Electronic Spectra of Electronic Spectra of
Metal ComplexesMetal Complexes
V. Santhanam
Department of Chemistry
SCSVMV
Color of the complexes
Observed
Color
Light
Absorbed
Abs Wavelength /
nm
Colorless UV <380
Yellow Violet 380-450
Orange Blue 450-490
Red Green 490-550
Violet Yellow 550-580
Blue Orange 580-650
Green Red 650-700
Colorless IR >700
Use of Electronic spectrum
•Electronicspectrum–duetod-d
transitions.
•Fromthetransitionenergy–energyofd•Fromthetransitionenergy–energyofd
e
-
canbegot.
•Theenergiesofdlevelsaffectedbymany
complicatedfactors.
Terms, States and Microstates
•Energy levels described by quantum numbers.
–Principal quantum number –n
–Azimuthalquantum number –l
–Magnetic quantum number –m–Magnetic quantum number –m
l
–Spin quantum number –m
s
•Butactual conditions are complicated .
•Quantum numbers alone are not sufficient
Why ?
Configuration
•Arrangement of e
-
s in an atom or ion
–Theelectronsarefilledintheincreasing
orderofenergy–Aufbauprinciple
–Ifmorethanoneorbitalishavingsame
energy,pairingofe-willnottakeplaceenergy,pairingofe-willnottakeplace
untilalltheorbitalsareatleastsingly
occupied–Hund’srule
–Notwoelectronscanhaveallthefour
quantumnumberssame–Pauli’sexclusion
principle
Jorgensen Relation
•We know that
•C.K.Jogensenderived a relationship for βas
(1-β) = h.k
B
B
where h-is ligandparameter
k-is metal ion parameter
β* = β–βh.k
•Thetotalorbital
angularmomentum
Lofagroupof
electronsinanatom
isgivenbyavector
sumoftheindividualsumoftheindividual
orbital angular
momental.
Spin-orbit coupling / L-S coupling
i) Russel-Saunders for
light atoms:
J = L + S, L + S –1, …, |L –S|
L = l
1+ l
2, l
1+ l
2–1, …, |l
1-l
2|
S = s
1+ s
2, s
1+ s
2–1, …, |s
1–s
2|
-Couple all individual orbital angular
momental to give a resultant total
orbital angular momentum L. (L = Ʃ l
i)
-Couple all individual spin angular
momentas to give a resultant total
spin angular momentum S. (S = Ʃ s)spin angular momentum S. (S = Ʃ s
i)
-Finally couple L and S to give the
total angular momentum J for the
entire atom.
Russel-Saunders
couplingworks
wellforthelight
elementsupto
bromine.
Spin-orbit coupling / L-S coupling
•The extent to which L and S are coupling is
called “Spin-Orbit Coupling constant”
•The perturbation created by L-S coupling is
sl.
Thisvalueisfor
singleelectronofa sl.
singleelectronofa
configurationand
alwayspositive
Spin-orbit coupling / L-S coupling
•For convenience, a parameter characteristic
of a term is used.
•L-S coupling constant for a term is
•The perturbation of energy of a term is
sl.•The perturbation of energy of a term issl.
s.2
•The valueispositiveforlessthanhalf
filledandnegativeiftheorbitalisgreater
thanhalffilled.
L-S Coupling continued
•Each term is split in to states by L-S coupling
specified by J values, which differing by unity.
•Each L value will have (2L+1) M
Lcomponents.
•Each S value will have (2S+1) Mcomponents.•Each S value will have (2S+1) M
Scomponents.
•So any term will have a degeneracy of
(2S+1).(2L+1)
L-S coupling in a p
2
system
Let us consider p
2
configuration
Maximum Lvalue possible is 2
Maximum Svalue possible is 0
So possible Jvalue is 2
Degeneracy of this state is
((2 x 2)+1) x ((2 x 0)+1) = 5
1
D
(9)
J=2
(9)
((2 x 2)+1) x ((2 x 0)+1) = 5
Next maximum Lvalue possible is 1
For this L value maximum Svalue
possible is 1
Possible Jvalues are 2, 1, 0
Degeneracy of this state is
((2 x 1)+1) x ((2 x 1)+1) = 9
3
P
(9)
J=2
(5)
J=1
(3)
J=0
(1)
2λ
λ
L-S coupling
•Each L and S combination is having a particular energy.
•Total angular momentum quantum number J
J = |L+S|, |L+S-1|, |L+S-2|,……|L-S|
•J values are of unit difference.
)1()1()1(
2
SSLLJJE
J
•J values are of unit difference.
