Ligand Field Theory

The ligand field theory (LFT) fleshes out the ideas of crystal field theory with molecular orbital theory concepts. It provides a method for understanding M–L bonding and antibonding orbitals; however, it has been strongly disputed by computational studies in favor of valence bond models that incorporate hypervalency. Still, LFT provides a more complete picture of complex bonding than crystal field theory, so we’ll discuss it here. Furthermore, the portions of LFT under dispute have nothing to do with CFT, so “no harm no foul.” Let’s take a look at the molecular orbitals of a hypothetical octahedral MLn complex to begin hashing out LFT.

The M–L bonding molecular orbitals mostly have ligand character, while the antibonding orbitals mostly reside on the metal.

Notice that the perturbations of the metallic d orbitals are consistent with the ideas of crystal field theory for the octahedral geometry. The labels dσ and dπ will be useful for us later, and indicate how the d orbitals overlap with the incoming ligand orbitals—the dσ orbitals overlap in a head-on, sigma-type manner, and the dπ orbitals overlap in a side-on, pi-type manner. The dσ orbitals are destabilized through electrostatic repulsion, while the dπ orbitals essentially remain non-bonding. What LFT adds to these CFT ideas is a description of the fate of the remaining unfilled metal valence orbitals and the filled ligand orbitals. According to the LFT picture, orbital overlap between the 4s, 4p and symmetry-matched 4d (i.e., the 4dσ) AOs and the six ligand HOMOs results in six bonding and six antibonding MOs. Notice that the math works out:

1 4s + 3 4p + 2 3dσ + 6 ligand HOMOs = 6 bonding MOs + 6 antibonding MOs

As necessary, the number of atomic orbitals in (12) equals the number of molecular orbitals out (12). But more important than the number of MOs is the character of each orbital—where is the electron density primarily located in these MOs? Ligand field theory provides a logical answer: the bonding MOs primarily possess ligand character, while the antibonding MOs are primarily localized on the metal. The guiding principle here is that MOs are composed primarily of those AOs to which they are closest in energy. In the end, we arrive at a critically important, albeit expected insight: the metal is electrophilic (owner of many unfilled MOs) while the ligands are nucleophilic (owners of filled MOs).

This may seem like a relatively simple picture at first glance. All we did, it seems, was tack the 4s, 4p and ligand orbitals on to crystal field theory. Yet this groundwork is necessary to build a truly powerful theory of the electronic structure of organometallic complexes. Until now, we’ve considered only the HOMO of the ligands. What happens when we take the HOMO and LUMO of the ligands into account? Read on…

Frontier Molecular Orbital Theory

The frontier molecular orbitals of a compound are at the “frontier” of electron occupation—the highest-energy occupied and lowest-energy unoccupied molecular orbitals (the HOMO and LUMO). The HOMO is logically viewed as nucleophilic or electron donating, while the LUMO is electrophilic and electron accepting. Furthermore, both chemical reactions and resonance can be explained by interactions (overlap) between a filled HOMO and an empty LUMO on one or more molecules. FMO theory uses these foundational ideas to explain the structure and reactivity of molecules, and at least in organic chemistry, the theory has been insanely successful.

When it comes to organometallic complexes, we can gain considerable insight by noticing that orbitals on the frontier are usually the metallic d orbitals. In the figure above, assuming that the metal is d6, the frontier falls between the dπ and dσ orbitals. Without a need for any quantum chemical calculations (only CFT and the number of d electrons are required), we can thus begin to anticipate the shapes of the frontier orbitals of organometallic complexes! Combined with a solid understanding of organic FMO theory and the frontier orbitals of organic ligands, we can recognize important orbital interactions in organometallic complexes that can explain a wide variety of observations.

Before digging in to an example, let’s introduce some terminology commonly used by organometallic chemists related to FMO theory.  The terms acidity and basicity are used in the context of FMO theory very broadly—prepare to expand your mind! Acidity refers to the desire of a ligand, metal center, or specific orbital to accept electron density (from any viable electron source, not just a Brønsted base). Basicity refers to the desire of a ligand, metal center, or specific orbital to donate electrons (to any viable electron sink, not just a Brønsted acid). FMO theory also distinguishes between σ-acids, σ-bases, π-acids, and π-bases. The first two, σ-X, want to accept or donate electrons in a σ-type fashion—that is, aligned head on with another orbital. The latter two, π-X, want to accept or donate electrons in a π-type fashion—that is, aligned side by side with another orbital. Most commonly, these terms are applied to ligands to

indicate their expected behavior in a fancy-schmancy but quite descriptive way, and the terms can also be applied to the metal center.

Sigma and pi acids and bases, with their most reactive frontier MOs and typical electronic behavior.

Why are these designations useful? Let’s look at an example now. Tungsten(0) hexacarbonyl and the other metal(0) carbonyls are interesting species. Let’s deconstruct W(CO)6 to see why…

Tungsten(0) hexacarbonyl. Why doesn't it fall apart?

We can make a very good argument that this compound ought to fall apart as soon as it’s made. The complex is neutral, but the deconstructed ligands and metal are all neutral too, and entropy loves the right-hand side of the figure above. Yet, the complex is reasonably stable. Why?!

FMO theory provides a good answer: there is more to the W–CO bond than meets the eye. Let’s zoom in on the tungsten–carbon bond and explore what might be going on here. Like almost all metal–carbon bonds, we expect W and C to be bound primarily through an interaction between a filled non-bonding orbital on carbon (HOMO) and an empty dσ orbital on the metal (LUMO). That’s depicted below, and it’s typical organometallic FMO theory. Interactions like this help explain why, for instance, extremely electron-poor ligands like PF3 make horrible ligands and bind to metals very weakly.

A typical orbital interaction between a metal and an organic ligand. Par for the course.

