before we did:
p2
ML & MS
Microstate
Table
States (S, P, D)
Spin multiplicity
Terms
3P, 1D, 1S
Ground state term
3P
For metal complexes we need to consider
d1-d10
d2
3F, 3P, 1G, 1D, 1S
For 3 or more electrons, this is a long tedious process
But luckily this has been tabulated before…
Free Ion terms for various d configurations.
We seek to understand
The “Same as “ relationships
What will happen when we put these free ion into a complex, octahedral
or tetrahedral.
The spectroscopy involved.
Explanation of the mirror symmetry behavior of the free ion GS terms.
Easy to understand using the hole formalism. dn has same terms as d10-n
Recall that we are ignoring “core” electrons. We only work with partially filled
subshells. Why?
Because the core, complete set of atomic orbitals within a subshell is wholly
occupied. All the orbital and spin angular momentum components cancel each
other out and add to 0.
Hole formalism: Consider d3. That configuration includes many different
orbital occupancies. Here is one of those occupancies. It will participate
on one or more of the terms.
-2
-1
0
1
2
- 1/2 x
x
x
1/2
Note that Lz is -3
We claim d3 to yield the same terms as d7
Explanation of the mirror symmetry behavior of the free ion GS terms.
Aim: d3 has same terms as d10-3 = d7
d7 allows for many different orbital occupancies. Here is one of them.
-2
-1
- 1/2 x
x
1/2
x
x
0
1
2
x
x
x
Note that Lz is -3
We claim this provides the same Lz and Sz as one of the d3 orbital occupancy
schemes. Consider the above to be formed from a completed d subshell and thee
holes to “create” the missing electrons.
The holes look like this
-2
- 1/2
1/2
-1
0
1
2
hole hole hole
Selection rules
(determine intensities)
Laporte rule
g  g forbidden (that is, d-d forbidden)
but g  u allowed (that is, d-p allowed)
Spin rule
Transitions between states of different multiplicities forbidden
Transitions between states of same multiplicities allowed
These rules are relaxed by molecular vibrations, and spin-orbit coupling
Group theory analysis of term splitting in octahedral.
Notice the correspondence to the symmetry analysis of the atomic orbitals when
they are split by an octahedral field.
High Spin Ground States
Recall that the set of d atomic electrons are split into a t2g set and an eg set by
repulsion with the negative octahedral ligands, similarly for tetrahedral environment.
So also are the free ion terms split by their environment.
dn
Free ion GS
Oct. complex
Tet complex
d0
1S
t2g0eg0
e0t20
d1
2D
t2g1eg0
e1t20
d2
3F
t2g2eg0
e2t20
d3
4F
t2g3eg0
e2t21
d4
5D
t2g3eg1
e2t22
d5
6S
t2g3eg2
e2t23
d6
5D
t2g4eg2
e3t23
d7
4F
t2g5eg2
e4t23
d8
3F
t2g6eg2
e4t24
d9
2D
t2g6eg3
e4t25
d10
1S
t2g6eg4
e4t26
We seek to understand the mirror like GS terms. dn and d10-n have the same
GS terms for the free ion.. How to explain this via the ligand splitting that we
know about (t2g and eg).
High Spin Ground States
Hole Formalism: The d10 electronic configurations is a spherical cloud of d electrons.
We can move to fewer electrons, d9, d8, etc. by creating holes in the spherical shell. Ex:
d9 = d10+ “d-hole”
Electronic configuration d1 has the term 2D. Ligand splitting occurs for 2D.
for electrons
for holes embedded in spherical cloud
2
Eg
eg
2
t2g
T2g
2
d
2
d
D
D
2
t2g
2
T2g
eg
Eg
d10-n electron configuration is equivalent to spherical d10 + holen It is the holen which
determines the orbital and spin angular momentum and the terms. Think positive
electrons. The term can be obtained from either the actual electronic configuration or the
hole configuration.
.
