Coordination Chemistry III:
Electronic Spectra
Chapter 11
Coordination Complexes
• Unlike most organic compounds, many
coordination compounds have vivid colors.
– These vivid colors are due to the electronic
transitions between the d-orbitals of the metal.
• The energy levels of d electron
configurations, however, are usually more
complicated than might be expected since
the electrons in the atomic orbitals interact
with each other.
Absorption of Light
• Complementary colors – if a compound
absorbed light of one color, the complement of
that color is observed.
– [Cu(H2O)6]2+ has a blue color. What is absorbed?
• Beer-Lambert Absorption Law
• log(Io/I)=A=lc (define variables)
• In a common absorption spectrum, the A is plotted versus
wavelength or cm-1 (1/).
Quantum Numbers of
Multielectron Atoms
• In order to understand energy transitions between states
with more than one electron, we need to understand in
more detail how these electrons interact with each other.
• Each conceivable set of individual ml and ms values
constitutes a microstate of the configuration.
– How many microstates in a d1 configuration?
– Examine the carbon atom (p2 configuration)
• Determine the electron configuration and quantum numbers.
• Independently, each of the 2p electrons could have one of six possible
combinations. The two electrons, however, are not independent of each
other.
Interaction between Electrons
• In any microstate, the individual orbital magnetic
moments (related to ml) and the spin magnetic moments
(related to ms) will interact resulting in an energy state or
term for the configuration.
– Commonly, a number of microstates will contribute to a single
term (degeneracies).
• This is an approximation using the one-electron wave
mechanical mode.
– Works well for 1st and 2nd row transition metals.
Term Designation (Free-Ion)
•
2S+1L
– L and S relate to the overall orbital and spin angular
momenta for the system (or state).
– L = overall orbital angular momentum quantum
number.
• Discuss possible values (analogous to the l quantum
number).
• L is related to the vectorial addition of vectors taken from
the l quantum number.
• Discuss the permitted ways in which the l values can
combine (Figure 20, Carter).
Overall Orbital Angular
Momentum Quantum Number, L
• The vectorial addition produces the possible terms for a
given configuration.
• For a given term, the magnitude of the resultant angular
momentum is fixed, [L(L+1)]1/2(h/2).
• The vector for the momentum can have a number of
allowed orientations.
– A partial lift of degeneracy by the magnetic field.
– Allowed orientations for a given term are associated with the
overall magnetic quantum number, ML.
• ML = L, L-1,….-L
Overall Orbital Angular
Momentum Quantum Number, L
• ML relates to the orbital multiplicity or orbital
degeneracy (Each has projection equal in magnitude
to ML(h/2)).
– What are the allowed orientations for a D term?
• In the Russell-Saunders scheme, ML = ml.
– An ML value can be assigned to each microstate.
• Therefore, a given L value must arise from a complete
set of microstates with the 2L+1 values.
– Examples
Overall Spin Quantum Number, S
• Spin state of the term.
• 2S+1 – spin multiplicity of the state.
– S=0, 1, 2
• Like L, S can be obtained by vectorial addition of the
spin angular momentum vectors (related to s or ms).
– Magnitude [S(S+1)]1/2(h/2)
• S is related to an overall spin quantum number, MS.
– MS = S, S-1,…-S (2S+1 values)
– Indicates the ‘allowed’ orientations of the vector relative to an
applied magnetic field. The produces the spin degeneracy in a
spin state.
Overall Spin Quantum Number, S
• Magnitude of the spin angular momentum is
[S(S+1)]1/2(h/2), but its projections on a
particular axis in the allowed orientation are
given by MS(h/2).
– Russell-Saunders scheme, MS=ms
In Summary
• Any given term with L and S values arises from
a set of microstates that has the necessary 2L+1
values of ML and also the necessary 2S+1
values of MS.
• Value of the term
– L = largest possible value of ML
• Let’s do the d2 configuration.
Determining Microstates
• Determine the number of microstates for a triplet P.
What are the allowed ML and MS values?
• Determine the number of microstates for a doublet D.
What are the allowed ML and MS values?
– It is sufficient to designate the microstates by x (only the
number is important).
• Number of microstates in a given free-ion term is
equal to (2L + 1)(2S +1).
– Row multiplied by column
Reducing a Microstate Table/Configuration
into Its Free-Ion Terms
• Notice that each term is composed of a rectangular
array of microstates.
– This process is shown for a p2 configuration in Table
11-4.
– The spin multiplicity is same as the number of columns
of microstates.
– The # of rows to include in the rectangular array is
equal to +L  -L (number of values).
• (2L+1)(2S+1)
Do the p2 and d2 configurations.
Term of the Lowest Energy
• The ground term (lowest energy) has the highest
spin multiplicity.
• If two or more terms share the same maximum
spin multiplicity, the ground term is the one
having the highest L value.
What is the term with the lowest energy for the d2
configuration?
Spin-Orbit Coupling
• There is also spin-orbit coupling wherein the
spin and orbital angular momenta couple with
each other.