The right-hand resonance structure, which depicts two bonds between tungsten and carbon, suggests that there is another bonding interaction between W and C…perhaps one where the LUMO of the ligand and the HOMO of the metal get involved? Can we draw a viable HOMO–LUMO interaction by reversing the typical roles of metal and ligand? The LUMO of the ligand is a π* orbital, and the HOMOs of the metal are the d orbitals not aligned along the octahedral axes (the dπ orbitals). The symmetry of these orbitals is just right! See the lower orbital interaction in the figure below.

There's a new interaction in town...and the metal is an electron donor in the pi-type interaction.

Please note that the tungsten atom is not rotated at all; two different d  orbitals are depicted in the two interactions. The second role-reversing interaction is important for π-acidic ligands like carbon monoxide, and is called backbonding. The interesting thing about backbonding is that when considering it, we must imagine the metal as electron donor and the ligand as electron acceptor. There is an important lesson here: consider the HOMO and LUMO of both the metal and ligands and their possible interactions. When it comes to tungsten(0) carbonyl, donation from the electron-rich (or π-basic)  metal center to the π-acidic CO ligands is what holds everything together!

Keep the fundamental ideas of FMO theory in mind as we move forward. Primarily, keep your mind open and remember that metal centers can act as electron donors toward acidic ligands, if the energetics and symmetry are right. We’re about to embark on an epic ligand survey, so we’ll have plenty of opportunities to apply these ideas!

CHM 401

 

Molecular orbital approach (Ligand Field Theory)

A Molecular Orbital Diagram using only sigma bonding can be constructed for a general Oh complex:

Bond Order = 6 - n(eg)/2

The sigma bonds are largely ligand based orbitals but take the appearance of d2sp3 hybrid orbitals about the metal.

The nb and lowest energy σ* orbitals are similar to the Crystal Field Theory description. The nb orbitals are strictly d like (dxy, dyz, dxz) and the σ* are mostly d orbital like (dx2–y2, dz2).

Inclusion of π bonding means using the t2g orbitals for overlap:

If the t2g orbitals are empty, then the ligand needs filled π orbitals to create a M-L π bond; the t2g orbitals become somewhat antibonding, are destabilized, and 10Dq is reduced.

If the t2g orbitals are occupied, then the ligand needs unfilled π* orbitals to create the M-L π bond; the t2g orbitals become somewhat bonding, are stabilized, and 10 Dq is increased.

This is known as π backbonding. It is the reason CO is such a strong ligand.

he valence-bond model and the crystal field theory explain some aspects of the chemistry of the transition metals, but neither model is good at predicting all of the properties of transition-metal complexes. A third model, based on molecular orbital theory, was therefore developed that is known as ligand-field theory. Ligand-field theory is more powerful than either the valence-bond or crystal-field theories. Unfortunately it is also more abstract.

The ligand-field model for an octahedral transition-metal complex such as the Co(NH3)63+

ion assumes that the 3d, 4s, and 4p orbitals on the metal overlap with one orbital on each of the six ligands to form a total of 15 molecular orbitals, as shown in the figue below.

Six of these orbitals are bonding molecular orbitals, whose energies are much lower than those of the original atomic orbitals. Another six are antibonding molecular orbitals, whose energies are higher than those of the original atomic orbitals. Three are best described as nonbonding molecular orbitals, because they have essentially the same energy as the 3d atomic orbitals on the metal.

Ligand-field theory enables the 3d, 4s, and 4p orbitals on the metal to overlap with orbitals on the ligand to form the octahedral covalent bond skeleton that holds this complex together. At the same time, this model generates a set of five orbitals in the center of the diagram that are split into t2g and eg subshells, as predicted by the crystal-field theory. As a result, we don't have to worry about "inner-shell" versus "outer-shell" metal complexes. In effect, we can use the 3d orbitals in two different ways. We can use them to form the

covalent bond skeleton and then use them again to form the orbitals that hold the electrons that were originally in the 3d orbitals of the transition metal.

Jump to main page content.

Your Tools

Help Log in

CommonView

Chem 416A - COSSAIRT Course Info

Answers to Your Questions

Reading and Handouts

Problem Sets

Exams

Answers to Your Questions

Questions Lectrure 11 10/17/12

1) Why haven't we addressed J (only L and S)?

We're getting there!

2) Why was e2'' lower in energy than e1' on the ferrocene MO diagram?

Because e2'' had an interaction with the metal d orbitals (a closer energy match and so a stronger interaction) and e1' only interacted with the metal p orbitals which are much higher in energy.

3) How do you determine the rectangular arrays from the microsates table?

You have to figure out all of your ML and MS combinations and put them in the table. You then literally draw a rectangle... each box within the rectangle should only contain one state, so that determines the number of rectangles you can draw. The rectangles have to be symmetric and span as much of the table as possible. You should always identify the biggest rectangle first and then see how many microstates are left and go from there. Check out the beginning of chapter 11 for another look!

4) Why does making the rectangular arrays work?

Using the microstate table the way we set it up, each term corresponds to a rectangular array. For each term, the spin multiplicity (the superscript 2S+1) is the same as the number of columns of microstates - for example a singlet term has one column, a doublet term has two columns, etc.

Announcements Lecture 11 Q&A and Road Map Posted

Oct 18, 2012 8:50 AM

Answers to questions from the lecture on 10/17/12 and the course road map I showed in class have ...

More

Problem Set 3 Posted

Oct 11, 2012 10:58 AM

Due at the start of class Friday October 19.

Lecture 8 Q&A

5) In solution how does ferrocene exist (staggered vs eclipsed)?

The cp ring is whizzing around... it can freely rotate on the iron.

6) How did you identify the symmetries of the group orbitals from the character table for ferrocene?

The easiest way is to see what orbital they interact with and then assign it based on that. So if the group orbital can interact with an s, p, or d orbital on the metal, it must have that symmetry... this can be done by inspection. Alternatively, you can assign the group orbital to an irreducible representation by seeing how it transforms under each of the operations of the point group like we've done before.

