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![d3 case
4T2g <--- 4A2g
4T1g <--- 4A2g
4T1g(P) <--- 4A2g
Calculate the value of B and Δ for the Cr3+ ion in [Cr(H2O)6)]3+ if
1=17000 cm-1, 2=24000 cm-1 and 3=37000 cm-1.
SOLUTION.
These values have been assigned to the following spin-allowed
transitions.
From the information given, the ratio 2 / 1 = 24000 / 17000 = 1.412
Using a Tanabe-Sugano diagram for a d3 system this ratio is found at
Δ/B=24.0
48](https://image.slidesharecdn.com/lecture2019exercises-190320122205/85/Transition-Metal-Electronic-Spectra-48-320.jpg)
![d3 case
4T2g <--- 4A2g
4T1g <--- 4A2g
4T1g(P) <--- 4A2g
Calculate the value of B and Δ for the Cr3+ ion in [Cr(H2O)6)]3+ if
1=17000 cm-1, 2=24000 cm-1 and 3=37000 cm-1.
SOLUTION.
These values have been assigned to the following spin-allowed
transitions.
From the information given, the ratio 2 / 1 = 24000 / 17000 = 1.412
Using a Tanabe-Sugano diagram for a d3 system this ratio is found at
Δ/B=24.0
48](https://image.slidesharecdn.com/lecture2019exercises-190320122205/75/Transition-Metal-Electronic-Spectra-48-2048.jpg)
Uploaded byChris Sonntag
Transition Metal Electronic Spectra
The document provides information about electronic spectra and terms for carbon p electrons and transition metal d electron configurations. It discusses: 1) Possible terms that arise from carbon's 2p electrons, including 1D2, 3P2, 3P1, 3P0 and 1S0 terms. Hund's rules are used to determine the ground state term. 2) Microstate tables that list all possible combinations of orbital angular momentum (L) and spin (S) for electron configurations. 3) Tanabe-Sugano diagrams that show the splitting of d electron terms in an octahedral ligand field and allow determination of transition energies. 4) Charge transfer transitions that can occur from the
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Transition Metal Electronic Spectra
- 1.
- 2. Free atom/ion terms Considerthe electron configuration of Carbon: 2s 2p 1 0 -1 Consider only the 2 p electrons, each has l = 1 with possible sums of mL = 2, 1 and 0 => D, P and S Terms (L = l1 + l2, l1 + l2 – 1, l1 + l2 -2, …. 0) The Spins can be parallel or antiparallel, so S = 1 or 0 So the multiplicities can be 2S+1 = 1 (singlet) or 3 (triplet) 2
- 3. microstates With L =2, 1, 0 and S = 1, 0 we may think that J = L + S , L + S – 1, …. would be 3, 2, 1, 0 -> BUT that is not possible because for L = 2 the two electrons must be in the same orbital and then they cannot have S = 1: 2p 1 0 -1 => L = 2 Possible combinations of L (as sum of mL) and S (as sum of spins) can be listed in a table, called “microstates” If we want to know all possible electrons states (terms) we have to go through this process 3
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- 5. 5 From that wecan find 1D2, 3P2, 3P1, 3P0 and 1S0 terms – the ground state is 3P0 (highest spin, highest L) 5 x 3P (9) 1S (1) 1S0 1D (5) 3P2 3P1 3P0 1D2
- 6. 6 Hund's rule (forgroundstate term symbol) 1. The term with maximum multiplicity lies lowest in energy 2. For a given multiplicity, the term with the largest value of L lies lowest in energy. 3. For atoms with less than half-filled shells, the level with the lowest value of J lies lowest in energy. When the shell is more than half full, the opposite rule holds (highest J lies lowest). 15 microstates for p2 Remember: This is the energy splitting of the 2 p electrons because of L-S coupling, NOT because of an external field !!
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- 12. Energies of Terms Thespin multiplicity rule is fairly reliable for predicting the ordering of terms, but the ‘greatest L’ rule is reliable only for predicting the ground term, the term of lowest energy; there is generally little correlation of L with the order of the higher terms. Thus, for d2 the rules predict the order 3F < 3P < 1G < 1D < 1S but the order observed for Ti2+ from spectroscopy is 3F < 1D < 3P < 1G < 1S 12
- 13. Term splitting (byel-el repulsion) Ground state (we get 3 groups of ground states which give similar splitting in Oh field) 13
- 14.
