Vibrational Spectroscopy

Once we have classified a molecule according to the methods described in the previous sections, we can use this information to deduce the properties of the IR spectrum of the molecule.

Example 1: Trans-[PtCl₂L₂]

We are going to look at the stretching vibrations of the Pt-Cl bonds. The analysis is exactly the same as in s-bonding.

The point group is D_2h.

The reducible representation reduces to A_g and B_1u. What this means is that the molecule has two modes of vibration, one with A_g symmetry and the other having B_1u symmetry.

The selection rule for absorpion in the IR spectrum is that the vibration must have the same symmetry as a p-orbital. If we look at the right-hand side of the group table, we can see that the A_g vibration has the same symmetry as an s-orbital, a d_{x² - y²} orbital or a d_z² orbital - i.e. (x², y², z²). Thus it is IR inactive.
The B_1u vibration has the same symmetry as the p_z orbital. This vibration is IR active.

The two molecular vibrations can be drawn like so:

When looking at molecular vibrations, remember that the centre of mass of the molecule must remain in the same place.

The selection rule for IR spectroscopy is that there must be a change in dipole moment during the vibration.
Trans-[PtCl₂L₂] has no dipole moment and the A_g vibration does not produce one. In contrast the B_1u vibration does produce a dipole moment - one Pt-Cl bond lengthens as the other shortens. Hence this vibration is IR active.

Raman Spectroscopy

There is a second type of vibrational spectroscopy, called Raman Spectroscopy. A laser is shone on the sample and energy corresponding to the vibration is absorbed. The scattering from the sample is observed at right angles to the laser beam. In addition to the frequency of the laser itself, a set of lines is observed at a lower energy - this energy corresponding to the frequency of the vibrations in the sample.
For Raman spectroscopy, the selection rules are different from IR - allowed vibrations must have the symmetry of s- and d-orbitals. For trans-[PtCl₂L₂], this is true for the A_g, B_1g, B_2g and B_3g vibrations. Hence the A_g vibration is observed in Raman, while the B_1u vibration is absent. The use of Raman spectroscopy complements the use of IR spectra in analysis.

There is further information available in a Raman spectrum. The fully symmetric vibration, with the same symmetry as an s-orbital, is polarised. If the laser beam is polarised, then the A_g vibration will be partially polarised. If a polarising filter is placed between the sample and the detector, then the A_g band wil decrease to zero intensity when the filter is crossed. The other bands will change to 6/7 of their original intensities.

Examples of other point groups

Cis-[PtCl₂L₂]

The point group for this molecule is C_2v.

The reducible representation is:

The symetries of the vibrations are A₁ and B₁.
The symetries of the p-orbitals are A₁, B₁ and B₂.
Hence both vibrations are allowed for IR spectroscopy and will be seen in the IR spectrum.
The symmetries of the s- and d-orbitals are A₁, A₂, B₁ and B₂. Hence both vibrations are allowed in the Raman spectrum, and the A₁ vibration will be fully polarised.

Explicit vibrations

The way to determine the explicit vibrations is exactly the same as the method used to find the explicit forms of the hydogen orbitals in the H₂O molecule. Simply look at what happens to the vibrations under A₁ and B₁ symmetry.

So the vibrations are:
n₁ + n₁ for the A₁ vibration
n₁ - n₁ for the B₁ vibration

Or in diagrammatic form:

Note that vectors are placed on the Pt atoms so that the centre of mass does not move.

CCl₄

The tetrahedron is a little more difficult. The trick is to set the axes correctly:

The molecule is placed inside a cube using alternate corners for the chlorine atoms. The edges of the cube define the x, y and z axes.

The reducible representation of the stretching vibrations is,

This reduces to A₁ and T₂.
The symbol T refers to a triply degenerate set of orbitals - i.e. three vibrations which are of exactly the same energy.
The p-orbitals are T₂, so only the T₂ vibrations are active in the IR spectrum -this gives rise to one band.
The s-orbital is A₁ and the d_xy, d_xz and d_yz orbitals are T₂, so both vibrations are Raman active with the A₁ vibration polarised.

Cr(CO)₆

We are only going to look at the carbonyl CºO stretching vibrations. This is because these vibrations are strong in the IR spectrum, in a clear region between 1800 and 2100 cm^-1. This makes them very distinctive.

The point group is O_h.

The reducible representation breaks down into A_1g, E_g and T_1u. The A_1g and E_g vibrations are Raman active with the A_1g strectch being fuly polarised. The T_1u vibration is IR active.

[Cr(CO)₅L]

It is possible to deal with all the stretches together, but it is generally easier to deal with the two types of carbonyls separately: the unique carbonyl trans to the L group, and the four carbonyls cis to L.

The point group is C_4v

The unique CO has a vibration which is A₁. This is always the case for a unique vibration. In this case, it is both IR and Raman allowed.

The CO stretches have symmetries A₁, B₁ and E. All are present in the Raman spectrum, but only the A₁ and E vibrations are IR allowed.

trans-[Cr(CO)₄L₂]

The symmetry is D_4h

The CO stretching vibrations are A_1g, B_1g, and E_u. The A_1g and B_1g vibrations are Raman active with the A_1g vibration fully polarised. The E_u vibration is IR active.

Explicit vibrations

A_1g: n₁ + n₂ + n₃ + n₄
B_1g: n₁ - n₂ + n₃ - n₄
E_u: n₁ - n₃ and n₂ - n₄

Comparison of [Cr(CO)₅L] and trans-[Cr(CO)₄L₂]

If we compare the four carbonyl groups which form a square plane in both complexes, we get

We have extra bands in the IR and Raman spectra in [Cr(CO)₅L] compared with trans-[Cr(CO)₄L₂]. In trans-[Cr(CO)₄L₂], the carbonyls are strictly planar, but in [Cr(CO)₅L] they can deviate from the plane a little. This means that for [Cr(CO)₅L], the intensity of the A₁ band in the IR and the E band in Raman will depend upon the deviation from the plane, and so in this case will be weak.

[Cr(CO)₅(pyridine)]

The presence of the pyridine rather than just L reduces the symmetry of the molecule from C_4v to C_2v at best.

The vibrations are A₁, A₂, B₁ and B₂. The A₁, B₁ and B₂ vibrations are IR active and al are Raman active, with the A₁ fully polarised.
The unique carbonyl is A₁ and both IR and Raman active, with the Raman band being fully polarised.

Comparison of [Cr(CO)₅L] and [Cr(CO)₅(pyridine)]

If we compare the four carbonyl groups which form a square plane in both complexes, we get

Note that the E band has split into two, the B₁ and B₂ bands, on reducing the symmetry from C_4v to C_2v.