You are on page 1of 23

Electronic Spectra

The colour of visible light absorbed by a TM complex depends on the

magnitude of o This depends on the ligands bonded to the metal.
some ligands induce large o high energy absorption
others induce small o low energy absorption
Hence different complexes of the same metal in the same oxidation state
may exhibit different colours have different absorption spectra.

The E of the light absorbed is used to promote an electron from a lower

to higher energy d orbital.

Note that CFT can only predict the splitting pattern of the d-orbitals.

The value of o is however derived from electronic spectra.

Types of transition

1. Charge transfer, either ligand to metal or metal to ligand.

these are often extremely intense
are generally found in the UV region but they may have a
tail into the visible spectrum.

2. d-d or ligand field transitions are transitions between orbitals that

are predominantly metal d orbitals
the separation is characterised by the strength of o.
can occur in both the UV and visible region.
transitions usually have small to medium intensities.
Electronic spectra of TM complexes
A single transition t2g2eg1
t2g3 has resulted into two
bands in the spectrum of the
d3 complex [Cr(NH3)6]3+ due
to electron-electron

Why study the electronic

Spectrum of the d 3 complex [Cr(NH ) ]3+.
Understand how to obtain 3 6
o from electronic
absorption spectrum of a The key point is that the presence of e-e
TM complex with more repulsions within d orbitals mean that
than one d electron when e- absorption frequencies
o because e-e
e repulsions are important. repulsions result in multiple absorptions.
Revision on Quantum Numbers (QN)

n = Principal QN size of orbital and its energy

l = Azimuthal/Orbital QN shape of orbital 0, 1, 2, 3..n-1
ml = Magnetic QN
ms = Spin QN for a single electron

L 0 1 2 3
orbital s p d f
ml 1 3 5 7

ml is a subset of l, where ml take values from l. 0.-l.

There are thus (2l +1) values of ml for each l value,

i.e. for the 1 s orbital (l = 0), 3 p orbitals (l = 1), 5 d orbitals (l = 2), etc.
The different ways in which electrons can occupy the orbitals specified
in configuration are called microstates.
For example, one microstate of a 2p2 configuration is (1+ , 1-), this
notation signifies that both the electrons occupy an orbital ml = +1
but do so with opposite spins.
+ indicates ms = + and indicates ms = -.
Another microstate of the same configuration is (-1+ , 0+).
In this microstate, both the electrons have ms = + but one occupies
the 2p orbital with ml = -1 and the other occupies the orbital ml = 0.
The microstates of a given configuration will have the same energy
only if electron-electron repulsions are negligible.
If we group together the microstates that have the same energy taking
into account e-e repulsions , we obtain energy levels called TERMS.
Russell-Saunders (R-S) coupling
The interactions that can occur are of three types:

spin-spin coupling S
orbit-orbit coupling L
spin-orbit coupling J

There are two principal coupling schemes used:

Russell-Saunders (or L - S) coupling
j - j coupling.
It follows that we can identify the terms of light atoms and put them in
order of energy by sorting the microstates according to their total spin
quantum number S and then according to their total orbital angular
momentum quantum number L.
This process called Russel-Saunders coupling.
Russell- Saunders Coupling

By analogy with the notation s, p, d, for orbitals with l = 0, 1, 2, ,

the total orbital angular momentum of an atomic term is denoted by
the equivalent upper case letter:
L = 0 1 2 3 4.
S P D F G.(omitting J)
The total spin is normally reported as the value of 2S + 1, which is
called the multiplicity of the term:

S= 0 1 3/2 2 5/2

2S + 1 1 2 3 4 5 6

singlet doublet triplet quartet quintet sextet

Russell Saunders term symbol
The multiplicity is written as a left superscript representing the value of L
and the entire label of a term is called a term symbol.

The Russell Saunders term symbol is then written as: (2S+1)L

For a d1 configuration:
S= +, hence (2S+1) = 2 ; L=2
and the ground term is written as 2D doublet dee

For an s1p1 configuration:

S= both +, hence (2S+1) = 3; or paired hence (2S+1) = 1 ; L=0+1=1
and the terms arising from this are 3P (triplet pee) or 1P (singlet pee)

For a p1d1 configuration: terms arising are 3F (triplet eff) or 1F (singlet eff)
Mulliken symbols for atomic and molecular orbitals
Symbol Meaning
a or A Non-degenerate orbital; symmetric to principal Cn
b or B Non-degenerate orbital; unsymmetric to principal Cn
e or E Doubly degenerate orbital or state
t or T Triply degenerate orbital or state
(subscript) g Symmetric with respect to centre of inversion
(subscript) u Unsymmetric with respect to centre of inversion
(subscript) 1 Symmetric with respect to C2 perp. to principal Cn
(subscript) 2 Unsymmetric with respect to C2 perp. to principal Cn

For splitting in a tetrahedral crystal field the components are similar,

except that the symmetry label g (gerade) is absent.
Energies of the terms- Hund's Rules
The Ground Terms are deduced by using Hund's Rules:
1) The Ground Term will have the maximum multiplicity.

