 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 1 + E 2 -1 0 G3H 3 0 1

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