7) How do you know which microstate each "x" in a rectangular array denotes?

It doesn't matter. Just the number of "x"s matters for this analysis.

8) Can rectangles of different dimensions be considered in the assignment of term symbols based on microstates?

Yes! In the example we did in class we had a 3x3 rectangle, a 1x5 rectangle, and a 1x1 rectangle. They all have to be considered.

9) Was I supposed to understand how the IR spectroscopy diagram of the three molecules described their CO stretches at this point?

Yes. You know how to figure out the number of CO stretches that are IR active given a molecule with a certain point group. The spectrum should have that number of peaks! We'll do an example or two on Friday.

10) How do you determine the atomic spin state value "S"?

Just a sum of +1/2 or -1/2 for the two electrons. So if I have two electrons that are both spin up (+1/2) then S = 1. The multiplicity in the term symbol (the superscript) is 2S+1.

11) Where does 2cos(2pi/5) come from in the D5h character table?

Ah, a C5 rotation is a rotation by 2pi/5. If you actually determined the characters by hand using a matrix, this is what the math would get you. For even rotations like C4 or C2, the math works out nicely to 1 or -1 or 0, so the tables for those look less complicated. But they all come from the same place.

posted.

Oct 10, 2012 2:00 PM

The answers to questions for the lecture on 10/10/12 have been posted.

Lecture 7 Q&A posted.

Oct 8, 2012 2:41 PM

The answers to questions for the lecture on 10/08/12 have been posted.

Problem Set 2 Posted

Oct 4, 2012 8:15 AM

Due at start of class Friday October 12.

Lecture 5 Q&A posted.

Oct 3, 2012 1:23 PM

The answers to questions for lecture on 10/03/12 have been posted.

Lecture 4 Q&A

12) What is electronic spectroscopy? Why do we have to determine the microstates?

Electronic spectroscopy allows us to see transitions between the ground state of a molecule to higher energy excited states, which corresponds to moving electrons to higher energy orbitals. In order to be able to assign spectra of this type, we have to determine term symbols, which comes easily from a microstate analysis.

13) How are you determining the relative energies in your MO diagrams?

Figuring out the ligand position relative to the metal d orbitals is qualitatively simple - for sigma/pi donors, the ligand orbitals are always lower in energy (and filled) than the metal d orbitals. For pi acceptors they are higher. The metal s and p orbitals should always be the highest in energy. Everything else in the middle is determined by how "close" in energy the interacting orbitals are. If the ligand orbital is interacting with a metal p orbital, there is very little interaction because they are far apart. Interactions with the d orbitals are much stronger.

Questions Lecture 10 10/15/12

1) Good readings on the isolobal analogy?

Roald Hoffman's Nobel Lecture: http://onlinelibrary.wiley.com/doi/10.1002/anie.198207113/abstract

Chapter 15.2 of Miessler and Tarr

Wikipedia also does a decent job: http://en.wikipedia.org/wiki/Isolobal_principle

2) How did we know we had an anionic ligand in Mo(NR2)3?

When we learned electron counting rules, NR2 was an XL-type ligand. The X is an indicator that it is anionic. Just like Cl is an X-type ligand (chloride anion).

3) For cyclic pi systems, do the diagrams only work for aromatic rings?

Yes. The polygon, vertex trick only works for aromatic rings.

4) Why does the metal change as the ligand is removed in the isolobal analogy?

posted.

Oct 1, 2012 2:05 PM

The answers to questions for the lecture on 10/01/2012 have been posted.

Lecture 3 Q&A posted.

Sep 30, 2012 5:11 PM

The answers to questions for the lecture on 09/28/2012 are posted.

Practice Assigning Point Groups

Sep 28, 2012 12:41 PM

http://symmetry.otterbein.edu/

Problem Set 1 Posted

Sep 27, 2012 6:00 PM

Due at start of class Wednesday October 3.

We were removing ligands as cations (+) with one electron because we were homolytically cleaving bonds like we did in methane - one electron stays on the metal and one goes with the ligand. The electron that stays on the metal makes the metal have a negative charge. Instead of drawing a negative charge, I drew the neutral molecule with the metal with one more d electron!

5) Is the reason we are finding the phase of the cyclic ring systems so that we can determine what metal orbital it could interact with?

Yes!

6) In the isolobal structure example, why doesn't the identity of the carbon atom change while the transition metal does?

We are homolytically cleaving bonds. When we break a C-H bond and give one to the carbon and one to the hydrogen, nothing changes about the charge because that is a normal covalent bond. When we homolytically cleave an M-L bond, the metal leaves with one more electron and the L (as in L-type, as in 2 electron donor) leaves with a positive charge and one less electron. We could draw the negative charge on the metal instead of changing its identity or we could do what I did and go to the metal with one higher d electron count.

7) Does the polygon/vertex trick work for sandwich compounds.

It only works for finding the energies and phases of the aromatic ligand itself.

8) What is a frontier orbital?

Frontier orbitals are the highest occupied and lowest unoccupied molecular orbitals. They are the ones you care about when considering the properties and reactivity patterns of the molecules.

9) How do you know if a ligand is sigma only or has pi contributions?

There are several ways. One is to look at the spectrochemical series - this is the easiest way. Another is to draw a lewis structure/and or look at the electron counting rules for that ligand (this can only give you an idea for the donors). The final way is to draw an MO diagram for the ligand itself.

10) Why is it that in doing the (sigma)v operations on Mo(NR2)3, the other two orbitals are not counted as changed?

Lecture Q&A is posted.

Sep 26, 2012 2:08 PM

The answers to questions for the lecture on 09/26/2012 are posted.

Lecture Q&A is posted.

Sep 24, 2012 12:59 PM

The answers to questions for the lecture on 09/24/2012 are posted!

Send questions about this workspace to Chemistry Fall 2012.

We have to look at all three orbital in determining the reducible representation. Two orbitals move, so there is no contribution. The other is bisected and changes phases, so it contributes -1. In total, the contribution is -1.

