arrangements of electrons in polyatomic atoms
DESCRIPTION
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 m l are possible. These arrangements are called microstates. - PowerPoint PPT PresentationTRANSCRIPT
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.