CHE00019I Core 6 Examination Outline Answers 2020-21
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CHE00019I
Core 6 Examination
Outline Answers 2020-21
Question 1
(a) (i)
There are 12 nuclei in this complex and therefore 3x12=36 degrees of freedom. As this is a non-linear molecule, there will be 3N-6 vibrational
(7)
degrees of freedom i.e. 36一6=30. [Application of knowledge to a problem Using the method of unshifted atoms
O
C
O C 
CO
Re
C 
C O
Br
|
C4v |
E |
2C4 |
C2 |
2 |
2 |
|
Number of unshifted atoms |
12 |
4 |
4 |
8 |
4 |
|
Contribution per unshifted atom |
3 |
1 |
一1 |
1 |
1 |
3N 36 4 一4 8 4
Reducing the reducible representation:
|
C4v |
I |
2C4 |
C2 |
2 |
2 |
|
|
|
|
36 |
4 |
一4 |
8 |
4 |
|
|
|
A1 |
1 |
1 |
1 |
1 |
1 |
|
|
|
A1x |
36 |
8 |
一4 |
16 |
8 |
64 |
8 |
|
A2 |
1 |
1 |
1 |
一1 |
一1 |
|
|
|
A2x |
36 |
8 |
一4 |
一16 |
一8 |
16 |
2 |
|
B1 |
1 |
一1 |
1 |
1 |
一1 |
|
|
|
B1x |
36 |
一8 |
一4 |
16 |
一8 |
32 |
4 |
|
B2 |
1 |
一1 |
1 |
一1 |
1 |
|
|
|
B2x |
36 |
一8 |
一4 |
一16 |
8 |
16 |
2 |
|
E |
2 |
0 |
一2 |
0 |
0 |
|
|
|
Ex |
72 |
0 |
8 |
0 |
0 |
80 |
10 |
i.e. 8a1+2a2+4b1+2b2+10e
Discarding translations (a1+e) and rotations (a2+e) yields 7a1+a2+4b1+2b2+8e. These 22 symmetries account for 30 vibrational modes, the e being doubly degenerate. 3N-6=30.
e
O
C a
O C 
CO
Re
C 
C O b
Br
|
C4v |
E |
2C4 |
C2 |
2Qv |
2Qd |
|
LCO |
5 |
1 |
1 |
3 |
1 |
2a1+b1+e.
4 symmetries accounting for 5 CO stretching modes, the e symmetry being doubly degenerate.
Inspection of the character table below reveals translational symmetry associated with a1 and e and polarisability tensors with a1, b1 and e. Thus: a1 IR and Raman; b1 Raman only; e IR and Raman.
There are therefore 4 IR active modes, two of which degenerate, yielding 3 fundamental bands in the IR spectrum and there are 5 Raman active modes yielding 4 fundamental bands in the Raman spectrum.
2022-08-04