4 ligand field theory
TRANSCRIPT
-
8/4/2019 4 Ligand Field Theory
1/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
1
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.
------------------------------------------------------------------------------------------------------------------
2p 2p
2s2s
O OO2
2s
*2s
*2p
*2p
2p
2plarger E
O2
Paramagnetic
Bond order of 2 (O=O)
2 MOs like this -
perpendicular to each other
------------------------------------------------------------------------------------------------------------------
2p 2p
2s2s
N NN2
2s
*2s
*2p
*2p
2p
2p
smaller E
N2
Diamagnetic
Bond order of 3 (N= N)
Very strong NN bond
Unlike O2, the 2p and 2p orbitals are
switched in energy (see next page)
HOMO = 2p, LUMO = *2p
-
8/4/2019 4 Ligand Field Theory
2/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
2
Ordering of MOs in O2 is2p, 2p, *2p, *2p but in N2 and CO, the order is 2p, 2p, *2p, *2p
Reason - O2 - larger E between s and p orbitals less orbital mixing
- larger E because across the period (B,C,N,O,F) more protons added to thenucleus, 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 2p and the *2s orbitals raises the energy of the 2p
above the 2p.
Changes in the energy of the MOs of homonuclear diatomic molecules in the 2nd
period
MO Diagram for CO
-
8/4/2019 4 Ligand Field Theory
3/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
3
HOMO (slightly antibonding)
LUMO
2p
2p
2s
2s
C OCO
2s
*2s
*2p
*2p
2p
2p
- CO is isoelectronic with N2 (C has one less electron than N, O has one more) MO diagram isvery similar to that of N2.
- Diamagnetic, Bond order of 3 (C= O)
- Important: HOMO = 2p (weakly antibonding), LUMO = *2p (strongly antibonding)
- Since O is far more electronegative than C, would expect a large dipole moment with - on O.
However, CO actually has only a small dipole moment (0.1 Debye) with the
- on Carbon!Usually electronegativities are a good indication of the direction and magnitude of the dipole ona molecule, but especially in molecules where orbitals with antibonding character are occupied,things are not so straightforward.
-
8/4/2019 4 Ligand Field Theory
4/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
4
Pictorial Representations of Possible Bonds
-
8/4/2019 4 Ligand Field Theory
5/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
5
MO theory ~ Ligand Field Theory
Constructing an MO diagram for an octahedral complex
Symmetry Adapted combinations of Ligand -orbitals in an Octahedral Complex
- any combination
between t2g andligands becomes
non-bonding
- In an octahedral environment, the metal orbitals (3d, 4s, 4p for a 1st row TM) divide bysymmetry into 4 sets: s = a1g, p = t1u, axial d = eg, inter-axial d = t2g.
- The orbitals of the six ligands can be combined to give six symmetry-adapted linearcombinations 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 metalorbitals and 6 ligand orbitals) combine to form six bonding and six anti-bonding MOs. Seenext page.
- This is a much more correct approach than crystal field theory, but is not as easy to use.
-
8/4/2019 4 Ligand Field Theory
6/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
6
MO energy levels for an octahedral complex
(only -bonding considered)
3d (eg + t2g)
4s (a1g)
4p (t1u)
O
a1g
eg*
t2g
t1u
eg
a1g*
t1u*
6 L
(a1g + t1u + eg)
TM
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 t2g and eg
* orbitals).
-
8/4/2019 4 Ligand Field Theory
7/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
7
-bonding ligands
(the above MO diagram does not take into account -bonding)
For -acceptor ligands, the bonding is SYNERGIC: -donation to the metal strengthens
-backbonding to the ligand, and -donation from the metal to the ligand strengthens the-donor component of bonding.
This is because -donation leads to increased electron density on the metal, which allows
increased -backdonation. Conversely, -backdonation reduces the amount of electron
density on the metal, which allows more -donation from the ligand to the metal.
Cr(CO)6: Octahedral complex with good -acceptor ligands
OCM
M C O
M C O
-interaction -interaction
lobe ofacceptor orbital(whole orbitalnot shown)
slightly antibondingHOMO of CO
axial d-orbitals strongly antibonding* orbitals of CO
M
L
L M L
-donor
-donor -acceptor
filled orbital empty orbital
NH3, CH3-, H-
Cl-, OH-, NR2-, OH2 CO, NO
+, CN-
M LM LM
-
8/4/2019 4 Ligand Field Theory
8/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
8
3d (eg + t2g)
4s (a1g)
4p (t1u)
O
a1g
eg*
t2g
t1u
eg
a1g*
t1u*
6 CO
large
(a1g + t1u + eg)
* orbitals of CO (t2g)
Cr
MO energy levels for an octahedral complex
with -acceptor ligands (e.g. [Cr(CO)6])
t2g*
-backdonation to CO from the t2g orbitals (which are non-bonding in the absence of-interactions between the metal and the ligands).
The 3 t2g orbitals and 3 high lying * orbitals of the CO ligands form 3 bonding MOs and
3 antibonding MOs.
Since the CO * orbitals are empty, the d-electrons occupy the bonding MO from thisinteraction.
The result is (1) a very large o, so the eg* orbital is likely to remain empty.
(2) the t2g orbital is strongly bonding (wants to be filled with 6 electrons)
complexes of strong -acceptor ligands are the most likely to obey the 18 electron rule
-
8/4/2019 4 Ligand Field Theory
9/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
9
Symmetry Adapted Linear Combinations of* orbitals in ML6 complexes
Q. Why are there only three ligand -acceptor orbitals shown in the MO diagram for CO when
there are 6 ligands, each with two empty * orbitals?
A. The 12 * 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 linearcombinations are of the correct symmetry to interact with the t2g orbitals (dxy, dxz, dyz) on themetal.
-
8/4/2019 4 Ligand Field Theory
10/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
10
Simplified picture of how -acceptor and -donor interactions
affect the MO diagram - Only the frontier orbitals are shown
O
eg*
t2g
large
* (t2g)
[Cr(CO)6]with -backdonation
from Cr to CO
eg*
t2g*
O
small
TM complex with -donation from
Ligand to Metal
eg*
t2g
eg*
t2g*
t2g
t2g-acceptor ligands
increase O
-donor ligands
decrease O
TM complex with-bonding only
L
L
TM complex with
-bonding only
-donor ligands
-donation from the ligands to the t2g orbitals the 3 t2g metal orbitals and 3 low lying, filledligand orbitals of -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 thisinteraction, and the d-electrons occupy the antibonding MO.
The result is (1) a small o
(2) the t2g orbital is weakly antibonding
-
8/4/2019 4 Ligand Field Theory
11/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
11
MO Diagram for a Tetrahedral Complex
e.g. [Ni(PPh3)4] (Ni0, d10, 18-electron complex)
3d (eg + t2g)
4s (a1g)
4p (t1u)
a1
e
t2*
t2
a1*
t2*
Ni
(a1g + t1u + eg)
t t ~ 4/9 o
4 PPh3
-
8/4/2019 4 Ligand Field Theory
12/12
Chemistry 3820 (Fall 2006)
2006 - Dr. Paul G. Hayes University of Lethbridge
12
Hard and Soft Acids and Bases