High Spin Ground States
Again, d10-n electron configuration is equivalent to spherical d10 + holen It is the holen
which determines the orbital and spin angular momentum and the terms. The term can
be obtained from either the actual electronic configuration or the hole configuration.
But note that ligand splitting of holes will yield inverted splitting relative to
electrons.
Principle: Electrons are repelled by ligands, but holes attracted to ligands.
Ex: d3 has t2g3eg0 electronic configuration (electrons repel ligands).
Ex: high spin d7 is either
electronic configuration of t2g5 eg2 (ligands destabilize electrons, favoring the t2g
.
orbitals)
or
hole3 with t2g1 eg2 hole configuration (ligands stabilize positive holes, favoring the eg)
High Spin Ground States
dn
Free ion GS
Oct. complex
Tet complex
d0
1S
t2g0eg0
e0t20
d1
2D
t2g1eg0
d2
3F
t2g2eg0
e2t20
d3
4F
t2g3eg0
e2t21
d4
5D
t2g3eg1
e2t22
d5
6S
t2g3eg2
e2t23
d6
5D
t2g4eg2
e3t23
d7
4F
t2g5eg2
e4t23
d8
3F
t2g6eg2
e4t24
d9
2D
t2g6eg3
d10
1S
t2g6eg4
= t2g
= eg
e1t20
e4t25
e4t26
GS has t2g
negative
electron,
repelling
the ligands.
Symmetry
is T2g for d1
octahedral
complex
GS has eg
positive
d-hole,
attracted to
ligands.
Symmetry
designation
of d9
octahedral
complex is
Eg
High Spin Ground States, d1 and d9 tetrahedral complexes
We have obtained the GS sym for octahedral d1 and d9. Now consider tetrahedral.
Recall the reversed tetrahedral orbital splitting. For tetrahedral ligand field e electron
destabilized less by ligands, t2 more so. Opposite for holes.
dn
Free ion GS
Oct. complex
Tet complex
d0
1S
t2g0eg0
e0t20
d1
2D
t2g1eg0
e1t20
d2
3F
t2g2eg0
e2t20
d3
4F
t2g3eg0
e2t21
d4
5D
t2g3eg1
e2t22
d5
6S
t2g3eg2
e2t23
d6
5D
t2g4eg2
e3t23
d7
4F
t2g5eg2
e4t23
d8
3F
t2g6eg2
e4t24
d9
2D
t2g6eg3
e4t25
d10
1S
t2g6eg4
e4t26
For d1 the
electron is e
GS.
=e
Symmetry is
thus E for
the d1 tet
complex.
= t2
For d9 or
d-hole1 the
d-hole
symmetry is
t2. The GS
complex
symmetry is
T 2.
High Spin Ground States: d4 and d6 octahedral and tetrahedral
The d6 is regarded as a single electron (t2g) on top of the spherical d5 half shell (t2g3eg2) .
For tetrahedral we have one electron (e) on top of half shell. But the spherical shell has
spin
which
must spin
be allowed
for.
Using
the high
multiplicities
from the free ion this means 5T2g for octahedral d6
and, similarly, 5dEn for the tetrahedral
d6GS
.
Free ion
Oct. complex
Tet complex
The GS of
the
octahedral
d4 complex
will have
symmetry of
the eg hole,
Eg.
d0
1S
t2g0eg0
e0t20
d1
2D
t2g1eg0
e1t20
d2
3F
t2g2eg0
e2t20
d3
4F
t2g3eg0
e2t21
d4
5D
t2g3eg1
d5
6S
t2g3eg2
d6
5D
t2g4eg2
Spin is
taken from
the free ion
yielding 5Eg.
d7
4F
t2g5eg2
e4t23
d8
3F
t2g6eg2
e4t24
d9
2D
t2g6eg3
e4t25
d10
1S
t2g6eg4
e4t26
= eg
e2t22
= t2
e2t23
= t2g
e3t23
=e
Reversed
splitting in
tetrahedral
leads to t2 for
hole in d4
tetrahedral
complexes.
Taking spin
from the free
ion GS
yields 5T2.
