APPLICATIONS OF ESR SPECTROSCOPY TO METAL COMPLEXES
APPLICATIONS OF ESR TOMETAL COMPLEXES V.SANTHANAM DEPARTMENT OF CHEMISTRY SCSVMV
METAL COMPLEXES – A SURVEY Metal complexes are important- Diverse biological roles Griffiths and Owen proved the M-L covalency by taking complexes (NH4)2[IrCl6] and Na2[IrCl6] The hyperfine splitting by Chloride ligands showed the covalent nature of M-L bond
Proved the back donation (pi-bonding) concept With the ESR data they were able to calculate ξ,ζ and λ of metal ions and the extent of delocalization In metal complexes the above said parameters were having lower values than the free metal ions.
THINGS TO BE CONSIDERED Nature of the metal Number of ligands Geometry No of d electrons Ground term of the ion
Electronic degeneracy Inherent magnetic field Nature of sample Energy gap between g.s and e.s Experimental temperature
NATURE OF THE METAL ION Since d metal ions have 5 d orbitals situations are complicated But the spectra are informative In 4d and 5d series L-S / j-j coupling is strong making the ESR hard to interpret
Crystal field is not affecting the 4f and 5f e- so the ESR spectra of lanthanides and actinides are quite simple. If ion contains more than one unpaired e- ZFS may be operative
GEOMETRY OF THE COMPLEX Ligands and their arrangement –CFS CFS in turn affect the electronic levels hence the ESR transitions The relative magnitude of CFS and L-S coupling is giving three situations.
If the complex ion is having cubic symmetry (octahedral or cubic) – g is isotropic Complexes with at least one axis of symmetry show two g values Ions with no symmetry element will show three values for g.
Symmetry of the complex ion- important – why? ESR is recorded in frozen solutions Spins are locked Lack of symmetry influences the applied field considerably.
Spin Hamiltonian of an unpaired e- if it is present in a cubic field is H = g β | Hx.Sx + Hy.Sy + Hz.Sz| If the system lacks a spherical symmetry and possess at least one axis ( Distorted Oh,SP or symmetric tops) then H = β |gxx Hx.Sx +gyy Hy.Sy + gzz Hz.Sz| Usually symmetry axis coincides with the Z axis and H is applied along Z axis then gxx = gyy = gL ; gzz = g||
If crystal axis is not coinciding with Z axis The sample is rotated about three mutually perpendicular axis and g is measured. g is got by one of the following relations for rotation aboutX axis - g2 = gyy2Cos2θ + 2gyz2 Cos2θ Sin2θ +gzz2 Sin2θY axis - g2 = gzz2Cos2θ + 2gzx2 Cos2θ Sin2θ +gxx2 Sin2θZ axis - g2 = gyy2Cos2θ + 2gxy2 Cos2θ Sin2θ +gyy2 Sin2θ
NUMBER OF d ELECTRONS Magnetically active nucleus cause hyperfine splitting. If more than one unpaired e- present in the ion, more no of transitions possible leads to fine structure in ESR spectrum. Here we have to consider two things Zero field splitting – due to dipolar interaction Kramer’s degeneracy
ZERO FIELD SPLITTING Considering a system with two unpaired e-s Three combinations S = +1 possible In absence of external ∆E2 field all three states are S=0 having equal energy ∆E1 With external field three S = -1 levels are no longer with same energy. Two transitions H≠0 possible; both with ZFS = 0 same energy
SPLITTING OF ELECTRONIC LEVELS EVEN IN ABSENCE OF EXTERNAL MAGNETIC FIELD IS CALLED ZERO FIELD SPLITTING (ZFS) The splitting may be assisted by distortion and L-S coupling also.
When there is a strong dipolar interaction the +1 level is raised in energy –Dipolar shift (D) This dipolar shift reduces the gap between S = -1 and S = 0 state Now the two transitions do not have same energy Results in two lines
EFFECT OF DIPOLAR SHIFT Ms = +1 D Ms = ±1 ∆E1 = ∆E2Ms = ±1,0 Ms = 0 Ms = 0 D Ms = -1ZFS = 0
KRAMER’S THEOREM Systems with even no. of unpaired e-s will contain a state with S = 0 But in the case of odd e- s no state with S = 0 since Ms = ½ In such cases even after ZFS the spin states with opposite Ms values remain degenerate which is called Kramer’s degeneracy
The levels are called Kramer’s doublets “ IN ANY SYSTEM WITH ODD NUMBER OF UNPAIRED e-s THE ZFS LEAVES THE GROUND STATE AT LEAST TWO FOLD DEGENERATE ”
EFFECT OF ZFS ON Mn(II) +5/2 ±5/2 +3/26 S +1/2 ±3/2 ±1/2 - 1/2 FREE ZFS AND RESULTING - 3/2 ION KRAMER’S DOUBLETS - 5/2
CONSEQUENCES OF ZFS Insome cases ZFS magnitude is very high than the splitting by external field. Then transitions require very high energy Some times only one or no transitions occur. Examples V3+ and Co2+
EFFECTIVE SPIN STATE - Co(II) Co(II) in cubic field has a ground term of 4F.Since it is a d8 system it have ±3/2 and ±1/2 levels. ZFS splits the levels by 200 cm-1 Since the energy gap is higher only the transition -1/2 to + 1/2 is seen. So it appears as if Co(II) has only one unpaired e- (Effective spin S’ = ½)
BREAK DOWN OF SELECTION RULE In some cases like V(III) the magnitude of ZFS very high. It exceeds the normal energy range of ESR transitions Normal transitions occur with ∆Ms = ±1 . But its energy exceeds the microwave region Then the transition from -1 to +1 levels with ∆Ms = ±2 occurs ,which is a forbidden one
+1 FORBIDDEN Ms= ±1 TRANSITION -1Ms = 0, ±1 NOT OCURRING Ms =0 0
MIXING OF STATES The magnitude of ZFS can be taken as originating from CFS. But orbitally singlet state 6S is not split by the crystal field even then Mn(II) shows a small amount of ZFS. This is attributed to the mixing of g.s and e.s because of L-S coupling
The spin – spin interaction is negligible. But for triplet states spin – spin terms are important and they are solely responsible for ZFS Naphthalene trapped in durene in diluted state shows two lines as if it has ZFS. Since there is no crystal field or L-S coupling this is attributed to spin – spin interaction of the πe- s in the excited triplet state
ESR AND JAHN-TELLER DISTORTION Jahn – Teller theorem : Any non-linear electronically degenerate system is unstable, hence it will undergo distortion to reduce the symmetry, remove the degeneracy and hence increase its stability. But this theorem does not predict the type of distortion Because of J-T distortion the electronic levels are split and hence the number of ESR lines may increase or decrease.
