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Transcript of 1 This molecular ion exits and has been experimentally measured ; its dissociation equals 2.791 eV...
![Page 1: 1 This molecular ion exits and has been experimentally measured ; its dissociation equals 2.791 eV and its H-H distance is 2.0 a 0 (1.06Å). There is no.](https://reader036.fdocuments.in/reader036/viewer/2022062516/56649da05503460f94a8c25c/html5/thumbnails/1.jpg)
1
This molecular ion exits and has been experimentally measured ; its dissociation equals 2.791 eV and its H-H
distance is 2.0 a0 (1.06Å).
There is no other 1 e - 2 nuclei stable system than H2+
Hydrogenoids exist even if they might be exoticHeH2+ is unstable relative to dissociation into He+ + H+.
H2+, the one electron system
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Write the Schrodinger Equation for H2+
Tell whether terms are simple or difficult
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3Simple, 1/R does not depend on the electron position !
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There are exact solutions of the equation.We will consider an approximate one, open to generalization
is a one-electron wave function: a molecular orbital
We will consider that is Linear Combination of Atomic Orbitals.
For H2+, it is possible to find them using symmetry.
Mirror or Inversion center: A single atomic function is not a solution
I a = b and I b = a The Molecular orbitals must have the molecular symmetry.
g = a + b and u = a - b are solutions:
I g = a + b = g
I u = a - b = u
g = a + b = g
u = a - b = u
Molecular orbitals -LCAO
gerade
ungerade
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Normalization
gIg 1sa+1sbI1sa+1sb1saI1sa1saI1sb1sbI1sa1sbI1sb2 + 2S
uIu 1sa-1sbI1sa-1sb1saI1sa- 1saI1sb- 1sbI1sa1sbI1sb2 - 2S
HBHASAB
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Normalization
gIg 1sa+1sbI1sa+1sb1saI1sa1saI1sb1sbI1sa1sbI1sb2 + 2S
uIu 1sa-1sbI1sa-1sb 1saI1sa- 1saI1sb- 1sbI1sa1sbI1sb2 - 2S
Neglecting S: g√1sa+1sb) and u√1sa-1sb)
With S: g√S1sa+1sb) and u√ S1sa-1sb)
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Density partition
gIg 1sa+1sbI1sa+1sb1saI1sa1saI1sb1sbI1sa1sbI1sb2 + 2S
¼ On atom A ¼ On atom B½ On the AB bond
½ On atoms
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Neglecting S:1g√1sa+1sb) With S: 1g√S1sa+1sb)
No node, the whole space is in-phaseSymmetric with respect to h v C∞ C2 and I
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Neglecting S:1g√1sa+1sb) With S: 1g√S1sa+1sb)
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Neglecting S:1u√1sa-1sb) With S: 1u√S1sa-1sb)
Nodal plane
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Charge, Bond index: Without S
gIg 1sa+1sbI1sa+1sb1saI1sa1sbI1sb1saI1sb1sbI1sa√
1/2 On atom A
D = C2 = 1/2
Q = 1 – D = +1/2
1/2 On atom B
By symmetry Half on each atoms
= 0 = 0
Square of the coefficient, square of amplitude
L = 1/√2 1/√2 = 1/2
LAB = CACB
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Charge, Bond index: With S
gIg 1sa+1sbI1sa+1sb1saI1sa1sbI1sb1saI1sb1sbI1sa√S
1/2 On atom A
DA = CA 2 + CACB SAB = 1/2
Q = 1 – D = +1/2
1/2 On atom B
1/(2+2S) 1/(2+2S) S/(2+2S) S/(2+2S)
Half of the contributionFor bonds
L = 1/√2 1/√2 S = S/2
LAB = CACB SAB
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Energies Eg and Eu
From the number of nodal planes, it follows that g is below u
Eg=()/(1+S) Eu=()/(1-S)
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Eg and Eu, bonding and antibonding states
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Eg and Eu, bonding and antibonding states
2s
2s 2s
1s1s
Rydberg States
Valence states
The bonding and antibonding levels are referred to “dissociation”
Not to the “free electron” ; an antibonding level could be higher than a bonding one if referred to a higher reference level.