11) When assigning symmetries of orbital combinations, do we do it is a whole?

Yes. You have to consider the symmetry of the entire set of orbitals (linear combination of molecular orbitals).

12) In the ferrocene example, how are the p-orbitals situated with respect to the metal?

The p orbitals are pointing at the metal (pz orbitals).

13) What chapters do I look at for MO diagrams?

Check out chapter 10. This has ligand field theory and AOM.

14) What are the limitations to the isolobal theory?

Well, they are many. The biggest one is that it is a very qualitative relative comparison. "Similar" not the "same".

Questions Lecture 9 10/12/12

1) In AOM how do you get the stabilization of the ligands in a sigma/pi donor?

You sum up all the d-orbital electron destabilization energy and then divide by the number of ligand electrons. In the case of the NbCl5 example, this would mean dividing the total destabilization energy by 10.

2) Why is e(pi) << e(sigma)? Can we assume this is always true for this class?

For this class we can indeed always assume this is the case. This is because the overlap between orbitals with pi symmetry is less than those with sigma symmetry interactions.

3) How do you know whether things are stabilized or destabilized using AOM?

In sigma and pi donor interactions you are always stabilizing the ligand

orbitals and destabilizing the metal d orbitals. For pi acceptors, the opposite is true.

4) Is AOM only used for figuring out d-orbital splitting?

Yes.

5) Does the sigma donor/pi acceptor affect the relative energies in the MO diagram?

Yes. Sigma donor (or pi donor) ligand orbitals will always be lower in energy than metal d orbitals. Pi acceptor ligand orbitals will always be higher in energy than metal d orbitals.

6) How do you calculate the stabilization or destabilization magnitude for ligand orbitals using AOM?

You treat all of the ligand orbitals the same and stabilize them (for donors) by the same amount you destabilized the metal orbitals. The net energy change should equal 0.

7) The nickel complex (Ni(CO)4) had xy below xy/yz but on the diagram for a square planar complex this was reversed. Why?

Pi effects! The MO we drew for a generic square planar complex was sigma only. CO ligands are pi acceptors and are going to stabilize orbitals they can interact with in a pi fashion, while the sigma only interactions will only be destabilizing.

8) Does the angular overlap model still apply when S(r) is not equal to 1?

Good question! Technically no, because then we would have to figure out how S(r) affects the picture, but I think qualitatively it would still work.

9) Lit reference for AOM?

Check out the reading page.

10) What is the difference between ligand field theory and AOM?

The main difference is that AOM doesn't care about the symmetry of the ligand orbitals. They are all treated equally. For that matter, it doesn't care about symmetry at all. AOM is a little more quantitative for calculating d-splitting, but I think ligand field theory is a little more flexible and useful.

Questions Lecture 8 10/10/12

1) Why does removing electron density have a stabilizing effect?

Removing electron density minimized electron-electron repulsions, which is a stabilizing force. It you remove electron density from an orbital, it stabilizes that orbital.

2) Could you explain "16 e-" rule for square planar complexes?

As we saw in class, the d splitting diagram for a square planar complex has one orbital that is much higher in energy than the others. Because it is so high in energy, square planar complexes tend to not have it filled, preferring to have only 8 d electrons (s2+p6+d8 = 16).

3) Are there transformation matrices to convert Oh symmetry into others, such as square planar?

I don't think there is a mathematical trick to directly convert Oh to square planar, but next class we will go over the "Angular Overlap Method" which is a completely mathematical way to get d-splitting. You will like this.

4) Why does going down a group (3d to 4d to 5d) increase (delta)o?

The larger 4d and 5d orbitals result in stronger metal - ligand interactions so this effect is similar to what is seen for more highly charged metal ions, which pull in the electron density from the ligands, making shorter metal - ligand bonds. This is a little confusing at first, but you can make sense of it by comparing delta for a Ni2+ ion vs a Pt2+ ion --- the Pt2+ ion will have stronger M-L bonds and a larger (delta)o.

5) Why is the character of (sigma)h for the py and px group orbitals in Oh equal to 0 in the reducible representation???

It turns out to explain the (sigma)h, you do actually have to use the coordinate system explanation we went over first (see figure 10.7 in the book). Two of the x vectors (or y vectors) are +1 and two are -1 (sum to 0).

6) What does weak field mean?

Weak field means small d orbital splitting!

7) Smaller radius is strong field, but 4d and 5d are strong field??

Close, but no! Higher charge means strong field.... a more highly charged metal has shorter M-L interactions and the larger 4d and 5d metals have stronger M-L bonds due to greater orbital overlap.

8) Do we see Jahn-Teller effects from differences in ligands or just from geometry?

We will see the Jahn-Teller effect when there is unequal occupation of degenerate orbitals - Cu(II) d9, for example... it is not necessarily related to the ligands. Where the ligands will matter is let's say you have a d6 metal center with strong field ligands in one case (so all electrons paired in t2g) and weak field ligands in the other (so 4 e- in t2g and 2e- in eg). The high spin complex (weak field) will have a Jahn Teller distortion because there is unequal occupation of the t2g set. One thing to note, however, is that the JT effect is stronger for unequal occupation of eg than t2g.

9) Is Jahn Teller only for d4 to d7?

Nope! It's for any d electron count where you could have unequal orbital occupation of degenerate orbitals. So for octahedral it's d1, d2, d4, d5, d6 (high spin), d7, d8, and d9!

10) Why does Co3+ have a larger (delta)o than Co2+?

The higher charge on the cobalt makes the metal hungrier for electrons and the metal - ligand bonds are shortened. This raises (delta)o.

11) How do I tell if a complex is low spin or high spin for two complexes with the same metal in the same oxidation state?

In that case, it is just ligand effects! You can use the spectrochemical series to help you decide.

12) Why is unequal occupation of degenerate orbitals so disfavored?

