1. The combination of infrared (IR) and Raman spectroscopies constitutes the field of vibrational spectroscopy, which is a powerful tool in analyzing molecular structure. Let us consider the vibrational spectroscopy of XY2. Let us propose that there are four candidate structures for XY2. I. Bent, with X central, but the X-Y bonds of different length (Cs symmetry); II. Bent, symmetric, with X central (C2v symmetry); III Linear, with X central (D∞h); IV. Linear with Y central (C∞v).
In each case, the vibrational representation would be:
I II III IV
Γvib 3a΄ 2a1 + b1 Σg+ +Σu- + Πu Σ+ + Σ- +Π
Referring to the appropriate character tables, and recalling that modes that transform as (x,y,z) are IR active, and that modes that transform as (x2,xy,…) are Raman active, we can summarize the number of each type of mode as:
# modes I II III IV
(or “bands”)
IR active 3 (3a΄) 3 (2a1 + b1) 1 (Πu) 2 (Σ+ + Π)
Raman active 3 (3a΄) 3 (2a1 + b1) 1 (Σg+) 2 (Σ+ + Π)
Polarized* 3 (3a΄) 2 (2a1) 1 (Σg+) 1 (Σ+)
Concident@ 3 (3a΄) 3 (2a1 + b1) 0 2 (Σ+ + Π)
* A polarized Raman band is essentially totally symmetric.
@ “Coincident” denotes that the same band is active in both IR and Raman.
Examination of the IR and Raman spectra for XY2 would reveal which candidate structure is correct.
(i) Fill in the table below for NH3 and H2CO in the style of the one above. (NOTE. You looked at both molecules in detail in a previous homework.)
Gvib(NH3) = ________ Gvib(H2CO) = ________
# modes:
NH3 (C3v) H2CO (C2v)
IR active ______ _____
Raman active _______ _____
Polarized* _______ _____
Coincident@ _______ _____
(ii) A molecule X2Y2 is suspected to have one of the following symmetries:
C2v, D2h, C2h or Cs
Working each possible geometry through, we find:
Γvib = 3a1 + a2 + 2 b1 (C2v)
Γvib = 3ag + au + 2 bu (C2h)
Γvib = 5a’ + a” (Cs)
Γvib = 2ag + b1g + b1u + b2u + b3u (D2h) (x axis through the X atoms.)
The vibrational spectroscopy gives us the following information. What is the geometry of the molecule?
IR active ___3___
Raman active ___3___
Polarized* ___2___
Coincident@ ___0___
(i) The principle of mutual exclusion reveals that the molecule having (i) the symmetry element i , the IR and the Raman modes would be mutually exclusive.
The following rule makes the question straightforward.
In the first problem the NH3 belongs to the C3v point group which doesnt have the i.
The no. of modes of vibration are 6. Among which all are both RAMAN and IR active.
Then H2CO has 6 modes of vibration and having point group of C2v( NO i).
All modes are IR and Raman active modes i.e coincident modes.
(ii) For the question we see that the Ir and Raman modes are not coincident. It says that the molecule doesnt have ANY (I) OR THE POINT OF INVERSION. Among the point groups given the C2h only has the (i) . THus the C2h point group would be taken.
Geomtry is C2h.
The no of modes will be 3* 4 - 6 (taking non-linear) = 6.
Hence the vibrational modes are also given 6.
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