Idea Transcript
5.61 2004
Lecture #33
page 1
BUILDING MOLECULES ONE ELECTRON AT A TIME H2 molecule: both electrons (with opposite spin) in σ1s MO Configuration is (σ1s)2 HOMONUCLEAR DIATOMICS He2 – four electrons - Configuration is (σ1s)2(σ∗1s)2 What's the net bonding energy?
Simplified Energy Level Diagram σ
u
E
1s
-1/2 au
1s
A
σ
∆Eel± ( R ) =
B
g
J '± K ' 1± S
2 electrons in a bonding MO ≡ “single bond” 2 electrons in an antibonding MO roughly cancel the bonding energy Bond order = ½[(# e- in bonding MOs) – (# e- in antibonding MOs)]
He2
- no covalent bond
He +2 ion - bond order ½ - stable Li2 – 6 electrons - Configuration (σ1s)2(σ∗1s)2(σ2s)2 - single bond Bond length 2.67 A
5.61 2004
Lecture #33
page 2
1s electrons hardly participate in bonding – could think of electrons in 1s AOs 1s ∼ e − Zr a0
Z=3 ⇒ exponential decays 3x closer than in H atom!
Configuration sometimes labeled KK(σ2s)2 – refers to filled “K” (n=1) shell Li2 has 6 electrons but only 2 valence electrons available for bonding
Be2
- configuration is KK(σ2s)2(σ∗2s)2 – no net bonding
B2 – what is correct configuration? KK(σ2s)2(σ∗2s)2.... (σ2pz)2?
.... (π2px) (π2py)? – with e- spins parallel or antiparallel? .... (π2px)2?
Antisymmetrize the spatial and spin parts of the wavefunction and see! Recall similar case for atoms, e.g. He atom 1s2s state (Lecture notes #27-28) In any case, B2 has a single bond C2 – several plausible configurations, all with double bond order e.g. KK(σ2s)2(σ∗2s)2(σ2pz)2(π2px)(π2py) N2 – configuration must be KK(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2 – triple bond O2 – several plausible configurations, all with double bond order e.g. KK(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)(π∗2py) F2 – several plausible configurations, all with single bond order e.g. KK(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)2(π∗2py)2 Ne2 – configuration is KK(σ2s)2(σ∗2s)2(σ2pz)2(π2px)2(π2py)2(π∗2px)2(π∗2py)2(σ∗2pz)2 – no net bonding, molecule not found in nature B2 – C2 – N2 – F2 – what are the actual configurations??
5.61 2004
Lecture #33
page 3
Our treatment can’t resolve σ2pz, π2px,y or σ∗2pz, π∗2px,y energies Better treatment: Use more AOs to construct MOs as LCAOs
ψ = c1 (1s A + 1sB ) + c2 (2 s A + 2 sB ) + c3 (2 pzA + 2 pzB ) + ... Do self-consistent field (SCF) Hartree-Fock determination of the coefficients Add more terms to LCAO until no further improvement – usually needs about 9 terms for 1st row homonuclear diatomics! When this is done the MOs can no longer be associated with individual AOs MOs are just numbered according to symmetry class RESULTS: same as LH side energy level picture below Two possible MO energy level diagrams for homonuclear diatomic molecules σ∗2pz σ∗2pz π∗2p π∗2p π2p σ2pz σ2pz π2p σ∗2s σ∗2s σ2s
σ2s
σ∗1s σ1s
σ∗1s σ1s
MOLECULAR SPIN STATES and TERM SYMBOLS Only spatial parts of molecular wavefunctions have been shown Generally, have to figure out z-components ML and MS and total angular momenta L, S, J as for atoms
5.61 2004
Lecture #33
page 4
M L = ml1 + ml 2 + ... with ml = 0 for σ orbitals, ml = ±1 for π orbitals, etc. M S = ms1 + ms 2 + ... for spin z-components (which don't depend on orbital states) Term symbols: orbital angular momentum states Σ, Π, ∆, Φ, ... for M L = 0,1,2,3,... Different from atomic case - S, P, D, F refer to orbital angular momentum, not its z-component Spin multiplicity 2S + 1 also indicated as for atoms 2 S +1
