Electron Spin Resonance (ESR) Spectroscopy
The phenomenon of electron spin resonance (ESR), also called electron paramagnetic resonance (EPR)/electron magnetic resonance, was discovered by the Soviet physicist Zavoisky. The phenomenon of electron spin resonance (ESR), also called electron paramagnetic resonance (EPR)/electron magnetic resonance, was discovered by the Soviet physicist Zavoisky. NMR is observed in radiofrequency (RF) region and ESR is observed in microwave (MW) region of the electromagnetic spectrum. ESR is observed primarily in systems containing only unpaired electrons. Thus, the systems which can be studied by ESR spectroscopy are organic or inorganic free radicals, triplet states and transition metal ions, their complexes, and metalloproteins with unpaired d- or f-electrons. Thus, the systems which can be studied by ESR spectroscopy are organic or inorganic free radicals, triplet states and transition metal ions, their complexes, and metalloproteins with unpaired d- or f-electrons.
For an isolated electron, without any outside forces, it has only an intrinsic angular momentum called “spin.” Because an electron is charged, the angular/spinning motion of this charged particle generates a magnetic field. In other words, the electron due to its charge and angular momentum, acts like a little bar magnet, or magnetic dipole, with a magnetic moment (m).
The electron with spin 1
2, the parallel state ( ) is designated as ms = + 1
2 and the antiparallel ( ) state is ms = - 1
2 (Fig. 1). The energy of each orientation is the product of μ and B0.
E = μ B0 where μ = ms g β ..1
β is a conversion constant called the Bohr magneton (also called μB) and ge is the spectroscopic g-factor of the free electron and equals 2.00. Therefore, the energies for an electron with ms = + 1
2 and ms = - 1
2 are, respectively, as follows:
E = ms g β B0 ..2
𝐸+1
2
= + 1
2g β B0 and 𝐸−1
2
= - 1
2g β B0
Figure 1 Electron Zeeman interaction
PRESENTATION OF ELECTRON PARAMAGNETIC RESONANCE SPECTRA In an EPR spectrometer, a paramagnetic sample is placed in a large magnetic field B0
which, as shown below, splits the energy levels of the ground state by an amount
∆E = hυ = g β B0 ..3
The value of g can be determined from the ∆E, the energy difference between the two spin levels. This can be done by both irradiating the sample either with MWs with a fixed frequency and sweeping the magnetic field or applying a constant magnetic field and scan the frequency of the electromagnetic radiation as in conventional spectroscopy.
In EPR, a klystron MW source is used. This X-band klystron has a narrow spectral band width of about 8.8-9.6 GHz. This makes it impossible to continuously vary the wavelength similar to optical spectroscopy. It is therefore in EPR necessary to vary or sweep the magnetic field, until the quantum of the MWs matches the energy difference between the field-induced energy levels. Thus, a peak (depending on energy levels) due to absorptions with the selection rule ∆ms = ±1 will be observed in the spectrum when the separation of two energy levels is equal to the quantum energy of the incident MW photon. At that instance, the absorption of energies will occur since the condition in Eq. 3 is satisfied. This field is called the “field of resonance,” the value of g is then calculated from υ (in GHz) and B0 (in gauss) using Eq. (3).
The direct detection of the absorption signal is possible only for samples containing a high concentration of unpaired electrons. Noise components over a wide range of frequencies appear with the signal, making its detection difficult. This is overcome by field modulation using a phase sensitive detector technique. This gives the signal as a first derivative as shown in Fig. 2.
Furthermore, due to short relaxation time of paramagnetic system, the absorption signals are usually broad and hyperfine splitting’s are buried and not seen (Fig. 2B). However, the derivative signal observed using field modulation exhibits well resolved hyperfine features.
Figure 2. Absorption and first derivative electron paramagnetic resonance spectra for single line (A) and multiple lines (B) systems.
