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Transcript
Magnetochimica AA 2011-2012 Marco Ruzzi Marina Brustolon 2. EPR in a nutshell Electron spin in a magnetic field e is proportional to S, meaning that e and S are vectors parallel to each other. They have opposite directions because the proportionality constant is negative. The latter one is written as the product of two factors:g and B: μe g B S e g B S g is a number called Landé factor, or simply g factor. For a free electron g = 2.002319. B eh 4me J T where me is the electron mass, e is the electron charge and h = 6.626 10-34 Js is the Planck constant. B is the atomic unit of magnetic moment, called Bohr magneton. The negative value of the electron charge is the reason why B 0 Zeeman effect Suppose to apply a constant magnetic field B to an electron spin. Since the energy of a magnetic moment e is given by the scalar product between e and B, the electron spin energy will depend on the orientation of e with respect to B E -μe B g | B | S B The dot product reduces to a single term if the direction of B coincides with one of the axes respect to which the B and S are represented. The choice of the reference frame is arbitrary and it can be chosen in such a way that the z axis is along the direction of B. In this case the equation for the energy becomes: Eg | B | B0 S z E ( 1/ 2 )g | B Β The splitting of the electron spin energy level into two levels in the presence of a magnetic field is called Zeeman effect. The resonance condition h E -E g | B | B0 In a magnetic field of 3.5 T, which is the standard magnetic field intensity used in many EPR spectrometers, for g=2.0023 we have = 9.5 GHz. Other regions of higher microwave frequencies used in commercial EPR spectrometers are Q-band (~ 34 GHz) and W-band (95 GHz). E 0 a Ea=+(1/2)gBB0 a,b E=gBB0 b B0=0 Eb=-(1/2)gBB0 B00 Finding the resonance condition E E a a hres 0 0 OK h0 b b Easier experimental arrangement Bres B0 constant res g B B0 h The microwave frequency should be changed to find the resonance condition 0 constant Bres h 0 g B The magnetic field should be changed to find the resonance condition Cavità Una tipica cavità EPR -30 -20 -10 0 10 20 30 -30 -20 -10 0 - 0 - 0 a b 10 20 30 Note: EPR spectra are recorded usually by modulating the magnetic field, and the amplitude of the modulated signal is then recorded. Therefore the EPR trace represents the derivative of the original signal. I(B) B B segnale EPR Field modulation coils I(B) B B Let us consider the simple case of an atom with a closed shell and an extra electron. In such an atom the electron spins are all coupled in pairs, except one. The electron angular momentum has two contributions: one arises from the electron spin, and another one arises from the orbital motion of the electron around the nucleus. The magnetic moment is the sum of two terms, referring to the two contributions: e B l g B S Contrary to atoms which are spherical, molecules are systems of low symmetry. For them the orbital angular momentum is suppressed (its average is zero), and the electron angular momentum, in the absence of spin orbit coupling, is only due to spin. The effect of spin orbit coupling is to restore a little amount of orbit contribution, which results in a deviation g =g-ge of the g factor (see the Table in the next slide for examples of the deviations.). Spin density on heavy atoms = larger deviation of the g factor from ge fattore- g Radicale e- 2,0023 CH 3 fattore-g Radicale But 2,0026 N Bu t 2,0061 O Bu t O 2,0029 S 2,0103 Bu t O 2,0047 HO O Bu t Bu t Bu t O Bu t 2,0052 2,0140 Hyperfine coupling 1 The