Electron Paramagnetic Resonance...
Transcript of Electron Paramagnetic Resonance...
Electron Paramagnetic Resonance Spectroscopy
Spectroscopy in Inorganic Chemistry
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Zavoisky in 1945
EPR
Electron Paramagnetic Resonance (ESR)
Electron Spin Resonance (ESR)
Electron Magnetic Resonance (EMR)EPR ~ ESR ~ EMR
same as NMRElectronic energy levels (EPR) GHz microwave frequencies
Nuclear energy levels (NMR) MHz
Applications
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Oxidation and reduction processes Reaction kinetics
Examining the active sites of metalloproteins• Kinetics of radical reactions• Spin trapping• Catalysis• Defects in crystals• Defects in optical fibers• Alanine radiation dosimetry• Archaeological dating• Radiation effects of biological compounds
Ex. Electrochemical oxidation or reduction
EPR
Instrument
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Common field strength 0.34 and 1.24 T9.5 and 35 GHzMicrowave region
EPR
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EPR
typical ESR spectrometer —a radiation source (klystron)a sample chamber between the poles of a magneta detection and recorder system
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quartz and very fragile.
standard: DPPH (diphenylpicrylhydrazyl radical)
g = 2.0036
pitch g = 2.0028
Bstdgsample = gstd ———
Bsample
for field-sweep, lower field (left-hand) than standard, higher g value
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One or more unpaired electron in molecule, ion, atom …
Unpaired electrons have spin and charge and hence magnetic
moment (Quantum mechanics)
Free radicals
Transition metal compounds
Electronic spin can be in either of two directions.
EPR
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EPR
a b
b Ms = -1/2
a Ms = +1/2
DE = gβB = hv
No magnetic field
B = 0
Magnetic field
B > 0
g g-factor (approximately 2.00232 free electron )
β Bohr magneton (9.2741 x 10-21erg.Gauss-1)
B magnetic field (Gauss or mT)
h Planck’s constant 6.626196 x 10-27erg.sec
ν frequency (GHz or MHz) (microwave )
Electron Zeeman Effect
hv = gβB
v = (gβ/h)B = 2.8024 x B MHz
for B = 3480 G ν= 9.75 GHz (X-band)
for B = 420 G ν= 1.2 GHz (L-band)
for B = 110 G ν= 300 MHz (Radiofrequency
B
E
E = -1/2gβB
E = 1/2gβB
Spectrometer frequencies used in EPR
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Microwave
band
B0
(Gauss)V0 (GHz) g=2 (kG)
L 300 1 0.35
S 1100 3 1.3
X 3400 9.5 3.4
X 3300 9.2
K 8600 24 8.5
Q 12500 35 12.2
W 35000 95 33.5
EPR9
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EPR
Better resolution
Find shoulder
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spin-lattice relaxation
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microwave radiation transferred from the spin system to its surroundings
long relaxation time ==> decrease in signal intensity
short relaxation time ==> resonance lines become wide
EPR
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Hyperfine Interactions EPR signal is ‘split’ by neighboring nuclei
Called hyperfine interactions
Can be used to provide information
Number and identity of nuclei
Distance from unpaired electron
Interactions with neighboring nuclei
E = gmBB0MS + aMsmI
a = hyperfine coupling constant
mI = nuclear spin quantum number
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Hyperfine Interactions
Coupling patterns same as in NMR
More common to see coupling to nuclei with spins greater
than ½
The number of lines:
2NI + 1
N = number of equivalent nuclei
I = spin
Only determines the number of lines--not the intensities
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couplings arise in two ways:
(i) direct dipole-dipole interaction
(ii) Fermi contact interaction
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Which nuclei will interact? Like as NMR selection rules.