ESR Absorption Positions: The g Factor
Eq. (3) shows that an ESR absorption will occur at a frequency v =∆E/h Hz. Thus, the position of an ESR absorption may be expressed in terms of absorption frequency. It is clear from Eq. (3) that the absorption frequency, i.e. the absorption position, varies with the applied field B0 . Since different ESR spectrometers operate at different magnetic fields, it is desirable to express ESR absorption positions in same form independent of the field strength. Thus, the ESR absorption positions are more conveniently expressed in terms of the observed g values.
Rearranging Eq. (3), we get
g = ∆E
β 𝐵0 = hυ
β 𝐵0
For measuring the g values of free radicals it is convenient to measure the field separation between the center of the spectrum of the unknown sample and that of a reference substance whose g value is accurately known. The most widely used reference is 1,1-diphenyl- 2-picrylhydrazyl free radical (DPPH) which is completely in free radical state and its g value is 2.0036. The reference substance is placed along with the unknown in the same dual resonant cavity.
LINE WIDTH OF ESR SIGNAL
The line width of ESR resonance depends on the relaxation time of the spin state under study. Out of the two possible relaxation processes, the spin-spin relaxation is very efficient, unless the sample is extremely dilute. The spin-spin relaxation time comes out to be 10–6 to 10–8 s, much shorter than the spin-spin relaxation time in case of NMR. The spin lattice relaxation is efficient at room temperature giving relaxation time of about 10–6 s, but becomes progressively less efficient at reduced temperatures. For most samples, therefore we can choose 10–7 s as a typical relaxation time and, using this in Heisenberg’s relation
∆E x ∆t = ℎ
2𝜋 we can calculate a frequency uncertainty (line width) as
∆υ = ℎ
4𝜋 ∙ 1
ℎ ∆t= 1
4𝜋 ∆t= 1
4 ×3.14 ×10−7 𝐻𝑧 = 0.796 𝑀𝐻𝑧
An even shorter relaxation time will increase this width, and line width of 10 MHz is not uncommon. Clearly, it is a much wider spectral line, than in case of NMR, where we find a normal line width for a liquid to be some 0.1 Hz.
A broad line is more difficult to observe and measure than a sharp line and, for this reason the ESR signals are recorded as the derivative of the absorption curve with respect to the magnetic field. An ESR absorption peak and its first derivative with respect to the magnetic field are shown in Fig. 3 (a) and Fig. 3 (b) respectively.
Fig. 3a Fig. 3b
The separation between maxima and minima of the derivative curve gives the line width.
INTENSITY OF ESR SIGNALS
The energy separation between the two spin states of an electron is much more than the energy separation between two nuclear spin states. The relative population of electrons in lower and higher energy spin states is governed by the Boltzman’s distribution law :
Thus, the lower energy spin state is more populated than the upper energy spin state of the electron. The difference in population between two electronic spin states is larger than that between two nuclear spin states due to the large value of ∆E in the former case. Consequently, the ESR signals are expected to be more intense than the NMR signals.
Hyperfine Structure
Hyperfine structure (HFS) occurs as a result of the magnetic interaction between the unpaired electronic spin S and the neighbouring nuclear spin I. The magnitude of this interaction is given by ‘a’ called isotropic hyperfine coupling constant.
The hyperfine interaction in ESR spectra is analogous to the fine structure, i.e., nuclear spin-spin interaction in NMR spectra. As a result of this interaction, the ESR signals or peaks are further split into several lines (HFS).
Characteristics of isotropic hyperfine splitting
1. It depends on the magnitude of magnetic moments of nuclear and electron spins.
2. The electron spin density in the immediate vicinity of the nucleus (Fermi contact) 3. It is independent of applied magnetic field
4. It obeys (n + 1) rule for I = 1
2 i.e., (2nI + 1)
5. The intensity ratio is obtained using Pascal triangle (analogous to NMR Spin-spin coupling) The Separation between the lines is usually of the order of 10-3-10-4 T (about 50 MHz) which is approximately 106 times larger than nucleus-nucleus coupling because an electron can approach a nucleus much more closer than another nucleus, and thus will interact more strongly with it. The biggest factor influencing the magnitude of electron-nucleus coupling is the
electron density at the coupled nucleus, i.e. the amount of time which the electron spends in the vicinity of the coupled nucleus.
