Download Section 1.6 - 1 1.6 Term Symbols A brief general review of atomic

1.6 Term Symbols
A brief general review of atomic configurations:
There are four different interactions in any atom that determine the relative energies of
atomic configurations (in order of importance):
a) Electrostatic attraction between electrons and the nucleus
Electrostatic energy ∝ 1/n2, where n = principal quantum number
b) Electron-electron repulsion
Leads to Hund's rule of maximum multiplicity
•
Because of Coulomb repulsion, pairing of electrons in one orbital costs energy:
→ Charge Correlation Energy
px
py
pz
E
ΔΕ = charge correlation energy
(e-e repulsion!)
py = φA
px
py
pz
ΔE = exchange energy
px = φB
px
py
pz
•
Exchange energy: Electrons of like spin in energy degenerate orbitals can
“exchange” leading to an overall lower energy (quantum mechanical resonance
phenomenon.)
•
Another way of looking at this is to consider a diagram of the px and py orbitals of an
atom. Although these orbitals are orthogonal (zero overlap integral), there is
nevertheless a region of actual orbital overlap in space (shown in red).
•
Consider one electron in each orbital. If they have opposite spin, their spatial wave
function can put them in exactly the same space at the same time (high electrostatic
repulsion). If they have the same spin, there must be a constraint on their spatial
wave function so as to not put them in exactly the same space at the same time
(therefore lower electrostatic repulsion if spin is the same)
Section 1.6 - 1
c) Spin-orbit coupling
i) Russel-Saunders for light atoms:
J = L + S, L + S –1, …, |L –S|
L = l1 + l2, l1 + l2 –1, …, |l1-l2|
S = s1 + s2, s1 + s2 –1, …, |s1 – s2|
ii) jj-coupling for heavy atoms (heavier than bromine)
j1 = l1 + s1, l1 + s1 – 1, …, |l1 – s1|
.
jn = ln + sn, ln + s1 – 1, …, |ln – sn|
J = j1 + j2, j1 + j2 –1, …, |j1 – j2|
d) Spin-spin interactions
Leads to the Pauli-Principle: No two Fermions in any spatially/energetically
confined system can have the same four quantum numbers (n, l, ml, ms).
SEE SUPP. INFO ON HUND’S RULE
•
In many-electron-atoms each electronic configuration can be described by a unique
term symbol.
E.g., for scandium
3d14s2:
2
D3/2, 2D5/2 (only one unpaired electron → easy to solve)
3d24s1:
2
S, 2D, 2G, 2,4P, 2,4F (more difficult to figure out)
•
The relative energies of the terms depend on spin their multiplicity (2S+1) and their
orbital angular momentum (L).
•
Based on Hund's rule, the lowest energy terms are the ones with the largest spin
multiplicity and among these, the term with the highest angular momentum is the
lowest.
Section 1.6 - 2
In polyelectronic atoms the motions and spins of the individual electrons are correlated
due to their electrostatic and spin-spin interactions.
Consequences:
a) Only electronic configurations that do not violate the Pauli Principle are allowed.
b) The different arrangements of x electrons occupying y orbitals are not all equal in
energy due to the different electron-electron repulsion energies in these microstates.
→ This leads to electronic fine structure of the atom.
E.g., the d2 configuration:
•
Two electrons can occupy any of the five d-orbitals depending on their quantum
numbers ml and ms.
One possible microstate would be:
ml
2
1
ms
+1/2
+1/2
0
-1
-2
… Obviously there are many more possible…
Deriving Spectroscopic Terms
•
Clearly, there is a finite number of (allowed) electronic configurations for 2 electrons
in 5 degenerate orbitals (and for any x electrons in y orbitals).
•
Each unique configuration, called a microstate, is defined by a unique combination
of quantum numbers.
•
Individual microstates may have different energies because each represents a different
spatial distribution of electrons within the atom resulting in different inter-electronic
repulsions.
•
Microstates of the same energy (degenerate) are grouped together into terms.
Section 1.6 - 3
•
A general term symbol that uniquely describes a specific electronic configuration
looks like this:
(2S+1)
LJ
where 2S + 1 is the spin multiplicity (and S is the total spin angular momentum.)
L is the total orbital angular momentum
J is the total angular momentum (spin + orbital)
S = 0 → “Singlet”
S = ½ → “Doublet”
S = 1 → “Triplet” etc.
•
NOTE: We use lower case letters to define single electron quantum numbers, and
upper case letters to define multiple electron terms.
•
Microstates can be visualized through the vector model of the atom.
→ Each electron has an orbital angular momentum l and a spin angular momentum s.
•
The single electron orbital angular momentum l (and hence the total orbital
angular momentum L) can only have certain orientations → quantization.
