Download Review Sheet on Determining Term Symbols

Review Sheet on Determining Term Symbols
Term symbols for electronic configurations are useful not only to the
spectroscopist but also to the inorganic chemist interested in understanding electronic and
magnetic properties of molecules. We will concentrate on the method of Douglas and
McDaniel (p. 26ff); however, you may find other treatments equal or superior to the D &
M method.
Term symbols are a shorthand method used to describe the energy, angular
momentum, and spin multiplicity of an atom in any particular state. The general form is
given as aTj where T is a capital letter corresponding to the value of L (the angular
momentum quantum number) and may be assigned as S, P, D, F, G, … for |L| = 0, 1, 2, 3,
4, … respectively. The superscript “a” is called the spin multiplicity and can be
evaluated as a = 2S +1 where S is the spin quantum number. The subscript “j” is the
numerical value of J, a new quantum number defined as: J = l +S, which corresponds to
the total orbital and spin angular momentum of the system. The term symbol 3P is read
as triplet – Pee state and indicates that there are two unpaired electrons in a state with
maximum orbital angular momentum, L=1.
The number of microstates (N) of a system corresponds to the total number of
distinct arrangements for “e” number of electrons to be placed in “n” number of possible
orbital positions.
N = # of microstates = n!/(e!(n-e)!)
For a set of p orbitals n = 6 since there are 2 positions in each orbital. Therefore, p2 (e =
2 and n = 6) so,
N = 6!/(2!(6-2)!) = (6*5*4*3*2*1)/((2*1) * (4*3*2*1)) = 15
We can introduce the idea of a hole formalism which states that for many
electronic properties one may consider systems with e or (n-e), the number of unoccupied
sites or “holes”, to be equivalent. Thus,
P4 (e = 4 and n = 6)
N = 6!/(4!(6-4)!) = 15
so P4 (2 holes) and P2 (4 holes) give equivalent values of N. The same is true for all p, d,
f, … systems such as d1/d9 or f2/f12.
If one should consider an atom with multiple unoccupied orbitals, the total
number of microstates will equal the product of the microstates for the individual orbitals.
(N = π N*a; a = 1 to i where a = orbital index). For an s1p2 configuration N = 2 (from S1)
* 15 (from p2) = 30.
We can define component ML in two ways. First,
M L = m1 + m2 + m3 + …
where ml corresponds to the l angular momentum quantum number for each electron
in a given orientation. Second,
ML = L, L-1, …, 0, …, -L
where L is the atomic orbital angular momentum. Thus, ML has 2L + 1 values.
Similar definitions for MS, representing the spin of the system may be made with
allowable values of MS = 2S+1
The total angular momentum quantum number, J, is given as J = L+S, L+S – 1,
…, |L-S|. It is the absolute magnitude of J that determines the total angular momentum.
Determination of Spectroscopic Terms
1.
2.
3.
4.
5.
Determine total number of microstates
Determine the possible values of ML and MS
Determine the Pauli – allowed microstates
Set up a chart of microstates
Resolve this group of microstates into atomic states
The following steps will be illustrated for a d2 configuration.
STEP 1 (Number of Microstates)
N = n!/(e!(n-e)! = 10!/(2!*8!) = 45
STEP 2 (Values of ML and MS)
For a d orbital, ml may range from +2 through –2 so there are 5 acceptable ml
values. The maximum value of ML for 2 electrons is when m1 = +2 and when m2 = +2 or
ML = 4 (this can only occur if MS = 0. Therefore, ML will range +4, +3, …, -4.
The allowed values for MS are 1, 0, -1 since two electrons may be spin aligned
(up or down) or paired.
STEP 3
The Pauli allowed microstates can be drawn as shown on the following
pages. Remember that certain combinations of quantum numbers are not allowed (e.g.,
ML = -4 and MS = +1 since this would require two parallel spins in the same orbital).
ML
MS
3
2
1
0
1
0
-1
-1
-2
-3
+1
+1
+1
+1
+1
+1
+1
+1
+1
+1
ml
=
+2
+1
↑
↑
↑
↑
↑
↑
↑
↑
ms=
+1/2
0
↑
↑
↑
↑
-1
↑
↑
↑
↑
-2
↑
↑
↑
↑
Ms
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
ml
=
+2
+1
↓
↓
↓
↓
↓
↓
↓
↓
ms=
-1/2
0
↓
↓
↓
↓
By symmetry an equivalent pattern must be observed for ms = -1/2 and MS = -1.
10 microstates for MS = +1
10 microstates for MS = -1
-1
↓
↓
↓
↓
-2
↓
↓
↓
↓
ML
MS
4
3
2
1
0
3
2
1
0
-1
2
1
0
-1
-2
1
0
-1
-2
-3
0
-1
-2
-3
-4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
ml
=
+2
↑
↑
↑
↑
↑
+1
ms=
+1/2
0
-1
-2
ml
=
+2
↓
↑
↑
↑
↑
↑
↓
↑
↑
↑
↑
↑
↓
↑
↑
↑
↑
↑
↓
↑
↑
↑
↑
↑
↓
+1
↓
↓
↓
↓
↓
ms=
-1/2
0
↓
↓
↓
↓
↓
-1
↓
↓
↓
↓
↓
These are the 25 microstates for MS = 0
Therefore,
10 (MS = +1) + 10 (MS = -1) + 25 (MS = 0) = 45 microstates as required by Step 1
-2
↓
↓
↓
↓
↓
STEP 4. Chart Microstates
On left column put range of ML values
On bottom row put range of MS values
Then enter the total number of microstates that have equivalent ML and MS values
Table 1.
