4 ligand field theory (1)

Chemistry 3820 (Fall 2006)

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MO Diagrams for O2 and N2
Drawing MO diagrams - always same number of MOs as AOs.
- more electronegative element on the right - lower in energy.
-----------------------------------------------------------------------------------------------------------------Paramagnetic

σ*2p

O2

Bond order of 2 (O=O)
π* 2p
2p

2p

π2p

2 MOs like this perpendicular to each other
σ2p

larger ∆E

σ*2s
2s

2s

σ2s
O

O2

O

-----------------------------------------------------------------------------------------------------------------σ*2p

N2

Diamagnetic
Bond order of 3 (N= N)

π*2p
2p

2p

σ2p

Very strong NN bond
Unlike O2, the σ2p and π 2p orbitals are
switched in energy (see next page)

smaller ∆E

π2p

HOMO = σ2p, LUMO = π*2p

σ*2s

2s

2s

σ2s
N

N2

N

© 2006 - Dr. Paul G. Hayes – University of Lethbridge


Chemistry 3820 (Fall 2006)

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Ordering of MOs in O2 is σ2p , π 2p, π*2p , σ*2p but in N2 and CO, the order is π 2p, σ2p , π*2p , σ*2p
Reason - O2 - larger ∆E between s and p orbitals à less orbital mixing
- larger ∆E because across the period (B,C,N,O,F) more protons added to the
nucleus, which pulls the electrons in closer to the nucleus (lower in energy). This
effect is felt more strongly by the s-orbitals than the p-orbitals, so the energy of
the s-orbitals drops more rapidly than that of the p-orbitals.
- N2 and CO - mixing between the σ2p and the σ*2s orbitals raises the energy of the σ2p
above the π 2p.

Changes in the energy of the MOs of homonuclear diatomic molecules in the 2 nd period

MO Diagram for CO

© 2006 - Dr. Paul G. Hayes – University of Lethbridge


Chemistry 3820 (Fall 2006)

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σ*2p

LUMO

π*2p

2p

σ2p

2p

HOMO (slightly antibonding)

π2p
2s

σ* 2s

2s

σ2s
C

CO

O

- CO is isoelectronic with N2 (C has one less electron than N, O has one more) – MO diagram is
very similar to that of N2.
- Diamagnetic, Bond order of 3 (C= O)
- Important: HOMO = σ2p (weakly antibonding), LUMO = π*2p (strongly antibonding)
- Since O is far more electronegative than C, would expect a large dipole moment with δ- on O.
However, CO actually has only a small dipole moment (0.1 Debye) with the δ- on Carbon!
Usually electronegativities are a good indication of the direction and magnitude of the dipole on
a molecule, but especially in molecules where orbitals with antibonding character are occupied,
things are not so straightforward.

© 2006 - Dr. Paul G. Hayes – University of Lethbridge


Chemistry 3820 (Fall 2006)

Pictorial Representations of Possible Bonds

© 2006 - Dr. Paul G. Hayes – University of Lethbridge

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Chemistry 3820 (Fall 2006)

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MO theory ~ Ligand Field Theory
Constructing an MO diagram for an octahedral complex
Symmetry Adapted combinations of Ligand σ-orbitals in an Octahedral Complex

- any combination
between t2g and
ligands becomes
non-bonding

-

In an octahedral environment, the metal orbitals (3d, 4s, 4p for a 1st row TM) divide by
symmetry into 4 sets: s = a1g , p = t1u , axial d = eg , inter-axial d = t 2g.

-

The orbitals of the six ligands can be combined to give six symmetry-adapted linear
combinations which are of the correct symmetry to interact with the s, 3 x p and 2 x axial-d
orbitals, but not the inter-axial d orbitals.

-

The result is that 3 orbitals (the inter-axial d-orbitals) are non-bonding, while the rest (6 metal
orbitals and 6 ligand orbitals) combine to form six bonding and six anti-bonding MOs. See
next page.

-

This is a much more correct approach than crystal field theory, but is not as easy to use.

© 2006 - Dr. Paul G. Hayes – University of Lethbridge


Chemistry 3820 (Fall 2006)

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MO energy levels for an octahedral complex
(only σ-bonding considered)
t1u*

4p (t1u)

a1g*

4s (a1g)
eg*
∆O
3d (eg + t2g)
t2g

(a 1g + t1u + e g)
eg

TM

6L
t1u
a1g

•
•

The six bonding orbitals are filled with 12 electrons from the six ligands
Orbitals shown in red (t2g and eg *) are the frontier orbitals where d-electrons reside
(which is why TM MOs are often simplified to show only t 2g and eg * orbitals).
© 2006 - Dr. Paul G. Hayes – University of Lethbridge


Chemistry 3820 (Fall 2006)

