Vibrational Spectroscopy In this part of the course we will look at the kind of spectroscopy which uses light to excite the motion of atoms. The forces required to move atoms are smaller than those required to move electrons (as nucleus and electron are bound by strong electrostatic forces). Thus the photon energies involved are somewhat lower; instead of UV/Vis radiation infra-red radiation is sufficient, hence the alternative title, infra-red spectroscopy. In the first section we will see that we can use classical arguments (springs again) to describe resonances in vibrational spectroscopy. We will learn how to relate the resonance frequency to the strength of a chemical bond and the mass of the nuclei. In the next section we will see how analysis of vibrational spectra can lead to information about molecular shapes. In the third section we will see that the infra-red spectrum can be used as a powerful analytical tool, because the spectrum yields a fingerprint characteristic of a particular molecule. Finally some quantum aspects of molecular vibrations will be investigated.
Absorption of IR radiation is sufficient to cause molecules to vibrate (the electronic state stays the same). The vibrational motions may be usefully sub-divided into stretching and bending. In general stretching involves separating nuclei whereas bending does not, thus stretching modes require higher energy photons than bending modes. In complex molecules vibrations are characteristic of functional groups, so IR spectra give an indication of what groups are inside a molecule, and are thus a useful analytical tool. As before we start with the simplest cases and progress to more complicated molecules. Diatomic molecules molecular springs For diatomics, there are no bending modes only a single vibrational mode. The vibrating molecule is often likened to an oscillating mass on the end of a spring: Typically the mass undergoes simple harmonic motion see physics courses.
For small oscillations the restoring force is thus proportional to the extension, so the oscillation and the displacement (x) follows a simple sinusoidal motion. Thus we can write generally: x x 0 sin( t) The frequency of the vibration is the number of waves per second and so is given by 1/T. Thus, 1 T The quantity is now seen to be related to the frequency of the motion. We call it the angular frequency and rearranging this equation we have
So by simple substitution the time dependent displacement is: x x 0 sin(t) Now we can examine the motion of the mass (m) according to Newton s laws. Provided displacement is not too large Hooke s law applies The constant here k is called the spring constant. A big k means that a large restoring force is exerted for a particular extension of the spring, i.e. we have a strong spring. Now, the equation of motion is given by Newton s second law, F ma Using Hooke s law and Newton s second law kx ma d x m dt d m dt x 0 sin t
Differentiating once: And again: So: kx m kx m d dt x cost 0 k m x sint m x 0 A simple rearrangement leads to the relation 1 Simply, this tells us that a strong spring (big k) vibrates at a higher frequency than a weak springy one. Also, if the mass is light the frequency will be high, so heavy masses vibrate at low frequencies. k has units kgs -, or Nm -1 k m Application to bonds the clamped diatomic molecule The idea is exactly the same as before a small extension of the bond leads to a proportional restoring force where k is now the force constant of the bond. In the same way as k relates to how strong the spring is, for molecules it relates to how strong the bond is. This makes it a useful quantity to know.
Problem: we can see that the vibrational frequency of HCl will depend on which end of the molecule we clamp because of m in our equation. In reality molecules are not clamped and will possess one unique vibrational frequency. Clearly the equation has to be modified for real molecules and the modification must be related to the mass. A detailed derivation gives us Where is the reduced mass 1 k m1m m m 1 Thus if the frequency of the diatomic vibrational mode is measured and the masses are known then the force constant (related to bond strength) can be determined. This is an important result.
Estimate or calculate exactly the force constant of the H 1 Cl 35 bond if the molecule absorbs at 3343.8 nm? [c = 3 x 10 8 ms -1, proton mass = 1.673 x 10-7 kg] 1. 300 400 Nm -1. 400 500 Nm -1 3. 500 600 Nm -1 convert tofrequency calc reduced mass Finally k 8 c 310 ms 3343.810 m1m m m 1 516.9Nm 1 1 9 8.9710 m (1 35) 1.67310 (1 35) 1.67310 7 7 0% 0% 0% 13 kg kg Hz 1.6610 7 kg 1 3
Similar experiments were performed for the other hydrogen halides. The results are shown below. Halide IR absorption / m k / Nm 1 band wavenumber / cm 1 HF.416 966 4139 HCl 3.344 xxx 990 HBr 3.774 41 650 HI 4.39 314 310 The table shows that photons of greater energy are required to stretch the HF bond compared to the HCl bond. This makes sense as we know that F atoms are smaller that Cl atoms and can therefore make shorter, stronger bonds. This trend continues down the table. We can also see this trend by looking at the force constants. Thus the HF bond is seen to be very difficult to stretch (966 Nm -1 ) being almost twice as difficult as the HCl bond (517 Nm -1 ).
