Question
Asked Nov 5, 2019
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Using a group theory approach and the CO bond vectors as your basis determine if Cr(CO)6 is IR-active

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Expert Answer

Step 1

The point group for the given molecule is Oh and the character table for this point group is shown below.

SC 6C
6C 3C, -(C)i
6S 8S 3060Linear functions, Quadratic
E
h
4
4
Rotations
Function
x2+Y+z
+1
+1
+1
+1
+1
+1
+1
+1
+1
+1
+1
A2
+1
+1
-1
-1
+1
-1
+1
+1
-1
(22 -x2 -y', x* = y*))
+2 0
+2
0
+2
-1
+2
(R R,R
+3 0
-1
+1
-1
+3
-1
-1
T.
+3 0
(xz, yz, xy
+1
-1
-1
+3
-1
+1
Au
+1
+1
+1
+1
+1
-1
-1
-1
-1
-1
A
А
+11
-1
-1
+1
-1
+1
-1
-1
+1
E
-2 0
+2
+2
+1
-2
T2
+3 0
-3-1
x, y,
-1
+1
-1
+1
+1
-31
+3 0
+1
-1
-1
+1
-1
o
O
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Image Transcriptionclose

SC 6C 6C 3C, -(C)i 6S 8S 3060Linear functions, Quadratic E h 4 4 Rotations Function x2+Y+z +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 +1 A2 +1 +1 -1 -1 +1 -1 +1 +1 -1 (22 -x2 -y', x* = y*)) +2 0 +2 0 +2 -1 +2 (R R,R +3 0 -1 +1 -1 +3 -1 -1 T. +3 0 (xz, yz, xy +1 -1 -1 +3 -1 +1 Au +1 +1 +1 +1 +1 -1 -1 -1 -1 -1 A А +11 -1 -1 +1 -1 +1 -1 -1 +1 E -2 0 +2 +2 +1 -2 T2 +3 0 -3-1 x, y, -1 +1 -1 +1 +1 -31 +3 0 +1 -1 -1 +1 -1 o O

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Step 2

For a bond vector approach, the contribution of the bond for each operation are taken as 1. So, the reducible representation for the given compound will be same as the number of unshifted bond for each symmetry operation.

E 8C6C 6C|3C, (C
6S 8S 3060
о,
No. of
6
2
4
2
unshifted
Bond
Contribu 11
11
1
1
1
1
1
1
tion Bond
Reducible 6 0
0 0
0
2
2
2
4
Represent
O
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Image Transcriptionclose

E 8C6C 6C|3C, (C 6S 8S 3060 о, No. of 6 2 4 2 unshifted Bond Contribu 11 11 1 1 1 1 1 1 tion Bond Reducible 6 0 0 0 0 2 2 2 4 Represent O

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Step 3

For the given molecule, A1g is 1, Eg is 1, a...

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