FACTORS AFFECTING THE g-VALUES Operating frequency of the instrument Concentration of unpaired e- Ground term of the metal ion present Direction and temperature of measurement Lack of symmetry Inherent magnetic field in the crystals Jahn – Teller distortion ZFS
SUSTAINING EFFECT The g value for a gaseous atom or ion for which L-S coupling is applicable is given by g = 1 +[J(J+1) + S(S +1) – L(L+1)] / 2J(J+1) For halogen atoms the g values calculated and experimental are equal. But for metal ions it varies from 0.2 -8
The reason is the orbital motion of the e- are strongly perturbed by the crystal field. Hence the L value is partially or completely quenched In addition to this ZFS and J-T distortion may also remove the degeneracy
The spin angular momentum S of e- tries to couple with the L This partially retains the orbital degeneracy The crystal field tries to quench the L value and S tries to restore it This phenomenon is called sustaining effect
Depending upon which effect dominate the L value deviates from the original value So L and hence J is not a good quantum number to denote the energy of e- hence the g value also
COMBINED EFFECT OF CFS AND L-S COUPLING Three cases arise depending upon the relative magnitudes of strength of crystal field and L-S coupling L-S coupling >>CFS CFS > L-S coupling CFS >> L-S coupling
L-S COUPLING >>CFS When L is not affected much by CFS, then J is useful in determining the g value Example rare earth ions 4f e- buried inside so not affected, g falls in expected region All 4f and 5f give agreeing results other than Sm(III) and Eu(III)
CFS > >L-S COUPLING IfCFS is large enough to break L-S coupling then J is not useful in determining g. Now the transitions are explained by the selection rule and not by g value The magnetic moment is given by μs = [n(n+2)] 1/2
All 3d ions fall in this category. Systems with ground terms not affected by CFS ie L=0 are not affected and the g value is close to 2.0036 There may be small deviations because of L-S coupling, spin – spin interaction and gs and es mixing
CFS >> L-S COUPLING In strong fields L-S coupling is completely broken and L= 0 which means there is covalent bonding. Applicable to 3d strong field , 4d and 5d series. In many cases MOT gives fair details than CFT.
Example1: Ni (II) in an Oh field For Ni(II) g calculation includes mixing of 3A2g(g.s) and 3T2g(e.s) g = 2 – [8λ/10Dq] For Ni (II) the g value is 2.25 hence 8λ/10 Dq must be - 0.25 From the electronic spectrum 10Dq for Ni(II) in an Oh field is known to be 8500 cm-1,λ is -270 cm-1
For free Ni(II) ion the λ is about -324 cm-1 the decrease is attributed to the e.s ,g.s mixing This example shows how λ and 10Dq can affect the g value
Example2: Cu (II) in a tetragonal field Cu (II) a d9 system. Ground term 2D 2 D Eg + 2T2g ( CFS) 2 Since Cu (II) is a d9 system it must undergo J-T distortion. So the Oh field becomes tetragonal.
T2g 2 Eg + 2B2g (J-T distortion) 2 Eg 2 B1g + 2A1g 2 The unpaired e- is present in 2A1g on applying the magnetic field the spin levels are split and we get an ESR line.
Cu (II) in various fields (E3) 2 Eg 2 T2g 2 B2g (E2) 2 D (E1) 2 B1g 2 Eg + 1/2 (E0) 2 A1g ESR - 1/2Free ion Oh field Tetragonal field H
The g value is given by g|| = 2 – 8 λ / (E2 – E0) g┴ = 2 – 2 λ / (E3 – E0) From electronic spectrum (E2 – E0) and (E3 – E0) can be calculated. From the above values λ can be calculated.
It is seen that when splitting by distortion is high g value approaches 2 If the distortion splitting is lower then resulting levels may mix with each other to give deviated g values.
d1 system ( Ti3+, VO2+)The energy gap is very less. The 2B2g may be furthervibrations mix these levels so T1 lowered by L-S couplingis very low-leading to broad lines which is not shown. 2 Eg 2 D 2 Eg 2 T2g ∆E + 1/2 B2g 2 ESR - 1/2 Free ion Oh field Tetragonal field H
d2 systems ( V3+ ,Cr4+) 3 A2g 3 A2g3 F 3 Eg +1 3 T1g ±1 3 A2g 0 0 -1Free ion Oh field J-T Distortion ZFS H
d3 systems ( Cr3+) 4 T1 4 T1 +3/2 ± 3/24 F +1/2 4 A2 4 B2 ± 1/2 - 1/2 + 3/2Free ion Oh field J-T Distortion ZFS H