These ideas are easily extended to give some clue as to the nature of chemical bonds in a molecule. Consider the following series of carbon oxygen bonds. bond C O typical band wavenumber / cm 1 1100 C O 1700 CO 170 We see that more energy is required to stretch a double bond than a single bond this makes sense as more electrons are found between the nuclei in a double bond and so the bond is stronger. The result for carbon monoxide suggests that the bond is even stronger than a double bond perhaps a triple bond? Such a structure can be envisaged: Thus IR spectroscopy has proved useful in elucidating a chemical structure. As hydrogen is a small atom a strong bond is expected in the H molecule. An IR spectrum was recorded over the range 400-5000 cm -1 but no absorption was observed. Independent measurements suggest the H H bond has a force constant of 575 Nm -1.
The HH bond is strong and the atoms light. The reason the vibration was not observed was that the 400 5000 cm -1 wavenumber range was not wide enough. 1. True. False reduced mass for homonuclear diatomic 1 k 1 575 kgs kg 1.3010 14 Hz m m m 0.836510 7 kg ~ 1.3010 8 c 310 14 4410 4 m 1 or 4400cm 1 0% 0% 1
Thus the spectrometer should/should not have observed a transition. We recall that just because absorption is in principle possible absorption may still be absent. This is another example of a selection rule; the vibrational transition in H is forbidden. It turns out that the selection rule associated with IR spectroscopy that stops H from absorbing IR radiation is associated with the symmetry of the molecule and the change in dipole moment during the vibration. There is another method, Raman Spectroscopy, which allows frequencies of H, and generally X molecules, to be determined. The selection rules for Raman are different because the interaction with the radiation is due to the polarisabiltiy, not the dipole moment. Selection Rule Vibrations may only show up in the IR spectrum if the vibration concerned causes a change in the molecular dipole moment There is an equivalent rule for Raman, requiring a change in the (shape of) the polarizability during a vibration.
It is easy to see how this selection rule arises. Recall that light has an oscillating electric field associated with it. If a vibration causes a change in the molecule s dipole moment then the vibrating molecule is really an oscillating electric field and can interact with light if there is no oscillating dipole then it cannot. HCl and H illustrate this nicely: This means that in general we can say for diatomics: homonuclear diatomic molecules IR inactive heteronuclear diatomic molecule IR active
For the HF molecule the vibration at 4139 cm -1 shows k = 966 Nm -1. At what wavenumber does the isotope DF absorb? Forceconstant independent of isotope so k calc reduced mass Finally or 1. 997cm -1. 069cm -1 3. 4139cm -1 4. 878cm -1 997cm 1 1 966Nm 7 m1m (19) 1.67310 kg 3.07 10 7 m m ( 19) 1.67310 kg 1 1 k 1 966 0% 0% 0% 0% 13 1 8.9910 s 7 6.83 3.07 10 1 3 4 ~ ~ HF Note DF 1 7 kg
Polyatomic Molecules In general a molecule will contain N atoms (N =, 3, ). How many vibrations can such a molecule possess? Each atom requires 3 coordinates to specify its location so that 3N numbers determine the entire molecule. For the whole molecule, three coordinates may be used to specify where the centre of mass is (during vibrations c.o.m. does not move) and three coordinates are required to specify the orientation of the molecule as a whole. This means that there are 3N - 6 coordinates left over and this is the number of vibrations that the molecule possesses. This is generally true for a non-linear molecule: if the molecule is linear then we only need two coordinates to specify its orientation. Overall: non-linear: 3N - 6 vib modes linear: 3N - 5 vib modes (e.g. diatomics: 3() - 5 = 1)
Example: N = 3 Linear, e.g. CO : modes: O=C=O should have 3(3) 5 = 4 modes. There are two stretching and two bending modes: Overall, two peaks appear in the IR spectrum.
