Question

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 XY_{2}. Let us propose that
there are four candidate structures for XY_{2}. I. Bent,
with X central, but the X-Y bonds of different length
(C_{s} symmetry); II. Bent, symmetric, with X central
(C_{2v} 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

Γ_{vib}
3a΄
2a_{1} +
b_{1}
Σ_{g}^{+} +Σ_{u}^{-} +
Π_{u}
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 (x^{2},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
(2a_{1} + b_{1})
1
(Π_{u}) 2 (Σ^{+} +
Π)

Raman
active 3
(3a΄) 3
(2a_{1} + b_{1})
1
(Σ_{g}^{+}) 2
(Σ^{+} + Π)

Polarized^{*} 3
(3a΄) 2
(2a_{1})
1
(Σ_{g}^{+}) 1
(Σ^{+})

Concident^{@} 3
(3a΄) 3
(2a_{1} + b_{1})
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 XY_{2} would
reveal which candidate structure is correct.

(i) Fill in the table below for NH_{3} and
H_{2}CO in the style of the one above. (NOTE. You looked at
both molecules in detail in a previous homework.)

G_{vib}(NH_{3}) =
________
G_{vib}(H_{2}CO) = ________

# modes:

NH_{3} (C_{3v})
H_{2}CO
(C_{2v})

IR active ______ _____

Raman active _______ _____

Polarized^{*}
_______
_____

Coincident^{@}
_______
_____

(ii) A molecule X_{2}Y_{2} is suspected to have
one of the following symmetries:

C_{2v}, D_{2h}, C_{2h} or C_{s}

Working each possible geometry through, we find:

Γ_{vib} =
3a_{1} + a_{2} + 2
b_{1} (C_{2v})

Γ_{vib} =
3a_{g} + a_{u} + 2
b_{u} (C_{2h})

Γ_{vib} =
5a_{’} +
a_{”}
(C_{s})

Γ_{vib} =
2a_{g} + b_{1g} + b_{1u} + b_{2u} +
b_{3u} (D_{2h}) (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___

Answer #1

(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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