Foundations of electronic transitions in atoms, molecules and especially transition metal compounds

1. Electronic Spectra
Dr. Chris, UP 2019

2. Between atoms
Emission Spectra

3. Double line in Na
The 3p level is split because of
spin-orbit coupling

4. Term symbols for atoms/ions
The electronic configuration 3p1 does not include the possible
energy levels:
We get so-called micro-states of the electron:
mL = 1 0 -1 mL = 1 0 -1 mL = 1 0 -1
L = 1 => P
S = ½ => multiplicity M = 2S+1 = 2
L + S = J = 3/2 and ½ (L+S, L+S-1. L+S-2 ….)

5. Angular Momentum L
The possible combinations of l for a pair of electrons
L = |l1 + l2| , |l1 + l2 ‐1| , … , |l1 ‐ l2|
What are the possible L values for 1s22s22p2 ?
Both open‐shell electrons are l = 1.
The possible combinations are 2,1,0
What about configuration [Xe] 6s2 4f1 5d1 ?

6. We have an l=3 and an l=2 electron. 3+2 = 5 and 3‐2 = 1.
The possible combinations are 5,4,3,2,1
Total spin quantum number S
S = |s1 + s2|, |s1 +s2 ‐1| , … ,| s1 ‐ s2|
Find S for 1s22s12p1
S = 1,0
Find S for 1s22s22p2
Multiplicity M = 2S + 1

7. S=1,0 the same as the previous problem. Notice that S is not affected
by which orbitals the electrons are in. S only cares about how many
open‐shell electrons there are, not about where they are
Find S for 1s22s22p3
We haven‟t done a three electron case yet, but they aren‟t hard. Find
all the combinations for a single pair first, and then factor in the third
electron.
For two electrons, we already know that the two possible S values are
S=1,0.
A third electron can either add or subtract ½ from these values, so the
final S is S = 3/2 and 1/2
An atom that only has closed shells will always be 1S
=

8. Total angular momentum J
J can range from L + S to |L – S|
What are the term symbols for 1s1 ?
Since there is only one electron, this is a simple problem. L=0
and M=1, so the only possible term symbol is 2S.
With only one electron, S = ½ , so J = 0 + ½ = ½.
The final term symbol is 2S½
What are the term symbols for 1s22s22p1?

9. There still only one open shell electron, so L=1, M=1 and S = ½.
We get a term symbol of the type 2P,. which gets split into separate
symbols because J = 3/2 and ½.
The term symbols are 2P3/2 and 2P1/2
What about 1s22s12p1?
Since l1 = 0 and l2 = 1, the only possible combination is L=1.
The possible combinations of S are: S=1,0.
This means that M = 3,1.
The term symbols will be of the form 1P and 3P.
For the 1P state, L=1 and S=0, so J=1.
For the second state, L=1 and S=1, so J=2,1,0.
The final term symbols are 1P1 and 3P2, 3P1, and 3P0

11. Energy Levels
1) High multiplicity values mean low energy
2) If there is a tie, high L values mean low energy
3a) If there is still a tie and the shell is less than
half full, then low J means low energy
3b) If the shell is more than half full, then high J
means low energy

12. example
s1p2 configuration
What is the ground state configuration ?
- Max spin is 3/2  M = 4
- L = 0 and L = 1  L = 1 = P
- J = 1 + 3/2, 3/2, ½ = 5/2, 3/2, ½
Less than half filled  lowest J is ground state
 4 P ½ is ground state

13. การจัดเรียงอิเล็กตรอน เทอม (terms)
d1, d9 2D
d2, d8 3F, 3P, 1G, 1D, 1S
d3, d7 4F, 4P, 2H, 2G, 2F, 22D, 2P
d4, d6 5D, 3H, 3G, 23F, 3D, 23P, 1I, 21G, 1F, 21D, 21S
d5 6S, 4G, 4F, 4D, 4P, 2I, 2H, 22G, 22F, 32D, 2P, 2S
For d- transition metals:
mL = 2 1 0 -1 -2
S = 1  M = 3
L = 3  F
J = 4,3,2,1  1
Ground state: 3 F 1
How would 1G look like ?

