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The molecule S2 is a bent molecule with C2y symmetry 1. С> (xz) S. aHowmany degrees of freedom are present in this molecule?

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Answer #1

a)

A molecule with N atoms has 3N degree of freedoms. Hence, since our molecule(SO2) has 3 atoms, the total number of degrees of freedom is 9.

b)

SO2 is a non-linear molecule. Hence, it has 3 translational and 3 rotational degrees of freedom. The remaining must be vibrational degrees of freedom.

Hence, the number of vibrational degrees of freedom in SO2 is

3-33.

Hence, the answer is 3.

c)

E represents the identity operation. Hence, when we apply E on SO2, both the bonds remain unchanged. Hence, we put 2 in the table below E.

When we do a C_2, through z axis, i.e rotate the molecule by 180 degrees through z axis, we exchange the positions of the S-O bonds. Hence, both the bonds are changed. Hence, we put 0 under C_2.

When we put a reflection plane in the xz molecular plane, the plane cuts all the three atoms in half. Hence, the bonds stay unchanged. Hence, we put 2 under σ(πε) operation.

When we put a reflection plane in the yz plane, it cuts the S atom in half and exchanges the position of the O atoms. Hence, the S-O bonds are exchanged in the process too. Hence, we put 0 under a(yz operation.

Hence, the character table with the reducible representation is:

C_{2v} E C_2 σ(πε) a(yz
\Gamma 2 0 2 0

d)

The character table of C_{2v} point group is

C_{2v} E C_2 σ(πε) a(yz
A 1 1 1 1
A_2 1 1 -1 -1
B_1 1 -1 1 -1
B_2 1 -1 -1 1

Now, the reducible representation we found in part c) is

\Gamma : 2 \ 0 \ 2 \ 0

Lets find out how many times the irreducible representation A occurs in our \Gamma.

The formula to find this is

\frac{1}{h} \sum_{all\ classes} \chi_R \times \chi_I \times N

Where h is the order of the group = 4 for C2v point group as it has 4 unique operations.

XR is the character of  the reducible representation, i.e 2, 0 , 2 and 0 for our reducible representations.

\chi_I is the character of the irreducible representation, i.e. for the first one, we are finding for A, hence, the characters are 1, 1, 1 and 1 (look at the character table for the row A1.)

N is the number of symmetry operations in the particular class of operations.

hence, the number of times A1 occurs in our reducible representation is

\frac{1}{h} \sum_{all\ classes} \chi_R \times \chi_I \times N \\ = \frac{1}{4}[2 \times 1 \times 1 + 0 \times 1 \times 1 + 2 \times 1 \times 1 + 0 \times 1\times 1 ] \\ \\ = \frac{1}{4} \times 4 = 1

Hence, the irreducible rerpesentation A1 occurs only 1 time in our reducible representation.

Similarly, the number of times A2 occur in our reducible representation is

\frac{1}{h} \sum_{all\ classes} \chi_R \times \chi_I \times N \\ = \frac{1}{4}[2 \times 1 \times 1 + 0 \times 1 \times 1 + 2 \times -1 \times 1 + 0 \times -1\times 1 ] \\ \\ = \frac{1}{4} \times [2+0-2+0] = 0

Hence, A2 does not appear in our reducible representation.

The number of times B1 appear is

Σ XR x X N h all classes 1 12 x 1 x 10 x -1 x 1 +2 x 1 x 1 +0 x -1 x 1] x [20 2 0] = 1 Η

hence, B1 appears one time in the reducible representation.

The number of times B2 appear is

\frac{1}{h} \sum_{all\ classes} \chi_R \times \chi_I \times N \\ = \frac{1}{4}[2 \times 1 \times 1 + 0 \times -1 \times 1 + 2 \times -1 \times 1 + 0 \times 1\times 1 ] \\ \\ = \frac{1}{4} \times [2+0-2+0] =0

Hence, B2 does not appear in our reducible representation.

Hence,the reducible representation can be written as

\Gamma = A_1 + B_1

Note that the basis we took to calculate the reducible representation are the S-O bonds, Hence, the reducible representation actually represent the vibration of the S-O bonds in SO2.

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