- 15. FREE ION TOCOMPLEX ION 15
- 16. Splitting in octahedralfield 16 L = 1 L = 2 L = 3
- 17. The same butin different symmetries: 17
- 18. Ground Terms ofd configurations Assign the Term symbols ! 18
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- 22. Splitting of termswith electron-hole analogy 22
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- 24. For d1,d9 andd4(HS), d6(HS) o can be directly measured: 24
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- 26. d2: from ionto complex 26
- 27. d2 case: Correlation betweenfree ion and complex terms: 27
- 28. excited state t2g 1eg 1 Including spin possibilities: 28
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- 31. Ground and excitedtriplet states d2 3F 3A2g 3T1g 3T2g Ground state 3P 3T1g 31
- 32. Why 2 T1states ? It requires more energy to bring an electron from the xy plane to the z2 orbital z2 x2-y2 xy xz yz Than from an xz or yz to the z2 orbital 32
- 33. Two triplet stateswith different energy mL = 2 1 0 -1 -2 Repulsion: LOW HIGH HIGH Repulsion: HIGH LOW LOW 3F 3P 33
- 34. Here o hasto calculated from Tanabe-Sugano: (d7) (d3) 34
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- 38. Compare d1, d2and d3 38
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- 40. Tanabe Sugano Diagrams Whatare the term symbols ? Which excited states are relevant ? https://chem.libretexts.org/Bookshelves/Physical_and _Theoretical_Chemistry_Textbook_Maps/Supplement al_Modules_(Physical_and_Theoretical_Chemistry)/S pectroscopy/Electronic_Spectroscopy/Electronic_Spe ctroscopy%3A_Interpretation 40
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- 42. Charge Transfer (CT)transitions -acceptor ligand MLCT From metal to ligand antibonding 42
- 43. -donor ligand LMCT From ligandto metal antibonding CT transitions are 1. more intense than d-d 2. at higher energy than d-d 43
- 44. d3 case The threeunpaired electrons are in t3 2g orbitals which give rise to 4A2g,2Eg,2T1g and 2T2g states. Which one is the ground state and why ? How does it look like ? If one electron is excited, the configuration is t2 2ge1 g which gives two quartet states 4T1g and 4T2g How do they look like ? 44
- 45. Analyse the firstexcited states for each configuration ! d4 case 45
- 46. Get ford2 case 46
- 47. The energy ratio ofthe 2 peaks is 1,49 That gives / B = 29 That gives me E2/ = 40.0 2 peaks in UV/VIS: at 25.600 cm-1 and 17.200 cm-1 http://wwwchem.uwimona.edu.jm/spectra/swingjs/TSd2.html 47
- 48. d3 case 4T2g <---4A2g 4T1g <--- 4A2g 4T1g(P) <--- 4A2g Calculate the value of B and Δ for the Cr3+ ion in [Cr(H2O)6)]3+ if 1=17000 cm-1, 2=24000 cm-1 and 3=37000 cm-1. SOLUTION. These values have been assigned to the following spin-allowed transitions. From the information given, the ratio 2 / 1 = 24000 / 17000 = 1.412 Using a Tanabe-Sugano diagram for a d3 system this ratio is found at Δ/B=24.0 48
- 49. The energy ratio2/1 is 1.4 /B = 24.0 E1/B = 24 17.000/B = 24 B = 17000/24= 708 cm-1 /708 = 24.0 = 708 cm-1* 24 = 17.000 cm-1 49
- 50. https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental _Modules_(Inorganic_Chemistry)/Crystal_Field_Theory/ Tanabe-Sugano_Diagrams • For ad7 metal ion determine the energy ratios for allowed transitions at Δoct/B of 20. • For a d6 metal ion of Δoct/B = 30 and B=530 cm-1 what would the energies of the 5 allowed transitions be? How many are in the UV-Vis range? How many are in the IR range? • Write out the allowed transitions for a d5 metal ion in a E/B> 28 ligand field. • A d4 complex exhibits absorptions at 5500 cm-1 (strong) and 31350 cm-1 (weak). What are the transitions that are being exhibited in the complex? What is the corresponding Δoct for the complex? • A spectrum of d7 metal complex seemingly exhibits only two intense transitions. What is the Δoct/B that this situation occurs? Please use reference to specific transitions and energy splitting. 50
- 51. d7 high andlow spin 51
- 52. • Δoct/B of20 yields E/B values of 38, 32, 18. Ratios then are 2.11 and 1.78 • Δoct/B = 30 yields E/B heights of 27, 40, 57, 65, 85. Energies are then 14310, 21200, 30210, 34450 and 45050 cm-1. All are in the UV-Vis range. *note you need to infer the E/B value for the last transition as the diagram does not extend that far up. • 2A2g<-2T2g, 2T1g<-2T2g, 2Eg<-2T2g, and 2A1g<-2T2g. • 31,350/5,500 gives a ratio of 5.7/1. The only Δoct/B value that matches is at 10. B value is then 550 cm-1. Δoctequals 5500 cm-1. • Three transitions are generated at low Δoct/B. However, at about a value of Δoct/B = 13 the transitions 4A2g<-4T1g, and 4T1g<-4T1g have the same energies which results in the appearance of only two absorptions. 52
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