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Reduction by formula - ex. 2 (PF5)


 

The vibrational modes associated with the P-F bonds in PF5

The PF5 molecule belongs to the D3h point group.

The molecule contains both axial and equatorial bonds, these are calculated separately as they are different. The number of bonds that remain unshifted by the operations are represented by the following:

 

E

2C3
3C2
sh

3S3

3sv

Gaxial

2

2
0
0

0

2

Gequatorial
3
0
1
3
0
1
 
G5F
5
2
1
3
0
3

 

Here the reduction will be performed on the representation of the 5 P-F although it could be done on the two sets separately.

There are three types of irrepresentations in this point group: A1, A2 and E.

The total number of symmetry operations is 12. There is one E element, three C2, one sh, two S3 and three sv planes: h = 1 + 2 + 3 +1 + 2 + 3 = 12.

 

The irrepresentations are calculated as follows using the formula (the colours indicate where the numbers came from)

 

The number of times the A1' occurs in the representation is:

nA1'

=

1/12

[

1*1*5

+

2*1*2

+

3*1*1

+

1*1*3

+

2*1*0

+

3*1*3

]

 

 

 

 

1E

 

2C3

 

3C2

 

1sh

 

2S3

 

3sv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

1/12 [5 + 4 + 3 + 3 + 0 + 9]

= 24/12 = 2

 

 



The number of times the A2' occurs in the representation is:

nA2'

=

1/12

[

1*1*5

+

2*1*2

+

3*-1*1

+

1*1*3

+

2*1*0

+

3*-1*3

]

 

 

 

 

1E

 

2C3

 

3C2

 

1sh

 

2S3

 

3sv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

1/12 [5 + 4 -3 + 3 + 0 -9]

= 0/12 = 0

 

 

 



The number of times the E' occurs in the representation is:

nE'

=

1/12

[

1*2*5

+

2*-1*2

+

3*0*1

+

1*2*3

+

2*-1*0

+

3*0*3

]

 

 

 

 

1E

 

2C3

 

3C2

 

1sh

 

2S3

 

3sv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

1/12 [10 -4 + 0 + 6 + 0 + 0]

= 12/12 = 1

 

 

 



The number of times the A1" occurs in the representation is:

nA1"

=

1/12

[

1*1*5

+

2*1*2

+

3*1*1

+

1*-1*3

+

2*-1*0

+

3*-1*3

]

 

 

 

 

1E

 

2C3

 

3C2

 

1sh

 

2S3

 

3sv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

1/12 [5 + 4 + 3 -3 + 0 -9]

= 0/12 = 0

 

 

 

 



The number of times the A2" occurs in the representation is:

nA2"

=

1/12

[

1*1*5

+

2*1*2

+

3*-1*1

+

1*-1*3

+

2*-1*0

+

3*1*3

]

 

 

 

 

1E

 

2C3

 

3C2

 

1sh

 

2S3

 

3sv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

1/12 [5 + 4 -3 -3 + 0 +9]

= 12/12 = 1

 

 

 



The number of times the E" occurs in the representation is:

nE"

=

1/12

[

1*2*5

+

2*-1*2

+

3*0*1

+

1*-2*3

+

2*1*0

+

3*0*3

]

 

 

 

 

1E

 

2C3

 

3C2

 

1sh

 

2S3

 

3sv

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

1/12 [10 -4 + 0 -6 + 0 + 0]

= 0/12 = 0

 

 

 

 



The representations reduces to 2 lots of A1', E' and A2":

G5F = 2A1' + E + A2"


This can be readily checked by adding up the characters for these rows: if correct, the representation should be regenerated:

 

 

E

2C3
3C2
sh

3S3

3sv

 

 A1'

 1

1

 1

1
1

+

 A1'

 1

1

 1

1
1

+

E'

 2

 -1

0

2

-1
0

+

 A2''

1

-1

-1

-1
1
 
 

 G5F

5

2

1

3

0
3

Therefore

G5F = 2A1' + A2" + E'

 



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