For the best answer, check out the original paper by Jahn and Teller, "Stability of Polyatomic Molecules in Degenerate Electronic States. I. Orbital Degeneracy" at:

http://rspa.royalsocietypublishing.org/content/161/905/220

13) In what cases would you compress in x,y rather than lengthen along z

for Jahn Teller of an octahedral complex?

I need to find an example....

Questions Lecture 7 10/08/12

1) Where should I read about the coordinate system from the end of class?

Chapter 10, figure 10.7. Go through the discussion, if you still don't understand, let me know!

2) How do you draw an internally consistent coordinate system?

The easiest way is what I did in class, pick the vector that always points along the bond, draw it in and then use a right handed system for everything. This should work. You just need to make sure that when you rotate the molecule and perform the symmetry operations, the coordinate system doesn't move.

3) What is a pi acceptor and a pi donor?

A pi accepting ligand has an empty set of p orbitals to accept electron density from the metal (stabilizing effect). A pi donating ligand contains a filled set of p orbitals that donate electron density to the metal center (destabilizing effect). This information is not from the character table. It comes from a combination of lewis structures, electron counting, and drawing an MO diagram for the ligand itself!

4) Is the order the sum of the kinds of symmetry elements or is it the total number of symmetry elements (C3 vs 3C3, for example)?

The order is equal to the total number of symmetry operations in the group, so 3C3 would count for 3.

5) Is there a difference between dashed and not dashed lines in the MO diagrams?

Nope.

6) In finding the reducible representation, does 0 mean no change?

No. You only get a contribution to the reducible representation if the atom (or orbital) doesn't move. So if I was looking at an atom and it doesn't move, it is a +1. That is why the number under E is always equal to the

sum of the things we are looking at - under E nothing moves.

7) How can a molecule satisfy the 18e- rule and still have a positive charge?

Perceptive question. The origin of 18e-s is from s2 p6 d10, but knowing that a molecule has 18e- doesn't tell you anything about how many d electrons it has. The number of d electrons comes from the oxidation state, which comes from looking at a combination of the ligand charges and the total charge on the molecule.

8) The Oh splitting we drew today was for a metal with six of the same ligands. Does having different ligands change things or can we always use this?

Indeed what we drew was for six of the same ligands. Things typically change if we have multiple kinds of ligands because the symmetry of the molecule changes! The amount of the change depends on the kinds of ligands. We will get to this.

9) How do you determine the oxidation state in Ti(NR2)3?

We said NR2 is a 3 electron, L-X type ligand. So there are three of these LX ligands, and thus three X's. So the titanium is Ti(III), d1.

10) What exactly is a "dative bond"?

A dative bond is a 2 electron bond where the electrons come entirely from the ligand.

11) Why does the ligand being in a macrocycle typically result in different electron counts on the metal in square planar complexes?

So, we said square planar complexes tend to be d8. The exception you see most frequently is if you have a ligand that is all tied together like a porphyrin (a macrocycle). The structure of the macrocycle forces the metal center to take on a certain geometry, in this case, square planar, even if it would prefer to have a different geometry based on electron count.

12) What is the best book for ligand field theory?

I personally like Albright, Burdett, and Wangbo - Orbital Interactions in Chemistry. Cotton's Chemical Applications of Group Theory is also good.

13) Where is the C2 axis in Oh?

Check out the otterbein website, or here http://www.huntresearchgroup.org.uk/teaching/teaching_MOs_year2/L7_Addn_Oh_point_group.pdf for good pictures.

14) To find the symmetries of the orbitals on the metal do we just look at the character table?

Yes! Look at the quadratic functions xz, yz, xy, z2, and x2-y2 on the character table.

15) Why does the pi acceptor push the t2g orbitals up in energy?

They don't. Pi donors destabilize t2g and pi acceptors stabilize t2g. We'll go over this again wednesday.

16) Are the ligands always lower in energy than the metal?

Not always, especially if you are talking about empty p orbitals on the ligands.

17) With pi donors in Oh symmetry is t2g ever higher in energy than eg?

Nope, always eg over t2g.

18) How do you get Re(I) d6 in the Re(CO)5(PF3) example?

It was because of the positive charge (which you forgot). All of the ligands bound to the metal are neutral 2 electron donors. The positive charge comes from the metal having one less electron than its valence count, so Re(I) d6.

Questions Lecture 6 10/05/12

1) Was I supposed to know what L and X mean?

They are just symbols for ligands that donate 2 electrons (L) and one electron (X). So an L2X ligand is a 5 electron donor (2+2+1).

2) How do you find the LUMOs and HOMOs of molecules by inspection of their lewis structures?

If only it were that easy. Typically if you have some extra information, knowing what the HOMO and LUMO look like can be simple. In the

example on F3B-O(C2H5)2, we knew we were talking about an acid base interaction. We know that the O(C2H5)2 has a high lying lone pair to donate (we just drew the MO for water) and I mentioned that boron has a low lying empty orbital to donate into. You would need a lot of practice before you could just figure this out from scratch.

3) Is electronegativity the only factor to consider when assigning relative energy levels in drawing MOs?

No, and it's not even the most important, just typically a reflection of orbital potential energies. What you really need to consider is orbital potential energies. Take a look at Figure 5.13 in your book. Notice, that the 2p and 2s of oxygen are both lower in energy than the 1s of hydrogen! That's why we drew it that way in class.

4) How do you know that an E representation is two orbitals?

We can tell how many orbitals an irreducible representation counts for by looking at the dimension (the character under E -identity in the table). The E irreducible representation always has a 2 and similarly, T always has a 3.

5) How does cyclopentadienyl ion interact with metals?

It typically sits atop a metal facially and it's typically drawn with a bond to the center of the aromatized ring. You will also see it drawn with bonds to all five carbons trying to represent how all interact equally by resonance. We count it as an L2X with interactions to the static nonaromatic structure through the two pi bonds (L2) and to the deprotonated carbon (X).