Let us illustrate the hyperfine structure resulting from the splitting of an ESR signal by considering an example of hydrogen atom having one proton and one electron. The two energy levels of a free electron in an applied magnetic field are shown in Fig. 4 (a) with ms = -1
2 aligned with the field and ms = + 1
2 aligned opposing the field. Thus, the ESR spectrum of a free electron would consist of a single peak (Fig. 4) corresponding to a transition between these energy levels.
Fig. 4 Effect of: (a) applied field on the spin energy states ms = ± 1
2 of an electron and (b) applied field and nuclear spins mI = ± 1
2 of proton on ms = - 1
2 and ms = + 1
2 energy states of an electron
Each of the two energy levels of the electron in hydrogen atom is split into two energy levels by the interaction with the nuclear spins of proton mI = ± 1
2 where m1 is the nuclear spin angular momentum quantum number. Thus, corresponding to the two energy states ms = - 1
2 and ms = + 1
2 four different energy levels are obtained as shown in Fig. 4(b). This is why ESR spectrum of hydrogen atom consists of two peaks of equal intensity (Fig. 5) corresponding to two transitions shown by two arrows in Fig. 4(b) rather than one corresponding to the arrow in Fig 4(a). The two peaks are of equal intensity because the probability of orientations of the nuclear spin of hydrogen atom causing different energy Ievels is equal. The selection rules in ESR are ∆mI = 0 and ∆ms = ±1 which allow us to decide between which levels transitions will give rise to spectral lines.
Fig. 5 First derivative ESR spectrum of hydrogen atom
Similarly, in the case of methyl radical, each of the two energy levels of the single unpaired electron on the carbon atom is split into four energy levels by the interaction with the nuclear spins of the three hydrogen nuclei (mI for a single hydrogen nucleus is ± 1
2 and that for the three hydrogen nuclei ± 3 x 1
2 = ± 3
2 ). Thus, corresponding to the two spin energy states of the unpaired electron, eight different energy levels are obtained as shown in Fig. 6. Four transitions are possible according to the ESR selection rules ∆mI = 0 and ∆ms = ±1. Thus, the ESR spectrum of methyl radical consists of four peaks (Fig. 7). The number of component peaks in the ESR signal of methyl radical can also be determined by the formula 2nI + 1 which also indicates (2 x 3 x 1
2) + 1 = 4 peaks. The observed relative intensities of the four component peaks are in the ratio 1 : 3 : 3 : 1. The relative intensities of the component peaks of a multiplet are directly proportional to the number of nuclear spin orientations of equivalent energy causing different energy levels (Fig. 6) and are given by coefficient of terms in the binomial expansion of (x + 1)n.
Fig. 6 Energy level diagram illustrating coupling between the spins of the unpaired electron and hydrogen nuclei in methyl radical
Fig. 7 First derivative ESR spectrum of methyl radical
Let us now discuss the ESR spectrum of 1,4-benzosemiquinone shown in Fig. 8. It exhibits five peaks with their intensities in the ratio 1 : 4: 6: 4: 1. In 1,4-benzosemiquinone the unpaired electron is delocalized over all the carbon and oxygen atoms. Thus, all the four protons are equivalent and the unpaired electron is coupled with four equivalent protons (for proton I = 1
2) resulting in the splitting of its ESR signal into five peaks according to the formula 2nI + 1. The ratio of relative intensities of these peaks is given by coefficients of the binomial expansion of (x + 1)n. In this case n = 4 (four equivalent protons are coupled with the unpaired electron). Hence, the intensity ratio comes to be 1:4:6:4:1.
Fig. 8 (a) Absorption ESR spectrum of 1,4-benzosemiquinone and (b) first derivative ESR spectrum of 1,4-benzosemiquinone
The ESR spectrum of 1,4-benzosemiquinone can also be explained graphically as shown in Fig. 9.