•The energy gap between any two successive J values is
•Forlessthanhalffilled,lowestvalueofJ,|L-S|hasthe
lowestenergy.(Normalmultiplet)
•FormorethanhalffilledhighestvalueofJ|L+S|hasthe
lowestenergy(Invertedmultiplet)
)1(
1,
JE
JJ
Normal and Inverted Multiplets
Normal Multiplet
Inverted Multiplet
Normal Multiplet
Inverted Multiplet
Tackling the Multi-ElectronProblem
Start by taking a stepback
•first we must treat the V
3+
‘free ion’ with no ligands in a spherically
symmetricenvironment
•since V
3+
is d
2
, we must consider e
–
-e
–
interactions, spin-spin
interactions, and spin-orbitalinteractions
Define a fewterms
•microstate –a specific valence electron configuration for amulti-•microstate –a specific valence electron configuration for amulti-
electron freeion
•atomic state –a collection of microstates with the sameenergy
•term symbol –the label for a free ion atomicstate
2S+1
L
L term symbollabel
0 S
1 P
2 D
3 F
4 G
“spin
multiplicity”
i! 10!
j!ij!2!10 2!
45# microstates
#e
–
#spaces
Multi-Electron QuantumNumbers
When we learn about atoms and electron configurations, we learn
the one-electron quantum numbers (n, l, m
l,m
s)
Multi-electron ions need multi-electron quantumnumbers
•L gives the total orbital angular momentum of an atomic state
(equal to the maximum value ofM
L)
•M
L is the z-component of the orbital angular momentum of amicrostate
•S gives the total spin angular momentum of an atomic state
(equal to the maximum value of M
S (for a given Lvalue)
•M
S is the z-component of the spin angular momentum of amicrostate
Spin-OrbitCoupling
So far we’ve considered only e
–
-e
–
(orbital angular momentum) and
spin-spin (spin angular momentum) interactions.
Orbitals and spins can also interact giving rise to spin-orbit
coupling
•J is the total angularmomentum
So for the
3
F ground state of a d
2
free ion, wehaveSo for the
3
F ground state of a d
2
free ion, wehave
To determine the lowest energy spin-orbit coupledstate:
1.For less than half-filled shells, the lowest J is the lowestenergy
2.For more than half-filled shells, the highest J is the lowestenergy
3.For half-filled shells, only one J ispossible
J L S 4
L S 2
so J 4, 3,2
So the
3
F ground state
electron configuration of
a d
2
free ion splits into
3
F4,
3
F3, and
3
F2.
Number of microstates for a system
)!2(!
!2
xyx
y
N
Where
N= number of microstates possible
y –number of orbitals
X-number of electrons
Number of microstates –p
1
system
+1 0 -1 y –number of orbitals -6
x -number of electrons –1
))!16.(2!.(1
)!6.(2
N
))!16.(2!.(1
N
))5.4.3.2.1(2.(1
)6.5.4.3.2.1.(2
N
N = 6
Number of microstates –p
2
system
y –number of orbitals -6
x -number of electrons –2
)!6.(2
N
))!26.(2!.(2
N
))4.3.2.1(2).(2.1(
)6.5.4.3.2.1.(2
N
N = 15
Term Symbols for p
2
•Maximum L value possible is 2
–Possible M
Lvalues are -L, -(L+1), …, 0 , …, +(L-1), +L
–i.e. -2 , -1 , 0 , +1 , +2
•Maximum S value possible is : 0
–Possible M
svalue is 0
•Considering all possible combinations
M
L +2 +1 0 -1 -2
Ms 0 0 0 0 0
Term
1
D
Term Symbols for p
2
•Next maximum L value possible is 1
–Possible M
Lvalues are -L, -(L+1), …, 0 , …, +(L-1), +L
–i.e. -1 , 0 , +1
•Maximum S value possible is : 1
–Possible M
svalue are +1 , 0 , -1
•Considering all possible combinations
M
S +1 0 -1
M
L+10 -1+10 -1+10 -1
Term
3
P
Term Symbols for p
2
•Next L value possible is 0
–Possible M
Lvalue is 0
•The only S value possible is : 0
–Possible M
svalue is 0
The term is
1
S
M
S 0
M
L 0
Term
1
S
Term Symbols for d
2
l
1= 2; l
2= 2
L = |l
1+l
2|, |l
1+l
2-1|, |l
1+l
2-2|…,0,…, |l
1-l
2|
So the possible values for L are
L = 4, 3, 2, 1, 0L = 4, 3, 2, 1, 0
s
1= ½ ; s
