Lecture 4Lecture 4

Electronic Microstates & Term Symbols

Suggested reading: Shriver and Atkins, Chapter 20.3 or Douglas, 1.4-1.5g ,

F b “ ” “l” “ ” d “ ”

Recap from last class: Quantum Numbers

Four quantum numbers: “n” , “l”, “ml”, and “ms”

Or, equivalently: “n” , “l”, “j”, and “mj”, q y , , j , j

Orbital angular momentum:

Spin angular momentum:

Total angular momentum:

Associated with f l

Associated with Vector sum of rotation of electron

cloud around nucleus

rotation/spin of electron around it’s

axis

orbital and spin angular

momentum

F b “ ” “l” “ ” d “ ”

Recap from last class: Quantum Numbers

Four quantum numbers: “n” , “l”, “ml”, and “ms”

Or, equivalently: “n” , “l”, “j”, and “mj”, q y , , j , j

1S

Orbital angular momentum:

Spin angular momentum:

Total angular momentum:

1 ssS

sz mS

1 L

mLz

1 jjJ

jz mJ

= 0, 1, 2, ….n1

m = 0 +

s = 1/2

m = ±1/2

j = -s,…,+s

m = j 0 +jm = -..., 0, … + ms = ±1/2 mj = -j,…0,…+j

Electronic Configurations and Microstates

• Electronic configurations tell us the number of electrons in each Electronic configurations tell us the number of electrons in each orbital, but they don’t tell us how the electrons occupy the

orbitals.

i.e., don’t always want to draw: , y

•“Terms”: the energy levels and configurations of atoms and ions (and molecules)

• Useful in understanding ionic and coordination compound spectra

General Form of Terms

naTb

•n: principle quantum number

•a = 2S+1 “multiplicity”p y(S is the total spin angular momentum of all electrons)

**a number**

For example, if two electrons are in two different orbitals we might have antiparallel spins:

S=s1+s2

= ½ - ½ =0

Or parallel spins: S=s1+s2 = ½ + ½ = 1

General Form of Terms

naTb

•n: principle quantum number of valence electrons

•a = 2S+1 “multiplicity”p y(S is the total spin angular momentum of all electrons)

**a number**

•T=L (total orbital angular momentum of all electrons; vector sum of , the orbital momentum of individual electrons)

l**a letter**L=0, 1, 2, 3, 4 S, P, D, F, G

•b=J (total angular momentum)**a number**

Examples of Term Symbols in the literature

Free Ions/ Atoms MoleculesFree Ions/ Atoms Molecules

Sundararajan Dalton Trans 2009 6021-6036

Co(PPH3)2Cl2 Used in palladium-catalyzed coupling reactions, 2010 Nobel Prize

Sundararajan, Dalton Trans., 2009, 6021 6036

Wang, Nature Materials 10 (2011)

Hydrogen (1 electron atom)

Ground state: 1s

l bln=1; =0

L= = 0

Only one term is possible (i.e., only one energy level for the one microstate)L= = 0

S=s= ½

)

2S½ (read “doublet S”)

J=L+S(vector sum) =0+½

In a magnetic field due to the Zeeman effect the 12S term yields two In a magnetic field, due to the Zeeman effect, the 12S1/2 term yields two closely-spaced energy levels (mj=½, -½)

Multiplicity terms

S= ½ 2S+1=a=2 doublet (electron can be spin up or spin down)

S= 1 2S+1=a=3 triplet (three different spin configurations with S= 1 2S+1=a=3 triplet (three different spin configurations, with spin wavefunction χ)

(both electrons spin up)

(electrons are indistinguishable))(2

1

(both electrons spin down)

Multiplicity terms

S= ½ 2S+1=a=2 doublet (electron can be spin up or spin down)

S= 1 2S+1=a=3 triplet (three different spin configurations with S= 1 2S+1=a=3 triplet (three different spin configurations, with spin wavefunction χ)

/ (f d ff f )S=3/2 2S+1=a=4 quartet (four different spin configurations)

)(3

1

3

)(3

1

3

Hydrogen (1 electron atom)

Excited state: 2p1

n=2; =1

Two terms are possible (i.e., two energy levels,

n=2; =1

L= = 1

depending on the orientations of the electron spin with respect to the

S=s= ½

spin with respect to the orbital angular momentum of the electron)

J=L+S= 1+½ or 1-½= ½ or 3/2

2P1/2 or 2P3/2

= ½ or 3/2

In a magnetic field due to the Zeeman effect the 22P1/2 term yields In a magnetic field, due to the Zeeman effect, the 2 P1/2 term yields two closely-spaced energy levels (mj=½, -½) while the 22P3/2 yields

four (mj=3/2, ½, -½, -3/2)

Helium (2 electron atom)