term symbol
M L = 0, L = 0, M S = 0, S = 0
H2
1
Σ "singlet sigma" state
M L = 0, L = 0, M S = ± 1 2, S = 1 2
H +2
2
Σ "doublet sigma" state
M L = 0, L = 0, M S = ± 1 2, S = 1 2
He +2
Li 2
ML
2
Σ "doublet sigma" state
M L = 0, L = 0, M S = 0, S = 0 1
Σ "singlet sigma" state
B2 - several possible configurations
First 8 electrons in filled σ orbitals – no net angular momentum Last 2 electrons in π bonding MOs 2 degenerate π bonding MOs, based on π2px,y AOs, available with L = 1, M L = ±1 If both e- in the same π bonding MO, M L = ±2, L = 2 Must have opposite spins, M S = 0, S = 0
5.61 2004
Lecture #33
page 5
If both e- in different π bonding MOs, then orbital angular momenta cancel, i.e. M L = 0, L = 0 Could have symmetric or antisymmetric spatial wavefunction π 2 p x (1) π 2 p y ( 2 ) ± π 2 p x ( 2 ) π 2 p y (1) Symmetric spatial wavefunction ψ + = π 2 px (1) π 2 p y ( 2 ) + π 2 px ( 2 ) π 2 p y (1) needs antisymmetric spin wavefunction α (1) β ( 2 ) − α ( 2 ) β (1) M S = 0, S = 0
Antisymmetric spatial wavefunction ψ − = π 2 px (1) π 2 p y ( 2 ) − π 2 px ( 2 ) π 2 p y (1) needs symmetric spin wavefunction α (1) β ( 2 ) , α (1) β ( 2 ) , or α (1) β ( 2 ) + α ( 2 ) β (1) M S = ±1 or 0, S = 1 3
Σ "triplet sigma" state
Which state is it? Recall that ψ − = 0 when both electrons at same point
ψ − has lower e--e- repulsions ⇒ 3 Σ is lower energy state Hund’s rule: HIGHEST SPIN MULTIPLICITY! 3 Σ wins! B2
M L = 0, L = 0, M S = 0, ±1, S = 1 3
Σ "triplet sigma" state
C2 - 4 e- in the two π bonding orbitals – orbital & spin angular momenta cancel
C2
M L = 0, L = 0, M S = 0, S = 0 1
Σ "singlet sigma" state
N 2 - two more e- go into σ2pz bonding MO – no additional angular momentum
N2
M L = 0, L = 0, M S = 0, S = 0 1
Σ "singlet sigma" state
5.61 2004
Lecture #33
page 6
O 2 has two electrons in the π antibonding orbitals – analogous to B2
2 degenerate π antibonding MOs, based on π∗2px,y, available with L = 1, M L = ±1 Just like B2 - highest spin multiplicity state Electrons in different spatial orbitals: M L = 0, L = 0
antisymmetric spatial wavefunction π * 2 p x (1) π 2 p y ( 2 ) − π * 2 p x ( 2 ) π 2 p y (1) symmetric spin wavefunction α (1) β ( 2 ) , α (1) β ( 2 ) , or α (1) β ( 2 ) + α ( 2 ) β (1) M S = ±1 or 0, S = 1 3
Σ "triplet sigma" state
O2
M L = 0, L = 0, M S = 0, ±1, S = 1 3
Σ "triplet sigma" state
This means that O 2 is paramagnetic – magnetic moment is attracted to, and oriented by, a magnetic field – exploited for medical O 2 monitoring! F2 - 4 e- in the two π antibonding orbitals – analogous to C2 - orbital and spin
angular momenta cancel F2
M L = 0, L = 0, M S = 0, S = 0 1
Σ "singlet sigma" state
5.61 2004
Lecture #33
page 7
SUMMARY OF HOMONUCLEAR DIATOMICS Relative MO energies (not to scale) for homonuclear diatomic molecules
σu2pz
σg2pz
σu2s σg2s
Species Li2 Be2 B2 C2 N2 O2 F2 Ne2
Ground-state electron configuration KK(σ2s)2 KK(σ2s)2(σ*2s)2 KK(σ2s)2(σ*2s)2(π2p)2 KK(σ2s)2(σ*2s)2(π2p)4 KK(σ2s)2(σ*2s)2(π2p)4(σ2pz)2 KK(σ2s)2(σ*2s)2(π2p)4(σ2pz)2(π*2p)2 KK(σ2s)2(σ*2s)2(π2p)4(σ2pz)2(π*2p)4 KK(σ2s)2(σ*2s)2(π2p)4(σ2pz)2(π*2p)4(σ*2pz)2
Bond Order 1 0 1 2 3 2 1 0
Bond Length (A) 267 245 159 124 110 121 141 310
Bond Energy (kJ/mol) 105 ≈9 289 599 942 494 154