APPLICATIONS OF ESR SPECTROSCOPY TO METAL COMPLEXES

1. APPLICATIONS OF APPLICATIONS OF
ESR ESR
TO TO
METAL COMPLEXESMETAL COMPLEXES
V.SANTHANAM
DEPARTMENT OF CHEMISTRY
SCSVMV

2. METAL COMPLEXES – A SURVEYMETAL COMPLEXES – A SURVEY
Metal complexes are important- Diverse biological Metal complexes are important- Diverse biological
rolesroles
Griffiths and Owen proved the M-L covalency by Griffiths and Owen proved the M-L covalency by
taking complexes (NHtaking complexes (NH
44))
22[IrCl[IrCl
66] and Na] and Na
22[IrCl[IrCl
66]]
The hyperfine splitting by Chloride ligands showed The hyperfine splitting by Chloride ligands showed
the covalent nature of M-L bondthe covalent nature of M-L bond

3. Proved the back donation (pi-bonding) conceptProved the back donation (pi-bonding) concept
With the ESR data they were able to calculate With the ESR data they were able to calculate ξξ,,ζζ
and and λλ of metal ions and the extent of delocalization of metal ions and the extent of delocalization
In metal complexes the above said parameters In metal complexes the above said parameters
were having lower values than the free metal ions.were having lower values than the free metal ions.

4. THINGS TO BE CONSIDEREDTHINGS TO BE CONSIDERED
Nature of the metalNature of the metal
Number of ligandsNumber of ligands
GeometryGeometry
No of d electronsNo of d electrons
 Ground term of the ionGround term of the ion

5. Electronic degeneracyElectronic degeneracy
Inherent magnetic fieldInherent magnetic field
Nature of sampleNature of sample
Energy gap between g.s and e.sEnergy gap between g.s and e.s
Experimental temperatureExperimental temperature

6. NATURE OF THE METAL IONNATURE OF THE METAL ION
Since d metal ions have 5 d orbitals situations are Since d metal ions have 5 d orbitals situations are
complicatedcomplicated
But the spectra are informativeBut the spectra are informative
In 4d and 5d series In 4d and 5d series L-S / j-jL-S / j-j coupling is strong coupling is strong
making the ESR hard to interpretmaking the ESR hard to interpret

7. Crystal field is not affecting the 4f and 5f e- so the Crystal field is not affecting the 4f and 5f e- so the
ESR spectra of lanthanides and actinides are ESR spectra of lanthanides and actinides are
quite simple.quite simple.
If ion contains more than one unpaired e- ZFS may If ion contains more than one unpaired e- ZFS may
be operativebe operative

8. GEOMETRY OF THE COMPLEXGEOMETRY OF THE COMPLEX
Ligands and their arrangement –CFSLigands and their arrangement –CFS
CFS in turn affect the electronic levels hence the CFS in turn affect the electronic levels hence the
ESR transitionsESR transitions
The relative magnitude of CFS and L-S coupling is The relative magnitude of CFS and L-S coupling is
giving three situations.giving three situations.

9. If the complex ion is having cubic symmetry If the complex ion is having cubic symmetry
(octahedral or cubic) – g is isotropic(octahedral or cubic) – g is isotropic
Complexes with at least one axis of symmetry Complexes with at least one axis of symmetry
show two g valuesshow two g values
Ions with no symmetry element will show three Ions with no symmetry element will show three
values for gvalues for g..

10. SYSTEM WITH AN AXIS OF SYMMETRY NO SYMMETRYSYSTEM WITH AN AXIS OF SYMMETRY NO SYMMETRY

11. Symmetry of the complex ion- important – why?Symmetry of the complex ion- important – why?
ESR is recorded in frozen solutionsESR is recorded in frozen solutions
Spins are lockedSpins are locked
Lack of symmetry influences the applied field Lack of symmetry influences the applied field
considerablyconsiderably..

