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example 1 - NH3


 

The vibrational modes associated with the N-H bonds in NH3 is found as follows:

The ammonia molecule belongs to the C3v point group.

The number of bonds that remain unshifted by the operations are represented by the following:

 

E

2C3

3sv

G3H

3

0

1



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

The total number of symmetry operations is 6. There is one E element, two C3 axes and three sv planes: h = 1 + 2 + 3 = 6.

 

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

=

1/6

[

1*1*3

+

2*1*0

+

3*1*1

]

=

1/6 [3 + 0 + 3]

=

1

 

 

 

 

1E

 

2C3

 

3sv

 

 

 

 

 

 

 

 

 

gE = 1

 

gC3 = 2

 

gsv = 3

 

 

 

 

 

 

 

 

 

cE(A1) = 1

 

cC3(A1) = 1

 

csv(A1) = 1

 

 

 

 

 

 

 

 

cE(3H) = 3

 

cC3(3H) = 0

 

csv(3H) = 1

 

 

 

 

 



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

nA2

=

1/6

[

1*1*3

+

2*1*0

+

3*-1*1

]

=

1/6 [3 + 0 -3]

=

0

 

 

 

 

1E

 

2C3

 

3sv

 

 

 

 

 

 

 

 

 

gE = 1

 

gC3 = 2

 

gsv = 3

 

 

 

 

 

 

 

 

 

cE(A2) = 1

 

cC3(A2) = 1

 

csv(A2) = -1

 

 

 

 

 

 

 

 

 

cE(3H) = 3

 

cC3(3H) = 0

 

csv(3H) = 1

 

 

 

 

 



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

nE

=

1/6

[

1*2*3

+

2*-1*0

+

3*0*1

]

=

1/6 [6 + 0 + 0]

=

1

 

 

 

 

1E

 

2C3

 

3sv

 

 

 

 

 

 

 

 

 

gE = 1

 

gC3 = 2

 

gsv = 3

 

 

 

 

 

 

 

 

 

cE(E) = 2

 

cC3(E) = -1

 

csv(E) = 0

 

 

 

 

 

 

 

 

 

cE(3H) = 3

 

cC3(3H) = 0

 

csv(3H) = 1

 

 

 

 

 



The representations reduces to A1 and E:

G3H = A1 + E



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

 

 

E

2C3

3sv

 

 A1

 1

 1

 +

 E

 2

 -1

 0

 

 

 

 

 

 

 G3H

 3

 0

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