14. TRANSITIONS IN ORGANIC
MOLECULES

15. Possible transitions

16. Example: trans-butadiene
(1) Determine the point group

17. (2) Determine the symmetry of the  orbitals
https://www.webqc.org

18. (3) Determine the change from HOMO with Bg symmetry to
LUMO with Au : has the product a x,y or z component ?
Is the transition 1(Bg) -> 2(Bg) allowed ?
The product is Bu
In the character table this has a
x and y component
=> HOMO-LUMO is allowed
https://www.webqc.org

19. example acetone
Is a transition possible between n and * ?
(1) Determine the point group
(2) The symmetry of the 2 orbitals
(3) Multiply the symmetries and check if the product
has x,y or z character

21. Exercise the same with the water molecule:
HOMO = non-bonding 1b1
LUMO = anti-bonding 4a1

22. Since B1 has the x-component, the transition is allowed

23. Selection rules
(1) An electronic transition must not change the multiplicity of the
system: S = 0.
This condition is called spin-selection rule
(2) Symmetry selection rule:
For centro-symmetric molecules with inversion center
(as for d-orbitals, typically in transition metal complexes)
the Laporte rule says that electronic transitions between “gerade” /
“ungerade” states are not allowed
( g -> g and u -> u are both forbidden).

24. Example: Benzene
Is the HOMO-LUMO transition allowed ?

25. The electronic ground state is 1A1g
(totally symmetric).
From the ground state, 3 transitions are symmetry allowed:
1A1g -> 1B1u , 1A1g -> 1B2u and 1A1g -> 1E1u

26. Transition Metal complexes
d-d transitions and CT

27. d-d spectra and MO theory:
3A2g →3T2g
3A2g →1Eg
υ, cm-1
UV
[Ni(NH3)6]2+
visible infrared

28. The electronic spectra of d-block complexes:
The features of electronic spectra that we need to be
able to master are:
1) naming of electronic states and d-d transitions,
e.g.3A2g, or 3A2g→1Eg
2) Explanation of relative intensities of bands in the
spectra of complexes of d-block metal ions. (The
Laporte and spin selection rules)
3) calculation of the crystal field splitting parameters
from energies of d-d bands

29. Naming of electronic states:
In names of electronic states, e.g. 4A2g, the labels A, E,
and T, stand for non-degenerate, doubly degenerate, and
triply degenerate, while the numeric superscript stands
for the multiplicity of the state, which is the number of
unpaired electrons plus one. Note that the electronic
states can be ground states (states of lowest energy) or
excited states:
4A2g
t2g
eg
Multiplicity =
3 unpaired electrons + 1
= 4
Non-degenerate
ground state =
„A‟
g = gerade
energy

30. eg
eg eg
t2g t2g
6A2g
3T2g
1A2g
Non-degenerate triply degenerate non-degenerate
Multiplicity
= 5 + 1
energy
t2g
Naming of electronic states (contd.):
NOTE: In determining degeneracy, one can re-arrange the electrons, but
the number of unpaired electrons must stay the same, and the number
of electrons in each of the eg and t2g levels must stay the same.
Multiplicity
= 2 + 1
Multiplicity
= 0 + 1

31. eg
eg eg
t2g t2g
5Eg
5T2g
2Eg
eg
eg eg
t2g t2g
3A2g
1Eg
3T2g
Naming of electronic states (contd.):
t2g
t2g
ground state excited state excited state
ground state excited state ground state
energy

32. Electronic transitions:
eg
eg
eg
eg
t2g t2g
t2g t2g
3A2g →3T2g
3A2g →1Eg
3A2g
3T2g
3A2g
1Eg
ground state excited state

33. visible infraredUV
green
3A2g →3T2g
3A2g →1Eg
[Ni(H2O)6]2+
The electronic spectrum of [Ni(H2O)6]2+:
λ,
The complex looks green, because it absorbs only weakly at 500 nm,
the wavelength of green light.