Further hole discussion. A d4 can be considered as spherical d5 with a positive hole
embedded in it. In octahedral complex the d-hole is eg symmetry. In tetrahedral the dhole is t2 symmetry.
Orgel diagram for d1, d4, d6, d9 in both octahederal and tetrahedral complexes.
Here is what we have: d1, d4, d6, d9 all have a d shaped object – either a d electron or d-hole – and
thus are designated as a D term for the free ion high spin GS. Ligand splitting causes t2(g) electron
to be lower than e(g), reversed in tetrahedral, reversed also for d-holes.
Now look at Orgel diagram for these configurations. Look at right part of diagram first
Both of these represent an
d1  d6
electron on top of a
spherical shell.
AnToctahederal
field splits
2g or T2
Energy
d4  d9
D
d4 into T2g and Eg (GS)
E
A tetrahedral splits
d4 into T2 (GS) and E
d1 into T2 and E (GS)
Eg or
d9 into
T2E(GS) and E
d6 into T2 and E (GS)
0
d-hole reversal
d9 into T2g and Eg (GS)
A tetrahedral splits
Both of these represent
d1, da6d-hole
tetrahedral
D
(reversal) in a tetrahedral
field
(a
d4, d9 octahedral
second reversal).
or
An octahederal field splits
d1 into T2g (GS) and Eg
d6 into T2g (GS) and Eg
Eg
d1, d6 octahedral
d4, d9 tetrahedral
ligand field strength
T2g or T2
Tetrahedral
reversal.
D
High Spin Ground States: d2, d3, d6, and d7
We have taken care of the d0, d1, d4, d5, d6, d9, and d10 configurations. Now have to do
d2, d3, d6, and d7 configurations. It turns out that all we have to do is solve d2.
d2 and d7 both are
electrons on top of
a spherical shell
yielding a splitting
pattern: 1, 2, 3
electrons
holes
1
3
d2
d7 = d5 + d2
2
d3=d5-hole2
2
d3 and d8 are both two
d-holes in a spherical
shell, yielding reversed
splitting: 3, 2, 1
3
d8=d10-hole2
1
But it is not so easy. Here is our approach:
We know the symmetry of the GS of the free d2 ion. How? We can get the terms
for d2 using the methods applied earlier to p2, etc. They are 3F, 1D, 3P, 1G, 1S.
We identify the GS as 3F. How?
We saw earlier that F in octahedral environment splits to A2g + T1g + T2g; in tetrahedral
we would get A2 + T1 + T2. Our problem is the energy ordering. Which is GS?
Thus the 3F GS for d2 splits into 3A2g + 3T1g + 3T2g. The 4F GS for d3 splits into 4A2g +
4T
4
1g + T2g. Where did the spin multiplicities come from??
But how do we decide on what becomes the GS after the splitting due to the ligands?
We use a correlation diagram. It shows the affect of increasing the ligand field
strength from zero (free ion) to very high where energy ordering is determined solely
by the occupancy of the t2g and the eg orbitals.
d2
F
r
e
e
I
o
n
t
e
r
m
s
We have two electrons in the
eg orbitals. It can be shown
that these give rise to 1A1g, 1Eg,
and 3A2g which have same
energy in strong ligand field.
Connect the terms of the same
symmetry without crossing.
Similarly, splitting occurs for these
occupancies.
Configurations
We have included
basedthe
on 3splitting
T1g originating
of d from
Splitting of free ion
electrons.
the 3P. We
Dominant
will need
initvery
immediately.
strong fields.
Same
Real complexes
terms.
symmetry as lower energy 3T1g from the 3F.
Orgel diagram for d2, d3, d7, d8 ions
This curvature will
complicate interpretation of
spectra. Same symmetry;
crossing forbidden
First look at
Energy
And now
d2 and d7 in tetrahedral
(reversed due to tetrahedral field)
T1 or T1g
7 in octahedral (2 elecs
Td12Tor
T1gT
and
d1g
1 or
on a spherical cloud)
P
and
and
T1 or T1g
T13 or T1g 8
d and d in tetrahedral (double F
reversal: d-holes and
tetrahedral)
T2 or T2g
All states shown are of the
same spin. Transitions occur
between them but weakly.
d3 and d8 in octahedral
(reversed due to d-holes).
