APPLICATIONS OF ESR SPECTROSCOPY TO METAL COMPLEXES
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APPLICATIONS OF ESR SPECTROSCOPY TO METAL COMPLEXES

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APPLICATIONS OF ESR SPECTROSCOPY TO METAL COMPLEXES APPLICATIONS OF ESR SPECTROSCOPY TO METAL COMPLEXES Presentation Transcript

  • 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.
  • SYSTEM WITH AN AXIS OF SYMMETRY NO SYMMETRY
  •  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)
  • +3/2 ±3/2 - 3/2±3/2,±1/2 ≈ 200 cm-1 +1/2 ±1/2 ONLY OBS.TRANSITION -1/2
  • 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
  • d4- system (weak field) 5 Eg (10) T2g 5 5 B2g +2 (15) (5) 5 A2g +1D5 (5)(25) 5 Eg ± 2 (2) ± 1 (2) (10) 5 B1g 0 (1) (5) 0 -1 -2