if you look at the peak shape in both IR-spectrum and UV/VIS spectrum of any molecules, you may notice two things. In IR spectrum of small molecules, the absorption peaks always tend to be very narrow, except for few bonds. On the other hand, In UV/VIS spectrum the absorption peaks always tend to be very wide compared to IR peaks, does anyone has an explanation for this?
There is a very simple explanation: Electronic excitations are usually accompanied by vibrational transitions which are usually not resolved. As a consequence broad bands result. This is pretty much the same situation as if you investigate the infrared spectra of gases. In this case, vibrational excitations are accompanied by rotational transitions. If your spectrometer is not able to resolve the individual bands, also comparably broad bands result. On the other hand, if you have liquids or solids, these rotational transitions are no longer possible and you see solely the band of the vibrational transition.
There is a very simple explanation: Electronic excitations are usually accompanied by vibrational transitions which are usually not resolved. As a consequence broad bands result. This is pretty much the same situation as if you investigate the infrared spectra of gases. In this case, vibrational excitations are accompanied by rotational transitions. If your spectrometer is not able to resolve the individual bands, also comparably broad bands result. On the other hand, if you have liquids or solids, these rotational transitions are no longer possible and you see solely the band of the vibrational transition.
It depends on characterization of functional groups with IR or subsbance with UV-Vis as well as scale of curve.
Mr. Shalan,
Basic principles of spectroscopy (Secondary school, structure of the atom and spectrum).
In case of UV/VIS spectrum,transitions occur from the ground vibrational level of the ground electronic state to many different vibrational levels of particular excited electronic states. Such transitions may give rise to vibrational fine structure in the main peak of the electronic transition. Since all the molecules are present in the ground vibrational level, nearly all transitions that give rise to a peak in the absorption spectrum will arise from the ground electronic state. If the different excited vibrational levels are represented as υ1,υ2, etc., and the ground state as υ0, the fine structure in the main peak of the spectrum is assigned to υ0 →υ0 ,υ0 →υ1,υ0→υ2 etc.,vibrational states. The υ0 →υ0 transition is the lowest energy (longest wave length) transition. Because spectrophotometer cannot resolve all transitions consists in electronic transition, it observe these types of transitions is that electronic spectrum of a molecules usually consists of a broad band of absorption centered near the wavelength of the major transition.
In IR spectroscopy,two of the most common Infrared band shapes are narrow and Broad . A typical example of a broad band is that displayed by O-H bonds, such as those found in alcohols and carboxylic acids. In any sample where hydrogen bonding occurs, the number and strength of intermolecular interactions varies greatly within the sample, causing the bands in these samples to be particularly broad. When intermolecular interactions are weak, the number of chemical environments is small, and narrow infrared bands are observed.Also,Compounds consisting of more than a handful of atoms have very many narrow infrared absorption bands.
The narrow character of IR bands is illusive. In fact, if we assume FWHM of the band to be due to fluctuations around the energy of the main transition, the correct consideration will be the ratio of fluctuation width to the main energy. So the higher the energy, the higher the absolute FWHM should be. For example, for C=O stretching (~1700 cm-1) FWHM is ~50 cm-1, and the above ratio is ~0.03. If we take a typical FWHM of 1000 cm-1 (free of vibronic replica) in UV-VIS at 500 nm, the above ratio will be 0.05.
@Oleg Dimitriev: At least for crystalline solids there are no energy fluctuations in the infrared. The width of the bands is determined by the oscillator strength and can therefore range from several to several hundred inverse centimeters...
The absorption bonds of FT-IR and UV-Vis are not related to each other and they have different mechanism for themselves.
Note two things: 1) IR graphs are wavenumbered while Uv-vis ones are nanometered. 2) The x-axis range of IR's is higher than what's normal for uv-vis'. It's not that simple to give justifications without referring to specific molecular eigenstates identification. -Zyad
It depends on the your ways. you should record UV spectrum at diluted concentration to absorbance in range from 0.2 to 0.6
Thomas Mayerhöfer provided a clear and concise answer. I dont understand why other people post incorrect or simply absurd answers.
In UV-Visible spectra Bonds will be in constant vibration, this variation will absorb nearby energies i.e, ΔE , for this reason UV peaks are broader.
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