12. Spin Hamiltonian of an unpaired e- if it is present in a Spin Hamiltonian of an unpaired e- if it is present in a
cubic field iscubic field is
H = g H = g ββ | H | H
xx.S.S
xx + H + H
yy.S.S
yy + H + H
zz.S.S
zz||
If the system lacks a spherical symmetry and possess at If the system lacks a spherical symmetry and possess at
least one axis ( Distorted Oh,SP or symmetric tops) thenleast one axis ( Distorted Oh,SP or symmetric tops) then
H = H = ββ |g |g
xxxx H H
xx.S.S
xx +g +g
yyyy H H
yy.S.S
yy + g + g
zzzz H H
zz.S.S
zz||
Usually symmetry axis coincides with the Z axis and H is Usually symmetry axis coincides with the Z axis and H is
applied along Z axis thenapplied along Z axis then
gg
xx xx = g= g
yyyy = g = g
LL ; g ; g
zzzz = g = g
||||

13. If crystal axis is not coinciding with Z axisIf crystal axis is not coinciding with Z axis
The sample is rotated about three mutually The sample is rotated about three mutually
perpendicular axis and g is measured.perpendicular axis and g is measured.
g is got by one of the following relationsg is got by one of the following relations
for rotation aboutfor rotation about
X axis - g
2
= g
yy
2
Cos
2
θ + 2g
yz
2
Cos
2
θ Sin
2
θ +g
zz
2
Sin
2
θ
Y axis - g
2
= g
zz
2
Cos
2
θ + 2g
zx
2
Cos
2
θ Sin
2
θ +g
xx
2
Sin
2
θ
Z axis - g
2
= g
yy
2
Cos
2
θ + 2g
xy
2
Cos
2
θ Sin
2
θ +g
yy
2
Sin
2
θ

14. NUMBER OF d ELECTRONSNUMBER OF d ELECTRONS
Magnetically active nucleus cause hyperfine Magnetically active nucleus cause hyperfine
splitting.splitting.
If more than one unpaired e- present in the ion, If more than one unpaired e- present in the ion,
more no of transitions possible leads to fine more no of transitions possible leads to fine
structure in ESR spectrum.structure in ESR spectrum.
Here we have to consider two things Here we have to consider two things
Zero field splitting – due to dipolar interaction Zero field splitting – due to dipolar interaction
Kramer’s degeneracyKramer’s degeneracy

15. ZERO FIELD SPLITTINGZERO FIELD SPLITTING
Considering a system Considering a system
with two unpaired e-swith two unpaired e-s
Three combinations Three combinations
possible possible
In absence of external In absence of external
field all three states are field all three states are
having equal energyhaving equal energy
With external field three With external field three
levels are no longer levels are no longer
with same energy.with same energy.
Two transitions Two transitions
possible; both with possible; both with
same energy same energy
∆E
2
∆E
1
S = +1
S = 0
S = -1
H ≠ 0
ZFS = 0

16. SPLITTING OF ELECTRONIC LEVELS EVEN IN SPLITTING OF ELECTRONIC LEVELS EVEN IN
ABSENCE OF EXTERNAL MAGNETIC FIELD IS ABSENCE OF EXTERNAL MAGNETIC FIELD IS
CALLED ZERO FIELD SPLITTING (ZFS)CALLED ZERO FIELD SPLITTING (ZFS)
The splitting may be assisted by distortion and L-S The splitting may be assisted by distortion and L-S
coupling also.coupling also.

17. When there is a strong dipolar interaction the +1 level When there is a strong dipolar interaction the +1 level
is raised in energy –Dipolar shift (D)is raised in energy –Dipolar shift (D)
This dipolar shift reduces the gap between S = -1 and This dipolar shift reduces the gap between S = -1 and
S = 0 stateS = 0 state
Now the two transitions do not have same energyNow the two transitions do not have same energy
Results in two linesResults in two lines

18. EFFECT OF DIPOLAR SHIFTEFFECT OF DIPOLAR SHIFT
Ms = ±1,0
Ms = ±1
Ms = 0
Ms = -1
Ms = 0
Ms = +1
∆E
1
= ∆E
2
ZFS = 0
D
D

19. KRAMER’S THEOREMKRAMER’S THEOREM
Systems with even no. of unpaired e-s will contain Systems with even no. of unpaired e-s will contain
a state with S = 0a state with S = 0
But in the case of odd e- s no state with S = 0 But in the case of odd e- s no state with S = 0
since Ms = ½since Ms = ½
In such cases even after ZFS the spin states with In such cases even after ZFS the spin states with
opposite Ms values remain degenerate which is opposite Ms values remain degenerate which is
called Kramer’s degeneracycalled Kramer’s degeneracy

20. The levels are called Kramer’s doubletsThe levels are called Kramer’s doublets
“ “ IN ANY SYSTEM WITH ODD NUMBER OF IN ANY SYSTEM WITH ODD NUMBER OF
UNPAIRED eUNPAIRED e
-s-s
THE ZFS LEAVES THE GROUND THE ZFS LEAVES THE GROUND
STATE AT LEAST TWO FOLD DEGENERATE STATE AT LEAST TWO FOLD DEGENERATE ””