T2 or T2g
Note the reversed ordering of the
splitting coming from F (T1/T2/A2).
The lower T1(g) now aims up and
should cross the upper T1(g) but T or T
1
1g
does not due to interaction with
the upper T1(g). Now have strong
curvature to avoid crossing.
Note the weak interaction of
A
A2gT1, the curvature.
the
2 ortwo
d2, d7 tetrahedral
A2 or A2g
0
d2, d7 octahedral
d3, d8 octahedral
d3, d8 tetrahedral
Ligand field strength (Dq)
Move to Tanabe-Sugano diagrams.
d1 – d3 and d8 – d9 which have only high spin GS are easier. Here is d2.
Correlation diagram for d2.
Convert to Tanabe-Sugano.
Tanabe-Sugano
Electronic transitions and spectra
d2 Tanabe-Sugano diagram
V(H2O)63+, a d2 complex
Configurations having only high spin GS
d2
d9
d1
d3
Note the two
lines curving
away from
each other.
d8
Note the two
lines curving
away from
each other.
Configurations having either high or low spin GS
The limit between
high spin and low spin
Determining Do from spectra
d1
Exciting
electron from t2g
to eg
d9
Exciting d-hole
from eg to t2g
Exciting d-hole
from eg to t2g
One transition allowed of energy Do
Exciting
electron from t2g
to eg
Determining Do from spectra
mixing
Here the mixing is not a problem since
the “mixed” state is not involved in the
excitation.
mixing
Lowest energy transition = Do
For d2 and d7 (=d5+d2) which involves mixing of the two T1g states, unavoidable problem.
Ground state and excited state
mixing which we saw earlier.
But note that the
difference in
energies of two
excitations is Do.
E (T1gA2g) - E (T1gT2g) = Do
Make sure you can identify the transitions!!
Can use T-S to calculate Ligand Field Splitting. Ex: d2, V(H2O)63+
Observed spectrum
u1: 17,800 cm-1
u2: 25,700 cm-1
Technique: Fit the observed energies to the diagram.
E/B
We must find a value of the splitting parameter, Do/B,
which provides two excitations with the ratio of
25,700/17,800 = 1.44
First, clearly u1 should correspond to 3T1  3T2 But note
that the u2 could correspond to either 3T1  3A2 or 3T1 
3T .
1
The ratio of u2/u1 = 1.44 is obtained at Do / B= 31
DO/B
Again, the root, basic problem
is that the two T1 s have
affected each other via mixing.
The energy gap depends to
some extent on the mixing!
Now can use excitation energies
For u1: E/B = 17,800 cm-1 /B = 29 yielding B = 610
cm-1
By using 31 = Do/B = Do/610 obtain Do = 19,000 cm-1
The d5 case
All possible transitions forbidden
Very weak signals, faint color
Jahn-Teller Effect found if there is an asymmetrically occupied e set.
octahedral d9 complex
b1g
x2-y2
z2
a1g
x2-y2
z2
b2g
xy
xy
xz
xz
yz
effect of octahedral field
yz
eg
elongation along the
x axis
Can produce two transitions.
This picture is in terms of the orbitals. Now for one derived from
the terms.
Continue with d9
Eg
2
T2g
B2g
2
D
A1g
2
Eg
B1g
Free ion term
for d9
effect of
octahedral field
GS will have d-hole
in either of the two eg
orbitals. ES puts dhole in either of the
three t2g orbitals.
effect of elongation
along z
For example, the GS
will have the d-hole in
the x2-y2 orbital which
is closer to the ligands.
Some examples of spectra
Charge transfer spectra
Metal character
LMCT
Ligand character
Ligand character
MLCT
Metal character
Much more intense bands