2) If there is more than 1 Term with maximum multipicity, then the

Ground Term will have the largest value of L.
Thus for a d2 the rules predict 3F < 3P < 1G < 1D < 1S

A simple graphical method for determining just the ground term alone
for the free-ions uses a "fill in the boxes" arrangement.

Note the idea of the "hole" approach:

A d1 configuration is treated as similar to a d9 configuration.

d1 has 1 electron more than d0 and

d9 has 1electron less than d10 i.e. a hole.
d-orbital ground terms
To calculate L, use the labels for each column to determine the value
of L for that box, then add all the individual box values together.

E.g. for a d7 configuration:

in the +2 box are 2 electrons, so L for that box is 2 x 2= 4
in the +1 box are 2 electrons, so L for that box is 1 x 2= 2
in the 0 box is 1 electron, L is 0
in the -1 box is 1 electron, L is -1 x 1= -1
in the -2 box is 1 electron, L is -2 x 1= -2
Total value of L is +4 +2 +0 -1 -2 = 3 F term.

To calculate S, simply sum the unpaired electrons using a value of

for each.
For maximum multiplicity S = 2s+1 = 2[3()] + 1 = 4 F
Also possible is S = 2s+1 = 2[1()] + 1 = 2 2P
Note that for 5 electrons with 1 electron in each box then the total
value of L is 0 L for a d1 configuration = L for a d9.
d-orbital ground terms
dn 2 1 0 -1 -2 L S Ground Term
d1 2 1/2 2D

d2 3 1 3F

d3 3 3/2 4F

d4 2 2 5D

d5 0 5/2 6S

d6 2 2 5D

d7 3 3/2 4F

d8 3 1 3F

d9 2 1/2 2D

The overall result shown in the Table is that:

4 configurations (d1, d4, d6, d9) give rise to D ground terms,
4 configurations (d2, d3, d7, d8) give rise to F ground terms
d5 configuration gives an S ground term.
Terms for 3dn free ion configurations
The Russell Saunders term symbols for the other free ion
configurations are given in the Table below.

Note that dn gives the same terms as d10-n

# of # of
Configu Ground
micro energy Excited Terms
ration Term
states levels
d1,d9 10 1 2D -
d2,d8 45 5 3F 3P, 1G,1D,1S

d3,d7 120 8 4F 4P, 2H, 2G, 2F, 2 x 2D, 2P

3H, 3G, 2 x 3F, 3D, 2 x 3P,

d4,d6 210 16 5D
1I, 2 x 1G, 1F, 2 x 1D, 2 x 1S

4G, 4F, 4D, 4P, 2I, 2H, 2 x

d5 252 16 6S
2G, 2 x 2F, 3 x 2D, 2P, 2S
What is the ground term of the configurations:
(a) 3d5 of Mn2+ and
(b) 3d3 of Cr3+?

If done correctly you should come up with (a) = 6S (a sextet ess term) and
(b) = 4F (a quartet eff term).

Remember the first rule for a ground term maximum multiplicity

S = 2 2S + 1 2 x 2 + 1 = 6 A sextet
L = 2 + 1 + 0 -1 -2 = 0 6S term

(b) S = 1 2S + 1 2 x 1 + 1 = 4 A quartet
L = 2 + 1 + 0= 3 4F term
The ligand field splitting of TM complex ground terms
Thus far our discussion refer only to free atoms, we will now take it a
step further to the domain of TM complex ions.

The treatment will be restricted to complex ions in an Oh crystal field.

The ground term energies for TM complexes are affected by the

influence of a ligand field.

The effect of a ligand field d orbitals split into t2g and eg subsets.

By analogy the D ground term states split into T2g and Eg.

f orbitals are split to give subsets known as t1g, t2g and a2g.

