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8. APPLICATIONS TO MOLECULAR ORBITAL THEORY

L TH S NHCHEMISTRY DEPARTMENT

NATURAL SCIENCES UNIVERSITY - HCM2010

CHEMICAL USES OF SYMMETRY AND GROUP THEORY

1

Introduction MO theory: study electronic structure,

bonding, and energy in molecules

There are many ways to treat the electronicstructure of molecules and their energy

In this chapter: emphasize the symmetryaspect of the MO theory

Symmetry dont provide the relative energiesof these AO, MO

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LCAO Molecular Method The simplest MO theory: molecular wave functions are

written as linear combinations of atomic wave functions(LCAO:Linear Combinations of Atomic Orbitals)

where: f i: atomic wave functions

Ci: coefficients

Note:

- Variation theorem: energy guessed from any Schrdingerfunction ( Y ) will be higher than the true energy

- Molecular electronic wave functions are not really linearcombinations of atomic electronic wave functions

LCAO Molecular Method The simplest MO theory: molecular wave functions

are written as linear combinations of atomic wavefunctions (LCAO: Linear Combinations of AtomicOrbitals)

Overlap of AOs MOs

Note:- Number of generated molecular orbitals equalsnumber of initial atomic orbitals

- The molecular orbitals must have the samesymmetries as the composed atomic orbitals

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Summary of condition for AO overlap

AOs involve in the overlap must have:

Similar energy (when big difference in AO energiesinsignificant bonding)

Same symmetry about the nuclear axis

AOs must be closed enough to provide good overlap of theorbitals

If: - AOs are same sign in the overlap region bonding

- AOs are different sign in the overlap region antibonding

MO for the H 2 molecule: D h group Two 1s orbitals of two H atoms involve in LCAO

Symmetry of these two 1s orbitals:

Symmetry of the two MO orbitals:

overlap of two 1s orbitals two MOs with s g+ and s u+ symmetry

D h E 2C s

v i 2S C2

G(1s) 2 2 2 0 0 0 = s g+ + s u+

D h E 2C s

v i 2S C2

Y b 1 1 1 1 1 1 = s g+

Y * 1 1 1 -1 -1 -1 = s u+

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Overlaps and energies of MOsof the H 2 molecule

E

AO overlaps in the H 2 molecule

)2( 2222 AB B Ab N YYYY

Y b = N ( Y A + Y B) Y* = N ( Y A Y B)

)2( 2222* AB B A N YYYY

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Experimental Observations:Photo e lectron Spectroscopy (PES)

Basis of the method: tobornbard an atomic ormolecular species withradiation of sufficientenergy to cause itsionization

M(g) + h n M+(g) + e(Photoelectron)

the energy balance being:

hn = IM + DEvib + DErot + Eke

Measure E ke raw PES

The PES of the H 2 molecule

The peak on the far right hand side of thespectrum ionization energy of the H 2moleculeOther additional peaks: due to vibration androtation of the molecule

Jack Barrett, Structure and Bonding, The Royal Society of Chemistry 2001

s Overlaps

s Overlap: (similar symmetry to the s orbital)- cylindrical symmetry to rotation about the nuclear axis- non-change sign when rotating about the nuclear axis- s bonding: no nodal plane

z is the nuclear axis

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p Overlaps

p Overlap: (similar symmetry to the p orbital)

- change sign which C 2 rotation about the nuclear axis

- AOs are perpendicular to the nuclear axis

- p bonding: 1 nodal plane containing the nuclear axis

d Overlaps

Overlap: (similar symmetry to the d orbital)

- change sign which C 4 rotation about the nuclear axis

- AOs plane are perpendicular to the nuclear axis

- d bonding: 2 nodal planes containing the nuclear axis

Note: antibonding always has an extra nodal plane perpendicular tothe nuclear axis

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Energies of 2s and 2p AOof the second row

- B, C, N: small 2s-2p energy difference large 2s-2p interaction

- O, F: big 2s-2p energy difference small 2s-2p interaction

Energy levels of Homonuclear Diatomic Molecules

of the Second Period

s

p

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N2 molecule: MO and PES spectrum

Figure 7.30 The PESs for the

diatomics N 2, O 2 and F 2.

