problems2 square planar mo diagram - hunt research group€¦ · · 2016-12-24hunt / problems...

Hunt / Problems Class 2 1

Problem Class 2: The Energy Diagram of a Square Planar Complex

dxy, dxz, dyz

dx2-y2, dz2 dx2-y2

dz2

L

L

L LL L

dxy

dxz, dyz Figure 1 dAO energy changes for a

tetragonal distortion

E

1 1 1 1 1 1 1 1 1 1A1g

A2g

Eg

(Tx, Ty)

Tz

B1g

B2g

1 1 1 -1 -1 1 1 1 -1 -1

1 -1 1 1 -1 1 -1 1 1 -1

1 -1 1 -1 1 1 -1 1 -1 1

2 0 -2 0 0 2 0 -2 0 0

1 1 1 1 1 -1 -1 -1 -1 -1A1u

A2u

B1u

B2u

1 1 1 -1 -1 -1 -1 -1 1 1

1 -1 1 1 -1 -1 1 -1 -1 1

1 -1 1 -1 1 -1 1 -1 1 -1

2 0 -2 0 0 -2 0 2 0 0Eu

x2+y2, z2

x2-y2

xy

(xz, yz)

Figure 2 character table for the D4h point

group

L

LL

LLL

L

L

y x

z

y

x

z(a) (b)

Figure 3 axial system is the same as for an

octahedral complex

ML

LLL

from above

ʹC2b(y )

ʹC2a(x)

C4(z),C2(z)

σνb(zy )σνa(zx)

M L

L

L

L

σ da, ʹ́C2a

σ db, ʹ́C2b

σνa(zx)

σνb(zy )

from the side

Figure 4 some of the symmetry elements for

D4h point group

Revision • square planar complexes can be derived by removing the

axial ligands from an octahedral complex • last year you learnt (based on crystal field theory)

roughly what happens to the key dAOs , Figure 1 o removing ligands from along the z-axis, reduces the e-

e interactions between the metal d electrons and the ligand σ-electrons, stabilising the dz2 dAO.

o there is a slight stabilisation of the dxz and dyz dAOs o the equatorial ligands slightly contract, raising the

energy of the dxy and dx2–y2 dAO • what can we say about square planar complexes using

MO theory? • the first step is to derive the energy diagram and some of

the MOs for such a complex. • by now you should be used to the procedure!

Point-Group • determine the molecular shape: square planar • identify the point group of the molecule: D4h o the character table is given in Figure 2

• define the axial system: Figure 3 o careful! I've noticed a proportion of students not

aligning the axis with the ligands o you can draw your molecule with the two ligands

coming towards you (Figure 3a), or you can draw them such that one set stays in the plane of the page (Figure 3b)

o my suggestion is to use the axis alignment shown in Figure 3a, but this is not a requirement

• find all of the symmetry operations on the molecule: o some of these are shown in Figure 4

The Fragment Orbitals • determine the fragments: the central metal and the

symmetry fragment consisting of the four in plane σ-ligands, put these on the bottom of your diagram

• 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


4s

4p

3d

px, py->pz->

s->

dx2-y2->dz2->dxy->dxz, dyz ->

spherical symmetry

octahedral symmetry

D4hsymmetry

Figure 5 working out the dAO symmetry

labels

t1ua1g

eg

Figure 6 Octahedral sigma MOs

uniqe orbitals

Figure 7 Square planar sigma MOs

Fragment Orbitals of the Metal • I recommend you write out all of the metal fragment

orbital symmetry labels, as shown in Figure 5 o this is because we will use this information later, and

it saves valuable time compared to looking these up each time we need them

o also labelled on Figure 5 are the spherical symmetry (s, p, d) labels and the ochahedral symmetry labels (a1g, t1u, eg, t2g,) , this is just to help you make the connection between all of these symmetry labels

o at this stage we assume the orbitals don't change just their labels

• AOs are easy as we can read their symmetry from the character table (remember the short-cuts!) o s->a1g o px, py-> eu and pz->a2u o dx2–y2 -> b1g and dz2 -> a1g o dxz, dyz -> eg and dxy -> b2g

Symmetry Adapted Fragment Orbitals for L4 • this is a bit harder, but we can do it • we already know what the octahedral symmetry

fragment orbitals should look like, Figure 6 o we can determine the square planar complex ligand

orbitals by removing the axial ligand AOs, from the octahedral orbitals, Figure 7

o we also know we should have 4 symmetry adapted orbitals, "4 orbitals in" means "4 orbitals out"

o you have also been given the symmetry adapted fragment orbitals in Lecture 5, but it is better to be able to generate them from knowledge you can bring to the problem rather than memorise them, this way you learn what to do when you don't have a pre-memorised set!

