This document discusses Raman spectroscopy and infrared spectroscopy techniques. It summarizes that Raman spectroscopy involves inelastic scattering of light from a sample, producing shifts in wavelength that are unique to different chemical bonds and functional groups. Infrared spectroscopy detects vibrational and rotational transitions in molecules by measuring the frequencies at which they absorb infrared radiation. The document then provides an example using infrared spectroscopy to determine the structure of two organic compounds based on their carbonyl stretching frequencies. Point group symmetry and character tables are used to determine the number of expected infrared-active vibrations.

1. Inorganic Spectroscopy
Part: RAMAN
https://www.youtube.com/watch?v=yQ1MctWU9Mg
(https://www.youtube.com/watch?v=TMLnUmbLwUI)
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3 possible transitions
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UV/VIS
Induced dipoles
https://www.youtube.com/watch?v=1_IqMY6t6w0
Elastic scattering
Rayleigh
(https://www.youtube.com/watch?v=1Q45PpodjJY)
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2. 5 6
• no sample preparation
• non-destructive
• can be used with a microscope
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3. 9 10
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 ?
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Find the point group of both molecules:
①
②
(Practise with:
http://symmetry.otterbein.edu/challenge)
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4. C2v
E C (z) (xz) (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
 8 0 4 0
Molecule ① : 8 CO groups
Reduce  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.
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C2v
E C (z) (xz) (yz)
 6 0 2 0
Molecule ②: also Point Group C2v
Reduce  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
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Fe(CN)6 complexes
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How can we explain the difference to Fe(III) ?
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5. Problem solving
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