34. On the previous slide we saw the two bands due to the
3A2g →3T2g and 3A2g →1Eg transitions. The band at λ =
1180 nm which is the 3A2g →3T2g transition shown below,
corresponds to Δ for the complex. This is usually
expressed as Δ in cm-1 = (1/λ(nm)) x 107 = 8500 cm-1.
The electronic spectrum of [Ni(H2O)6]2+:
eg
eg
t2g t2g
3A2g →3T2g3A2g
3T2gΔ
= Δ
= 8500
cm-1

35. Note the weak band at 620 nm that corresponds to the
3A2g →1Eg transition. The electron that is excited moves
within the eg level, so that the energy does not involve Δ,
but depends on the value of P, the spin-pairing energy.
The point of interest is why this band is so weak, as
discussed on the next slide.
The electronic spectrum of [Ni(H2O)6]2+:
eg
eg
t2g t2g
3A2g →1Eg3A2g
1EgΔ
= 16100
cm-1

36. The electronic spectrum of [Ni(H2O)6]2+:
The two peaks at higher energy resemble the 3A2g→3T2g transition, but
involve differences in magnetic quantum numbers of the d-orbitals,
and are labeled as 3A2g→3T1g(F) and 3A2g→3T1g(P) to reflect this:
3A2g →3T2g
3A2g →3T1g(F)
3A2g →3T1g(P)
3A2g →1Eg
λ,
[Ni(H2O)6]2+

37. The Selection rules for electronic transitions
There are three levels of intensity of the bands that we
observe in the spectra of complexes of metal ions.
These are governed by two selection rules, the Laporte
selection rule, and the spin selection rule. The Laporte
selection rule reflects the fact that for light to interact with
a molecule and be absorbed, there should be a change
in dipole moment. When a transition is „forbidden‟, it
means that the transition does not lead to a change in
dipole moment.
The Laporte Selection rule: This states that transitions
where there is no change in parity are forbidden:
g→g u→u g→u u→g
forbidden allowed

38. All transitions within the d-shell, such as 3A2g→3T2g are
Laporte forbidden, because they are g→g. Thus, the
intensity of the d-d transitions that give d-block metal
ions their colors are not very intense. Charge transfer
bands frequently involve p→d or d→p transitions, and so
are Laporte-allowed and therefore very intense.
The Spin Selection rule: This states that transitions that
involve a change in multiplicity (or number of unpaired
electrons) are forbidden. This accounts for why
transitions within the d-shell such as 3A2g→1Eg that
involve a change of multiplicity are much weaker than
those such as 3A2g→3T2g that do not.
The Selection rules for electronic transitions

39. The Selection rules for electronic transitions
3A2g →3T2g
Charge-transfer band – Laporte and spin allowed – very intense
[Ni(H2O)6]2+
a
b c
3A2g →1Eg Laporte and spin forbidden – very weak
a, b, and c, Laporte
forbidden, spin
allowed, inter-
mediate intensity

40. The three types of bands present in e.g. [Ni(H2O)6]2+ are:
1) Laporte-allowed plus spin allowed charge transfer
bands of very high intensity
2) Laporte-forbidden plus spin-allowed d→d transitions
(e.g. 3A2g→3T2g) of moderate intensity
3) Laporte forbidden plus spin-forbidden d→d transitions
(3A2g→1Eg) of very low intensity.
The Intensity of bands in complexes of d-block ions:

41. The MO view of electronic transitions in an
octahedral complex
t1u*
a1g*
eg*
t2g
t1u
eg
4p
4s
a1g
3d
t2g→t1u*
M→L Charge transfer
Laporte and spin
allowed
t1u→t2g
L→M Charge transfer
Laporte and spin
allowed
t2g→eg
d→d transition
Laporte forbidden
Spin-allowed or
forbidden
The eg level in CFT
is an eg* in MO
In CFT we consider
only the eg and t2g
levels, which are a
portion of the over-
all MO diagram
σ-donor orbitals
of six ligands

42. There are two mechanisms that allow „forbidden‟
electronic transitions to become somewhat „allowed‟.
These are:
1) Mixing of states: The states in a complex are never
pure, and so some of the symmetry properties of
neighboring states become mixed into those of the
states involved in a „forbidden‟ transition.
2) Vibronic Coupling: Electronic states are always
coupled to vibrational states. The vibrational states may
be of opposite parity to the electronic states, and so help
overcome the Laporte selection rule.
Why do we see ‘forbidden’ transitions at all?

