What type of molecules are excited by uv radiation

This arises from increased solvation of the lone pair, which lowers the energy of the n orbital. Often but not always , the reverse i. This is caused by attractive polarisation forces between the solvent and the absorber, which lower the energy levels of both the excited and unexcited states. This effect is greater for the excited state, and so the energy difference between the excited and unexcited states is slightly reduced - resulting in a small red shift.

Many inorganic species show charge-transfer absorption and are called charge-transfer complexes. For a complex to demonstrate charge-transfer behaviour, one of its components must have electron donating properties and another component must be able to accept electrons. The extent of the delocalization is shown in red.

Notice that there is delocalization over each of the three rings - extending out over the carbon-oxygen double bond, and to the various oxygen atoms because of their lone pairs. But the delocalization doesn't extend over the whole molecule. The carbon atom in the centre with its four single bonds prevents the three delocalized regions interacting with each other. The rearrangement now lets the delocalization extend over the entire ion.

This greater delocalization lowers the energy gap between the highest occupied molecular orbital and the lowest unoccupied pi anti-bonding orbital.

It needs less energy to make the jump and so a longer wavelength of light is absorbed. You will know that methyl orange is yellow in alkaline solutions and red in acidic ones.

The structure in alkaline solution is:. In acid solution, a hydrogen ion is perhaps unexpectedly picked up on one of the nitrogens in the nitrogen-nitrogen double bond. This now gets a lot more complicated! The positive charge on the nitrogen is delocalized spread around over the structure - especially out towards the right-hand end of the molecule as we've written it.

The normally drawn structure for the red form of methyl orange is. But this can be seriously misleading as regards the amount of delocalization in the structure for reasons discussed below after the red warning box if you are interested. Let's work backwards from the absorption spectra to see if that helps. The yellow form has an absorption peak at about nm. That's in the blue region of the spectrum, and the complementary color of blue is yellow. That's exactly what you would expect.

The red form has an absorption peak at about nm. That's at the edge of the cyan region of the spectrum, and the complementary color of cyan is red. Again, there's nothing unexpected here. Notice that the change from the yellow form to the red form has produced an increase in the wavelength absorbed. An increase in wavelength suggests an increase in delocalisation. That means that there must be more delocalization in the red form than in the yellow one.

Here again is the structure of the yellow form:. If you use the normally written structure for the red form, the delocalization seems to be broken in the middle - the pattern of alternating single and double bonds seems to be lost. The real structure is somewhere between the two - all the bonds are identical and somewhere between single and double in character. That's because of the delocalization in benzene.

The two structures are known as canonical forms, and they can each be thought of as adding some knowledge to the real structure. For example, the bond drawn at the top right of the molecule is neither truly single or double, but somewhere in between. Similarly with all the other bonds. The two structures we've previously drawn for the red form of methyl orange are also canonical forms - two out of lots of forms that could be drawn for this structure. We could represent the delocalized structure by:.

These two forms can be thought of as the result of electron movements in the structure, and curly arrows are often used to show how one structure can lead to the other.

In reality, the electrons haven't shifted fully either one way or the other. Just as in the benzene case, the actual structure lies somewhere in between these. You must also realize that drawing canonical forms has no effect on the underlying geometry of the structure.

Bond types or lengths or angles don't change in the real structure. For example, the lone pairs on the nitrogen atoms shown in the last diagram are both involved with the delocalisation. For this to happen all the bonds around these nitrogens must be in the same plane, with the lone pair sticking up so that it can overlap sideways with orbitals on the next-door atoms.

The fact that in each of the two canonical forms one of these nitrogens is shown as if it had an ammonia-like arrangement of the bonds is potentially misleading - and makes it look as if the delocalization is broken. The problem is that there is no easy way of representing a complex delocalized structure in simple structural diagrams.

It is bad enough with benzene - with something as complicated as methyl orange any method just leads to possible confusion if you aren't used to working with canonical forms. It gets even more complicated! If you were doing this properly there would be a host of other canonical forms with different arrangements of double and single bonds and with the positive charge located at various places around the rings and on the other nitrogen atom.

When sample molecules are exposed to light having an energy that matches a possible electronic transition within the molecule, some of the light energy will be absorbed as the electron is promoted to a higher energy orbital. An optical spectrometer records the wavelengths at which absorption occurs, together with the degree of absorption at each wavelength.

The resulting spectrum is presented as a graph of absorbance A versus wavelength, as in Figure 4 shown below. Figure 4: Simplified UV-Visible absorption spectrum for 1,3-butadiene.

The absorption peaks at nm, which is in the ultraviolet range and not in the visible range. The region from nm to nm is called the near ultraviolet or simply the ultraviolet , and from to about nm the visible. Light in these regions has less energy and is only able to excite relatively loosely bound electrons, such as those in pi bonds. Note the absorption spectrum of benzene, phenanthrene, and naphthacene in Fig. Colorless benzene does not absorb in the visible part of the spectrum, while the yellow naphthacene absorbs in the violet and blue part of the visible spectrum.

The light we see when we look at a sample of naphthacene is white light from which the violet and blue portions have been removed by absorption; that is, yellow light is reflected. It is important to remember this complementary nature of color: The color observed on reflection is always white light from which the absorbed color has been removed.

An object that appears green actually absorbs purple light, for example, while a compound that absorbs green light will appear red. The relationship between absorbed and observed color is shown in Fig. As a general rule, molecules containing conjugated systems of pi electrons absorb light closer to the visible region than saturated molecules or those with isolated double or triple bonds.

The longer the conjugated system, the longer the wavelength of the light absorbed.