2= ½
So the possible values for S are
S = 1, 0
Microstates of d
2
Maximum L value is 4
So the possible M
Lvalues are +4, +3, +2, +1, 0, -1, -2, -3, -4
Maximum S value possible is 0
So Mcan be only 0
M
L+4+3+2+10 -1-2-3-4
M
S0 0 0 0 0 0 0 0 0
Term
1
G (9)
So M
scan be only 0
Microstates of d
2
Next maximum L value is 3
So the possible M
Lvalues are +3, +2, +1, 0, -1, -2, -3
Maximum S value possible is 1
So Mcan be +1, 0, -1So M
scan be +1, 0, -1
M
L +3 +2 +1 0 -1 -2 -3
M
S
+10-1+10-1+10-1+10-1+10-1+10-1+10-1
Term
3
F (21)
Microstates of d
2
Next maximum L value is 2
So the possible M
Lvalues are +2, +1, 0, -1, -2
Maximum S value possible is 1
So Mcan be +1, 0, -1So M
scan be +1, 0, -1
M
L +2 +1 0 -1 -2
M
S 0 0 0 0 0
Term
1
D (5)
Microstates of d
2
•Next maximum L value possible is 1
–Possible M
Lvalues are +1 , 0 , -1
•Maximum S value possible is : 1
–Possible M
svalue are +1 , 0 , -1
•Considering all possible combinations
M
S +1 0 -1
M
L+10 -1+10 -1+10 -1
Term 3
P (9)
Microstates of p
2
•Next L value possible is 0
–Possible M
Lvalue is 0
•The only S value possible is : 0
–Possible M
svalue is 0
The term is
1
S
M
S 0
M
L 0
Term
1
S (1)
Inter-electronic repulsions
The Crystal Field Splitting of
Russell-Saunders termsin weak o
hcrystal fields
Russell-Saunders Terms Crystal Field Components
S(1) A
1g
P(3) T
1g
D(5) E
g+T
2gD(5) E
g+T
2g
F(7) A
2g+ T
1g+ T
2g
G(9) A
1g+E
g+ T
1g+T
2g
H(11) E
g+ T
1g+ T
1g+ T
2g
I(13) A
1g+ A
2g+ E
g+ T
1g+T
2g+T
2g
MullikenSymbols
MullikenSymbol Explanation
A Non-degenerate orbital; symmetric to principal C
n
B Non-degenerate orbital; unsymmetricto principal C
n
E Doubly degenerate orbital
T Triply degenerate orbital
(subscript) g Symmetric with respect to center of inversion(subscript) g Symmetric with respect to center of inversion
(subscript) u Unsymmetricwith respect to center of inversion
(subscript) 1 Symmetric with respect to C
2perp. to principal C
n
(subscript) 2 Unsymmetricwith respect to C
2perp. to principal C
n
(superscript) 'Symmetric with respect to σ
h
(superscript) " Unsymmetricwith respect to σ
h
Transition Moment Integral
dM
es
ex
gs
ex
dM
es
ey
gs
ey
dM
esgs
dM
es
ez
gs
ez
•Probability of any transition is proportional to M
i
2
•Allowed transition M
i≠ 0
•Forbidden transition M
i= 0
Transition Moment Integral
gs es gs es
i o i o s s
M d d
soe
•Foranyallowedelectronictransition,theM
ivalueshouldbe
non-zero.
•soeitherfirstintegralorsecondoneshouldbezeroifM
iis
zero.
•Thedipolemomentoperatordealswiththedisplacementof
atomsandassociateddipolemomentchanges,sothespin
partcanbeisolated.
( ) ( )f x f x
( ) ( )
( ) ( )
f x f x
f x f x
f
1 operator
f
2 Product
x
5
x
3
x
5
x
13
Zero
x
5
x
3
x
2
x
10
Non-zero
x
2
x
3
x
5
x
10
Non-zero
x
2
x
3
x
2
x
7
Zero
productfunction dx
ψ
e
gs
µ ψ
e
es
Product Mi
Odd Odd Odd Odd Zero
Odd Odd Even Even Non-zero
Even Odd Odd Odd Non-zero
Even Odd Even Odd Zero
The Selection rules for electronictransitions
Charge-transfer band –Laporte and spin allowed –veryintense
[Ni(H
2O)
6]
2+
a
3
A
2g →
1
E
g Laporte and spin forbidden –veryweak
a, b, and c,Laporte
forbidden, spin
allowed, inter-
mediateintensity
b
3A
2g→
3
T
2g
c
mediateintensity
Why the‘forbidden’ transitions occur?
The orbital selection rule or the Laporteselection
rule may be relaxed by any one of the following
Departurefromcubicsymmetry
d-pmixing
Vibroniccoupling
Mixing of states
•Thestatesinacomplexareneverpure.
•Whentwostatesareenergeticallytooclose
theymaybemixedbythermalvibrations
itself.itself.
•Someofthesymmetrypropertiesof
neighboringstatesbecomemixed.
•Thismakesthetransitionspartiallyallowed
whicharenormallyforbidden.