’
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From the number of nodal planes, it follows that g is below u
The atomic energy level
Remember !
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This term represents the difference between Hmol and Hat.• Either the electron is close to A:
R and rb are nearly the same and [1/R - 1/rb] is small• Or the electron is far from A and 1sa
2 is small
<1saIHmolecularI1sa> ~ <1saIHatomicI1sa>
or Haa, the atomic level
-13.6 eV for H
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This is a natural reference for a bond formation.For a system involving similar AOs, = 0
This is not the usual reference (free electron)For conjugated systems of unsaturated hydrocarbon It is the Atomic energy of a 2p orbital
or Haa, the atomic level
-11.4 eV for C (2p level)
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or Hab,the bond interaction From the number of nodal planes, it follows that g is below u
represents the interaction energy between A and B
2 represents half of the energy gap (Eg - Eu )
is the resonance integral (~ -3 eV) negative
It should be roughly proportional to the overlap
Eg = ()/(1+S) Eu = ()/(1-S) Eg - Eu = (1-S)/(1-S2) - ()(1+S)/(1-S2) Eg - Eu = (- 2S) /(1-S2) ~
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This is a natural unit for a bond formation.For a system involving similar bonds, is the unit
We define the unit including the negative sign.For conjugated systems of unsaturated hydrocarbon It represents half of a C=C bond (2 electrons gain the energy of the splitting)
or Hab, the value of the splitting
A C=C bond is 2
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Eg and Eu, with S
S
S
The gap is ~2; the average EM value is close to above it.
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The average EM value is above
Emean = (Eg+Eu)/2= [()/(1+S) +()/(1-S)]/2Emean = [ ()(1-S)/(1-S2) +()(1+S)/(1-S2)]/2
Emean =(S)/(1-S2)
S > 0 The mean value corresponds to <0 >0 a destabilization (energy loss)
The antibonding level is more antibondingthan the bonding level is bonding!
Small
Emoyen
1+S
1-S
Emean
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g
udistanceinternucléaire
EnergieEnergy
A-B distance
The bonding level is stable for the equilibrium distance
Electron in the antibonding level should lead to dissociation
At small d, e2/R dominates
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The molecule with several electrons
The orbitalar approximation: Molecular configurations.
H2
1g2Ground state
1u2
1g1u
2g2
Excited states
Rydberg states
Three rules: Pauli, Stability and Hund diagram of states
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The molecule with several electrons
The orbitalar approximation: Molecular configurations.
H2
1g2Ground state
1u2
1g1u
2g2
Excited states
Rydberg states
Three rules: Pauli, Stability and Hund diagram of states
2s1s 2s2s
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Diagram of orbitals
S
S
AO
right
AO
left
MO center
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Diagram of orbitals
2e : best situation
1e
2e
3e
4e 0 (-4S)
# e energy gain
Positive (4e - repulsion)
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Etatdiexcité
Etat excitésinguletou triplet
Etat Fondamental
*
*
*
udiexcited state S=0 E=
First excited states: gu ↓ ± ↓ ; and ↓↓ E=
One is alone =singlet state S=0 E=S
3 are degenerate = triplet spin S=1
gGround state S=0 E=
orbitals:
Diagram of States:
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Mulliken charge, Bond index: ground state
gIg 1sa+1sbI1sa+1sb21saI1sa21sbI1sb21saI1sb21sb
I1sa
1 On atom A
DA = ii CA 2 +i i CACB SAB =
1
Q = 1 – D = 0
1 On atom B
Half of the contributionFor bonds
L = 2 1/√2 1/√2 S = 1 S
L = 2 1/√2 1/√2 = 1
LAB = i i CACB SAB
i : occupancy of orbital i
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Mulliken charge, Overlap population: excited states
DA = ii CA 2 +i i CACB SAB =
1
Q = 1 – D = 0
First Excited States
OP = 2 1/√2 (-1)/√2 S = -1 S
OPAB = i i CACB SAB
i : orbital i occupancy
diexcited State
DA = ii CA 2 +i i CACB SAB =
1
Q = 1 – D = 0
OP = 1 1/√2 1/√2 S + 1 1/√2 (-1)/√2 S = 0
OPAB = i i CACB SAB
i : orbital i occupancy
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Rydberg states, from 2s and 2p
To find M.O.s First construct Symmetry orbitals
Each atom A or B does not have the molecular symmetry,
It is necessary to pair atomic orbitals between symmetry related atoms.