energy of nuclear spins is influenced by a magnetic field. This effect is called nuclear Zeeman effect. E g N | N | B0 I z In the presence of a nuclear spin the electron spin experiences an additional magnetic field provided by the nuclear magnetic moment, which affects the resonance conditions. The electron nucleus spin interaction is called hyperfine interaction. It gives rise to a splitting of the resonance EPR lines into several components, two components for interaction with a nuclear spin I = 1/2, three components for a I = 1 nucleus, and in general 2I+1 components for the interaction with a spin I nucleus. Hyperfine coupling 2 The hyperfine energy contribution is called contact (or Fermi) contribution and it is: Ehf a S I where h.c.c.) a is a constant (hyperfine coupling constant, which depends on |(0)|2, the square of the wave function which describes the electron motion calculated in the point where the nucleus is. | a | g | B | B0 Under these conditions (high field approximation) the energy term En due to the nuclear spin represents a small perturbation on the electron spin energy, which becomes: Etot g | B | B0 S z g N | N | B0 I z a S z I z Hyperfine coupling 3 For a free radical (S = 1/2) containing a single magnetic nucleus with I = 1/2, there are four possible values of the total energy, corresponding to the electron and nuclear spin components Sz = ±1/2 and Iz = ±1/2 as illustrated in the next slide. The corresponding energies are: E1 1 / 2 g | B | B0 1 / 2 g N | N | B0 1 / 4a E2 1 / 2 g | B | B0 1 / 2 g N | N | B0 1 / 4a E3 1 / 2 g | B | B0 1 / 2 g N | N | B0 1 / 4a E1 1 / 2 g | B | B0 1 / 2 g N | N | B0 1 / 4a EPR consists of transitions between pairs of energy levels characterized by different values of Sz, but the same Iz value. Energy levels for S=1/2 coupled to I=1/2 E Nuclear Zeeman E Electron Zeeman E E1 a/2 E2 EPR transitions 0 E3 S z 1 ; I z 0 a/2 E Hyperfine E4 Scheme of the energy levels for an external magnetic field B0 S=1/2 coupled to I=1/2: the EPR spectrum E a mS mI (4) (3) 1/2 1/2 1/2 -1/2 (2) -1/2 -1/2 (1) -1/2 1/2 B Accoppiamento con più nuclei equivalenti L’intensità relativa delle righe spettrali dovute all’accoppiamento con più nuclei equivalenti con I=1/2 è data dai coefficienti dello sviluppo binomiale (a+b)n. Per calcolare questi ultimi si può utilizzare il triangolo di Tartaglia. Nuclei (I=1/2) Equivalenti Intensità relativa 0 1 1 1 2 1 3 1 4 1 5 1 6 1 7 1 8 9 1 1 8 9 10 35 1 5 15 35 70 126 1 4 20 56 84 3 10 21 1 6 15 28 36 3 5 7 2 4 6 1 6 21 56 126 1 1 7 28 84 1 8 36 1 9 1 H Radicale Metile H C H z 23 G B Mi + 3 2 + 1 2 1 2 3 2 EPR spectrum of Naphtalene radical anion 1 2 5 Gauss A quintet of quintets, each 1:4:6:4:1 Nitroxide radicals Some stable radicals O CH3 H3C H3C CH3 N CH3 O H3C H3C CH3 N H3C H3C Tempone N H3C NO 2 N CH3 CH3 O Nitronyl Nitroxide radical CH3 O O 2N N CH3 di-t-butyl-nitroxide R O CH3 O CH3 O Tempo N CH3 ONa N T T T = NO 2 H3C T 2,2-diphenyl-1picrylhydrazyl DPPH H3C Trityl S S S S CH3 CH3 H3C H3C N CH3 CH3 O a a 10 Gauss 3315 Gauss 14N; I = 1; a.n. = 99.63 L’intensità relativa delle righe spettrali dovute all’accoppiamento con più nuclei caratterizzati da I>1/2 si può ricavare dal corrispondente “diagramma a piramide”. Esempio: 2 nuclei equivalenti con I=1 B0 Intensità 1 relativa 2 O O N N 3 N N O O 1 8.0 G Ph Ph 2 Le bande EPR Campi di risonanza e l per g = 2 alle frequenze a W comunemente usate in EPR. Banda W Frequenza (GHz) L S X K Q W 1.1 3.0 9.5 24.0 35.0 94.0 B0 (Gauss) l (mm) 392 1070 3389 8560 12485 33600 330 100 32 12.5 8.6 3.2 N.B. g = h/BB0 alti valori di g danno righe a campi più bassi! 22