•
The total orbital angular momentum L of a group of electrons in an atom is given
by a vector sum of the individual orbital angular momenta l.
•
Two simple examples are p2 and d2:
(Source: “Molecular Symmetry and Group Theory”, R.L. Carter, Wiley, 1998)
Section 1.6 - 4
•
The total angular momentum J is related to the energy, i.e. different combinations
of l and s will result in different energies or terms.→ SPIN-ORBIT COUPLING!
•
There are two ways of defining J:
1) Russel-Saunders coupling:
-
Couple all individual orbital angular momenta l to give a resultant total orbital
angular momentum L. (L = Σl)
-
Couple all individual spin angular momenta s to give a resultant total spin angular
momentum S. (S = Σs)
-
Finally couple L and S to give the total angular momentum J for the entire atom.
-
Russel-Saunders coupling works well for the light elements up to bromine.
2) j-j coupling
- Couple individual orbital l and spin s angular momenta first to the complete electron
angular momentum j. (j = l + s)
- Couple all j to give the total angular momentum J. (J = Σj)
- j-j coupling is much more complicated to treat, but should be used for elements heavier
than bromine.
Applying the Russel-Saunders Scheme:
•
Need to know the values of L and S:
L =
total orbital angular momentum quantum number associated with collection of
microstates with L.
S=
total spin angular momentum quantum number associated with collection of
microstates with S.
→ L and S define the spectroscopic term.
L = maximum ML
ML = 0, ±1, ±2, …, L
S = maximum MS
MS = S, S-1, S-2, …, -S
and
ML = Σ ml
MS = Σ ms
where ml and ms are values for individual electrons in a given microstate
Section 1.6 - 5
In order to find the terms L and S we have to sum up ml and ms of all possible
microstates.
•
There are 2L+1 possible orientations of L and 2S+1 possible orientations of S.
Therefore, the total number of microstates in one term given L and S will be
(2L + 1) × (2S + 1)
•
This must be so as the possible values of ML and MS are:
ML = 0, ± 1, ± 2, …, ± L
and L = max. ML
•
MS = S, S –1, S-2, …, -S
and
S = max. MS
By convention the atomic term symbols are assigned as follows (function of L and S):
L=
S=
2S+1
0
S
0
1
1
P
½
2
2
D
1
3
3
F
3/2
4
4
G
2
5
5
H
5/2
6
6
I
6
7
Example of a Many Electron Atom: Carbon in its ground-state [He]2s22p2
First an important realization:
•
Closed (sub)shells make zero contribution to angular momentum. If all orbitals are
filled with 2 electrons, S = 0 and L = 0 → i.e., all ang. momenta CANCEL OUT!
for [He]2s22p2 that leaves the p2 configuration to be considered.
Each p electron can have:
n = 2, l = 1
ml = 1, 0, -1
ms = ½, - ½
(2p orbitals)
(3 possible values: px, py, pz)
Determining the number of microstates:
# microstates =
n!
e! h!
n = total # sites available (i.e., 2 x # of orbitals)
e = number of electrons
h = number of holes (i.e., n – e)
For p2, all possible combination of 2 electrons gives 6!/2!4! = 15 combinations, which
are best collected and visualized in a table:
Section 1.6 - 6
ml
ms
½
-½
0
½
-½
-1
½
-½
ML = Σml
MS = Σms
1
x
x
x
x
x
x
x
x
x
x
x
1
1
x
x
x
x
x
2
0
x
1
0
0
1
x
x
0
0
1
0
1
-1
x
x
x
x
x
x
0
-1
0
0
0
0
x
-1
0
-1
1
-1
0
x
x
x
-2
0
x
-1
-1
•
The total angular momentum quantum numbers L and S are the largest possible
values of ML and MS
•
We now have to consider that just as for l and ml, ML and MS can have
ML = L, L-1, L-2, …, -L
i.e. ML = 0, ± 1, ± 2, …, ± L; for any given L there are 2L+1 microstates
and
MS = S, S-1, S-2, … -S
i.e. MS = 0, ±1, ± 2, …, ± S; for any given S there are 2S+1 microstates
•
Now start with the maximum ML: We see from the table that the maximum ML here
is 2, and that it only occurs in combination with MS = 0.
→ Therefore, we must have a term with L = 2 and S = 0: 1D
This term accounts for (2L + 1)(2S + 1) = 5 (ML,MS) microstates (2,0), (1,0), (0,0), (-1,0),
(-2,0) and leaves ten microstates to be accounted for.