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Rydberg states
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The drawings or the symmetry labels are unambiguous
Mathematic expression is Ambiguous; it requires defining S
+ for positive S, good for pedagogy- for similar direction on the z axis, better for generalizationbetter for computerization.
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Sigma overlap
S-S S-d
p-ds-p
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u g
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u g
overlap is lateral; it concerns p or d orbitals that have a nodal plane
p-d in-phased-d in-phase
p-d out-of-phased-d out-of-phase
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Bonding and antibonding d-d orbitals
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overlap for d orbitals
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Rydberg states
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Rydberg states
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There is no interaction no overlap no mixing between orbital of different symmetry
S = 0Les recouvrementsse compensent deux à deux
S > 0
z
2px
2pz
s is symmetric relative to z
p is antisymmetric relative to z
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and separation
Linear molecule: symmetry relative to C∞ : SYM and
Planar molecule: symmetry relative to : SYM and
orbitals in linear molecule: 2 sets of degenerate Eg and Eu orbitals.
Degenerate for H, not for C∞ not for V ; appropriate combination
shows symmetry.
WARNING! Do not confuse and orbitals and and overlaps.
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orbitals in linear molecule: 2 sets of degenerate Eg and Eu orbitals.
*
Bonding Antibonding
Rea
l
C
ompl
ex
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Euler transformationComplex real
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Orbitale antiliante
g
Orbitale liante
u
Bonding Antibonding
orbitals.
The overlap (the resonance integral) is weaker than the one.
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Bonding Antibonding
orbitals: Lateral overlap.
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orbitals.
2S and 2PZ mix
3u
3g
2g
antiliant
niveaux non liants
liant
Symétrie u
Symétrie g
Orbitales Moléculaires Orbitales de Symétrie NIVEAUX 2S-2P
2u
non bonding
Bonding
Antibonding
M. O. Symmetry Orbitals g are bondings-p hybrization u are antibonding
Symmetry g
Symmetry u
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hybridization
Mixing 2s and 2p: requires degeneracy to maintain eigenfunctions of AOs.
Otherwise, the hybrid orbital is an average value for the atom, not an exact solution.
This makes sense when ligands impose directionality: guess of the mixing occurring in OMs.
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Hybrid orbitals on A
non bonding
Bonding
Antibonding
M. O.
Hybrid orbitals on B
The non bonding hybrids Can be symmetryzed
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Method to build M.O.s
• Determine the symmetry elements of the molecule
• Make the list of the functions involved (valence atomic orbitals)
• Classify them according to symmetry (build symmetry orbitals if necessary by mixing in a combination the set of orbitals related by symmetry)
• Combine orbitals of the same symmetry (whose overlap is significant and whose energy levels differ by less than 10 eV).
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LCAO
a
This is a unitary transformation;
n AO → n MO
MO AO
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Combination of 2 AOs of same symmetry
They mix to generate a bonding combination and an antibonding one.
The bonding orbital
is the in-phase combination
looks more like the orbital of lowest energy
(larger coefficient of mixing)
has an energy lower than this orbital
The antibonding orbital is the out-of-phase combination looks more to the orbital of highest energy (larger coefficient of mixing) has an energy higher than this orbital
A B
A B
B
A
Niveau liant
Niveau AntiliantAntibonding
Bonding
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Combination of 3 AOs of same symmetry
One bonding combination, one non-bonding and one antibonding.
Either 2 bonding and 2 antibonding, or 1 bonding, 2 non-bonding and an antibonding
In general,
Combination of 4 AOs of same symmetry
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Populating MOs
1. Fill the in increasing order, respecting the Pauli principle.
2. Do not consider where the electron originate ! This is a different problem « correlating » the « initial distribution » to the final one. To determine the ground state just respect rule 1!