21 L

21 ssS

G d 1 2Ground state: 1s2

n=1; 1=2=0

L= 1+2= 0 1S0

S=s1+s2= ½-½ (Pauli) = 0

J L S( t ) 0J=L+S(vector sum)=0

Helium (2 electron atom)

21 L

21 ssS

G d 1 12 1Ground state: 1s12p1

1=0; 2=11P0

L= 1+2= 13P0

“singlet P”

“triplet P”

S=s1+s2= ½-½ or ½+½= 0 or 1

P0 p

J=L+S(vector sum)=0

Carbon (6 electron atom)

654321 L

654321 ssssssS

G d 1 22 22 2 The electrons in the 1s and Ground state: 1s22s22p2

1=0; 2=0; 3=0; 4=0; 5=1; 6=1

The electrons in the 1s and 2s states will be spin paired. The two electrons 3 4 5 6

L= 2, 1, 0 D, P, Sin p can be either in px, py, or pz orbitals, and either paired or unpaired:

S= 0 or 1 2S+1 = 1, 3

J L S( t ) 3 2 1

paired or unpaired:

m=-1 m=0 m=+1

J=L+S(vector sum)=3, 2, 1

Microstates

Microstates: the different ways in which electrons Microstates: the different ways in which electrons can occupy certain orbitals

Grouping together the microstates that have the same energy when electron-electron repulsions are taken

into account, yields the terms (i.e., the , y ( ,spectroscopically distinguishable energy levels)

!)!2(!2

0

0

ee NNNN

Number of microstates:

No=# degenerate orbitals (i.e., three degenerate p orbitals)

Ne=# electrons

Carbon, continued: [He]2p2

15!2)!232(

)!3(2!)!2(

!2

0

0

ee NNN

N

No=# degenerate orbitals for p=3N =# electrons in p=2Ne # electrons in p 2

What are the microstates and which has the lowest What are the microstates, and which has the lowest energy?

Carbon, continued: [He]2p2

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

ml +1 0 ‐1 +1 0 ‐1 ML=ml MS=ms

1 1 11 1 1

2 0 1

3 ‐1 1

4 1 ‐1

5 0 ‐1

6 ‐1 ‐1

7 2 0

8 1 0

9 0 0

10 1 0

11 0 0

12 ‐1 0

13 0 0

14 ‐1 0

15 ‐2 0

Carbon, continued: [He]2p2

M

+1 0 ‐1

MS=ms

‐2 0 1 0

‐1 1 2 1ML=ml 0 1 3 1

+1 1 2 1

2 0 1 0

ML=ml

+2 0 1 0

Note: the array is symmetric about the lines through ML=0 and MS=0, providing a check of the tabulation

ML=2 1 0 -1 - 2 L=2 a D termML 2, 1, 0, 1, 2 L 2 a D termThe values in the array occur only for MS=0 S=0

**a 1D term!**

Subtracting out the 1D term:

M

+1 0 ‐1

MS=ms

‐2

‐1 1 1 1ML=ml 0 1 2 1

+1 1 1 1

2

ML=ml

+2

ML=1, 0, -1 L=1 a P termThe values in the array occur MS=1, 0, -1 S=1

**a 3P term!**a P term!

Subtracting out the 3P term:

M

+1 0 ‐1

MS=ms

‐2

‐1 0 0 0ML=ml 0 0 1 0

+1 0 0 0

2

ML=ml

+2

ML=0 L=0 a S termMS=0 S=0

**a 1S term!**a S term!

The energies of the terms

1 Hund’s rule: For a given configuration the term with 1. Hund s rule: For a given configuration, the term with the greatest multiplicity lies lowest in energy the triplet term of a configuration (if one is permitted) will have a lower energy than a singlet term

2. L Rule: For a term of a given multiplicity, the term with _ g p y,the greatest value of L lies lowest in energy if L is high, the electrons can effectively avoid each otherother

3. J-Rule: if subshell is less than half filled, lowest J is llowest energy.

If greater than half filled, highest J is lowest energyIf equal to half-filled, only one J possibleq , y J p

Energy Diagram for Carbon

Ground state of Carbon: 3P0 (“triplet P0”)er

gyE

ne

Beyond Russell-Saunders coupling

For light atoms and the 3d series (Z30) the energies For light atoms and the 3d series (Z30), the energies of the microstates are determined first by the electron

spin (S), then their orbital angular momentum (L).

i.e, as before, total momentum J is determined by summing first the orbital angular momenta, then the g g ,

spins, and then combining both: Russel-Saunders coupling

For heavier atoms, must consider spin-orbit coupling: l h l f d h“jj-coupling”. For each electron, find j=+s, then sum j’s

of each electron to find total J of atom/ion

For this class, knowledge of Russell-Saunders coupling is sufficient