• work out the new symmetry labels for these fragments, (you can use the information already written down for the metal orbitals) o totally symmetric-> a1g o px, py phase pattern-> eu o dx2–y2 phase pattern -> b1g

• how do we know the relative energy ordering? We look at the orbital overlap and guestimate the energies o clearly the totally symmetric a1g is lowest o the eu are long range (2 bonds and ultimately with the

metal in between!) and essentially non-bonding o the b1u has closer through space antibonding and will

be highest in energy • put the FO energy levels onto your diagram


4s

4p

3d

ML

L

L

L xy

zL

L

L

LM

px, py->pz->

s->


b1ga1g

eu

b2geg

a1g

eg

b2g

eu

b1g

eu

a2u

a1g

a2ueu

Figure 8 Basic energy level diagram, non-

bonding and weak interactions

Figure 9 a1g and b1g MOs

4s

4p

3d

ML

L

L

L xy

zL

L

L

LM

px, py->pz->

s->


b1ga1g

1eu

b2geg

a1g

1eg

1b2g

eu

b1g

2eu

1a2u

a1g

a2ueu

2a1g

2b1g

1b1g1a1g

a1g

b1g

8e 8e16e

3a1g

Figure 10 Basic energy level diagram, including a1g and b1g MO energy levels

Forming the MO Diagram • combine fragment orbitals of the same symmetry,

estimate the splitting energy and draw in the MO energy levels and key MOs:

• start with the weakly and non-interacting orbitals first o remember that we don't draw the long lines joining the

energy levels as this makes the diagrams messy. o the metal pAOs are easy to assign, the eu interact with

the ligands and the a2u remains non-bonding o the metal dAOs of b2g and eg symmetry are non-

bonding, Figure 8 • next are the b1g and a1g FOs o the b1g interaction is straight forward, except we may

not know exactly how high in energy it goes! (we will come back to this shortly)

• but what about the a1g FOs, there are two of these on the metal side, one from the 4s and one from the dz2! o we have two options, combine the lowest energy

orbitals, or to follow the basic TM-Oh format, in this case we can either route! (I will show you both)

o the easiest case is to assume the two lowest energy orbitals combine, so the ligand a1g will form a MO with the metal dz2 a1g FO.

o except we may not know exactly how high in energy it goes!

• to determine how high the anti-bonding a1g and b1g MOs go we need to draw out and evaluate the orbital character, Figure 9 o always draw out difficult orbitals BEFORE you assign

where the energy level goes. I see many students trying to put the energy level in first and getting frustrated because they not know where it should go, ALWAYS draw the MO first!

o the negative equatorial torus of the dz2 FO is anti-bonding (through bond) with the equatorial ligands

o however, there is good through space overlap of the ligand FOs and positive overlap of the equatorial ligands with the axial component of the dz2 FO

o the anti-bonding a1g MO is actually not very anti-bonding, it will not be destabilised by much

o the anti-bonding b1g MO is has strong directed anti-bonding overlap with the ligand FOs, this MO is strongly antibonding, Figure 10

o should the b1g go above or below the 4s? The b1g is strongly anti-bonding and thus we can predict it will go above the 4s, we also know from analogy with the Oh diagram that this "eg" orbital goes above the 4s