6) Why is I- a better base than Br-?

It's not!!! It's a softer base than Br-. Softer, not better. This means it will form stronger bonds with late metals like Hg.

7) What was the MO for triangular H3 trying to show and why is it important?

It was trying to demonstrate how you can use fragments of molecules to draw MO diagrams, not just atoms themselves. It also introduced symmetry adapted linear combinations of atomic orbitals!!

8) What is a dative interaction?

It is a two electron interaction where the ligand is donating the two electrons to the metal. So all of the L-type interactions we talked about could be considered "dative". Another term is "coordinate covalent".

9) If we don't count electrons using the XL formalism is that okay?

Sure! I am just teaching the way I think makes the most sense, but if you have already learned another method or you prefer to use the book, feel free. You will always (if you do it correctly) get the same answer.

10) Is there a more analytical definition of hard/soft?

Yes. Pearson wrote a later paper about this. Here's the reference: Robert G. Parr and Ralph G. Pearson (1983). "Absolute hardness: companion parameter to absolute electronegativity". J. Am. Chem. Soc. 105 (26): 7512–7516. doi:10.1021/ja00364a005

11) What exactly is polarizability?

Electric polarizability is the relative tendency of a charge distribution, like the electron cloud of an atom or molecule, to be distorted from its normal shape by an external electric field, which is applied typically by inserting the molecule in a charged parallel-plate capacitor, but may also be caused by the presence of a nearby ion or dipole. Generally, polarizability increases as volume occupied by electrons increases. In atoms, this occurs because larger atoms have more loosely held electrons in contrast to smaller atoms with tightly bound electrons. On rows of the periodic table, polarizability therefore increases from left to right. Polarizability increases down on columns of the periodic table. Likewise, larger molecules are generally more polarizable than smaller ones.

12) In H3 the nonbonding and antibonding orbitals are degenerate?

Yes, but only in triangular H3 with D3h symmetry. If we had an isosceles triangle they wouldn't be!

13) What is the benefit of reviewing the neutral ligand formalism? How do we use this tool?

Ah! You will see soon. Being able to quickly count electrons and determine oxidation states is essential to understanding reactivity and electronic properties of transition metal complexes.

14) More practice on neutral ligand formalism?

Check out chapter 13 of your book!

15) In H2O why is there an a1 relatively antibonding to O and bonding to H?

So when you interact two orbitals you generate a bonding and an antibonding level. What happens with H2O (and many other molecules) is that three orbitals are interacting at once. So qualitatively one is bonding to everything, one is antibonding to everything and one is in between. The in between one should be bonding with respect to the highest energy orbital, but antibonding with respect to the lowest energy orbital it comes from.

16) If the lowest energy atomic orbitals are not from the same irreducible representation (eg. a1 and a1) do the first compatible set make the sigma molecular orbital?

To determine if something is sigma bonding or pi bonding (or delta bonding) the symmetry of the interaction is important. If there is no node in the interaction then it is a sigma bond (think sphere), if there is one node then it is pi (think dumbbell) and if there are two nodes then it is delta (think flower). The lowest energy interactions don't have to be sigma bonds, but they usually are.

17) Could you go over how to use lewis structures to get number of electron donated in the X/L model?

Yes, next class!

18) Inequivalent lone pairs... Is our understanding of pKa solely based on the nonbonding lone pair?

Yes. This is the lone pair that get's involved in acid-base chemistry. The other one can be thought of as relatively inert. Water only has one pKa! I don't think anyone has ever measured the pKa of the inert lone pair (H4O2+).

Questions Lecture 5 10/03/12

1) Lots of questions about the pi bonding set being lower in energy than the sigma bonding set in interacting the C and O 2p orbitals and assignment of the HOMO as the carbon lone pair. Here is the complete MO drawn a bit more clearly:

As can be seen from this diagram, the carbon 2s and 2p orbitals mix strongly with the oxygen 2p orbitals because of the close energy match. There is a complete description of this in the book on page 146-147 along with figure 5.14. For the HOMO being the lone pair, since the oxygen 2p electrons are lower in energy they are used to fill the pi bonding levels, leaving the higher energy p electrons of carbon to fill the HOMO sigma level!

2) There were also several questions on interacting different orbitals like pz and dxz. When you do this you need to picture things in 3D and the centers of the atoms (or orbitals) need to line up! This will help you to see when things do or do not have overlap. There is a useful example in section 5.1.3 of the book.

3) For D3h, what are the two C3's?

When operations such as C3 are in the same class, the listing in the character table is 2C3, indicating that the results are the same, whether rotation is in the clockwise or counterclockwise direction (or, alternatively, that C3 and C3

2 give the same result). In either case, this is equivalent to two columns in the table being shown as one. Similar notation is used for multiple reflections.

4) What's the difference between upper case and lower case symmetry labels (irr reps)?

Upper case letters are used for the representations in character tables.

When we assign molecular orbitals to irreducible representations, the convention is to use lower case.

5) How do we assign symmetry labels to molecular orbitals?

We look at the interactions and see which they belong to... We will finish up H3 next class and we will be doing other examples for several more classes so you will get the idea soon.

6) Does the longer bond length show less stability?

Yes.

7) Will we talk about hybrid orbitals?

Not exactly.... but orbital mixing in MO diagrams is important as we saw for CO. This is a more accurate representation than hybrid orbitals.

8) What does the 2e' orbitals of H3 look like?

I'll draw them next class!

9) Why does greater electronegativity correspond to lower energy orbitals?

Orbital energy is really determined by orbital potential energy (table 5.2 in the book). These values steadily become more negative from left to right within a period as the increasing nuclear charge attracts all the electrons more strongly. In general this value tracks qualitatively well with electronegativity, so we will use that periodic trend for ease of understanding.

10) Is the oxygen lone pair in CO as unreactive as the bonding electrons?