2 CH2 → H2C=CH2 may be a fragment analysis to build ethene in the ground state, not an easy reaction leading directly to the ground state! C-C
p
C-C
p
CH2 H2C=CH2 CH 2
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A-A Homonuclear diatomic molecules
Generalization of the LCAO approach:Build the symmetry orbitals and classify them by symmetryIf E2s-E2p < 10 eV combine orbitals of same symmetry
If E2s-E2p > 10 eV do not
0
Energy
Z
NLi-C O-F
2p
2s
10 eV
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For Homonuclear diatomics
E2s(A) = E2s(B) E2p(A) = E2p(B)
Making symmetry orbitals, we combine symmetry related orbital first!
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57Orbitales : les niveaux 2 u et 3u se combinent
si les niveaux 2s et 2p sont proches en énergie
3u
3g
2u
2g
Symmetry orbitals type
Hybridization: 2sg and 3sg may mix; 2su and 3su may mix
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Symmetry orbitals 2
Place relative des Orbitales 3 g et des Orbitales u:
L'importance du relèvement de 3 g décroît
avec l'électronégativité de l'atome
3u
3g
1u
1g
Due to hybridization, 3g goes upThe relative order of E3g and E1u may change
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Cas du lithium à L'azote. 3g en dessous de 1u.
Li-N: 3g above 1u
.
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Lithium: Li2 2 valence electrons : one occupied MOConfiguration: (core)2g
2. Li-Li single bond . Beryllium: Be2 4 valence electrons configuration : Configuration: ((core)2g
22u2. 2 occupied MOs
no bond (excepting weak polarization). Boron: B2 6 valence electrons configuration: (core)2g
22u22u2’u.
2 occupied MOs + 2 unpaired electrons (Hund’s rule). B2 is paramagnetic. Bonding equivalent to a single bondCarbon: C2 8 valence electrons configuration: (core)2g
22u22u
22’u2. 4 occupied MOs 3 bonding, one antibondingStrong bonding :C=C: 2 bonds.
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Nitrogen N2 10 valence electrons : 5 occupied MOConfiguration: (core)2g
22u22u
22’u23g2.
1 bond and 2 bonds: This is the most stable case with the maximum of bonding electrons.It corresponds to the shortest distance and to the largest dissociation energy. It is very poorly reactive, inert most of the time (representing 80% of atmosphere). 3g close but above 1u . A N-N elongation weakens the bonding and the hybridization; 3 3g passes below 1u
.
N N
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Cas de l'oxygène et du fluor. 3g en dessous de 1u.
O-F: 3g below 1u
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Oxygen O2 10 valence electrons : 5 occupied MO plus 2 unpaired electrons Configuration:1 s bond and 1 p bond (2 halves). paramagnetic.
Fluor F2 12 valence electrons : 7 occupied MO (core)2g
22u22u
22’u23g23g
23’g2 A single bond Neon Ne2. No bond
.
O O° °
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Li-H
Li H
Li H
1sH
2sLi
H
Li
Niveau liant
Niveau Antiliant
-5.4 eV
-13.6 eV
Li Li-H H
Antibonding
Bonding
= 0.9506 (2sLi) - 0.3105(1sH). = 0.3105 (2sLi) +0.9506 (1sH). dH= 1.807 QH= -.807 dLi= 0.193 QLi= +.807 Li-H is 80.7% ionic, 19.7% covalent. There is a dipole momentLi+-H-
.
Large coefficient on Li in antibonding
Large coefficient on H in bonding
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HF
FH-FH
2s
2p
1s
*
Only one s bond
Large coefficient on 2pZ(F)
DipoleH+–F-
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CO
-11.4 eV
-21.4 eV
-14.8 eV
C CO O
2p
2s
2p
E(eV)
C OCO
N2CO
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orbitals of CO
• Antibonding: 2/3 2pZ(C) -1/3 2pZ(O)
• Non bonding: 1/3 2pZ(C)-1/3 2pZ(C)+1/3 2pZ(O)
It accounts for the electron pair on C
• Bonding: 2/3 2s(C) +1/3 2pZ(O)
+ =C O C O
+ =C O C O
+ =C O C O