Another Example Non-linear Molecule, e.g. water: should have 3(3) 6 = 3 modes. There are two stretching modes: and one bending mode: Thus three peaks appear in the IR spectrum of water. These examples illustrate one way in which IR spectroscopy can be used to determine the structure of a molecule the IR spectra of carbon dioxide and water alone will tell us that CO is linear and water is bent.
70 60 Infrared (Intensity) 50 40 30 0 10 0-10 1500 000 500 3000 3500 4000 wavenumber (cm -1 )
Calculate the number of vibrational modes in acetylene (C H ) then draw them. Label each for IR/Raman activity.
Larger Molecules - Fingerprinting Obviously as N gets bigger the number of vibrational modes becomes large. Spectra of such molecules will be complex. However, to a good approximation the vibrational frequency of a particular bond or functional group is independent of the surrounding bonds. This means that we can create a table of stretching and bending modes associated with different bonds these are called correlation tables, an example of which is shown in the attached chart and table
IR Spectra: Formaldehyde Certain types of vibrations have distinct IR frequencies hence the chemical usefulness of the spectra The gas-phase IR spectrum of formaldehyde: (wavenumbers, cm -1 ) Tables and simulation results can help assign the vibrations! Formaldehyde spectrum from: http://www.cem.msu.edu/~reusch/virtualtext/spectrpy/infrared/infrared.htm#ir Results generated using B3LYP//6-31G(d) in Gaussian 03W.
Group Frequencies GROUP C-H Alkynes 3333-367(s) stretch 700-610(b) bend C=C Alkenes 1680-1640(m,w)) stretch CC Alkynes 60-100(w,sh) stretch C=C Aromatic Rings 1600, 1500(w) stretch C-O Alcohols, Ethers, Carboxylic acids, Esters 160-1000(s) stretch C=O Aldehydes, Ketones, Carboxylic acids, Esters 1760-1670(s) stretch Monomeric -- Alcohols, Phenols 3640-3160(s,br) stretch O-H Hydrogen-bonded -- Alcohols, Phenols 3600-300(b) stretch Carboxylic acids 3000-500(b) stretch N-H Amines 3500-3300(m) stretch 1650-1580 (m) bend C-N Amines 1340-100(m) stretch CN Nitriles 60-0(v) stretch NO Nitro Compounds 1660-1500(s) asymmetrical stretch 1390-160(s) symmetrical stretch v - variable, m - medium, s - strong, br - broad, w - weak
The Raman Effect The incident radiation excites virtual states (distorted or polarized states) that persist for the short timescale of the scattering process. Polarization changes are necessary to form the virtual state and hence the Raman effect Virtual state Virtual state hv 1 hv 1 hv Anti-Stokes line This figure depicts normal (spontaneous) Raman effects hv 1 hv 1 hv Stokes line Scattering timescale ~10-14 sec (fluorescence ~10-8 sec) Ground state (vibrational) H. A. Strobel and W. R. Heineman, Chemical Instrumentation: A Systematic Approach, 3 rd Ed. Wiley: 1989.
More on Raman Processes The Raman process: inelastic scattering of a photon when it is incident on the electrons in a molecule When inelastically-scattered, the photon loses some of its energy to the molecule (Stokes process). It can then be experimentally detected as a lower-energy scattered photon The photon can also gain energy from the molecule (anti-stokes process) Raman selection rules are based on the polarizability of the molecule Polarizability: the deformability of a bond or a molecule in response to an applied electric field. Closely related to the concept of hardness in acid/base chemistry. P. W. Atkins and R. S. Friedman, Molecular Quantum Mechanics, 3 rd Ed. Oxford: 1997.
More on Raman Processes Consider the time variation of the dipole moment induced by incident radiation (an EM field): Induced dipole moment ( t) ( t) ( t) polarizability EM field If the incident radiation has frequency and the polarizability of the molecule changes between min and max at a frequency int as a result of this rotation/vibration: t) mean polarizability 1 cos t cos t ( int 0 = max - min Expanding this product yields: 1 ( t) 0 cost 0 cos( int ) t cos( int 4 ) t Rayleigh line Anti-Stokes line Stokes line P. W. Atkins and R. S. Friedman, Molecular Quantum Mechanics, 3 rd Ed. Oxford: 1997.