21. EFFECT OF ZFS ON Mn(II)EFFECT OF ZFS ON Mn(II)
+5/2
+3/2
+1/2
- 1/2
- 3/2
- 5/2
±5/2
±3/2
±1/2
6
S
FREE
ION
ZFS AND RESULTING
KRAMER’S DOUBLETS

22. CONSEQUENCES OF ZFSCONSEQUENCES OF ZFS
In some cases ZFS magnitude is very high In some cases ZFS magnitude is very high
than the splitting by external field.than the splitting by external field.
Then transitions require very high energyThen transitions require very high energy
Some times only one or no transitions occur.Some times only one or no transitions occur.
Examples VExamples V
3+3+
and Co and Co
2+2+

23. EFFECTIVE SPIN STATE - Co(II)EFFECTIVE SPIN STATE - Co(II)
Co(II) in cubic field has a ground term of Co(II) in cubic field has a ground term of
44
F.F.Since it is a Since it is a
dd
88
system it have system it have ±3/2 and ±1/2 levels.±3/2 and ±1/2 levels.
ZFS splits the levels by 200 cmZFS splits the levels by 200 cm
-1-1

Since the energy gap is higher only the transition -1/2 Since the energy gap is higher only the transition -1/2
to + 1/2 is seen.to + 1/2 is seen.
So it appears as if Co(II) has only one unpaired e- So it appears as if Co(II) has only one unpaired e-
(Effective spin S’ = ½)(Effective spin S’ = ½)

24. ±3/2,
±1/2
±3/2
±1/2
+3/2
- 3/2
+1/2
-1/2
ONLY
OBS.TRANSITION
≈ 200 cm
-1

25. BREAK DOWN OF BREAK DOWN OF
SELECTION RULESELECTION RULE
In some cases like V(III) the magnitude of ZFS very In some cases like V(III) the magnitude of ZFS very
high.high.
It exceeds the normal energy range of ESR transitionsIt exceeds the normal energy range of ESR transitions
Normal transitions occur with Normal transitions occur with ∆Ms = ±1 . But its energy ∆Ms = ±1 . But its energy
exceeds the microwave regionexceeds the microwave region
Then the transition from -1 to +1 levels with Then the transition from -1 to +1 levels with ∆Ms = ±2 ∆Ms = ±2
occurs ,which is a forbidden oneoccurs ,which is a forbidden one

26. Ms = 0, ±1
Ms= ±1
Ms =0
+1
-1
0
FORBIDDEN
TRANSITION
NOT OCURRING

27. MIXING OF STATESMIXING OF STATES
The magnitude of ZFS can be taken as originating The magnitude of ZFS can be taken as originating
from CFS.from CFS.
But orbitally singlet state But orbitally singlet state
66
S is not split by the S is not split by the
crystal field even then Mn(II) shows a small crystal field even then Mn(II) shows a small
amount of ZFS.amount of ZFS.
This is attributed to the mixing of g.s and e.s This is attributed to the mixing of g.s and e.s
because of L-S couplingbecause of L-S coupling

28. The spin – spin interaction is negligible.The spin – spin interaction is negligible.
But for triplet states spin – spin terms are But for triplet states spin – spin terms are
important and they are solely responsible for ZFSimportant and they are solely responsible for ZFS
Naphthalene trapped in durene in diluted state Naphthalene trapped in durene in diluted state
shows two lines as if it has ZFS.shows two lines as if it has ZFS.
Since there is no crystal field or L-S coupling this Since there is no crystal field or L-S coupling this
is attributed to spin – spin interaction of the is attributed to spin – spin interaction of the ππee
- -
s s
in the excited triplet statein the excited triplet state

29. ESR AND JAHN-TELLER DISTORTIONESR AND JAHN-TELLER DISTORTION
Jahn – Teller theorem :Jahn – Teller theorem :
Any non-linear electronically Any non-linear electronically
degenerate system is unstable, hence it will degenerate system is unstable, hence it will
undergo distortion to reduce the symmetry, undergo distortion to reduce the symmetry,
remove the degeneracy and hence increase its remove the degeneracy and hence increase its
stability.stability.
But this theorem does not predict the type of But this theorem does not predict the type of
distortiondistortion
Because of J-T distortion the electronic levels are Because of J-T distortion the electronic levels are
split and hence the number of ESR lines may split and hence the number of ESR lines may
increase or decrease.increase or decrease.

