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MO Diagrams for More Complex Molecules Chapter 5 Wednesday, October 22, 2014
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### Transcript of MO Diagrams for More Complex Molecules Chapter 5 Wednesday, October 22, 2014.

MO Diagrams for More Complex Molecules

Chapter 5

Wednesday, October 22, 2014

Boron trifluoride1. Point group D3h

2.

3. Make reducible reps for outer atoms

Γ2s 3 0 1 3 0 1

Γ2pz 3 0 -1 -3 0 1

Γ2px 3 0 -1 3 0 -1

Γ2py 3 0 1 3 0 1

4. Get group orbital symmetries by reducing each Γ

Γ2s = A1’ + E’

Γ2pz = A2’’ + E’’

Γ2px = A2’ + E’

Γ2py = A1’ + E’

xy

z

xy z

x yz

Boron trifluoride

Γ2s = A1’ + E’

Γ2pz = A2’’ + E’’

Γ2px = A2’ + E’

Γ2py = A1’ + E’

What is the shape of the group orbitals?

2s:

A1’ E’(y)

? ?

E’(x)

The projection operator method provides a systematic way to find how the AOs should be combined to give the right group orbitals (SALCs).

Which combinations of the three AOs are

correct?

BF3 - Projection Operator Method

A1’ = Fa + Fb + Fc + Fa + Fc + Fb + Fa + Fb + Fc +

Fa + Fc + Fb

Fa

In the projection operator method, we pick one AO in each set of identical AOs and determine how it transforms under each symmetry operation of the point group.

AO E C3 C32 C2(a) C2(b) C2(c)

σh S3 S32 σv(a) σv(b) σv(c)

Fa Fa Fb Fc Fa Fc Fb Fa Fb Fc Fa Fc Fb

Fb Fc

The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results.

A1’ 1 1 1 1 1 1 1 1 1 1 1 1

A1’ = 4Fa + 4Fb + 4Fc

A1’

BF3 - Projection Operator Method

A2’ = Fa + Fb + Fc – Fa – Fc – Fb + Fa + Fb + Fc – Fa –

Fc – Fb

Fa

In the projection operator method, we pick one AO in each set of identical AOs and determine how it transforms under each symmetry operation of the point group.

AO E C3 C32 C2(a) C2(b) C2(c)

σh S3 S32 σv(a) σv(b) σv(c)

Fa Fa Fb Fc Fa Fc Fb Fa Fb Fc Fa Fc Fb

Fb Fc

The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results.

A2’ 1 1 1 -1 -1 -1 1 1 1 -1 -1 -1

A2’ = 0

There is no A2’ group orbital!

BF3 - Projection Operator Method

E’ = 2Fa – Fb – Fc + 0 + 0 + 0 + 2Fa – Fb – Fc + 0 + 0 + 0

Fa

In the projection operator method, we pick one AO in each set of identical AOs and determine how it transforms under each symmetry operation of the point group.

AO E C3 C32 C2(a) C2(b) C2(c)

σh S3 S32 σv(a) σv(b) σv(c)

Fa Fa Fb Fc Fa Fc Fb Fa Fb Fc Fa Fc Fb

Fb Fc

The group orbital wavefunctions are determined by multiplying the projection table values by the characters of each irreducible representation and summing the results.

E’ 2 -1 -1 0 0 0 2 -1 -1 0 0 0

E’ = 4Fa – 2Fb – 2Fc

E’(y)

Γ2s = A1’ + E’

Γ2pz = A2’’ + E’’

Γ2px = A2’ + E’

Γ2py = A1’ + E’

What is the shape of the group orbitals?

2s:

A1’ E’(y)

? ?

E’(x)

We can get the third group orbital, E’(x), by using normalization.

BF3 - Projection Operator Method

∫𝜓 2𝑑𝜏=1 Normalization condition

BF3 - Projection Operator Method

∫𝜓 2𝑑𝜏=1𝜓 𝐴1

′=𝑐a ¿¿

𝑐a2∫ ¿¿¿ nine terms, but the six

overlap (S) terms are zero.

A1’ wavefunction

Normalization condition for group orbitals

𝑐a2 ¿

𝑐a2 [1+1+1 ]=1 𝑐a=

1

√3

𝜓 𝐴1′=1

√3¿¿So the normalized A1’ GO is:

Let’s normalize the A1’ group orbital:

BF3 - Projection Operator Method

𝜓𝐸 ′( 𝑦)=𝑐a ¿¿

𝑐a2∫ ¿¿¿ nine terms, but the six

overlap (S) terms are zero.

