Problem Class 2: The MO Diagram of a Square Planar Complex
L
• square planar complexes can be derived by removing the
axial ligands from an octahedral complex
L
L
• last year you will have learnt (based on crystal field
L
L
theory) roughly what happens to the key dAOs in this
L
case, Figure 1
dx2-y2
dx2-y2, dz2
o removing ligands from along the z-axis, reduces the eeg
e interactions between the metal d electrons and the
dz2
ligand σ-electrons, stabilising the dz2 dAO.
dxy
o there is a slight stabilisation of the dxz and dyz dAOs
t2 g
o the equatorial ligands slightly contract, raising the
dxz, dyz
dxy, dxz, dyz
energy of the dxy and dx2–y2 dAO
Figure 1 dAO energy changes for a tetragonal
o note that the dz2 dAO can be stabilised by so much
distortion
that it can lie lower than the dxy AO
z
• what can we say about square planar complexes with
L
MO theory?
L
L
• the first step is to derive the MOs for such a complex.
x
L
L
• by now you should be used to the procedure!
y
L
Figure 2 axial system is the same as for an
octahedral complex
C4, C2
σv
σv
L
L
L
C2′
• define the axial system: Figure 2
• find all of the symmetry operations on the molecule:
Figure 3
C2′
L
σ d C2′′
σ v C2′
σd
σv
from above
Figure 3 symmetry elements for D4h point
group
D4 h
E
2C4
C2
C2′
C2′′
i
2S4
A1g
1
1
1
1
1
1
1
σ h 2σ v
Point-Group
• determine the molecular shape: square planar
• identify the point group of the molecule: D4 h
2σ d
1
1
1
A2g
1
1
1
-1
-1
1
1
1
-1
-1
B1g
1
-1
1
1
-1
1
-1
1
1
-1
B2g
1
-1
1
-1
1
1
-1
1
-1
1
Eg
2
0
-2
0
0
2
0
-2
0
0
A1u
1
1
1
1
1
-1
-1
-1
-1
-1
A2u
1
1
1
-1
-1
-1
-1
-1
1
1
B1u
1
-1
1
1
-1
-1
1
-1
-1
1
B2u
1
-1
1
-1
1
-1
1
-1
1
-1
Eu
2
0
-2
0
0
-2
0
2
0
0
The Fragment Orbitals
• determine the fragments: the central metal and the
symmetry fragment consisting of the four in plane σligands
• determine the energy level and symmetry labels of the
fragment orbitals:
o this will be considered in two parts
o the metal orbitals which is relatively easy
o and the symmetry adapted orbitals of the four ligands
Fragment Orbitals of the Metal
• the dAOs are easy as we can read their symmetry from
the character table,
• in addition these will be very similar to the octahedral
fragment MOs (minus the contributions from the ligands
on the z-axis)
o s-> a1g
Tz
(Tx, Ty)
Figure 4 character table for the D4h point
group
o px, py-> eu and pz-> a2u
o dx2–y2 -> b1g and dz2 -> a1g
o dxz, dyz -> eg and dxy -> b2 g
Hunt / Problems Class 2
1
Symmetry Adapted Fragment Orbitals for L4
• then by analogy with the octahedral symmetry fragment
orbitals (Figure 5) it is fairly easy to determine square
planar complex orbitals (Figure 6) by removing the
axial ligand AOs.
• you have also been given the symmetry adapted
fragment orbitals in Lecture 5
• the symmetry of these fragments is
o totally symmetric-> a1g
a1g
t1u
eg
Figure 5 Octahedral sigma MOs
eu
o px, py phase pattern-> eu
o dx2–y2 phase pattern -> b1g
eu
a1g
b1g
Figure 6 Square planar sigma MOs
2eu
2b1g
3a1g
eu
4p
1a2u
t1u
Forming the MO Diagram
• combine fragment orbitals of the same symmetry,
estimate the splitting energy and draw in the MO energy
levels and MOs: Figure 7
• the dAOs in this diagram look nothing like those we
expected based on crystal field theory, Figure 9
a2u
2b1g
dx2-y2
2a1g
2a1g
2b1g
3a1g
4s
a1g
a1g b1g
3d
eg t 2 g
1b2 g
1b2 g
1b2 g
dz2
dxy
1eg
1eg
1eg
dxz, dyz
2a1g
a1g
b2 g eg
Figure 9 Comparing the expected dAO configuration and that obtained
from the basic energy level diagram
b1g
eu
a1g
1eu
1b1g
1a1g
L
M
M
L
L
L
L
L
L
L
Figure 7 Basic energy level diagram
2a1g
un-mixed
o the three non-bonding b2g and eg orbitals lie as
expected however the 3a1g looks out of place and
there is an unoccupied metal s AO below the highest d
antibonding levels!
o this configuration is therefore highly unfavourable as
it stands
Electronic configuration
• the four σ-ligands will provide 2e each giving 8e
• the metal will provide 7 or more electrons, I've used 8e
• giving a total 0f 16e
3a1g
Mixing
• determine if any MO mixing occurs, form the mixed
2a
3a
orbitals and redraw the MO diagram with the mixed
mixed
MOs
Figure 8 diagram showing the orbitals to be
1g
1g
mixed and the resulting MOs
Hunt / Problems Class 2
2
• mixing will only occur if the 2a1g orbital is occupied
o for the 2a1g orbital to be occupied there must be 8 "d" electrons on the metal, and hence this
structure is unstable unless there are at least 7 dAO electrons present, at which point
stabilisation of the square planar structure occurs
o this is why square planar complexes primarily occur only for d8 and sometimes d7 or d9
complexes
• the 2a1g and 3a1g MOs can mix, if the 2a1g orbital is formally occupied then mixing is likely as
this will lower the energy of this MO and thus lower the total energy of the complex the key a1g
orbitals, and the resultant mixed orbitals are shown in Figure 8
• the final energy level diagram is given in Figure 10
2eu
eu
4p
t1u
1a2u (pz)
3a1g (s)
2b1g (dx2-y2)
a2u
3a1g
1b1g
4s
a1g
a1g
b1g
3d
eg t 2 g
b2 g
2a1g(dz2)
a1g
1b2 g (dxy)
1eg (dyz & dxz)
eg
b1g
eu
a1g
1eu
1b1g
z
1a1g
x
y
M
L
L
M
L
L
L
L
L
L
Figure 10 final energy level diagram
MO Diagram Checklist!
Analysis
• the relative ordering of the dAO based MOs now
Pd(NH3)4]2+
coincides with the earlier treatment
M(CN)4]2- M=Ni, Pd
o there is a single high energy orbital, which is the
[PtCl4]22b1g which is the dx2-y2 M based MO
[AuBr4]2
o moreover although the 2a1g MO is called a dz
Ir(CO)(Cl)(PPh3)2] Vaska's complex
orbital it will have a large 4s AO contribution due
Rh(Cl)(PPh3)3 Wilkinson's Catalyst
to mixing
o the remaining dAO based MOs remain as expected
• d8 square planar complexes are not uncommon, some examples are in the box to the right.
Hunt / Problems Class 2
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