30. FACTORS AFFECTING FACTORS AFFECTING
THE g-VALUESTHE g-VALUES
Operating frequency of the instrumentOperating frequency of the instrument
Concentration of unpaired e-Concentration of unpaired e-
Ground term of the metal ion presentGround term of the metal ion present
Direction and temperature of measurementDirection and temperature of measurement
Lack of symmetryLack of symmetry
Inherent magnetic field in the crystalsInherent magnetic field in the crystals
Jahn – Teller distortionJahn – Teller distortion
ZFSZFS

31. SUSTAINING EFFECTSUSTAINING EFFECT
The g value for a gaseous atom or ion for which L-S The g value for a gaseous atom or ion for which L-S
coupling is applicable is given bycoupling is applicable is given by
g = 1 +[J(J+1) + S(S +1) – L(L+1)] / 2J(J+1)g = 1 +[J(J+1) + S(S +1) – L(L+1)] / 2J(J+1)
For halogen atoms the g values calculated and For halogen atoms the g values calculated and
experimental are equal.experimental are equal.
But for metal ions it varies from 0.2 -8But for metal ions it varies from 0.2 -8

32. The reason is the orbital motion of the e- are The reason is the orbital motion of the e- are
strongly perturbed by the crystal field.strongly perturbed by the crystal field.
Hence the L value is partially or completely Hence the L value is partially or completely
quenchedquenched
In addition to this ZFS and J-T distortion may also In addition to this ZFS and J-T distortion may also
remove the degeneracyremove the degeneracy

33. The spin angular momentum The spin angular momentum SS of e- tries of e- tries
to couple with theto couple with the L L
This partially retains the orbital This partially retains the orbital
degeneracydegeneracy
The crystal field tries to quench the L The crystal field tries to quench the L
value and S tries to restore it value and S tries to restore it
This phenomenon is called sustaining This phenomenon is called sustaining
effecteffect

34. Depending upon which effect dominate the L value Depending upon which effect dominate the L value
deviates from the original value deviates from the original value
So L and hence J is not a good quantum number So L and hence J is not a good quantum number
to denote the energy of e- hence the g value alsoto denote the energy of e- hence the g value also

35. COMBINED EFFECT OF CFS AND L-S COUPLINGCOMBINED EFFECT OF CFS AND L-S COUPLING
Three cases arise depending upon the relative Three cases arise depending upon the relative
magnitudes of strength of crystal field and L-S magnitudes of strength of crystal field and L-S
couplingcoupling
L-S coupling >>CFSL-S coupling >>CFS
CFS > L-S couplingCFS > L-S coupling
CFS >> L-S couplingCFS >> L-S coupling

36. L-S COUPLING >>CFSL-S COUPLING >>CFS
When L is not affected much by CFS, then J is When L is not affected much by CFS, then J is
useful in determining the g valueuseful in determining the g value
 Example rare earth ionsExample rare earth ions
4f e- buried inside so not affected, g falls in 4f e- buried inside so not affected, g falls in
expected regionexpected region
All 4f and 5f give agreeing results other than All 4f and 5f give agreeing results other than
Sm(III) and Eu(III)Sm(III) and Eu(III)

37. CFS > >L-S COUPLINGCFS > >L-S COUPLING
If CFS is large enough to break L-S If CFS is large enough to break L-S
coupling then J is not useful in determining coupling then J is not useful in determining
g.g.
Now the transitions are explained by the Now the transitions are explained by the
selection rule and not by g valueselection rule and not by g value
The magnetic moment is given by The magnetic moment is given by
μμ
ss = [n(n+2)] = [n(n+2)]
1/21/2

38. All 3d ions fall in this category.All 3d ions fall in this category.
Systems with ground terms not affected by CFS ie Systems with ground terms not affected by CFS ie
L=0 are not affected and the g value is close to L=0 are not affected and the g value is close to
2.00362.0036
There may be small deviations because of L-S There may be small deviations because of L-S
coupling, spin – spin interaction and gs and es coupling, spin – spin interaction and gs and es
mixingmixing

39. CFS >> L-S COUPLINGCFS >> L-S COUPLING
In strong fields L-S coupling is completely broken and In strong fields L-S coupling is completely broken and
L= 0 which means there is covalent bonding.L= 0 which means there is covalent bonding.
Applicable to 3d strong field , 4d and 5d series.Applicable to 3d strong field , 4d and 5d series.

In many cases MOT gives fair details than CFTIn many cases MOT gives fair details than CFT..

