Re: Color of Transition metal solutions

Charles wrote:
On Sat, 28 Mar 2009 13:45:21 -0700 (PDT), terry
<tfmann@xxxxxxxxxxxxxx> wrote:

Q. I understand transition metal complexes to be colored because of
absorbtion of radiation in the visible region when electron are
promoted to higher energy d orbitals by photons of light. But there
is something bugging me. If you have a beaker with a finite number of
these molecules, each one can only be excited once so it cant keep on
absorbing the light for ever, unless of course spontaneous de
excitation is occuring at the same rate, but in that case doesnt
quantam physics dictate that de- excitation should result in the
release of a photon of exactly the same energy or wavelength and
cancel out the absortpion?. Can anybody see the error in my
arguement? thanks


de-excitation can be in steps, the total energy to be released must be
the same, but some can be transferred to vibration, heat. some can be
released at lower frequencies, longer wavelength. this is what
happens with fluorescence. some energy can be coupled to other
molecules, if they are able to accept it at appropriate levels.

Most of the energy goes to heat eventually. It occurs most rapid;y in liquids, where close contact among molecules facilitates the energy transfer. A small fraction of the energy is re-emitted as scattered light, including Raman scattering. as well as Rayleigh scattering. Fluorescence also happens, but it is vanishingly small for most transition metal complexes. But it has been said, "Everything fluoresces to some extent." In a well-designed spectrophotometer, the absorption cell is close to the detector to increase the fraction of scattered light that is included in the measurement. That design feature is most important in solutions that are somewhat turbid.
.

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