•
ML
MS
Cross off the five microstates that we have accounted for and we are left with:
1
1
0
1
1
0
1
-1
0
0
0
-1
0
0
-1
1
-1
0
-1
-1
• The maximum value of ML is now 1, and it occurs with a maximum MS = 1:
→ Therefore, we must have a term with L = 1 and S = 1: 3P
This term accounts for nine microstates: (1,1), (1,0), (1,-1), (0,1), (0, 0), (0,-1), (-1,1),
(-1,0), (-1,-1) and leaves only one microstate (0,0) to account for:
→ Therefore, we must have a term with L = 0 and S = 0: 1S
Each term (1D, 3P, and 1S) defines a state (group of microstates of the same energy.)
Section 1.6 - 7
Total Angular Momentum in Many Electron Atoms: Finding J and MJ
Applying the Russel-Saunders Scheme: to our C example…
• As the carbon atom is a light atom we can now use the Russell-Saunders coupling
scheme to account for spin-orbit coupling and its effects.
• Considering only the 9 microstates of the 3P term we find the values of J. Again we
will use a table to visualize the possible combinations:
ML
MS
MJ
1
1
1
0
0
0
-1
-1
-1
1
0
-1
1
0
-1
1
0
-1
2
1
0
1
0
-1
0
-1
-2
The largest MJ = 2, i.e. the largest J = 2; there are 2J+1 = five states: MJ = 2, 1, 0, -1, -2
If we cross these MJ values off the above table, we are left with:
•
The largest remaining MJ =1, therefore J=1: there are three MJ states: 1, 0, -1
•
Finally, all that is left is MJ = 0, therefore J=0: there is only one MJ state: 0
→ The term 3P consists of three terms with different total angular momenta:
3
P2
3
3
P1
P0
Each of the terms is degenerate by (2L+1)(2S+1) = (2J+1) giving the total 15 microstates
we started with:
(2J+1)
1
So
D2
3
P2
3
P1
3
P0
1
Total
1
5
5
3
1
15
Section 1.6 - 8
The graphs above give a graphical summary of the whole process.
Source: Purcell & Kotz, Inorganic Chemistry, 1977; P.W. Atkins, Physical Chemistry,
3rd Edition, 1987
**HOMEWORK: Derive ALL the term symbols for the Ti2+ ion in the gas phase.
The correct answer is given in the table below.
1
G
F
1
D
3
P
1
S
Total # of microstates in d2
3
9
21
5
9
1
45
Q. Which term represents the ground state?
A. This can be determined using Hund’s Rules:
1) The term with the highest multiplicity (= microstates with highest number of unpaired
electrons) is lowest in energy.
2) For a term of given multiplicity, the greater the value of L, the lower the energy.
Section 1.6 - 9
Classical explanation: The higher L, the better correlated the orbital motion of the
electrons, the less repulsion, the lower the energy.
→ For the d2 configuration term energies we find the theoretical order:
•
3
F < 3P < 1G < 1D < 1S
3
F < 1D < 3P < 1G < 1S
Experimentally found is:
Hund’s rules are not always reliable regarding the largest L term,
→
but are always reliable in establishing the ground state.
•
How to quickly find the ground state:
-
Find the microstate with the highest multiplicity.
-
Find the highest possible ML for that microstate
→ Ground state term.
•
Finding the ground state in free ions – e.g. d4 and d7:
ms = +1/2
ml =
+2
+1/2
+1/2
+1/2
+1
0
-1
MS = 1/2 + 1/2 + 1/2 +1/2 = 2
ML = +2 + 1 + 0 + -1 =2
so... L = 2; S = 2
5
D
ms = +1/2 -1/2 +1/2 -1/2 +1/2
-2
ml =
+2
+1
0
+1/2
+1/2
-1
-2
MS = 1/2-1/2 + 1/2-1/2 + 1/2 +1/2 +1/2 = 3/2
ML = +2 + 2 + 1 + 1 + 0 + -1+ -2 =3
so... L = 3; S = 3/2
4
F
NOTE:
Rule #3. For less than half-filled sub-shells, the term with the lowest value of J is lowest
in energy.
Section 1.6 - 10
Applications of term symbols:
a) Term symbols allow a quick energetic ordering of atomic microstates (Hund’s Rules):
b) Spectroscopic selection rules tell us which transitions are expected to have zero
intensity based on the harmonic oscillator approach.
•
Selection rules for electronic transitions can be expressed using term symbols:
ΔS = 0
(electrons are Fermions with half-integral spin; photons are bosons with
integral spin → light cannot affect spin)
ΔL = 0, ± 1 with Δl = ± 1
(the orbital angular momentum of an electron must change, but this does
not necessarily affect the overall momentum)
ΔJ = ± 1, 0
(but J = 0 cannot combine with J =0)
Section 1.6 - 11