ML
+4
+3
+2
+1
0
-1
-2
-3
-4
1
1
2
2
2
1
1
+1
1
2
3
4
5
4
3
2
1
0
MS
1
1
2
2
2
1
1
-1
The sum of the values in this chart must total 45, the total number of microstates.
STEP 5. Resolve Microstates
We can immediately see that there is a microstate which must correspond to ML =
-4 and MS = 0. This must be a G state since L = 4 and it must be an orbital singlet since
MS = 0. Therefore, the first term is 1G (We will assign J values later.). A 1G
representation corresponds to a 9 x 1 matrix (ML x MS) since L can range from +4
through –4 (9 elements) and S can only equal 0 (1 element). Removing 9 microstates
from Table 1 yields Table 2.
Table 2.
ML
+3
+2
+1
0
-1
-2
-3
1
1
2
2
2
1
1
+1
1
2
3
4
3
2
1
0
MS
1
1
2
2
2
1
1
-1
Elements Removed: 1G (9 microstates)
(Notice that the elements removed for 1G came from the MS = 0 column, with one from
each ML row as required.)
Now go to the upper left corner (italicized and red) element. The ML and MS value for
this element yield the next microstate representation. ML=3 therefore F, MS =1 therefore
triplet, so the element is 3F.
3
F is a 7 x 3 matrix = 21 microstates that are removed to give Table 3.
Table 3.
ML
+2
+1
0
-1
-2
1
1
1
+1
1
2
3
2
1
0
MS
1
1
1
-1
Elements Removed: 1G, 3F (30 microstates)
Now got to the single element in row +2 (red) which gives 1D (ML = 2, MS = 0).
This is a 5 x 1 matrix yielding Table 4.
Table 4.
ML
+1
0
-1
1
1
1
+1
1
2
1
0
MS
1
1
1
-1
Elements Removed: 1G, 3F, 1D (35 microstates)
Upper left gives ML = 1, MS = 1 or a 3P which is a 3 x 3 matrix. This leaves table 5.
Table 5.
ML
0
1
0
MS
Elements Removed: 1G, 3F, 1D, 3P (44 microstates)
This table must correspond to 1S so the d2 configuration gives:
1
G, 3F, 1D, 3P, 1S (45 microstates)
Assigning J Values
J = L + S: number of microstates = 2J + 1
G ! 1G4
(L = 4, S = 0)
9 microstates
S ! 1S0
(L = 0, S = 0)
1 microstate
1
1
D ! D2
(L = 2, S = 0)
5 microstates
3
F ! 3F4
(L = 3, S = 1)
9 microstates
3
3
P ! P2
(L = 1, S = 1)
5 microstates
Notice that the total number of microstates does not equal 45 at this point! All of the
microstates associated with 1G, 1S, and 1D are accounted for, while 12 3F microstates and
4 3P microstates are missing. This is because we have only calculated the state of
maximum multiplicity! Both the 3F and 3P states can be factored. We do this by lowering
the J value by one until all microstates have been accounted for. Thus,
For
1
1
3
F4 (9 microstates)
F3 (7 microstates)
3
F2 (5 microstates)
3
21 microstates
3
P2 (5 microstates)
P1 (3 microstates)
3
P0 (1 microstate)
3
9 microstates
We now have accounted for all 45 microstates:
1
G4, 1S0, 1D2, 3F4, 3F3, 3F2, 3P2, 3P1, 3P0
We must now arrange these in a predicted orbital energy scheme.
Rules for Assigning Energies (Hund’s Rules)
1. Ground state will have the largest spin multiplicity i.e. 3P is lower in energy than
1
P
2. Two states with the same spin multiplicity can be distinguished by L values. The
state with the largest L is of lowest energy, i.e. 1D < 1S
3. After checking 1 and 2 above: If the subshell is less than 1/2 full, the lowest J
corresponds to the lowest energy. If the subshell is greater than 1/2 full, the
highest J corresponds to the lowest energy.
1
S
1
1
1
D
G
1
1
S0
D2
1
1
G4
S
G
3
P
s2p6d2
1
3
P
P2
P1
3
P0
3
3
F
D
3
3
F
3
F4
F3
3
F2
3
Actual
Predicted ground state is 3F2.
CAUTION! The energy difference between levels can be less than or equal to the
splitting of states. Therefore, the only configuration we are certain of is the ground state.
The actual order is give at the far right.
Miscellaneous Information:
1. It usually is not productive to associative a particular electron configuration with a
specific term symbol since there can be considerable overlap.
2. Orbitals with L = odd number have odd parity (ungerade). Orbitals with L = even
number have even parity (gerade).
3. Selection rules for transitions
∆l = ±1
∆S = 0
∆j = ±1 or 0
3
i.e.
P2 ! 3D3 allowed
1
S0 ! 3P1 not allowed (spin forbidden)
4. Filled, 1/2 filled, or empty orbitals represent orbital singlets.
5. For lighter elements, L and S coupling (Russell – Saunders) is a good description.
Spin – orbit coupling (j – j) can have an effect, especially for heavy elements.