Adapted with permission from

Brundle C.R. and Baker A.D.

(1977) Electronic Spectroscopy:

Theory, Techniques and

Applications, vol. 1, Academic

Press (ISBN 0-12-137801-2).

Copyright Elsevier.

D.J. Willock, Molecular Symmetry (Electronic version)

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Energy difference of ns-np of the elements Li-Kr


MO: The Ligand Group Orbital approach (LGO) For polyatomic molecules with a central atom and

other outer atoms (ligands): H 2O, NH3, SF6,[Co(NH3)6]2+

Approach: molecular orbitals are constructed from:(1) atomic orbitals of the central atom, and (2)symmetry a dapted c ombinations of ligand orbitals

(SAC or Ligand group orbitals , LGOs). LGOs are formed by linear combination of the ligand

AOs:

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LGOs

MO: H 2O molecule - C 2v group

H2O: two s bonds only Step 1: Find the

symmetry of the two sbonds:

Reducible representation of thetwo s bonds of H 2O

Gs

s v

s v

0 2

Gs = A1 + B2

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Step 2: Find AOs of oxygen to match with the symmetryof the s bonds

From the character table:

A1: Os AO 2s, 2p z or a 1 orbitals

B2: 2p

yb

2orbital

Note: the 2p x of the oxygen does not involve in the constructionof the two s bonds.

Step 3: Construct the LGOs to match with the symmetryof the s bonds

Linear combination of the two1s of two H atoms 2 LGOswith the A 1 and B 2 symmetry

LGO with A 1 symmetry:LGO

1= 1/2 ( f

1+ f

2)

LGO with B 2 symmetry:LGO2= 1/2 ( f 1 - f 2)+

+

++ +

_

_

_

J. Barrett, Structure and Bonding (electronic version)

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Pictorial overlap of a 1 orbitals

a 1 orbital ofOxygen

LGO 1 orbital ofH (A1 symmetry)


Pictorial overlap of b2

orbitals

b1 orbital ofOxygen

LGO 2of H

B 2

B2


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Step 4:

The MOdiagram forH2O

H2O


MO diagram for H 2O

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PES of H2O

MO for NH 3: C3v group

Symmetry of the three sbonds:

Irreducible representationof the three s bonds:

Gs =

C3v E 2C3 3s vGs

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The valence AOs of N involvein the construction of MOs:

From the character table:

A1: Ns AO 2s, 2p

zor a

1orbitals

E: 2p x, 2p y e orbitals

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Construct the LGOs of H:

Take symmetry operations of s1

orbital of H1

in theC3v group:

Dropping the repeating

C3v E C3 C32 s v s v s v

s1 s1 s2 s3 s1 s2 s3

C3 E C3 C32

s1 s1 s2 s3

Construct the LGOs of H: Projecting out the a 1 and e SALCs of the s 1 (H1 atom):

LGOs of Hs orbitals:LGO1 = 1/3 (s 1 + s2 + s3)

LGO2 = 1/ 6 (2s 1 s2 s3)

LGO 3 = 1/ 2 (s 2 s3)

C3 E C3 C32

A1T js1 s1 s2 s3 (s1 + s2 + s3)

ET js1 2s 1 s2 s3 (2s 1 s2 s3)

ET j(s2 s3) 2(s 2 s3) (s3 s1) (s1 s2) 2(s 2 s3)

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Figure 7.22The MOdiagram for theC 3v moleculeNH3. For clarity,the dotted lines

for the a 1orbitals aredrawn in adifferent style tothose for the

others.