Fig. 9 Energy level diagram illustrating coupling between the unpaired electron and the four protons in 1,4-benzosemiquinone,
The deuterium atom 12𝐻 is a simple example of a system having I = 1. In this case, there are three values of mI i.e. mI = +1, 0 and -1 corresponding to ms =+ 1
2), and three values of mI
corresponding to ms = - 1
2., i.e. mI = -1, 0 and + 1. Thus, each of the two energy levels of the electron in deuterium is split into three energy levels by coupling with the nuclear spins of deuterium resulting insix different energy levels as shown in Fig. 10. According to the ESR selection rules (∆ms = ±1 and ∆mI = 0), there are three allowed transitions. Thus. ESR spectrum of deuterium atom consists of three peaks of equal intensity (Fig. 11) corresponding to three transitions shown by arrows in Fig. 10. The three peaks are of equal intensity because the probability of the orientations of the nuclear spin of deuterium atom causing different energy levels is equal.
Fig. 10 Energy level diagram illustrating coupling between the unpaired electron and deuterium nucleus
Fig. 11 First derivative ESR spectrum of deuterium atom
ESR SPECTRUM OF BENZENE ANION (𝑪𝟔̇ 𝑯𝟔−)
Benzene anion can be formed by the reaction of an alkali metal with benzene in a solvent such as tetrahydrofuran (THF). Benzene anion results when the alkali metal atom transfers an electron to the benzene molecule. The observed ESR spectrum of 𝑪𝟔̇ 𝑯𝟔− (Fig. 12) is symmetrical with seven peaks indicating that the unpaired electron density is distributed equally among the six protons in the system.
Fig. 12. ESR spectrum of benzene anion.
As there are six equivalent protons in benzene anion, the expected number of ESR signals = (2n I + 1) = 2 x 6 x 1
2 +1 = 7. Seven lines with intensity ratio 1 : 6 : 15 : 20 : 15 : 6 : 1, are actually observed in the ESR spectrum of 𝑪𝟔̇ 𝑯𝟔−. The spacing between the lines gives the hyperfine coupling constant (a) of 3.76 gauss.
Cyclopentadienyl Radical
This radical has one unpaired electron (S = 1
2) interacting with five equivalent protons.
Hence, the number of peaks observed is calculated as (2nI + 1 = 2 x 5 x 1
2 +1 = 6). Therefore, the ESR spectrum (Fig. 13) contains a sextet i.e., six equally spaced lines.
Figure 13 Electron spin resonance spectrum of the cyclopentadienyl radical.
HYPERFINE SPLITTING RULES FOR PREDICTING THE NATURE OF ESR SPECTRUM FOR MORE THAN ONE MAGNETIC NUCLEI
In general an ESR signal is split by n equivalent magnetic nuclei of spin I into (2nI + 1) signals. When the splitting is caused by both a set of n equivalent nuclei of spin In and a set of m equivalent nuclei of spin Im , the number of ESR lines is given by (2n In + 1) (2n Im + 1).
The following specific cases illustrate the use of these general rules :
When the unpaired electron is delocalized over an non-equivalent protons (I = 1
2), a spectrum consisting of 2n lines will be observed.
When the unpaired electron is delocalised over n equivalent protons, a total of (n+1) lines, corresponding to (2n x 1
2+ 1 ) will be observed. Thus, methyl free radical 𝑪𝑯̇ 𝟑 , in which the unpaired electron is delocalised over three equivalent protons will give rise to four ESR signals with intensity ratio 1 :3 :3 : 1. Similarly, the benzene anion (𝑪̇ 𝑯𝟔 𝟔−) displays seven ESR signals as its unpaired electron is delocalised over six equivalent protons. The intensity ratio of these signals being 1 : 6 : 15 : 20 : 15 : 6 : 1.
The intensity ratio of (n + 1) signals arising out of the delocalisation of an unpaired electron over n equivalent protons is given by the coefficients of the binomial expression (1 + x)n, For, two protons, n = 2
Thus, the intensity ratio of three signals will be 1 : 2 : 1.