40. Example1: Ni (II) in an OExample1: Ni (II) in an O
hh field field
For Ni(II) g calculation includes mixing of For Ni(II) g calculation includes mixing of
33
AA
2g2g(g.s) (g.s)
and and
33
TT
2g2g(e.s)(e.s)
g = 2 – [8g = 2 – [8λλ/10Dq]/10Dq]
For Ni (II) the g value is 2.25 hence For Ni (II) the g value is 2.25 hence 88λλ/10 Dq must /10 Dq must
be - 0.25be - 0.25
From the electronic spectrum 10Dq for Ni(II) in an From the electronic spectrum 10Dq for Ni(II) in an
OO
hh field is known to be 8500 cm field is known to be 8500 cm
-1-1
,,λλ is -270 cm is -270 cm
-1-1

41. For free Ni(II) ion the For free Ni(II) ion the λλ is about -324 cm is about -324 cm
-1-1
the the
decrease is attributed to the e.s ,g.s mixingdecrease is attributed to the e.s ,g.s mixing
 This example shows how This example shows how λλ and 10Dq can affect and 10Dq can affect
the g valuethe g value

42. Example2: Example2: Cu (IICu (II) in) in
a tetragonal field a tetragonal field
Cu (II) a dCu (II) a d
99
system. Ground term system. Ground term
22
DD

22
D D
22
EE
g g + +
22
TT
2g 2g ( CFS) ( CFS)
Since Cu (II) is a dSince Cu (II) is a d
99
system it must undergo J-T system it must undergo J-T
distortion.distortion.
So the OSo the O
hh field becomes tetragonal. field becomes tetragonal.

43. 
22
TT
2g 2g
22
EE
gg + +
22
BB
2g 2g (J-T distortion) (J-T distortion)

22
EE
gg
22
BB
1g1g + +
22
AA
1g1g
The unpaired e- is present in The unpaired e- is present in
22
AA
1g1g
 on applying the magnetic field the spin levels are on applying the magnetic field the spin levels are
split and we get an ESR line.split and we get an ESR line.

44. Cu (II) in various fieldsCu (II) in various fields
2
D
2
T
2g
2
E
g
2
E
g
2
B
2g
2
B
1g
2
A
1g
+ 1/2
- 1/2
ESR
Free ion O
h
field Tetragonal field H
(E
3
)
(E
2
)
(E
1
)
(E
0
)

45. The g value is given byThe g value is given by
gg
|||| = 2 – 8 = 2 – 8 λλ / (E / (E
22 – E – E
00))
gg
┴ ┴ = 2 – 2 = 2 – 2 λλ / (E / (E
33 – E – E
00))
 From electronic spectrum (EFrom electronic spectrum (E
22 – E – E
00) and (E) and (E
33 – E – E
00) )
can be calculated. can be calculated.
From the above values From the above values λλ can be calculated. can be calculated.

46. It is seen that when splitting by distortion is high g It is seen that when splitting by distortion is high g
value approaches 2value approaches 2
If the distortion splitting is lower then resulting If the distortion splitting is lower then resulting
levels may mix with each other to give deviated g levels may mix with each other to give deviated g
valuesvalues..

47. dd
11
system ( Ti system ( Ti
3+3+
, VO, VO
2+2+
))
2
D
2
T
2g
2
E
g
2
B
2g
+ 1/2
- 1/2
ESR
Free ion O
h
field Tetragonal field H

2
E
g
The
2
B
2g
may be further
lowered by L-S coupling
which is not shown.
The energy gap is very less.
vibrations mix these levels so T
1

is very low-leading to broad lines
∆E

48. dd
22
systems ( V systems ( V
3+3+
,Cr ,Cr
4+4+
))
3
F
3
A
2g
3
A
2g
3
T
1g
3
E
g
3
A
2g
± 1
0
0
+ 1
- 1
Free ion O
h
field J-T Distortion ZFS H

49. dd
33
systems ( Cr systems ( Cr
3+3+
))
4
F
4
T
1
4
T
1
4
A
2
± 3/2
± 1/2
+3/2
+1/2
- 1/2
+ 3/2
4
B
2
Free ion Oh field J-T Distortion ZFS H

50. dd
44
- system (weak field)- system (weak field)
5
D
(25)
5
T
2g
(15)
5
E
g
(10)
5
E
g
(10)
5
B
2g
(5)
5
A
2g
(5)
5
B
1g

(5)
± 2 (2)
± 1 (2)
0 (1)
+2
+1
0
-1
-2