MO diagram for NH 3

CH4: Td group

Symmetry of the four s bonds:

Gs

Irreducible representation of the

four s bonds:

Gs =

The orbitals of C with suitable

symmetry:

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Symmetry Adapted Linear Combinations of H orbitals(LGOs)

A1 symmetry T 2 symmetry

LGO of H:

(a 1 ) = s 1 + s 2 + s 3 + s 4

1(t) = s 1 s 2 + s 3 s 4 2 (t) = s 1 s 2 s 3 + s 4

3(t) = s 1 + s 2 s 3 s 4

13

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Pictorial overlap of the orbitals

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MO diagram for CH 4

Bond order = 4

For C H bond: 1

Figure 7.20: The MO diagram for CH 4 in the T d point group.


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PES of the valence orbital region of CH 4

Figure 7.21: The PES of the valence orbital region of CH4. The originalspectra were reported in two sections (from 12 to 16.3 eV and from 21.9to 24.2 eV ) and are bought together here on a single energy scale. Thevertical axis is photoelectron intensity in arbitrary units. Adapted fromPotts A.W., Price W.C. (1972) Proceedings of the Royal Society of London,Series A 326: 165 179. With permission from the Royal Society.


Hypervalent molecules

Hypervalency: molecules (PF 5, SF6 and XeF 4) which

the central main group element appears to expand

octet rule

Hypervalency has remained a controversial subject.

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Hypervalent molecule: SF 6

Symmetry of the six s bonds: O h group

Irreducible representation of the xis sbonds:

Gs = GF =

The orbitals of C with suitable symmetry:

Character Table of the O h group

Note:

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Central atom (S) LGOs of outer atoms

a 1g

t1u

Symmetry Adapted Linear Combinations ofs orbitals of F (LGOs)

Symmetry Adapted Linear Combinations ofthe s orbitals of F (LGOs)

e g orbitals

(non-bonding)

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MO diagram for SF 6

Electron configuration of SF 6: (1a 1g )2 (1t 1u )6 (1e g)4

The bond order: bond order = (8 0) = 4

or 4/6 = 2/3 per S-FD.J. Willock, Molecular Symmetry (Electronic version)

Transition metal complexes:-bonding in octahedral ML 6 complexes

LGOs of s bond of ligands:

AO of central atom:

a 1g - s

e g - (d z2, d x2-y2 )

t 1u - (p x, p y, p z)

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MO diagram for ML 6 complexes (s bonds only)

-bonding in octahedral complexesTwo kinds of p bonding incomplexes:

donors (halides, OH - and O 2-ligands): occupied AOs, usuallylone pairs, lower energy than themetal orbitals

acceptors (such as CO, CN -,C2H4): empty AOs, higher energythan the metal orbitals

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Construct SALCs of the ligand -orbitals

Symmetry of the p bonding:Gp = T1g + T2g + T1u + T2u

(Can be calculated from: Gp = Gbond - Gs )

AOs of metal can adapt the symmetry:

T2g AOs of M: (d xy, d xz, d yz): t 2g

T1u - (p x, p y, p z): t 1u

LGOs: t 2g and t 1u : form p bondst 1g and t 2u : from p non-bonds

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Examplesof the poverlaps

MO diagram for ML 6 complexes: p donor

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MO diagram for ML 6 complexes: p acceptor

The frontier orbitals of Benzene

The highest energy atomic orbitals of benzene are the porbitals of benzene ring

Assume: the frontier MOs are composed of LCAO of theC 2pp orbitals in D 6h group:

Symmetry of the six C 2p p orbitals:

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D6h character Table

Construct the LCAOs by projecting outthe appropriate linear combination

D6h E C6 C3 C2 C32 C65

A2u T jf 1 f 1 f 2 f 3 f 4 f 5 f 6B2gT jf 1 f 1 f 2 f 3 f 4 f 5 f 6E1gT jf 1 2f 1 f 2 f 3 2f 4 f 5 f 6E1gT j(f 2 + f 3) 2f 2 + 2f 3 f 3 + f 4 f 4 f 5 2f 5 f 6 f 6 f 1 f 1 + f 2E2u T jf 1 2f 1 f 2 f 3 2f 4 f 5 f 6E2u T j(f 2 f 3) 2f 2 2f 3 f 3 + f 4 f 4 + f 5 2f 5 2f 6 f 6 + f 1 f 1 + f 2