43. Mixing of states: Comparison of [Ni(H2O)6]2+ and [Ni(en)3]2+:
[Ni(H2O)6]2+
[Ni(en)3]2+
3A2g →3T2g
3A2g →3T2g(F)
The spin-forbidden 3A2g →1Eg is close to the spin-allowed
3A2g →3T2g(F) and „borrows‟ intensity by mixing of states
The spin-forbidden 3A2g →1Eg is not close
to any spin allowed band and is very weak
3A2g →1Eg
Note: The two spectra are
drawn on the same graph
for ease of comparison.

44. Electronic transitions are coupled to vibrations of various
symmetries, and the latter may impart opposite parity to
an electronic state and so help overcome the Laporte
selection rule:
Vibronic coupling:
electronic ground
state is „g‟
electronic excited
state is „g‟
g→g transition
is forbidden
g→(g+u) transition
is allowed
energy
coupled vibration
υ4’ is „u‟
Electronic transitions, as seen
in the spectra of complexes of
Ni(II) shown above, are always
very broad because they are
coupled to vibrations. The
transitions are thus from ground
states plus several vibrational
states to excited states plus
several vibrational states (υ1, υ2, υ3),
so the „electronic‟ band is actually
a composite of electronic plus
vibrational transitions.
υ5
υ3
υ1
υ5’
υ3’
υ1’

45. Symmetry of vibrational states, and their
coupling to electronic states:
T1u
symmetry
vibration
A1g
symmetry
vibration
(symbols have same meaning for
vibrations: A = non-degenerate,
T = triply degenerate, g = gerade,
u = ungerade, etc.)
The band one sees in the
UV-visible spectrum is the
sum of bands due to transitions
to coupled electronic (E) and
vibrational energy levels (υ1, υ2, υ3)
observed
spectrum
E E- υ1
E- υ2
E- υ3
E + υ1‟
E + υ2‟
E + υ3‟

46. The spectra of high-spin d5 ions:
6A2g →4T2g
energy
For high-spin d5 ions all possible d-d transitions are spin-forbidden. As a
result, the bands in spectra of high-spin complexes of Mn(II) and Fe(III)
are very weak, and the compounds are nearly colorless. Below is shown
a d-d transition for a high-spin d5 ion, showing that it is spin-forbidden.
eg
eg
t2g t2g
Complexes of Gd(III) are colorless, while those of other lanthanide
M(III) ions are colored, except for La(III) and Lu(III). Why is this?

47. The spectra of complexes of tetrahedral
metal ions:
As we have seen, a tetrahedron has no center of symmetry,
and so orbitals in such symmetry cannot be gerade. Hence
the d-levels in a tetrahedral complex are e and t2, with no „g‟
for gerade. This largely overcomes the Laporte selection
rules, so that tetrahedral complexes tend to be very intense
in color. Thus, we see that dissolving CoCl2 in water
produces a pale pink solution of [Co(H2O)6]2+, but in alcohol
tetrahedral [CoCl2(CH3CH2OH)2] forms, which is a very
intense blue color. This remarkable difference in the spectra
of octahedral and tetrahedral complexes is seen on the next
slide:

48. The spectra of octahedral [Co(H2O)6]2+ and
tetrahedral [CoCl4]2- ions:
[CoCl4]2-
[Co(H2O)6]2+
The spectra at left
show the very intense
d-d bands in the blue
tetrahedral complex
[CoCl4]2-, as compared
with the much weaker
band in the pink
octahedral complex
[Co(H2O)6]2+. This
difference arises
because the Td com-
plex has no center of
symmetry, helping to
overcome the g→g
Laporte selection rule.