Structure & Reactivity
Structure Determination with Spectroscopy
UV2. UV-Visible Spectroscopy: Metal Ions.
Transition metals are often associated with
brightly-coloured compounds. Because of their relatively low
electronegativity, transition metals are frequently found as positively-charged
ions, or cations. In solution, metal ions would not swim around by
themselves, but would attract other molecules to them. These molecules
bind to the metal ions, forming coordination complexes.
These transition metal complexes or coordination
complexes have lots of electrons, and they can often interact with lots of
different photons. When they do, there are a number of possible electronic
transitions that can result. These transitions might involve the metal ion
itself, or the ligands -- those molecules or ions that bind to the
metal ion.
One very common transition is called a charge transfer
transition. It involves the excitation of an electron from the ligand to
the metal, or vice versa. Sometimes the former case is referred to as a
ligand-to-metal charge transfer, or LMCT; the latter case would be a
metal-to-ligand charge transfer, or MLCT. However, these two acronyms are
sometimes used interchageably to suggest some sort of transition that involves
both the ligand and the metal, without worrying too much about the direction.
In addition, sometimes electrons can be excited from
one level to another, just on the metal ion. For transition metals, these
electronic excitations are called d-d transitions. For transition
metals, the valence electrons are in the d sub-shell, and in a d-d transition,
the electron is excited from one d level to another. That sounds like it
might be pretty easy -- the electron isn't going very far, after all -- but d-d
transitions are actually quite inefficient. The electronic transitions
involved just are not very good at capturing photons.
By comparison, it might sound like it would be difficult to move an
electron from the metal all the way to the ligand, but it's actually pretty
easy. These electronic transitions interact with photons very efficiently.
As a result, there is a kind of counterintuitive
relationship in the UV-visible spectra of transition metal complexes: d-d
transitions require very little energy but occur relatively infrequently,
meaning they give very weak absorbances in the spectrum. MLCT transitions
require much more energy but they happen frequently, leading to stronger
absorbances in the spectrum.
We aren't going to worry about why different
electronic transitions occur with greater or lesser efficiency; that has to do
with symmetry and group theory, and some mathematics that we aren't equipped to
handle right now. We just need to understand that these distinctions can
influence what UV spectra look like.
Problem UV2.1.
In the following spectra, assign
the observed "peaks" or absorbance bands to MLCT or d-d transitions.
Problem UV2.2.
In
the preceding
spectra, predict the observed colour of each of the complexes.
The whole idea of UV spectroscopy is that different compounds might absorb
photons of different wavelengths based on their electronic structures. We
might be able to look at the UV spectrum of a compound and tell its identity or
structure; that task would be especially straightforward if we had a few
different options to choose from.
Now we see that the y axis also matters to some extent. A specific
compound might absorb at 250 nm and at 450 nm; however, an additional
characteristic of that compound might be that it absorbs very strongly
at 250 nm and only weakly at 450 nm. Knowing that information
might help us distinguish the spectrum of that compound from the spectrum of
another that has similar absorption wavelengths, but that shows peaks in the UV
spectrum of very different sizes.
How strongly a compound absorbs photons at a particular wavelength is described
by a quantity called the extinction coefficient (or alternatively, in
different variations, the absorption coefficient or the molar absorptivity
coefficient).
For example, it can be shown that aqueous potassium permanganate, KMnO4,
at its absorption maximum of 530 nm, has an extinction coefficient ε =
14 L g-1 cm-1. How much light is absorbed by a
sample of potassium permanganate at that wavelength depends on a couple of
additional factors, however. First of all, how concentrated is the sample?
The more potassium permanganate there is dissolved in the sample, the more light
it can absorb. Also, the longer the light travels through the cuvette, the
more permanganate it will encounter, and so the more light it will absorb.
So absorbance can be described as
A = ε c l
in which ε = extinction coefficient, l = path length, c =
concentration.
Most cuvettes are 1 cm wide; if the extinction coeffieicnt contains a unit of cm-1,
we can just use b = 1 cm and that part is taken care of. From there, the
absorbance will just depend on the concentration of the permanganate.
Problem UV2.3.
Calculate the absorbance at 530 nm for the following concentrations of potassium
permanganate in a 1 cm cuvette.
a) 0.025 g /L b) 0.011 g/L
Problem UV2.4.
Calculate the concentration of potassium permanganate (in g/L) given the following
absorbance readings at 530 nm in a 1 cm cuvette.
a) A = 0.75 b) A = 0.21
Problem UV2.5.
Although the maximum absorption is around 530 nm, the visible absorption band in
potassium permanganate actually stretches from 500 to 600 nm. What colour is potassium permanganate?
Frequently, extinction coefficients have units of L mol-1 cm-1
or M-1 cm-1. In these cases, the concentration is
measured in mol L-1 rather than g L-1. The molecular weight (molar mass)
of the compound, in g mol-1, is the conversion factor between
one unit and the other. For example, the extinction coefficient for the
530 nm peak of potassium permanganate is approximately 2 200 L mol-1
cm-1.
Problem UV2.6.
Confirm that the two extinction coefficients for potassium permanganate, given
with different units, are actually equivalent.
Problem UV2.7.
Calculate the absorbance at 530 nm for the following concentrations of potassium
permanganate in a 1 cm cuvette.
a) 1.5 x 10-4 mol /L
b) 9.0 x 10-5 mol/L
Problem UV2.8.
Calculate the concentration of potassium permanganate (in mol/L) given the following
absorbance readings at 530 nm in a 1 cm cuvette.
a) A = 0.47 b) A = 0.89
See additional discussion of
coordination complexes.
This site is written and maintained by Chris P. Schaller, Ph.D., College of Saint Benedict / Saint John's
University (with contributions from other authors as noted). It is freely
available for educational use.
Structure & Reactivity in Organic, Biological and Inorganic Chemistry by Chris Schaller is licensed under a Creative Commons Attribution-NonCommercial 3.0 Unported License.
Send corrections to cschaller@csbsju.edu
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