Chem 104A, UC, Berkeley. Example: Z eff. Chem 104A, UC, Berkeley. Be:

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1 Example: 4 O: 1s s p p Z eff (0.85) 5(0.35) 4.55 Z Be: B: C: 1s s 1 1s s p 1s s p Hund s rule: For any set of orbitals of equal energy (degenerate orbitals), the electron configuration with the maximum number of parallel spins is the ground state. 1

2 Now, 3d orbitals: K: Ca: [ Ar] [ Ar] 1 For neutral, many-electron atoms, the orbital is filled before the 3d orbital. How to write the electron configuration? In the sequence of ionization energy from left to right, those first be removed, in the right. Example: Fe (Z=6): 6 See p50 DG [ Ar]3d Note the aufbau process gives you: 6 [ Ar] 3d Fe: 3p 3d s s p 3s 3p 3d 3 p 7(0.35) 8(0.85) (1.00) d 5(0.35) 18(1.00) (0.35) 14(0.85) 10(1.00).5 Z Z Z eff eff eff (3p) (3d ) 6.5 () 3.75 E(3p)<E(3d)<E() electrons will be first removed!

3 Question: [ Ar]3d 8 Since E(3d) <E(), why not?? 1. Electron-electron repulsion. Spin-pairing energy Compare the total energy a m b n Binding energy m( m 1) E( A) mw ( a) nw ( b) J ( a, a) n( n 1) m nj ( a, b) J ( b, b) e-e repulsion 3

4 Example: Sc Z=1 1 [ Ar]3d E=-W(d)-W(s)+J(ds)+J(ss) E(3d 3 [ Ar]3d ) E(3d 1 3 ) E=-3W(d)+3J(dd) W ( d) 3J ( dd) W ( s) J ( ds) J ( ss) 5.68eV Ground state: 1 [ Ar]3d W(d)=4.75eV W(s)=1.60eV J(ds)=8.38eV J(dd)=11.78eV J(ss)=6.6eV e-e repulsion in d orbital: ground state 1 [ Ar]3d Stronger shielding for, smaller effective Z: electron will be first removed. Ground state for Sc 1 1 :[ Ar]3d 4

5 Exceptions: Special stability of half-filled and filled shell: Cr :[ Ar]3d Cu :[ Ar]3d Mo :[ Kr]4d Ag :[ Kr]4d Au :[ Xe]4 f s 5s 5d 10 6s C : 1s s p These microstates are defined by a set of unique quantum numbers and correspond to electronic states of a specific energy. 5

6 Atomic Energy States Term Symbol Reading: DG 1 Atomic energy states Total orbital angular momentum L Total spin angular momentum S Total angular momentum J S+1 L J Term Symbol S1/ Double S one-half 6

7 n electron system L= l 1 +l +..+l n, l 1 +l +..+l n, -1, l 1 +l +..+l n, -, If all l are equal, the minimum is zero, if one l is larger than the others, the minimum is that given by orienting the other angular momentum to oppose it as long as L 0. (vector sum of all vector l I ). Designation: L =0, 1,, 3, 4, 5. S, P, D, F, G, H. Example: p (l1=1, l=1): L=,1, 0 p f 1 (l1=1, l=1, l3=3) L=5, 4, 3,, 1 7

8 Total spin angular momentum S: S=n/, n/-1, n/-, 0, for n even S=n/, n/-1, n/-,.1/, for n odd Vector sum of all of the vector C : 1s s p s i S=0 S=1 S=0 Spin multiplicity=s+1 S=0, S+1=1, singlet S=1/, S+1=, Doublet S=1, S+1=3, Triplet. Total angular momentum J: J=L+S, L+S-1,. L-S Allowed vector sum for L S 8

9 Hydrogen atom Ground state vs. Excited States s S1/ p P3/ p P1/ Energy 1 S s 1/ 1s S1/ Electron configuration: Term Symbols: 1 1 p 3p S L D 1 P1 1 S0 3 D3,,1 3 P,1,0 3 S1 9

10 C : 1s s p X C : 1s s p S??? L D 1 P1 1 S0 3 D3,,1 3 P,1,0 3 S1 10

11 M L : total orbital-orientation quantum No. M L m 0, 1,,... l L L=max(M L ) L + 1 possible values For a given value of S, there will be S+1 spin states characterized by M S. M S = m s Ms=S, S-1, S-,.., -S S=max (MS) 11

12 C : 1s s p S=0 Ms= 0 S=1 Ms= 1 Ms= -1 Ms= 0 S=0 Ms= 0 Atomic energy states Total orbital angular momentum L Total spin angular momentum S Total angular momentum J S+1 L J M L : total orbital-orientation quantum No. M L m 0, 1,,... L MS= ms l L=max(M L ) S=max(MS) 1

13 N (y)! x!(y x)! In general, the total No. of ways N that x electrons can be placed into y orbitals of equivalent energy with either spins is given by C : 1s s p 6! 15!4! microstates 4!=Factorial 4=1xx3x4 Step I: Sum all m s and m l for each microstates, Determine M S, M L for each MS= ms ML= ml 13

14 m l MS= ms ML= ml M s M L Step II: Starting with maximum M S, partition microstates into sets. Note: A term S+1 L corresponds to (L+1) (S+1) microstates S=max (M S ) L=max(M L ) 14

15 M s M L max( M ) 1 S=1 1,0, 1 For M s 1, max( ) 1 L s M s M L=1 1,0, 1 M L J=,1,0 3 P 3 3, P 1, P0 9 microstates M s M L max( M ) 0 S=0 0 S L= M s max( M ), 1,0,1, L M L J= 1 D 5 microstates 15

16 M s M L max( max( M ) 0 S=0 0 S M s M ) 0 L=0 M L 0 L J=0 1 S0 1 microstate E 1 S0 1 D C : 1s s p Electron configuration 3 P 3 P1 3 P0 Atomic Energy States 16

17 Ground State Terms for many-electron atoms: determine which of several energy states will be lowest in energy. 1. The ground state term always has maximum spin multiplicity. therefore, for the carbon atom, the 3 P,1,0 terms will be more stable than the 1 D and 1 S 0. When comparing two states of the same spin multiplicity, the state with the higher L value is usually more stable. This is the case with the 1 D and 1 S 0 of the carbon atom. 3. For given S and L values, the minimum J value is most stable if there is a single open subshell that is less than halffull, and the maximum J is most stable if the subshell is more than half-fill. Thus for the carbon atom, the 3 P 0 state is the ground state term. Example: Ta 14 3 :[ Xe]4 f 5d 6s Rule 1. S=3/ S+1 =4 Rule. max (M L ) L=3 Rule 3. Jmin=3/ 4 F3/ Ground state 17

18 Identifying Ground State: 3d 5 6 S 5/ 3d 3 F 3d 8 3 F 4 Term Symbol: example 3 Er [ Xe]4 f 11 18

Chapter 6 Part 3; Many-electron atoms

Chapter 6 Part 3; Many-electron atoms Read: BLB 6.7 6.9 HW: BLB 6:59,63,64,67,71b-d,74,75,90,97; Packet 6:10 14 Know: s & atoms with many electrons Spin quantum number m s o Pauli exclusion principle o