• I will come back and discuss the a1g and b1g MOs further shortly, but first let's complete this MO diagram


un-mixed

mixed

3a1g

3a1g2a1g

2a1g

Figure 11 mixing of the of the a1g MOs

4s

4p

3d

ML

L

L

L xy

zL

L

L

LM

px, py->pz->

s->


b1ga1g

1eu

b2geg

a1g

1eg

1b2g

eu

b1g

2eu

1a2u

a1g

a2ueu

2a1g

2b1g

1b1g1a1g

8e 8e16e

3a1g

2a1g

3a1g

Figure 12 Stage 2 energy level diagram,

including mixing of the a1g MOs

Electronic configuration • the four σ-ligands will provide 2e each giving 8e • we know from last year that d8 complexes form square

planar TMs, so we know the metal will be d8 • thus the total electron count is 16e • put the electrons into the energy diagram, filling energy

levels sequentiall, Figure 10

Mixing • determine if any MO mixing occurs, form the mixed

orbitals and redraw the MO diagram with the mixed MOs

• remind yourself of the "rules" for mixing, Lecture 2 o necessary conditions for MO mixing to occur are that

only MOs of the same symmetry (but not the same bonding anti-bonding pair) can mix and mixing must stabilise the total energy of the molecule

o mixing tends to be large when at least one of the following criteria are met; MOs are close in energy, one of the MOs is non-bonding or unoccupied, MOs are in the HOMO-LUMO region

o take one MO and then "add" the other MO once "as is", and once with the phase reversed, then examine the resultant MOs to determine which is lowered in energy and which is raised in energy

• the 2a1g and 3a1g MOs fulfil these criteria, they are in the HOMO-LUMO region, the 2a1g is occupied and the 3a1g is non-bonding and not occupied, Figure 12 o we mix the 2a1g and 3a1g MOs, the results are quite

strange! o make the 4s AO quite large. We know this should be

the case because (from Foundation course material) the radial extent of the 3d is small and the 4s is larger and more diffuse.

• it is important to remember that all MOs of the same symmetry can mix slightly, so the MOs of a real molecule can be more complex. In a qualitative MO diagram we estimate the largest mixing interactions, and provide a starting point for understanding the more complex MOs of a real molecule.

• how far up does the 3a1g MO go? o this is hard to guess it may go above or slightly below

the 2b1g MO, this will depend on the exact system studied and is best determined by a calculation

• how far down does the 2a1g MO go? o this is hard to guess, mixing is very "quantum

mechanical" and so very sensitive to the exact system being studied. The 2a1g MO could potentially even go below the non-bonding dAOs!


4s

4p

3d

ML

L

L

L xy

zL

L

L

LM

px, py->pz->

s->


b1ga1g

1eu

b2geg

a1g

1eg

1b2g

eu

b1g

2eu

1a2u

a1g

a2ueu

2b1g

1b1g1a1g

b1g

8e 8e16e

3a1g2a1g

2a1g

Figure 13 Reordering of the a1g MOs

4s

4p

3d

ML

L

L

L xy

zL

L

L

LM

px, py->pz->

s->


b1ga1g

1eu

b2geg

a1g

1eg

1b2g

eu

b1g

2eu

1a2u

a1g

a2ueu

2b1g

1b1g1a1g

8e 8e16e

2a1g3a1g

2a1g2a1g

Figure 14 Based on TM-Oh template

un-mixed

mixed

3a1g

3a1g2a1g

2a1g

Figure 15 Mixing of the new a1g MOs.

MO Diagram Checklist • don't forget the checklist, have you got everything you

should have on the diagram? • I haven't been adding annotations, but describing orbital

overlaps and energies in the text, in an exam you would be expected to include annotations

Discussion of a1g and b1g energy levels • earlier I said I would come back and discuss the a1g and

b1g MOs further • first, what would happen if I thought the dz2-L a1g MO

should lie above the 4s based a1g MO? Figure 13 o we can see that mixing will still occur between the

2a1g and 3a1g MOs, the 2a1g would be lowered in energy, and we would end up with the same diagram as given in Figure 12

• second, what would happen if I used the Oh-TM diagram as my template and interacted the ligand a1g FO with the 4s based FO? Figure 14 o in this case we could still have mixing, however as the

MOs involved are further apart we could expect this to be much weaker. Nevertheless, the same final mixed MOs would result, Figure 15.

o however the problem now is that the 3a1g (which is already high in energy) would go up even further and the 2a1g (which is already non-bonding) would drop down below the 1b2g and 1eg MOs! This is possible in a real system.

What about a real complex? • in an exam situation you would normally be expected to

draw the ENERGY level diagram and show key orbitals such as the 2a1g, 3a1g and 2b1g MOs. (READ the question carefully!) Alternatively you might be asked to interpret the real MOs.

