Electron Paramagnetic Resonance Spectroscopyr.takjoo.profcms.um.ac.ir/imagesm/1006/stories/Spectroscopy/93/section 23.pdfNo magnetic field B = 0 Magnetic field B > 0 g g-factor (approximately

Electron Paramagnetic Resonance Spectroscopy

Spectroscopy in Inorganic Chemistry

Ferdowsi University of Mashhad2

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


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


Common field strength 0.34 and 1.24 T9.5 and 35 GHzMicrowave region

EPR


EPR

typical ESR spectrometer —a radiation source (klystron)a sample chamber between the poles of a magneta detection and recorder system


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


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


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

Ferdowsi University of Mashhad

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


EPR

Better resolution

Find shoulder

Ferdowsi University of MashhadEPR

11

spin-lattice relaxation


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

Ferdowsi University of MashhadEPR

13


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


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


couplings arise in two ways:

(i) direct dipole-dipole interaction

(ii) Fermi contact interaction


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, …


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


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


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


Relative Intensities for I = ½


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


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



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


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


Relative Intensities for I = 1

pyrazine radical anion


(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

25


For N I=1 (+1, 0,-1) ↑ → ↓

+2 ↑↑+1 ↑→0 ↑↓

+1 →↑0 →→-1 →↓0 ↓↑-1 ↓→-2 ↓↓

12321

probability

↑

→

↓

N N


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


(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

25


EPR


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



Example:

VO(acac)2

Interaction with vanadium nucleus

V: I = 7/2

2nI + 1 = 2(1)(7/2) + 1 = 8 line expected


EPR spectrum of vanadyl acetylacetonate



EPR

11 line


EPR


EPR


EPR


EPR

Concentration effect


EPR


EPR


EPR

anisotropic systems


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


system with an axis of symmetry no symmetry


Li+ – 13CO2- in CO2 matrix

large 13C and small 7Li (I = 3/2) hyperfine splitting


Cr porphyrin→ oxidation → radical I=3/2 Abundance 9.5%

EPR

radical





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



trans-[Cr(pyridine)4Cl2]+

frozen solution in DMF/H2O/MeOH


trans–[Rh(pyridine)4Cl2]Cl·6H2O

powder


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


zero-field splitting

in the absence of magnetic field, 2S + 1 energy states split depends on the structure of sample, spin-orbit coupling



Copper(II) acetylacetonate (Cu(acac)2)


• 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.


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


• 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]



62

Ex. 12 d9 system

CuII(TPP) complex (frozen solution in CCl3H)

Cu(acac)2 frozen solution


64

multiple resonanceENDOR (electron-nuclear double resonance)

Ex. 13 [Ti(C8H8)(C5H5)] in toluene (frozen solution)

(a) ESR spectrum (b) 1H ENDOR spectrum


BH4- + •C(CH3)3 → [BH3•]- + HC(CH3)3


g = 2.005A(N) = 0.45 mT


S(=NBut)2 • - g = 2.0071

A(N) = 0.515 mT


(MeO)3PBH2•


CrIII(porphyrin)Cl


EPR


Documents

Electron Paramagnetic Resonance Spectroscopyr.takjoo.profcms.um.ac.ir/imagesm/1006/stories/Spectroscopy/93/section 23.pdfNo magnetic field B = 0 Magnetic field B > 0 g g-factor (approximately