By analogy, the F ground term when split by a crystal field will give
states known as T1g, T2g, and A2g.
Correlating the splitting of spectroscopic terms in Oh ligand field

R-S term Number of states Terms in Oh symmetry

S 1 A1g
P 3 T1g
D 5 Eg , T2g
F 7 A2g , T1g , T2g
G 9 A1g , Eg , T1g , T2g

Note: The ground term for first-row transition metal ions is either:
D, F or S which in high spin Oh A, E or T states.

Now back to the spectrum of [Cr(NH3)6]3+ & e-e repulsions.

Due to d-d transitions the spectrum shows two central bands with
intermediate intensities.
How do we account for two bands in a single electron transition?
Spectroscopic terms
The terms of an octahedral complex are labelled by the symmetry species
of the overall orbital state; a superscript prefix shows the multiplicity of
the term:

The ground term of the d 3 Cr3+ discussed above is 4F which gives rise
to 4A2g , 4T1g , 4T2g in the octahedral crystal field.

The two eg t2g transitions of the excited state are thus labeled:
4T 4A
2g and T2g A2g.
4 4

These are called molecular term symbols serves a function similar to

atomic term symbols discussed earlier on.


For ALL octahedral complexes, except high spin d5, simple CFT would
predict that only 1 band should appear in the electronic spectrum and
that the energy of this band should correspond to the absorption of
energy equivalent to .

In practice however, ignoring spin-forbidden lines, this is only

observed for those ions with D free ion ground terms i.e., d1, d9 as well
as d4, d6.

The observation of 2 or 3 peaks in the electronic spectra of d2, d3, d7

and d8 high spin octahedral complexes requires further treatment
involving electron-electron interactions.

Using the Russell-Saunders (LS) coupling scheme, these free ion

configurations give rise to F free ion ground states which in octahedral
and tetrahedral fields are split into terms designated by the symbols
A2g, T2g and T1g.
Ligand field transitions

Ground state configuration of the octahedral d 3 complex [Cr(NH3)6]3+

t2g3 absorption near 25 000 cm-1 excitation t2g2eg1 t2g3.
There are 2 eg t2g transitions possible:

28 500 cm-1 21 550 cm-1

The dz2 dxy transition, promotes The dz2 dzx transition merely
an electron from the xy-plane into relocates an electron that is
the already electron-rich z-direction already largely concentrated
(the axis is electron-rich because along the z-axis.
both dyz and dzx are occupied
large e-e repulsion). Here there is no e-e repulsion.
Selection rules for electronic spectra of TM complexes.

The Selection Rules governing transitions between electronic energy

levels of transition metal complexes are:

1. The Spin Rule S = 0

2. The Orbital Rule (Laporte) l = +/- 1

The first rule states that allowed transitions must involve the promotion
of electrons without a change in their spin.

The second rule states that if the molecule has a centre of symmetry,
transitions within a given set of p or d orbitals (i.e. those which only
involve a redistribution of electrons within a given subshell) are
Selection rules for electronic spectra of TM complexes.

Relaxation of the Rules can occur through:

a) Spin-Orbit coupling gives rise to weak spin forbidden bands

b) Vibronic coupling - an octahedral complex may have allowed

vibrations where the molecule is asymmetric
Absorption of light at that moment is then possible.

c) -acceptor and -donor ligands can mix with the d-orbitals

transitions are no longer purely d-d.
Electronic Spectra
Expected Values:

For M2+ complexes, expect = 7500 - 12500 cm-1 or = 800 - 1350 nm.

For M3+ complexes, expect = 14000 - 25000 cm-1 or = 400 - 720 nm.

Extinction coefficients for octahedral complexes are expected to be

around 50-100 times smaller than for tetrahedral complexes.

For a typical spin-allowed but Laporte (orbitally) forbidden transition in

an octahedral complex, expect < 10 m2mol-1.
Electronic Spectra
Example: The complex [(Co(NH3)6]+3 absorbs light with a wavelength
of 437 nm. What is the splitting energy ?

Solution: The complex absorbs light at the given frequency, so must

be equal to the light energy absorbed.

E = hc/
h = 6.626 x 10-34 Js ;
c = 3.00 x 108 m/s and
= 437 nm = 4.37 x 10-7 m.

E = (6.626 x 10-34 Js x 3.00 x 108 m/s) /4.37 x 10-7 m = 4.55 x 10-19 J

This is equal to 4.55 x 10-19 J/atom x 6.022 x 1023 atoms/mole

Therefore = 274 kJ/mole.