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![The C–O stretching wavenumbers are shifted to lower values when
there are changes in the extent of back-bonding in the compound.
Removing positive charge from the metal causes the shift of
electrons from the metal to the CO π∗ orbitals causes the CO
wavenumber values to decrease.
The highest excess of negative charge on the metal occurs in the
*V(CO)6 +− complex and so more back-bonding occurs than in the
other complexes.
The next highest excess of electron density is in Cr(CO)6 , and then
[Mn(CO)6 ]+.
5](https://image.slidesharecdn.com/spectroscopicmethodsirpart2-180904065115/75/Spectroscopic-methods-IR-part-2-5-2048.jpg)
Uploaded byChris Sonntag
Spectroscopic methods IR part 2
This document discusses spectroscopic methods for analyzing organometallic compounds, focusing on IR and Raman spectroscopy of metal carbonyl complexes. It explains that the C-O stretching wavenumbers in metal carbonyl complexes shift to lower values when there is more back-bonding from the metal to the carbonyl π* orbitals. The number of IR peaks expected for related complexes can be determined from the point group and character tables. It also briefly mentions using IR, Raman, UV-Vis, and NMR spectroscopy to study other types of inorganic complexes.
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Spectroscopic methods IR part 2
- 1. Spectroscopic Methods in InorganicChemistry Dr.Chris UP August 2018 PART 2: IR of Organo-Metal compounds M(CO)x and RAMAN 1
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- 5. The C–O stretchingwavenumbers are shifted to lower values when there are changes in the extent of back-bonding in the compound. Removing positive charge from the metal causes the shift of electrons from the metal to the CO π∗ orbitals causes the CO wavenumber values to decrease. The highest excess of negative charge on the metal occurs in the *V(CO)6 +− complex and so more back-bonding occurs than in the other complexes. The next highest excess of electron density is in Cr(CO)6 , and then [Mn(CO)6 ]+. 5
- 6. Electron density onthe metal Higher C-O strength Lower C-O strength M=C=O character 6
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- 8. CO Substitution patterns Wecompare cis- and trans-ML2(CO)2 complexes in IR: What are the point groups ? 8
- 9. Tetrahedral Td OctahedralOh Linear: D∞h for A-B-A ( i ) C ∞h for A-B http://en.wikibooks.org/wiki/Introduction_to_Mathematical_Physics/N_body_problem_in_quantum_mechanics/Molecules 9
- 10. Character Tables forcis and trans 10
- 11. Representations of 2C-O groups Which contains the irreducible representations : Which contains the irreducible representations : Conclusion: Number of IR peaks for cis and trans complex: 11
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- 13. How do youdistinguish whether the structure of transition metal complex molecule M(CO)4L2 is cis or trans by inspection of the CO stretching region of the IR spectra? -> determine the symmetry group: M(CO)4L2 13
- 14. -> Check thecharacter tables: c2v d4h 14
- 15. Reducing stretching motions 4stretching vectors 4 0 0 2 4 0 0 0 0 0 0 4 2 0 15
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- 19. Exercise 19 How many IRpeaks do we expect from these ML(CO)3 compounds ? Step 1: determine the point groups of each molecule Step 2: Get the characters for the 3 stretching vectors in the character table of this point group Step 3: find the irreducible representations that are included in this rep -> how many have x, y or z ?
- 20. Tetrahedral Td OctahedralOh Linear: D∞h for A-B-A ( i ) C ∞h for A-B http://en.wikibooks.org/wiki/Introduction_to_Mathematical_Physics/N_body_problem_in_quantum_mechanics/Molecules 20
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- 24. M(II)hexamine complexes N-H stretch N-H bend M-NH3 bend M-NH3 rock 24 Which compoundhas the weakest/strongest M-N bond ?
- 25. The spectra presentedin Figure 5.6 show a trend in the wavenumber shifts for the three hexamine complexes; the N–H bands shift to lower wavenumbers from Co to Cr to Ni. This indicates that the N–H bond order (bond strength) decreases as the metal– N bond order increases in the stability order mentioned 25
- 26. ATR Method –attenuated total reflection 26
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- 29. 3 possible transitions 29 UV/VIS Induceddipoles https://www.youtube.com/watch?v=1_IqMY6t6w0 Elastic scattering Rayleigh
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- 33. • no samplepreparation • non-destructive • can be used with a microscope 33
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- 37. Figliola, et. al.Organometallics 2014, 33, 4449 Compound 1 A = S 2061 2021 1976 1955 1878 Compound 2 A = Se 2054 2014 1970 1950 1875 We should decide which structure is more likely based on IR: 5 peaks, higher frequencies for A = Sulfur ① or ② ? 37
- 38. Find the pointgroup of both molecules: ① ② (Practise with: http://symmetry.otterbein.edu/challenge) 38
- 39. C2v E C (z)s(xz) s(yz) Linear f, rotations Quadratic f A1 1 1 1 1 z x, y, z A2 1 1 -1 -1 R xy B1 1 -1 1 -1 x, R xz B2 1 -1 -1 1 y, R yz G 8 0 4 0 Molecule ① : 8 CO groups Reduce G: ¼ (8x1x1 + 0x1x1 + 4x1x1 + 0x1x1) = 3 A1 ¼ (8x1x1 + 0x1x1 + 4x-1x1 + 0x-1x1) = 1 A2 ¼ (8x1x1 + 0x-1x1 + 4x1x1 + 0x-1x1) = 3 B1 ¼ (8x1x1 + 0x-1x1 + 4x-1x1 + 0x1x1) = 1 B2 The A2 can be ignored since it does not contain x, y or z and is therefore not IR active. This gives 7 IR active CO vibrations. 39
- 40. C2v E C (z)s(xz) s(yz) G 6 0 2 0 Molecule ②: also Point Group C2v Reduce G: ¼ (6x1x1 + 0x1x1 + 2x1x1 + 0x1x1) = 2 A1 ¼ (6x1x1 + 0x1x1 + 2x-1x1 + 0x-1x1) = 1 A2 ¼ (6x1x1 + 0x-1x1 + 2x1x1 + 0x-1x1) = 2 B1 ¼ (6x1x1 + 0x-1x1 + 2x-1x1 + 0x1x1) = 1 B2 The A2 can be ignored since it does not contain x, y or z and is therefore not IR active. This gives 5 IR active CO vibrations. A1 1 1 1 1 z x, y, z A2 1 1 -1 -1 R xy B1 1 -1 1 -1 x, R xz B2 1 -1 -1 1 y, R yz 40
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- 42. How can weexplain the difference to Fe(III) ? 42
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