It is buried with the bonding electrons, so conceptually this is a fair way to look at it.

11) In ionic bonding the difference in electronegativity is great yet the bonds are strong. What's the MO diagram for that look like?

For the most part MO diagrams give us good representations of bonding interactions where electrons are shared, and in this class that is the way we will use them. The book does go over an example of ionic bonding in section 5.3.2, but it even cautions that this is not necessarily the best way

to describe bonding in these molecules. A different model is needed.

Questions Lecture 4 10/01/12

1. How do we know that A1 and B1 from C2v are IR active?

A vibrational mode is IR active if it corresponds to an irreducible representation that has the same symmetry (or transforms) as the Cartesian coordinates x, y, z, because a vibrational motion that shifts the center of charge of the molecule in x, y, or z results in a change in the dipole moment. To find how an irreducible representation transforms, just look at the right hand portion of the character table.

2. How do we know if something transforms a certain way?

We look at the character table and ask ourselves, how does that thing move under the symmetry operations of the point group (define your z axis as the principle rotation axis). For example, the reason the Cartesian axis y transforms as B2 in C2v is because under E the vector is unchanged, under C2 it flips direction, under (sigma)v it flips directions, and under (sigma)v' it is unchanged.

3. Why are x, y, and z IR and Raman active?

They are not. Only x, y, and z are IR active. Raman active vibrations transform as x2, y2, z2, xy, xz, yz, and combinations of these.

4. How do you count the number of operations in a class?

The number of operations in a class is the number of that type of equivalent operation in the molecule. For example, in octahedral (Oh) symmetry the top of the character table says 6C4, meaning there are six symmetry equivalent C4 axes or 6 operations in the C4 class.

5. How does (sigma)v'(yz) for H2O flip the x vector if Oxygen is on the z axis?

If we look at the mirror plane that is in the yz plane, it contains the y and z axes and doesn't change them. However, it reflects the x axis, giving it a -1 contribution to our analysis.

6. All questions on axes with the cis and trans dicarbonyl complexes will be addressed at the start of next class!

7. What is the significance of the results of the formula for determining

irreducible representations from a reducible one?

A reducible representation by itself doesn't tell us anything about the molecule - in order to use the power of the character table, we have to find what collection of irreducible representations it corresponds to, which is what the formula does.

8. Is there a way to predict where IR stretches will be using group theory?

Sadly, no.

9. Why is A1 the symmetric stretch for H2O and B1 antisymmetric?

The vibrations transform as A1 for the symmetric stretch (and bend) and as B1 for the antisymmetric stretch. That's why they're drawn that way.

10. Why does D2h ML2(CO)2 only have one IR active mode?

When we did the CO stretching analysis, we found two stretching modes, Ag and B3u. Only one of these is IR active because only one of them transforms as x, y, or z.

11. Does antisymmetric mean "not symmetric"?

Yes.

12. What does "off the diagonal" mean?

We will normally not be working with matrices in this class, but I demonstrated where the characters in the character table came from last class. When we had block diagonalized matrices, the numbers on the diagonal corresponded to the characters in the character table. When we do a transformation that moves an atom, it's contribution is "off diagonal" meaning it doesn't contribute to the trace.

13. How do we define the z axis?

The z axis always corresponds to the highest order rotational axis, Cn.

14. What's the difference between (sigma)v and (sigma)v'?

They are both parallel to the principle rotation axis and just correspond to two different reflection planes.

15. Does it matter how we define the axes or will answers be okay if we

are consistent?

We should always label the z axis as the principle rotation axis. If you don't do this, but are consistent you will always get abundant partial credit, but will loose a couple points for not properly defining z.

16. What does the 3 in B3u stand for?

I still need to find a good answer for this...

 

Questions Lecture 3 09/28/2012

1. What is a (sigma)d exactly - it's not the same as a (sigma)v!

Indeed it is true. They are not the same. Here are the full definitions:

σv: Vertical mirror plane - Mirror plane that is parallel to, and encompasses, the principal rotation axis. σd: Dihedral mirror plane - A σv that bisects angle between C2 axes ⊥ to principal axis.

2. Why could you take Rpi (which only dealt with x and y) and just add the 1 for the z? Is it because z is unaffected by a C2 rotation?

That is exactly correct! 

3. How do we incorporate other irreducible representations that have A, B, E, and T?

We will see this next class.

4. What class would be needed to learn the mathematical aspects of character tables?

A simple class on linear algebra would be helpful, and if you wanted to go deeper, a math class on group theory.

5. (sigma)h, (sigma)v, (sigma)d - I'm lost!

These are the three kinds of mirror planes. (sigma)h is a mirror plane perpendicular to the principle rotation axis and (sigma)v is a mirror plane parallel to the principle rotation axis. (sigma)d is a subset of (sigma)v - see answer to question 1.

6. I don't understand the rules for having a complete group.

We'll go over this once more next class.

7. The sum of the dimensions squared is equal to the order???

The dimensions are the characters under E, the order is the number of classes (E, C2, sigmah, i). So in the example in class, the order was 4 and of we squared the characters under E and then added them up we need to get 4 to have a complete set of irreducible representations!

8. What does irreducible representations are orthogonal mean?

This means that if you multiply the characters for any pair of irreducible representations and sum them up, you should get 0!

9. What's the difference between inversion and an improper rotation?

Inversion is a subset of improper rotation - an S2 is technically an inversion.

10. If you matrix multiply 2 symmetry operations, should you get a new symmetry operation in the group?

YES!

11. What are reducible and irreducible representations exactly?

 

Representations: these are the set of matrices that can be used to do the mathematics of group theory and label objects

There are two types of representations: reducible and irreducible:

Reducible representation are linear combinations of irreducible reps.

Irreducible reps can be used to generate groups, just as symmetry elements

 

11. Can we look a little closer at Sn symmetry operations?

Sn is the symmetry operation that combines proper rotation and a reflection through a perpendicular reflection plane. The axis about which the rotation occurs is called an improper axis. We'll do some examples on PS 2.