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Construct the LCAOs by projecting outthe appropriate linear combination

D6h LCAO

A2u T jf 1 f 1 +f 2 + f 3 + f 4 + f 5 + f 6B2gT jf 1 f 1 f 2 + f 3 f 4 + f 5 f 6E1gT jf 1 2f 1 + f 2 f 3 2f 4 f 5 + f 6E1gT j(f 2 + f 3) 3( f 2 + f 3 f 5 f 6)

E2u T jf 1 2f 1 f 2 f 3 + 2f 4 f 5 f 6E2u T j(f 2 f 3) 3( f 2 f 3 + f 5 f 6)

Normalized LCAOs

Y A2u 1/6 (f 1 +f 2 + f 3 + f 4 + f 5 + f 6) Y B2g = 1/6 (f 1 f 2 + f 3 f 4 + f 5 f 6)

Y E1ga = 1/12 (2f 1 + f 2 f 3 2f 4 f 5 + f 6) Y E1gb = 1/2 (f 2 + f 3 f 5 f 6)

Y E2u a = 1/12 (2f 1 f 2 f 3 + 2f 4 f 5 f 6) Y E2u b = 1/2 ( f 2 f 3 + f 5 f 6)

Pictorial presentation of LCAO

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The energy of the p MOs of benzenebased on the Hckel approximation

Energy diagram:

Etotal = 2(2 b ) + 4(b ) = 8 b (from quantum mechanics calculation)

The Ethene molecule: D 2h group

Figure 7.33: The axis system and basisused to obtain SALCs for the H(1s)orbital set in the D 2h molecule ethene.

Character Table of the D 2h group


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Ethene: Symmetry of the four H(1s)


Ethene: Symmetry of the four H(1s)

Problems: Use the projection operator to obtain thenormalized SALC H(1s) functions and show that your answerscorrespond to the orbital phase patterns above.


a g b1g b2u b3u

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Ethene: Symmetry of the C(2s,2p) orbitals

G(C(2s,2p)) = 2A g + B1g + B2g + B1u + B2u + 2B3uD.J. Willock, Molecular Symmetry (Electronic version)

Ethene: Symmetry of the C(2s,2p) orbitals

G(C(2s,2p)) = 2A g + B1g + B2g + B1u + B2u + 2B3u

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Figure 7.34: The MO diagram for the ethene molecule

This figure shows the C

C energy levels in the D 2h

point group and in the

absence of the four H

atoms.

In this hypothetical

situation, the 1b 1u and

1b 2u MO have the same

energy, as do the 1b 1g

and 1b 2g SALCs (similar

to the homonuclear

diatomics of C 2)D.J. Willock, Molecular Symmetry (Electronic version)


Figure 7.35 shows the energy scheme for the C C bond in the ethene molecule with

the sp-hybridization formed the matched symmetry 2s and 2p x orbitals of C

Figure 7.35: The MO diagram for the ethene moleculewith the s-p hibridization

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Figure 7.36 The MO diagram for the D 2hmolecule ethene, using hybrid orbitals.


Figure 7.36 brings together the C-C and

H(1s) SALCs to form the MO diagram ofethane. It is found that the C-C orbitalsderived from p y have partners in theH(1s) SALCs but those from p z orbitalsdo not. Hence, the degeneracy expectedfor the bonds in Dh is lost.

- In the MO diagram of Figure 7.36 the H orbitalsare placed higher in energy than the bondingSALCs of the C-C set since the electronegativity ofC is greater than H.- Now there are 12 AO functions, and so there willbe 12 MOs formed. However, to keep the diagramsimple, only the six occupied valence orbitals andtwo lowest unoccupied levels are shown.- The first five orbitals contain bonding C-C and C-Hinteractions, while the highest occupied level is thebonding C-C 1b 1u that is nonbonding for C-H.

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