L

m=2

m=1

m=0

m=-1

m=-2

ML

arrangements of electrons in polyatomic

atoms• for an atom with several valence electrons, a number of arrangements of these electrons in orbitals of different l and ml are possible. – These arrangements are called microstates.

• Some of these microstates have the same energy (are degenerate) whereas others have different energy, presenting different energy states.

L

m=2

m=1

m=0

m=-1

m=-2

ML

microstates

• each valid arrangement of electrons (specified by n, l, ml and ms for each electron) is called a microstate.

• Some have different energies and some the same

• how many can we have?• n, number of electron sites; e, number of electrons; n-e, number of "holes"

N =n!

[e!(n−e)!]

L

m=2

m=1

m=0

m=-1

m=-2

ML consider a d1 configuration

• the number of microstates is:– n = 10, e = 1

• N = 10!/[1!(9!)] = 10N =n!

[e!(n−e)!]

L

m=2

m=1

m=0

m=-1

m=-2

ML

pigeonhole diagrams of the

10 d1 microstatesml= 2 1 0 -1 -2

L

m=2

m=1

m=0

m=-1

m=-2

ML

the net orbital angular

momentum L• find the value of ML and Ms

for each microstate– ML = ∑ml

– Ms = ∑ms

• group by values of ML and Ms

• arrange in a table

L

m=2

m=1

m=0

m=-1

m=-2

ML

like this: one X for each microstate

Ml\Ms +1/2 -1/2

2 X X

1 X X

0 X X

-1 X X

-2 X X

L

m=2

m=1

m=0

m=-1

m=-2

ML

These microstates

correspond to one

spectroscopic term

• the orbital an-gular momentum L is the same, the projection, ML is different

• Each microstate corresponds to some orientation of L = 2 and Ms = ±1/2

ML

Ms

ML

Ms

L

Ms

L ML L

L

m=2

m=1

m=0

m=-1

m=-2

ML

L and S?

• L for a given set of microstates is the maximum ML

– ML= ∑ ml = L, L -1, L -2, ..., - L

– (2 L + 1 values of ML)

• S is the maximum Ms

– S = ∑ ms and the different values of MS

– MS = S, S-1,...,0,..- S

• so for the d1 case we could arrange the microstates by ML and MS to obtain these values for different terms.

L

m=2

m=1

m=0

m=-1

m=-2

ML

10 microstates

ML for S = 1/2 ML for S = -1/2

2 2

1 1

0 0

-1 -1

-2 -2

L

m=2

m=1

m=0

m=-1

m=-2

ML

assigning terms from

microstates• have a groups of related microstates

• #microstates in a term = (2L +1)(2S+1)

– L = 2, S = 1/2 (10 microstates)• L = 0 1 2 3 4 5• notation S P D F G H

(then by alphabet, omitting J)• Have a D term with spin multiplicity (2S+1) = 2

(doublet)• 2D term (pronounced doublet-dee)

L

m=2

m=1

m=0

m=-1

m=-2

ML a p2 system is more complex

• number of valid (remembering pauli exclusion, etc.) microstates is 15

• there are microstates which cannot be described by a single value of L and Ms

• a systematic treatment of the microstates is given on the next slide

• the notation: ml =1, ms= 1/2 is represented by 1+

L

m=2

m=1

m=0

m=-1

m=-2

ML

EXAMINATION of permutations of

ml and ms

(1+,1+) (1+, 1-) (1+, 0+) (1+, 0-) (1+, -1+) (1+, -1-)

Pauli

(1-,1+) (1-, 1-) (1-, 0+) (1-, 0-) (1-, -1+) (1-, -1-)

same Pauli

(0+, 1+) (0+, 1-) (0+, 0-) (0+, 0+) (0+, -1+) (0+, -1-)

repeat

(0-, 1+) (0-, 1-) (0-, 0-) (0+,0-) (0-, -1+) (0-, -1-)

(-1+, 1+) (-1+, 1-) (-1+, 0+) (-1+, 0-) (-1+, -1-)

(-1-, 1+) (-1-, 1-) (-1-, 0+) (-1-, 0-) (-1-, -1-)