J = L+S, L+S-1, L+S-2,….L-S| (no negative values)
J is a subscript on the right side of the L quantum #
– Determine the spin-orbit coupling in the 1S and 3P
free-ion terms. Designate these with J.
• Page 388
Lowest Energy Term Including
Spin-Orbit Coupling
• For subshells (such as p2) that are less than
half-filled, the state having the lowest J value
has the lowest energy. For subshells that are
more than half-filled, the state having the
highest J value has the lowest energy.
– Determine the lowest energy state for the p2
configuration.
Electronic Spectra of
Coordination Compounds
• Absorption spectra in most cases involve the d orbitals
of the metal.
• Identifying lowest energy term (quickly)
– Sketch the energy levels showing the d electrons.
– Spin multiplicity of lowest-energy state equal highest
possible number of unpaired electrons +1.
– Determine the maximum possible value of ML (ms) for the
configuration.
– Combine steps.
High- and low-spin d6 configurations in octahedral symmetry.
Reducing the Symmetry of the
Free Ion by a Ligand Field
• The orbital term symbols for the free
atom/ion are identical to the symbols for the
appropriate symmetry species in the
spherical group, R3 (show Table).
– There are no inherent symmetry restrictions on
possible orbital degeneracies.
• D – fivefold degeneracy (ML = ?)
• F – ???
Correlation Table
Reducing the Symmetry of the
Free Ion by a Ligand Field
• When a ligand field is imposed, however, there are
restrictions placed on the maximum orbital degeneracy.
– In an octahedral complex, the maximum degeneracy is 3 (Ttype IR’s, look at table). Orbital degeneracies present in the
free ion must be split.
• All term symbols will be redefined by the new
symmetry.
• Upon reducing the symmetry, the free-ion term can be
treated as a reducible representation composed of
irreducible representations in the appropriate character
table.
Reducing the Symmetry of the
Free Ion by a Ligand Field
• The total number of microstates, Dt, remains the
same.
• Do the d1 configuration.
– What does this split into when an octahedral field is
imposed?
– How many microstates are possible when imposing the
ligand field?
The new ligand-field terms will retain the original spin
multiplicities found in the free-ion terms.
Examine
Correlation Diagrams – Relating Electron
Spectra to Ligand Field Splitting
• Examine the correlation diagram for a d2 configuration
in an octahedral ligand field.
– Far left (absence of ligand field) – the free-ion terms. On this
side, the ligand field has very little influence.
• Examine the previous table.
– Far right (strong ligand field) – the states are largely
determined by the ligand field.
• Observed before in the crystal field and ligand field discussion.
• In real compounds, the situation is somewhere in the
middle.
Splitting of Free-Ion Terms
• Irreducible representations are produced.
– Examine the correlation diagram and the Oh table.
• Table 11-6
– Each state has symmetric characteristics of that IR.
– IRs are also obtained from the strong-field limit
configurations (right side of table).
• The IRs on the right and left sides must
‘correlate’.
Selection Rules for Transitions
• Transitions between states of the same
parity are forbidden.
– Laporte Selection Rule
• Transitions between states of different spin
multiplicities are forbidden.
– 4A2  4T1 and 4A2  2A2
Discuss how the rules are relaxed on the expected
absorption intensities.
Tanabe-Sugano Diagrams
• Special correlation diagrams useful in the interpretation
of electronic spectra.
– Lowest energy state is plotted along the horizontal axis.
• o/B (field strength)
– Vertical axis is the measure of the energy above the ground
state
• E/B
B = Racah parameter, a measure of the repulsion between terms of the
same multiplicity.
– Lines connecting states of the same symmetry cannot cross.
2
Ad
Configuration, [V(H2O)6
3+
]
Electron Configurations
• The diagrams of d2-d8 are illustrated in Figure 11-7.
• There is a vertical line near the center of d4-d7 diagrams.
What does the vertical line represent? What happens
once the line is crossed?
• In some cases, the absorption bands are off-scale (x-axis)
or obscured by the charge-transfer bands (discussed
later).
• Discuss the terms for the configurations. They are not
the same as observed for the orbital terms.
Jahn-Teller Distortions
• Spectrum of the [Ti(H2O)6]3+ complex.
– The complex has a d1 configuration (Fig. 11-8).
– Go over terms for the configurations.
– Why are there two overlapping bands?
• If degenerate orbitals are asymmetrically occupied a distortion will
occur to remove the degeneracy.
• Jahn-Teller distortion is usually only significant for
asymmetrically occupied eg orbitals.
– The others cannot be resolved.
– The most common distortion is along the z-axis (elongation).
Symmetry Labels for Configurations
• Symmetry labels for electron configurations
match their degeneracies.
– T = triply degenerate asymmetrical state
– E = doubly degenerate asymmetrical state
– A or B = nondegenerate
Do a few without Jahn Teller splitting/distortion.
• Effect of a D4h field.