Every isotope of every element has a ground state nuclear spin quantum number, I
has value of n/2, n is an integer
Isotopes with even atomic number and even mass number have I= 0, and have no EPR spectra
12C, 28Si, 56Fe, …
Isotopes with odd atomic number and even mass number have neven
2H, 10B, 14N, …
Isotopes with odd mass number have n odd
1H, 13C, 19F, 55Mn, …
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a b
b
a
B
E
Ms = +1/2
Ms = -1/2
E = gmBBoMS + amBMSmI
MI = +1/2
MI = -1/2
MI = -1/2
MI = +1/2
∆mI = 0
2nI+1 → 2*1*½+1=2
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Hyperfine Interactions Relative intensities determined by the number of interacting
nuclei
If only one nucleus interacting
All lines have equal intensity
If multiple nuclei interacting
Distributions derived based upon spin
For spin ½ (most common), intensities follow binomial distribution
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Relative Intensities for I = ½
N Relative Intensities
0 1
1 1 : 1
2 1 : 2 : 1
3 1 : 3 : 3 : 1
4 1 : 4 : 6 : 4 : 1
5 1 : 5 : 10 : 10 : 5 : 1
6 1 : 6 : 15 : 20 : 15 : 6 : 1
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Relative Intensities for I = ½
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CH3•
a b
b
a
B
E
Ms = +1/2
Ms = -1/2
MI = +3/2
MI = -3/2
MI = -3/2
MI = +3/2
MI = +1/2
MI = -1/2
MI = -1/2
MI = +1/2
∆mI = 0
2nI+1 → 2*1*½+1=4
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CH3•
MI = +3/2
MI = -3/2
MI = -3/2
MI = +3/2
MI = +1/2
MI = -1/2
MI = -1/2
MI = +1/2
∆mI = 0
2nI+1 → 2*1*½+1=4
+3/2 ↑↑↑+1/2 ↑↑↓ ↑↓↑ ↓↑↑
-1 /2 ↑↓↓ ↓↑↓ ↓↓↑-3/2 ↓↓↓
1331
probability
1 3 3 1
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Hyperfine Interactions
Example:
Radical anion of benzene [C6H6]•-
Electron is delocalized over all six carbon atoms
Exhibits coupling to six equivalent hydrogen atoms
2NI + 1 = 2(6)(1/2) + 1 = 7
7 lines
relative intensities 1:6:15:20:15:6:1
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Relative Intensities for I = 1
N Relative Intensities
0 1
1 1 : 1 : 1
2 1 : 2 : 3 : 2 : 1
3 1 : 3 : 6 : 7 : 6 : 3 : 1
4 1 : 4 : 10 : 16 : 19 : 16 : 10 : 4 : 1
5 1 : 5 : 15 : 20 : 45 : 51 : 45 : 20 : 15 : 5 : 1
6 1 : 6 : 21 : 40 : 80 : 116 : 141 : 116 : 80 : 40 : 21 : 6 : 1
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Relative Intensities for I = 1
pyrazine radical anion
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(a) coupling to 2 14N nuclei (1:2:3:2:1 quintet)split by 4 H atoms further into 1:4:6:4:1 quintet
(b) Na+ salt, further splitting into 1:1:1:1 quartet
2nI+1 → 2*2*1+1=5
2nI+1 → 2*4*½+1=5
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For N I=1 (+1, 0,-1) ↑ → ↓
+2 ↑↑+1 ↑→0 ↑↓
+1 →↑0 →→-1 →↓0 ↓↑-1 ↓→-2 ↓↓
12321
probability
↑
→
↓
N N
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For H I=1/2 ↑
+2 ↑↑+1 ↑→0 ↑↓
+1 →↑
0→→
-1 →↓0 ↓↑-1 ↓→-2 ↓↓
12321
probability
↑
→
↓
+2 ↑↑↑↑+1 ↑↑↑↓ ↑↑↓↑ ↑↓↑↑ ↓↑↑↑0 ↑↑↓↓ ↓↑↑↓ ↓↑↓↑ ↑↓↑↓ ↑↓↓↑ ↓↓↑↑-1 ↓↓↓↑ ↓↓↑↓ ↓↑↓↓ ↑↓↓↓
-2 ↓↓↓↓
14641
probability
N N H H H H
pyrazine radical anion
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(a) coupling to 2 14N nuclei (1:2:3:2:1 quintet)split by 4 H atoms further into 1:4:6:4:1 quintet
(b) Na+ salt, further splitting into 1:1:1:1 quartet
2nI+1 → 2*2*1+1=5
2nI+1 → 2*4*½+1=5
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EPR
Hyperfine Interactions
Coupling to several sets of nuclei
First couple to the nearest set of nuclei
Largest a value
Split each of those lines by the coupling to the next closest
nuclei
Next largest a value
Continue 2-3 bonds away from location of unpaired electron
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Hyperfine Interactions
Example:
VO(acac)2
Interaction with vanadium nucleus
V: I = 7/2
2nI + 1 = 2(1)(7/2) + 1 = 8 line expected
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EPR spectrum of vanadyl acetylacetonate
Hyperfine Interactions
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EPR
11 line
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EPR
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EPR
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EPR
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EPR
Concentration effect
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EPR
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EPR
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EPR
anisotropic systems