L

m=2

m=1

m=0

m=-1

m=-2

MLsummary of microstates

ML=

+2 0 -2 1 0 -1 1 0 -1 1 0 -1 1 0 -1

+1 ↑↓ ↑ ↑ ↓ ↓ ↑ ↑ ↓ ↓

0 ↑↓ ↑ ↑ ↓ ↓ ↓ ↑ ↑ ↓

-1 ↑↓ ↑ ↑ ↓ ↓ ↓ ↓ ↑ ↑

Ms=

0 0 0 +1 +1 +1 -1 -1 -1 0 0 0 0 0 0

L

m=2

m=1

m=0

m=-1

m=-2

ML Today’s DJ question

• Write all the microstates for the neutral fluorine atom.

L

m=2

m=1

m=0

m=-1

m=-2

ML

Group Work

• groups of 3: and

write the microstates for a d2 electronic configuration

L

m=2

m=1

m=0

m=-1

m=-2

ML

energies and angular momentum

• there are several components to the energy of the atom, excluding the stable core electrons

• the n values of the valence electrons are pretty much the same

• the angular momentum of the orbitals added to the net electron spin lead to different energy levels called states.

L

m=2

m=1

m=0

m=-1

m=-2

ML

Russell-Saunders

(L-S) coupling• in L-S coupling, the total angular momentum of the electronic configuration, J, is the sum of the orbital angular momentum, L (ML (max), and the spin, S (∑ms).

J = L + S

ML

Ms

L

J

L

m=2

m=1

m=0

m=-1

m=-2

ML What are L and S?

• L for a given set of microstates is the maximum ML

– ML= ∑ ml = L, L -1, L -2, ..., - L

– (2 L + 1 values of ML)

• S is the maximum Ms

– S = ∑ ms and the different values of MS

– MS = S, S-1,...,0,..- S

• so for the p2 case we can arrange the microstates by ML and MS to obtain these values for different terms.

L

m=2

m=1

m=0

m=-1

m=-2

ML

Microstates for p2 arranged by

Msand MLMS

1 0 -1

+2 x

+1 x xx x

ML 0 x xxx x

-1 x xx x

-2 x

L

m=2

m=1

m=0

m=-1

m=-2

ML See L = 2 and S = 0 term

MS

1 0 -1

+2 x

+1 x x x x

ML 0 x x xx x

-1 x x x x

-2 x

L

m=2

m=1

m=0

m=-1

m=-2

ML L = 1, S = 1 term

MS

1 0 -1

+2 x

+1 x x x x

ML 0 x x x x x

-1 x x x x

-2 x

L

m=2

m=1

m=0

m=-1

m=-2

ML L = 0, S = 0 term

MS

1 0 -1

+2 x

+1 x x x x

ML 0 x x x x x

-1 x x x x

-2 x

L

m=2

m=1

m=0

m=-1

m=-2

ML

assigning terms from

microstates• have several groups of related microstates

•#microstates = (2L +1)(2S+1)

– L = 2, S = 0 (5 microstates) (D term)– L = 1, S = 1 (9 microstates) (P term)– L = 0, S = 0(1 microstate) (S term)

• L = 0 1 2 3 4 5• notation S P D F G H (then by

alphabet, omitting J)

L

m=2

m=1

m=0

m=-1

m=-2

ML indicating Spin Multiplicity; S

• the spin is shown by the numerical superscript value = 2S + 1 preceding the letter term symbol – L = 2, S = 0 (2S + 1 = 1) 1D term– L = 1, S = 1 (2S +1 = 3) 3P term– L = 0, S = 0 (2S + 1= 1) 1S term

L

m=2

m=1

m=0

m=-1

m=-2

ML Ground State Term?

• lowest E term that has highest S: – here the 3P or 3 F

• for two terms with same S, that with greater L will be the ground state

L

m=2

m=1

m=0

m=-1

m=-2

ML

Develop the table for the d2 configuration

• Make a matrix with rows for ML

and MS.

• Put an X for each microstate in each box with corresponding ML

and MS.

• It better be symmetrical!

L

m=2

m=1

m=0

m=-1

m=-2

ML

Now…

• Find the largest value of ML and MS and assign the L and S values and the term symbol.

• Eliminate those microstates and repeat until all microstates are eliminated.