• in a research situation, you want to know what all of the orbitals look like. Finding a nice simple square planar ML4 complex with four σ-donor ligands is not easy!

• a full MO diagram (of the σ-MOs) is given on the next page for [PdCl4]2- . On-line you can also find the checkpoint file for [PdCl4]2- which will allow you to view the real 3D MOs.

• remember Cl is actually a π-donor ligand and so there are more MOs than you would see for a pure σ-donor ligand.

• this should be giving you a hint that we are covering only the beginnings of MO theory here, real life (as usual) is much more complicated! However, we do need these simple models to build from otherwise we would be lost.


Figure 16 Full MO diagram.

• there are a few things to note in Figure 16

o the 1b1g MO is lower than the 1a1g MO, this can be rationalised by the good σ-overlap of the FOs

o the 1eu contains some additional contributions due to the Cl ligand being a π-donor.

o the 2a1g lies below the dAO manifold of formally non-bonding MOs.

o our qualitative MOs are surprisingly accurate! • d8 square planar complexes are not uncommon, some

examples are in the box to below:

Pd(NH3)4]2+

M(CN)4]2- M=Ni, Pd[PtCl4]2- [PdCl4]2-

[AuBr4]-

Ir(CO)(Cl)(PPh3)2] Vaska's complexRh(Cl)(PPh3)3 Wilkinson's Catalyst

o many of these structures are not highly symmetric,

however despite the symmetry lowering we can still use the simple MO diagrams developed here to understand the bonding in these complexes!


4s

4p

3d

ML

L

L

L xy

zL

L

L

LM

px, py->pz->

s->


b1ga1g

1eu

b2geg

a1g

1eg

1b2g

eu

b1g

2eu

1a2u

a1g

a2ueu

2a1g

2b1g

1b1g1a1g

8e 8e16e

3a1g

2a1g

3a1g

Figure 17 Stage 2 energy level diagram,

including mixing of the a1g MOs

dx2-y2

dz2

dxy

dxz, dyzb2g eg

a1g

a1g

eg

b2g

a1g

b1gb1g

crystal field theoryMO theory

4s

Figure 18 comparing MO theory to crystal

field theory

Analysis and Interpretation • for analysis we will return to the more standard diagram,

reproduced here again as Figure 17 • for mixing to occur we need the HOMO 2a1g to be

occupied o the HOMO could be occupied by only 1 electron, thus

it is possible for a d7 system to exhibit mixing. o however mixing is dependent on the occupation, and

will be less overall if there is only 1e • The predicted "dAO" energies crystal field theory (CFT)

and MO theory are shown in Figure 18 o Should we compare CFT and MO theory? The short

answer is NO! o CFT is much older than MO theory and was

developed before we could study TM complexes to the level of detail we can today (both experimentally and computationally!)

o CFT is based on an IONIC model of bonding, while qualitative MO theory is based on a mostly COVALENT model of bonding

o in reality both ionic and covalent effects are present o in TM-complexes the covalent description dominates,

and the best way to think of this is as the MO model modified slightly by the ionic contributions ie CFT.

o of course, a calculation takes ALL of these effects into account, in addition to quantum mechanical based interactions such as electron exchange and correlation

• what can we say about both theories? o in CFT the explanation for the stability of the dz2

orbital is that there is less repulsion from the negative charge on L thus stabilising the "full" dz2 orbital.

o in MO theory we say that there has been mixing of the a1g MOs, you might have noticed that the ligands don't actually effect the outcome, and that the mixing has essentially changed the "nature" of the dz2 AO by mixing in some of the 4s AO.

o the theories are apparently contradictory! o the stability of the dz2 MO is primarily driven by the

mixing and stabilisation of the a1g MO, there will be some ionic effects on the dz2 FO making a small contribution

o based on the MO description the b2g and eg remain degenerate, while in CFT these MOs have different energies. In the real system we can expect different energies for these orbitals, due to ionic interactions.

o we have seen that in a real system that the 2a1g MO can drop below the dxy, dxz and dx2-y2 MOs, based on the CFT this could also happen however it is more likely based on MO theory. We have seen this in the [PdCl4]2- example shown here.

problems2 square planar mo diagram - hunt research group€¦ · · 2016-12-24hunt / problems...

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