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solids, frozen solutions, radicals prepared by irradiation of crystalline materials, radical trapped in host matrices, paramagnetic point defect in single crystals
for systems with spherical or cubic symmetry g factors
for systems with lower symmetry, g ==> g‖ and g┴ ==> gxx, gyy, gzz
ESR absorption line shapes show distinctive envelope
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system with an axis of symmetry no symmetry
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Li+ – 13CO2- in CO2 matrix
large 13C and small 7Li (I = 3/2) hyperfine splitting
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Cr porphyrin→ oxidation → radical I=3/2 Abundance 9.5%
EPR
radical
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HMn(CO)5 /solid Kr matrix at 77 Khu-→ •Mn(CO)5
A‖(55Mn) = 6.5 mT
A┴(55Mn) = 3.5 mTA┴(83Kr) = 0.4 mT
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trans-[Cr(pyridine)4Cl2]+
frozen solution in DMF/H2O/MeOH
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trans–[Rh(pyridine)4Cl2]Cl·6H2O
powder
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transition metal complexes• the number of d electrons• high or low spin complex• consequence of Jahn-Teller
distortion
• zero-field splitting and Kramer’s degeneracy ESR spectra of second and third row transition metal complexes are often hard to observed, however, rare-earth metal complexes give clear, useful spectra short spin-lattice relaxation times
==> broad spectral lines
low temperature experiments will be needed to observe spectra
d3 system
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zero-field splitting
in the absence of magnetic field, 2S + 1 energy states split depends on the structure of sample, spin-orbit coupling
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Copper(II) acetylacetonate (Cu(acac)2)
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• Copper has two nuclear magnetically active isotopes. Both isotopes have a nuclear spin of I=3/2, but they vary in their natural abundance.
• The 63Cu isotope has a natural abundance of 69% while the 65Cu isotope has a natural abundance of 31%.
• Since the nuclear magnetogyric ratios are quite similar with 7.09 for 63Cu and 7.60 for 65Cu, the hyperfine coupling to each isotope is nearly identical.
• As a result, the ESR spectrum shows four resonances as it couples to the one nuclear spin I=3/2 in each molecule.
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Mo2O3dtc4
• The complex is dinuclear and containsmolybdenum(V)
• The strong centerline is due to the molecules with the 96Mo isotope. This isotope has a nuclear abundance of 75 % with a nuclear spin I=0. Because of the spin of zero, only a single resonance is observed.
• The 95Mo isotope is 15.72 % and the 97Mo isotope is 9.46 % abundant, both with a spin of I=5/2 with similar magnitudes of the magnetogyric ratio (but opposite signs). As a result, about 25% of the EPRsignal is split into a sextet of lines.
[G]3 4 0 0 3 4 5 0 3 5 0 0 3 5 5 0 3 6 0 0 3 6 5 0 3 7 0 0 3 7 5 0
-1 2 0
-1 0 0
-8 0
-6 0
-4 0
-2 0
0
2 0
4 0
6 0
8 0
1 0 0
1 2 0
1 4 0[*1 0 ̂3 ]
Fe(NO)dtc2
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• The nitrosyl group has an unpaired
electron
• The electron is located at the nitrogen
atom and therefore couples with the
nucleus
(14N: 99.638 % abundance, I=1)
• A three line spectrum is observed for
this compound (=2*1+1)
[G]3390 3400 3410 3420 3430 3440 3450 3460
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
[*10^ 3]
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Ex. 12 d9 system
CuII(TPP) complex (frozen solution in CCl3H)
Cu(acac)2 frozen solution
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multiple resonanceENDOR (electron-nuclear double resonance)
Ex. 13 [Ti(C8H8)(C5H5)] in toluene (frozen solution)
(a) ESR spectrum (b) 1H ENDOR spectrum
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BH4- + •C(CH3)3 → [BH3•]- + HC(CH3)3
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g = 2.005A(N) = 0.45 mT
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S(=NBut)2 • - g = 2.0071
A(N) = 0.515 mT
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(MeO)3PBH2•
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CrIII(porphyrin)Cl
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EPR
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