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π-bonding ligands
(the above MO diagram does not take into account π-bonding)
L

M

L

M

L

M

empty orbital

filled orbital
σ-donor
M

NH3 , CH3 -, H -

•
•

M

L

L

π-donor

π-acceptor

Cl-, OH -, NR 2- , OH 2

CO, NO+, CN -

For π-acceptor ligands, the bonding is SYNERGIC: σ-donation to the metal strengthens
π-backbonding to the ligand, and π-donation from the metal to the ligand strengthens the
σ-donor component of bonding.
This is because σ-donation leads to increased electron density on the metal, which allows

increased π-backdonation. Conversely, π-backdonation reduces the amount of electron
density on the metal, which allows more σ-donation from the ligand to the metal.

Cr(CO)6: Octahedral complex with good π-acceptor ligands
σ-interaction
C

M
lobe of
acceptor orbital
(whole orbital
not shown)

π-interaction
O
M

C

O

M

C

O

slightly antibonding
HOMO of CO

axial d-orbitals

strongly antibonding
π* orbitals of CO

© 2006 - Dr. Paul G. Hayes – University of Lethbridge


Chemistry 3820 (Fall 2006)

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MO energy levels for an octahedral complex
with π-acceptor ligands (e.g. [Cr(CO)6])
t1u*

4p (t1u)

a1g*

4s (a1g)

t2g*

π* orbitals of CO (t2g)

eg*

large
∆O

3d (eg + t2g)

t2g
(a1g + t1u + eg)
eg

Cr

6 CO
t1u
a 1g

•

π-backdonation to CO from the t2g orbitals (which are non-bonding in the absence of
π-interactions between the metal and the ligands).

•

The 3 t2g orbitals and 3 high lying π* orbitals of the CO ligands form 3 bonding MOs and
3 antibonding MOs.

•

Since the CO π* orbitals are empty, the d-electrons occupy the bonding MO from this
interaction.

•

The result is (1) a very large ∆o , so the e g* orbital is likely to remain empty.
(2) the t2g orbital is strongly bonding (wants to be filled with 6 electrons)
à complexes of strong π-acceptor ligands are the most likely to obey the 18 electron rule
© 2006 - Dr. Paul G. Hayes – University of Lethbridge


Chemistry 3820 (Fall 2006)

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Symmetry Adapted Linear Combinations of π* orbitals in ML6 complexes
Q. Why are there only three ligand π-acceptor orbitals shown in the MO diagram for CO when
there are 6 ligands, each with two empty π* orbitals ?
A. The 12 π* orbitals of the ligands can be combined to form 12 symmetry adapted linear
combinations of atomic orbitals (3 x T1u , 3 x T2g , 3 x T1g and 3 x T2u). Only the three T2g linear
combinations are of the correct symmetry to interact with the t2g orbitals (dxy , dxz, dyz) on the
metal.

© 2006 - Dr. Paul G. Hayes – University of Lethbridge


Chemistry 3820 (Fall 2006)

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Simplified picture of how π-acceptor and π-donor interactions
affect the MO diagram - Only the frontier orbitals are shown
t 2g *

π* (t2g)

eg *

eg*

L
large
∆O
t 2g
TM complex with
σ-bonding only

π-acceptor ligands
increase ∆O

t2g
[Cr(CO)6]
with π-backdonation
from Cr to CO

eg *

π-donor ligands
decrease ∆O

eg *
small
∆O
t 2g*

t 2g

TM complex with
σ-bonding only

t 2g

L

TM complex with πdonation from
Ligand to Metal

π-donor ligands
•

π-donation from the ligands to the t2g orbitals à the 3 t2g metal orbitals and 3 low lying, filled
ligand orbitals of π-symmetry form 3 bonding MOs and 3 antibonding MOs (thus t2g is
lowered and ∆o increases).

•

Since the interacting ligand orbitals are full, these electrons occupy the bonding MO from this
interaction, and the d-electrons occupy the antibonding MO.

•

The result is (1) a small ∆o
(2) the t2g orbital is weakly antibonding
© 2006 - Dr. Paul G. Hayes – University of Lethbridge

Chemistry 3820 (Fall 2006)

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MO Diagram for a Tetrahedral Complex
e.g. [Ni(PPh3)4] (Ni0 , d10 , 18-electron complex)
t2*

4p (t 1u)

a 1*

4s (a1g)
t2*

∆t

∆t ~ 4/9 ∆o

3d (eg + t2g)
e

(a1g + t1u + eg)
a1

4 PPh 3

Ni

t2

© 2006 - Dr. Paul G. Hayes – University of Lethbridge


Chemistry 3820 (Fall 2006)

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Hard and Soft Acids and Bases

© 2006 - Dr. Paul G. Hayes – University of Lethbridge