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### Transcript of 1.6 Term Symbols A brief general review of atomic ... · Section 1.6 - 1 py = φA px = φB 1.6 Term...

• Section 1.6 - 1

py = A

px = B

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

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)

= charge correlation energy (e-e repulsion!)

E = exchange energy

E

px py pz

px

px

py pz

pzpy

• Section 1.6 - 2

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 HUNDS RULE In many-electron-atoms each electronic configuration can be described by a unique

term symbol. E.g., for scandium 3d14s2: 2D3/2, 2D5/2 (only one unpaired electron easy to solve) 3d24s1: 2S, 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 - 3

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:

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.

ml 2 1 0 -1 -2

ms +1/2 +1/2

• Section 1.6 - 4

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 = 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 - 5

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 = l) - Couple all individual spin angular momenta s to give a resultant total spin angular

momentum S. (S = s) - 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 = j) - 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 S = maximum MS ML = 0, 1, 2, , L MS = S, S-1, S-2, , -S

and

ML = ml MS = ms where ml and ms are values for individual electrons in a given microstate

• Section 1.6 - 6

!!!#hensmicrostate =

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 MS = S, S 1, S-2, , -S

and L = max. ML and S = max. MS

By convention the atomic term symbols are assigned as follows (function of L and S): L = 0 1 2 3 4 5 6

S P D F G H I S = 0 1 3/2 2 5/2 6

2S+1 1 2 3 4 5 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 (2p orbitals) ml = 1, 0, -1 (3 possible values: px, py, pz) ms = , -

Determining the number of microstates: 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 - 7

ml ms

1 x x x x x - x x x x x 0 x x x x x - x x x x x -1 x x x x x - x x x x x ML = ml 2 1 1 0 0 1 1 0 0 0 -1 -1 -1 -1 -2 MS = ms 0 1 0 1 0 0 -1 0 -1 0 1 0 0 -1 0

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. Cross off the five microstates that we have accounted for and we are left with: ML 1 0 1 1 0 0 0 -1 -1 -1 MS 1 1 0 -1 0 -1 0 1 0 -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 - 8

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 2 1 0 1 1 -1 0 0 1 1 0 0 0 0 -1 -1 -1 1 0 -1 0 -1 -1 -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:

3P2 3P1 3P0

Each of the terms is degenerate by (2L+1)(2S+1) = (2J+1) giving the total 15 microstates we started with:

(2J+1) 1So 1 1D2 5 3P2 5 3P1 3 3P0 1

Total 15

• Section 1.6 - 9

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.

1G 9 3F 21 1D 5 3P 9 1S 1

Total # of microstates in d2 45 Q. Which term represents the ground state? A. This can be determined using Hunds 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 - 10

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:

3F < 3P < 1G < 1D < 1S

Experimentally found is:

3F < 1D < 3P < 1G < 1S

Hunds 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:

NOTE:

Rule #3. For less than half-filled sub-shells, the term with the lowest value of J is lowest in energy.

ml = +2 +1 0 -1 -2

MS = 1/2 + 1/2 + 1/2 +1/2 = 2ML = +2 + 1 + 0 + -1 =2 so... L = 2; S = 2 5D

ms = +1/2 +1/2 +1/2 +1/2

ml = +2 +1 0 -1 -2

MS = 1/2-1/2 + 1/2-1/2 + 1/2 +1/2 +1/2 = 3/2ML = +2 + 2 + 1 + 1 + 0 + -1+ -2 =3 so... L = 3; S = 3/2 4F

ms = +1/2 -1/2 +1/2 -1/2 +1/2 +1/2 +1/2

• Section 1.6 - 11

Applications of term symbols:

a) Term symbols allow a quick energetic ordering of atomic microstates (Hunds 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:

S = 0 (electrons are Fermions with half-integral spin; photons are bosons with integral spin light cannot affect spin)

L = 0, 1 with l = 1

(the orbital angular momentum of an electron must change, but this does not necessarily affect the overall momentum)

J = 1, 0 (but J = 0 cannot combine with J =0)