1. The document discusses principles of spectroscopy, including absorption spectroscopy which measures light absorbed by a sample (e.g. UV-Vis, IR, NMR) and emission spectroscopy which measures light emitted from an excited sample (e.g. fluorescence, Raman spectroscopy). 2. It covers electronic transitions in molecules, character tables, determining symmetry of molecular orbitals, and which transitions are allowed. Singlet and triplet states are also discussed. 3. Advantages of fluorescence spectroscopy are highlighted, including high sensitivity down to the microgram level and specificity from characteristic excitation and emission wavelengths. Examples of applications to inorganic analysis of metals and anions are provided.

1. Inorganic Spectroscopy
Dr. Chris, UP 2019
Part 1: Principles
1 2
Interaction of EMR
with Matter
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Electromagnetic Radiation and
Applications in Spectroscopy
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2. Two Principles of Spectroscopy
1. Absorption Spectroscopy:.
• e.g. UV (185 - 400 nm) / Visible (400 - 800 nm),
IR (0.76 - 15 μm) and NMR (m range)
2. Emission Spectroscopy:
• Measure the light that comes out of a sample when it
is excited by a light source
• e.g.Raman- and Fluorescence Spectroscopy (in both
cases excitation by a laser source)
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https://phet.colorado.edu/en/simulation/molecules-and-light
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Check out the behaviour in all 4 lights:
--> water vs CO2 vs Ozone
What are the differences ?
Electronic
Transitions
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Questions
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3. 9
http://slideplayer.com/slide/9514618/
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• How much is the absorption A if half of the light
intensity is absorbed by a sample ?
• How much light intensity is absorbed when A = 1 ?
• What is the difference in light absorption between
A = 2.5 and A =3 ?
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4. Possible electronic transitions
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(a) MO diagram Carbonyl Group
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(b) Assign symmetries to MOs
The symmetry of the whole molecule:
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5. Character Table of C2v:
(https://www.webqc.org)
Which symmetry do the n- , HOMO and LUMO orbitals have ?
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 1 -1 -1 1
HONO (n) 1 -1 1 -1
LUMO (*) 1 -1 -1 1
Determine symmetry of the 3 frontier orbitals:)
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Determine allowed transitions:
HOMO –> LUMO = b2 x b2 = a1 = z allowed
n –> LUMO = b1 x b2 = a2 not allowed
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Electronic states of molecules
b2
b1
b2
Symmetry of MOs:
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6. State of the molecule:
b2 x b2 = A1
b2
b2
b1
b1 x b2 = A2Molecules with all
electrons paired are
always totally symmetric
Symmetry of the whole molecule:
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Instead of saying:
One electron moved from b1 (HOMO) to b2 (LUMO)
[which is not allowed since b1 x b2 = a2]
We can also say:
The molecule changed its state from A1 (ground state)
to A2 (excited state)
or: A1 -> A2*
Electronic Transitions:
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Example 2: Butadiene
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Which transitions are allowed ?
(1) Determine the symmetry of the molecule
(2) Look up the character table
(3) Determine the characters for the HOMO and the two
LUMOs
(4) Multiply the characters for HOMO with each LUMOs
(5) Look up if these characters contain x,y or z
component
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7. Compare with
Particle in the Infinite Potential Well
The bigger L, the lower are the energy levels !
h2
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• π → π* transition
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Summary of usual electron transitions
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http://www.chemguide.co.uk/analysis/uvvisible/theory.html
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8. http://sammypiccolo.com/australian-capital-
territory/solvent-effect-in-uv-spectroscopy-pdf.php
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Solvent influence
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Polar solvents could form hydrogen-bonds and create
molecule aggregations – a non-polar solvent instead has
less interactions and should give us a more detailed
spectrum:
Example:
UV spectrum of PHENOL in
Ethanol and I-octane
Polar molecules like acetone are stabilized by a polar
solvent (water).
Therefore it needs more energy to reach the excited state.
31 32
hexane
MeOH
hexane
MeOH
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9. One form has  max at 280 nm, the other at 203 nm.
Which one is which and why ?
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FLUORESCENCE (LUMINISCENCE)
SPECTROSCOPY
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Triplet
States
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3 kinds of electronic transitions
Which has highest and lowest energy ?
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10. Flouresence emission always
occurs from the LOWEST
vibrational excited state
BECAUSE:
the relaxation process is fast !
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http://micro.magnet.fsu.edu/primer/techniques/fluorescence/fluorescenceintro.html
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Absorption
Emission (A)
Det
Det
Emission (B)
Det
Det
Excitation
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We measure the excitation spectrum in order to find out the
best wavelength, which we should use for the emission
spectrum – example:
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11. Excitation vs Emission Spectrum
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Advantages
• More sensitive when compared to other
absorption techniques. Concentrations as low
as μg/ml or ng/ml can be determined.
(One molecule can emit light many times, but
in absorption only one time)
• Precision up to 1% can be achieved easily
• As both excitation & emission wave lengths
are characteristic it is more specific than
absorption methods.
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Inorganic Analysis
• Is in competition to AAS for metal cations
• Especially useful for uranium salts
• And for certain anions like (a) cyanide:
Absorption at 400 nm,
emission at 480 nm
0.2 – 50 ug/L
(b) Phosphate:
An ion association complex between molybdophosphate and rhodamine B
provides the basis for an assay for phosphorous at 0.04 to 0.6 µg. The
fluorescence is measured at 575 nm with an excitation at 350 nm, after first
extracting excess of the rhodamine with chloroform.
Fluorescence Spectroscopy in Inorganic Analysis
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12. Special metal-ion + chelate ligand combinations give
fluorescence signals typical for the metal:
Probably the first use of metal chelates was in the analysis of
aluminium using the pentahydroxyl-flavone called morin
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