2. 5 6
• no sample preparation
• non-destructive
• can be used with a microscope
7 8
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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 ?
11
Find the point group of both molecules:
①
②
(Practise with:
http://symmetry.otterbein.edu/challenge)
12
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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.
13
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
14
Fe(CN)6 complexes
15
How can we explain the difference to Fe(III) ?
16
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