E’(y) wavefunction

𝑐a2 ¿

𝑐a2 [4+1+1 ]=1 𝑐a=

1

√6

𝜓𝐸 ′( 𝑦)=1

√6¿¿So the normalized E’(y) GO is:

Now let’s normalize the E’(y) group orbital:

BF3 - Projection Operator Method

𝜓𝐸 ′( 𝑦)=1

√6¿¿

𝜓 𝐴1′=1

√3¿¿

the probability of finding an electron in in a group orbital, so for a normalized group orbital.

𝜓𝐸 ′(𝑥)=1

√2¿So the normalized E’(x) GO is:

Γ2s = A1’ + E’

Γ2pz = A2’’ + E’’

Γ2px = A2’ + E’

Γ2py = A1’ + E’

What is the shape of the group orbitals?

2s:

A1’ E’(y)

?

E’(x)

Now we have the symmetries and wavefunctions of the 2s GOs.

BF3 - Projection Operator Method

We could do the same analysis to get the GOs for the px, py, and pz orbitals (see next slide).

notice the GOs are orthogonal

(S = 0).

2s:

A1’ E’(y) E’(x)

BF3 - Projection Operator Method

2py:

A1’ E’(y) E’(x)

2px:

A2’ E’(y) E’(x)

2pz:A2’’ E’’(y) E’’(x)

boron orbitals

A1’

A2’’

E’

E’

2s:

A1’ E’(y) E’(x)

BF3 - Projection Operator Method

2py:

A1’ E’(y) E’(x)

2px:

A2’ E’(y) E’(x)

2pz:A2’’ E’’(y) E’’(x)

boron orbitals

A1’

A2’’

E’

E’

little overla

p

F 2s is very deep in energy and won’t interact with boron.

H

He

Li

Be

B

C

N

O

F

Ne

BC

NO

F

Ne

NaMg

Al

SiP

SCl

Ar

AlSi

PS

ClAr

1s2s

2p 3s3p

–18.6 eV

–40.2 eV

–14.0 eV

–8.3 eV

Boron trifluoride

π*

Boron Trifluoride

π

nb

nb

σ

σ

σ*σ*

A1′

E′

A2″

A2″

A1′

A2″

E′

Energ

y

–8.3 eV

–14.0 eV

A1′ + E′

A1′ + E′

A2″ + E″

A2′ + E′

–18.6 eV

–40.2 eV

d orbitals

• l = 2, so there are 2l + 1 = 5 d-orbitals per shell, enough room for 10 electrons.• This is why there are 10 elements in each row of the d-block.

σ-MOs for Octahedral Complexes1. Point group Oh

2.

The six ligands can interact with the metal in a sigma or pi fashion. Let’s consider only sigma interactions for now.

sigmapi

2.

3. Make reducible reps for sigma bond vectors

σ-MOs for Octahedral Complexes

4. This reduces to:Γσ = A1g + Eg + T1u

six GOs in total

σ-MOs for Octahedral Complexes

Γσ = A1g + Eg + T1u

Reading off the character table, we see that the group orbitals match the metal s orbital (A1g), the metal p orbitals (T1u), and the dz2 and dx2-y2 metal d orbitals (Eg). We expect bonding/antibonding combinations.

The remaining three metal d orbitals are T2g and σ-nonbonding.

5. Find symmetry matches with central atom.

σ-MOs for Octahedral Complexes

We can use the projection operator method to deduce the shape of the ligand group orbitals, but let’s skip to the results:

L6 SALC symmetry label

σ1 + σ2 + σ3 + σ4 + σ5 + σ6 A1g (non-degenerate)

σ1 - σ3 , σ2 - σ4 , σ5 - σ6 T1u (triply degenerate)

σ1 - σ2 + σ3 - σ4 , 2σ6 + 2σ5 - σ1 - σ2 - σ3 - σ4 Eg (doubly degenerate)

12

34

5

6

σ-MOs for Octahedral Complexes

There is no combination of ligand σ orbitals with the symmetry of the metal T2g orbitals, so these do not participate in σ bonding.

L

+

T2g orbitals cannot form sigma bonds with the L6 set.

S = 0.T2g are non-bonding

σ-MOs for Octahedral Complexes

6. Here is the general MO diagram for σ bonding in Oh complexes:

SummaryMO Theory

• MO diagrams can be built from group orbitals and central atom orbitals by considering orbital symmetries and energies.

• The symmetry of group orbitals is determined by reducing a reducible representation of the orbitals in question. This approach is used only when the group orbitals are not obvious by inspection.

• The wavefunctions of properly-formed group orbitals can be deduced using the projection operator method.

• We showed the following examples: homonuclear diatomics, HF, CO, H3

+, FHF-, CO2, H2O, BF3, and σ-ML6