Mixing of states: Comparison of [Ni(H
2O)
6]
2+ and[Ni(en)
3]
2+:
[Ni(H
2O)
6]
2+
[Ni(en)]
2+
3A
2g 2g→
3
T(F)
The spin-forbidden
3
A
2g →
1
E
g is close to thespin-allowed
3
A
2g →
3
T
2g(F) and ‘borrows’ intensity by mixing ofstates
The spin-forbidden
3
A
2g →
1
E
g is not close
to any spin allowed band and is veryweak
3A
2g→
1
E
g
Note:Thetwospectraare
drawnonthesamegraph
foreaseofcomparison.
[Ni(en)
3]
2+
3A
2g→
3
T
2g
VibronicCoupling
•The vibrational
statesmaybeof
oppositeparityto
the electronic
g
u
g
the electronic
states,andsohelp
overcome the
Laporteselection
rule.
g
g
Symmetry of vibrational states, and their
coupling to electronicstates:
observed
spectrum
E-υ
1
E-υ
2
E-υ
3
1
E + υ ’
E
E +υ
2’
E +υ
3’
T
1u
symmetry
vibration
A
1g
symmetry
vibration
(symbols have same meaningfor
vibrations: A =non-degenerate,
T = triply degenerate, g =gerade,
u = ungerade,etc.)
The band one sees inthe
UV-visible spectrum isthe
sum of bands due to transitions
to coupled electronic (E)and
vibrational energy levels (υ
1, υ
2,υ
3)
T
d
vsO
h
•Atetrahedronhasnocenterofsymmetry,
andsoorbitalsinsuchsymmetrycannot
begerade.Hencethed-levelsina
tetrahedralcomplexareeandt
2.
•ThislargelyovercomestheLaporte•ThislargelyovercomestheLaporte
selectionrules,sothat tetrahedral
complexestendtobeveryintenseincolor.
•DissolvingCoCl
2inwaterproducesa
palepinksolutionof[Co(H
2O)
6]
2+,butin
alcoholforms,whichisatetrahedral
[CoCl
4]
2-
havingveryintensebluecolor.
[Co(H
2O)
6]
2+
and [CoCl
4]
2-
[CoCl
4]
2-
The spectra at left
show the very intense
d-d bands in the blue
tetrahedral complex
[CoCl
4]
2-
, as compared
with the much weaker
[Co(H
2O)
6]
2+
with the much weaker
band in the pink
octahedral complex
[Co(H
2O)
6]
2+
.This
differencearises
becausetheTcom-d
plex has no centerof
symmetry, helping to
overcome the g→g
Laporte selectionrule.
Shape of Electronic bands
•Variation of 10Dq
•Lower symmetry components
•Vibrationalstructure & F.C Principle
•Jahn-Teller distortion
•Spin-Orbit coupling
•Ligands possess σ, σ*, π, π*, and nonbonding (n) molecularorbitals.
•Iftheligandmolecularorbitalsarefull,chargetransfermayoccur
fromtheligandmolecularorbitalstotheemptyorpartiallyfilledmetal
d-orbitals.
•LMCTtransitionsresultinintensebands.Forbiddend-dtransitions
mayalsotakeplacegivingrisetoweakabsorptions.
• Ligand to metal charge transfer results in the reduction of themetal.
MLCT
•Ifthemetalisinalowoxidationstate(electronrich)and
theligandpossesseslow-lyingemptyorbitals(e.g.,COor
CN−).
•LMCTtransitionsarecommonforcoordination
compounds having π-acceptorligands.
•Upontheabsorptionoflight,electronsinthemetal•Upontheabsorptionoflight,electronsinthemetal
orbitals are excited to the ligand π*orbitals.
•MLCT transitions result in intense bands. Forbidden d –d
transitions may alsooccur.
•Thistransitionresultsintheoxidationofthemetal.
Metal-to-Ligand Charge-
Transfer transition.
Effect of Solvent Polarity on CTSpectra
You are preparing a sample for a UV/Vis experiment and youdecide
to use a polar solvent. Is a shift in wavelength observedwhen:
Both the ground state and the excited state are neutral.
The excited state is polar, but the ground state isneutralThe excited state is polar, but the ground state isneutral
The ground state and excited state ispolar
The ground state is polar and the excited state isneutral
The excited state is polar, but the
ground state isneutral
The ground state and excited state
is polar
The ground state is polar and the
excited state isneutral
The MO view of electronic transitions inan
octahedralcomplex
t*
1u
a1g*
4p
t
2g→t
1u*
M→L Chargetransfer
Laporte and spin
allowed
t
2g→e
g
d→d transition
Laporteforbidden
Spin-allowed or
forbidden
eg*
t
2g
4s
eg
a
1g
t
1u
3d
t
1u→t
2g
L→M Chargetransfer
Laporte andspin
allowed
The e
g level inCFT
is an e
g* inMO
In CFT weconsider
only the e
g and t
2g
levels, which are a
portion of the over-
all MOdiagram
σ-donor orbitals
of sixligands