For three protons, n = 3, and (1 +x)3 = 1 + 3x + 3x2 +x3
So, the intensity ratio of the four signals will be 1 : 3 : 3 : 1.
If the odd electron is delocalised over two sets of non-equivalent protons, the number of signals expected is the product of the number expected for each set, i.e., (2nIn +1) (2mIm + 1)
Naphthalene Anion Radical
The naphthalene anion 𝑪𝟏𝟎̇ 𝑯𝟖−, consists of an unpaired electron delocalised over the entire naphthalene ring. It has two different sets ( & ), each of four equivalent protons. The ESR spectrum would thus show, i.e., (2nIn +1) (2mIm + 1) = (4 + 1) x (4 + 1) = 25 lines. This radical is formed when naphthalene insolution in 1,2 dimethoxyethane is reduced with potassium metal. A green colored solution is obtained whose EPR spectrum is shown below.
By considering the pattern and coupling constants of a1 = 4.90 G and a2 = 1.83 G of hyperfine splittings, the species formed is consistent with the naphthalene radical anion.
Figure 14 Electron spin resonance spectrum of the naphthalene anion radical (K+ is the counterion) at 298K.
p-Nitrobenzoate Dianion Radical
The structure, experimental spectrum and its stick diagram for the reconstruction of the spectrum for the p-nitrobenzoate dianion radical are shown in Fig. 15. The unpaired electron
in the molecule first interacts with the 14N (I = 1) with large splitting (AN) to give a triplet signal of intensity ratio 1:1:1. Furthermore, it consists of two sets of two equivalent protons. So, each splits again into a triplet with intensity ratio 1:2:1 and again each signal splits into triplet to give overall 27 lines as shown in Fig. 15. However, there is no way based on the spectrum alone to decide which splitting corresponds to which set of protons. A comparison was made with the calculated results and with spectra of similar radicals for which deuterium substituted radicals have been studied. It indicates that the smallest splitting, AH·, is associated with the protons closer to the CO2, while the relatively large splitting, AH, corresponds to the protons near to the NO2 group.
Figure 15 The structure stick diagram and electron spin resonance spectrum of p- Nitrobenzoate dianion radical.
Diphenylpicrylhydrazyl Radical
In the case of DPPH, the unpaired electron is delocalized on to two nitrogen nuclei (I = 1 and n = 2). Hence, the two nitrogen nuclei are equivalent and the spectrum (Fig. 16) exhibits 2nI + 1 = 5, five lines with intensity ratio: 1:2:3:2:1. However, in its solid polycrystalline form, it exhibits a single-EPR line that is narrowed due to Heisenberg spin exchange.
Figure 16 The structure of diphenylpicrylhydrazyl radical and its electron paramagnetic resonance spectrum.
Problem I. Which of the following will show electron spin resonance (ESR) spectrum? Give reason for your choice.
Solution:
(a) Hydrogen atom (H) has electronic configuration ls1, i.e. it has one unpaired electron.
Thus, it will show ESR spectrum.
(b) Hydrogen molecule (H2) has electronic configuration (σ1s)2, i.e. it has no unpaired electron. Thus, it will not show ESR spectrum.
(c) (c) Chlorine atom (Cl) has electronic configuration 1s22s22p63s23px23py22px1. i.e. it has one unpaired electron. Thus, it will show ESR spectrum.
(d) The electronic configuration of Na+ is ls22s22px22py22pz2. Since it has no unpaired electron, it will not show ESR spectrum.
(e) The electronic configuration of cuprous ions (Cu+) is 3d10. Since it has no unpaired electron, it will not show ESR spectrum.
Problem 2. Calculate the ESR frequency of an unpaired electron in a magnetic field of 3000 G (0.30 T).
Solution: we have
Problem 3. Calculate the g value if the methyl radical shows ESR signal at 3290 G (0.3290 T) in a spectrometer operating at 9230 MHz.
