Question
Asked Nov 5, 2019

Using a group theory approach and the CO bond vectors as your basis determine if MnCl(CO)5 is IR-active

check_circleExpert Solution
Step 1

The point group of MnCl(CO)5 is C4V.

The character table of C4V is given below.

quadratic
2C 4 (z) C2
С *
20y204near functions,rotations
E
functions
x+y2
+1
A1
+1
+1
+1
+1
Z
+1
+1
-1
+1
-1
х* -у?
В,
+1
+1
+1
-1
-1
+1
B2
+1
-1
+1
-1
ху
(x.y) R .R,)
(xz.yz)
Е
-2
0
+2
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quadratic 2C 4 (z) C2 С * 20y204near functions,rotations E functions x+y2 +1 A1 +1 +1 +1 +1 Z +1 +1 -1 +1 -1 х* -у? В, +1 +1 +1 -1 -1 +1 B2 +1 -1 +1 -1 ху (x.y) R .R,) (xz.yz) Е -2 0 +2

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Step 2

The irreducible representation is given below.

Е 2C, С, | 20у
20
Number of unshifted bonds 6
2
2
2
Contribution of bonds
1
1
1
1
г
6
2
2
4
2
41
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Е 2C, С, | 20у 20 Number of unshifted bonds 6 2 2 2 Contribution of bonds 1 1 1 1 г 6 2 2 4 2 41

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Step 3

The reduction formula is given as equation (1) in which N is the number of times a symmetry species occurs in the reducible representation, h is the ‘order of the group, Χr is the character for a particular class of operation in the reducible representation Χi is the character of the i...

N--Σχn'..... ()
e
A6x1+2xx2+2x1x1+2x4x1+2x2x1]
=3
(2)
A26x1x1+ 2x1x2+2x 1x 1+ 4x-1x2 + 2 x-lx2
=0
B, 6x1x1+2x-1x2+2x1x1+4x2x1+2x2x-1]
=1
(3)
B2 6x1x1+2x-1x2+2x1x1+4x-1x2+2x1x2]
=0
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N--Σχn'..... () e A6x1+2xx2+2x1x1+2x4x1+2x2x1] =3 (2) A26x1x1+ 2x1x2+2x 1x 1+ 4x-1x2 + 2 x-lx2 =0 B, 6x1x1+2x-1x2+2x1x1+4x2x1+2x2x-1] =1 (3) B2 6x1x1+2x-1x2+2x1x1+4x-1x2+2x1x2] =0

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