new to this).

Ex 11.8    [V(H2O)63+    17,800 and 25, 700 cm–1  Ratio 1.44,

0 /B = 31                    ∆0 = 19,000 cm-1  B = 610 cm-1  (ok)

11.14d   [VF6]3–           14,800 and 23,250 cm–1  Ratio 1.57,

0 /B = 26                       ∆0 = 16,100 cm-1  B = 621 cm-1  (ok)

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11.14e   VCl3(CH3CN)3 14,409 and 21413 cm–1  Ratio 1.49,

Book: ∆0 /B = 34.5       Why not between the other two if ratio is?   How did they come up with this??

0 = 16,970 cm-1  B = 492 cm-1

Mine following example         :

ratio of 1.49 gives : ∆0 /B = 28 not 34.5 (makes sense?)

0 = 23,440 cm-1  B =  (655+579)/2 = Bave617  cm-1

0 is not between spectra peaks, does that make sense?

Nicole Crowder / University of Mary Washington

Hi Lary,

I think what is going on here is differences in the spectrochemical properties of the ligands. H2O and F- are both low field ligands, meaning that delta o is small. The acetronitrile is higher in the spectrochemical series, meaning that its impact on d orbital splitting and ΔO will be greater.

Although you can find the appropriate ratio of v2/v1 in many places on the TS diagram (especially given the highly parallel lines for the allowed transitions in the d2 diagram), you need to take into account the higher ΔOcaused by the ligands, which is why the book has placed it in the 30's.

When I give problems on this, I often tell my students to assume that the ratio of the first two allowed transitions occurs at a ΔO/B value of x, and then give them an appropriate x value based on the ligand field strength.

Hope that helps!

Lary / Wright State University

In reply to by Nicole Crowder / University of Mary Washington

Hi Nicole,

I had thought it was likely due to the field strength and there fore higher ∆o/B.  I didn't see how to get that higher ∆/B from the method in the text which causes us to think we were doing something wrong.  I can see how your solution gets around the situation.  I'll try to be very careful on writing this week's exam.  I've seen some comments that people don't teach TS much or at all.  Do you have an opinion?  (I teach the second semester of the "senior" Advanced inorganic course).

Thanks for being so quick

Lary

Matt Whited / Carleton College

Hi Lary,

You're right that the v2/v1 ratio is 1.49, corresponding approximately to a delta/B value of 28. This should allow you to get an answer that is in the ballpark of what Miessler reports, although it's always tough to be precise if you are reading off a chart rather than working through the equations. Honestly, it isn't so important to be exact, since the diagrams will never correspond exactly to your spectra due to the choice of Racah parameters used to generate the diagrams not matching yours exactly. So take your answers more as guidelines. I think your problem above is a miscalculation of B values or delta. When I solve I get delta/B = 28.7, E1/B = 26.4 (so B = 545 cm-1) and E2/B = 39.3 (so B = 596 cm-1). If you average those B values (570 cm-1) and plug back in to find delta, you get delta = 16,400 cm-1 (even if you use the larger value of B that you report above, it should correspond to a delta value of around 17,000 cm-1). If you or your students are having trouble working from diagrams, I recommend the JavaScript T-S applets (only available for a few of them, but quite useful): http://wwwchem.uwimona.edu.jm/courses/Tanabe-Sugano/TSapplets.html.

My first clue that there was a math error in your solution was that you found a value for delta that is greater than the energy of your second transition. If you look at the T-S diagram for d2, you see that there is no point at which delta/B > E2/B. That sort of check can be useful, in addition to the spectrochemical series intuition check that Nicole mentions.

Hope this helps.

- Matt