Solution: we have
Problem 4. Predict the number of lines in the ESR spectrum of each of the following radicals:
(a) [CF2H]· (b) [13CH3]· (c) [CF2D]· (d) [CClH2]· (e) [13CF2H]·
Solution: The number of lines in an ESR signal = (2nIn +1) (2mIm + 1) where n and m are the number of different kinds of coupled nuclei having spin In and Im, respectively.
(a) The spin of F = 1
2, i.e. In = 1
2 and H = 1
2, i.e. Im = 1
2; n and m for the radical [CF2H]· are 2 and 1, respectively. So, (2 x 2 x 1
2 + 1) (2 x 1 x 1
2 + 1) = 6 lines (b) [13CH3]· Here n = 1, In = 1
2; m = 3, Im = 1
2. So, (2 x 1 x 1
2 + 1) (2 x 3 x 1
2+ 1) = 8 lines (c) [CF2D]· Here n = 2, In = 1
2;; m = 1, Im = 1 (spin of D = 1). So, (2 x 2 x 1
2 + 1) (2 x 1 x 1 + 1) = 9 lines
(d) [CClH2]· n = 1, In = 3
2; (spin of Cl = 3
2); m = 2, Im = 1
2;. So, (2 x 1 x 3
2 + 1) (2 x 2 x 1
2 + 1) = 12 lines
(e) [13CF2H]· Here n = 1, In = 1
2; m =2, Im = 1
2; o = 1, Io = 1
2; So, (2 x 1 x 1
2 + 1) (2 x 2 x 1
2 + 1) (2 x 1 x 1
2 + 1) = 12 lines.
Problem 5. A free electron is placed in a magnetic field of strength 1.3 T. Calculate the resonance frequency, if g = 2.0023.
Solution: Resonance condition for ESR absorption is
Problem 6. How many ESR peaks would you get for sodium atom (I = 3
2 ) Solution: Sodium atom has one unpaired electron (S = 1
2 ) . The electron spin will couple with nuclear spin of sodium (I = 3
2 ) to give a hyperfine structure. The number of ESR peaks ESR signal = (2nI +1) = (2 x 1 x 3
2 + 1) = 4 For, I = 3
2 ; mI = +3
2 , +1
2, −1
2 −3
2 and for S = 1
2 , mS = +1
2, −1
2
Fig. 17. ESR spectrum of sodium atom.
Four transitions are allowed according to the selection rule, mS = 1 and mI = 0.
Thus, four peaks are observed in the ESR spectrum of sodium atom.
Problem 7. Which of the following will show ESR spectra and why? (a) Benzene, C6H6 (b) Benzene anion, C6H6- (c) Cyclopentadienyl cation, C5H5+ (d) Cyclopentadienyl anion, C5H5-
Solution: (a) Benzene, C6H6 Hückel molecular-orbital (HMO) calculations on benzene, the n- electron distribution may be represented as
Since there is no unpaired electron, benzene will not show ESR spectrum.
(b) Benzene anion, C6H6- From HMO calculations on C6H6-, the n-electron distribution can be represented as
Since there is one unpaired electron, C6H6- will show ESR spectrum.
(c) Cyclopentadienyl cation, C5H5+; The n-electron distribution in the molecular orbitals of C5H5+is as follows:
Since there are two unpaired electrons, C5H5+ will show ESR spectrum.
(d) Cyclopentadienyl anion, C5H5-.The n-electron distribution in the molecular orbitals of C5H5- is
Since there is no unpaired electron, C5H5- will not show ESR spectrum.
Problem 8. Trichloromethyl radical [CC13]· shows ten lines in the hyperfine structure of its ESR spectrum. Calculate the spin of the Cl nucleus.
Solution: The radical [CC13]· has one unpaired electron which interacts with the nuclear spin of Cl. We know that the number of lines are equal to 2nI + 1. Hence, 2nI + 1 =10 (n is the number of interacting nuclei and I is their spin).
Or, (2 x 3 x I + 1) = 10 I = 9
6 = 3
2
therefore, spin of the Cl nucleus = 3
2