 

Questions Lecture 2 09/26/2012

1) What is the interaction between (electrons in) orbitals that makes one singlet lower in energy than another?

It is due to magnetic repulsion effects. Electrons of the same spin magnetically repel each other and want to be far apart. In oxygen, the 1(delta)g state (electrons of opposite spins paired in a single orbital) is lower in energy than the 1(sigma)g

+ state (electrons of opposite spins separated in two orbitals). This is because there is an magnetic attraction between electrons of opposite spins.

2) Not sure how to follow the electron filling diagram with the arrows.

Here it is in a nicer form:

3) Why would we expect Zn2+ to be most stable?

Having a filled d-shell (a d10 species) is very favorable.

4) What does it mean to share lone pairs with more electronegative ligands?

Electrons can be and are shared unevenly in real molecules. More electronegative ligands have the bulk of the charge localized on them.

5) Will we be considering hybrid orbitals with transition metals?

Not generally. We will see few examples of real molecules with s,d mixing but the effect is somewhat small.

6) What decides the z-axis in a complex?

Oooo. Great question. We will learn this next class - the z axis is the principle rotation axis (group theory).

7) Are there exceptions to lp-lp > bp-bp repulsions?

Not for our purposes of learning vsepr, but because multiple bonds are more repulsive, and because in transition metals we can have even quintuple bonds, there may come a point where an exception would arise.

8) What is the rationale for the electropositive character of transition metals?

Check out the electronegativity chart. Most things that transition metals bond with just happen to be more electronegative.

9) Do we have to memorize the different ideal bonding configurations?

NO!

10) In :C(triple bond)O: why is there a partial negative charge on carbon and a partial positive charge on oxygen?

If we count one electron from each bond plus the lone pair, carbon has 5 electrons and so does oxygen. Neutral carbon has 4 electrons in its valence shell (-1) and neutral oxygen has 6 (+1).

11) Do multiple bonds take up more space because of the shape of the bonding orbitals compared with single bonds?

Kind of - if you think about the orbitals in multiple bonds, you can think of sp hybridization. I typically just think about the number of electrons involved in the interaction. A double bond has 4 electrons, so it needs

more space (is more repulsive) than a single bond that only has 2 electrons.

12) What is the name for electrons in bonds that are shared evenly?

Covalent (as opposed to ionic, shared unequally).

13) How do ligand types and the 18e- rule come into play?

We will get there in a couple classes!!

14) How much variation in bond angles do we see due to differences in ligand electronegativities?

Here's a good example:

Molecule               X - P - X bond angle

PF3                        97.8

PCl3                      100.3

PBr3                      101

15) What are the differences in electron counting according to ionic and covalent methods?

In one, you have to assign oxidation states first, in the other you can use only neutral ligands. We'll do a lot more of this in a couple classes.

16) When drawing SO42- do you just pull the electrons off the sulfur and put them on oxygen?

Yes. Sulfur ends up as S(6+, no electrons) and each O ends up as O(2-, 8 electrons). It doesn't matter what resonance form you use.

17) Is the strong lp-lp repulsion due more with electron density delocalization or greater electron density?

I like to use the localization argument. But technically, electron density and electron localization are pretty much the same argument.

18) Why do double and triple bonds have a greater repulsive effect?

More electrons in a small space. As you mention, it also does shorten the

M-L bonds which does indeed contribute.

19) How does shielding affect the electronegativity at the metal.

D orbitals are less penetrating and are more greatly shielded by s and p orbital electrons.

20) Why does the energy of a d orbital on a metal change as ligands are added?

Electron-electron repulsive forces (for now we're just thinking about ligands as point charges).

21) Confused about delatE and pairing energy. What is deltaE?

deltaE, as we will learn about a lot more in a few classes, is the separation in energy between d orbital energy levels. If we're talking about an octahedral complex, it's the separation between the Eg levels and the T2g levels. If this separation is large, then it is likely that the complex will have its electrons paired (low spin) and if the separation is small, then it is more likely to have the electrons in singly occupied orbitals (high spin). We'll learn a lot more about this in days to come.

Questions Lecture 1 09/24/2012

1) What is singlet vs triplet oxygen?

Many people asked this and it is an excellent question. Normal ground state oxygen is called "triplet" because it has two unpaired electrons with the same spin in its highest occupied molecular oribitals (degenerate e set). When oxygen is excited, it can pair those spins in one of the orbitals, or it can leave the two electrons separate, but flip one of the spins. The picture below shows what I mean:

 

2) What is an MO diagram?

Molecular orbital diagram. It shows how atomic (or group) orbitals combine to make molecular orbitals.

3) Are the lecture slides online?

Typically the lectures will be on the chalk board. Today was a special case. I'll take a poll next class to see if we need the notes online.

4) Does "group" in "group theory" refer to the groups in the periodic table?

No. It refers to a mathematical group. In mathematics, a group is an algebraic structure consisting of a set together with an operation that combines any two of its elements to form a third element. You will see we will get there.

5) Carbon with six bonds???

Crazy I know. Perhaps a better way to think of them is "interactions". I refer you to: http://www.nature.com/nchem/journal/v4/n2/full/nchem.1257.html

6) What is the mechanism of the de-excitation in heme?

When the iron is present, the deactivation pathway is loss as heat from molecular vibrations.

7) Do you have a group theory worksheet?

If you still need extra notes once we go over the group theory sections, let me know! Check out Chapter 4 in the book for a sneak preview.

8) What is the degradation mechanism of Schrock's TREN-HIPT catalyst?

The ligand arms get protonated and the ligand falls off :(

9) How does the puckering affect the MO diagram of metal-bound porphyrins?

It destabilizes dz2 and dxz mostly. The broken symmetry is indeed very significant in the properties of those kinds of molecules.

10) What/who inspired you to be an inorganic chemist?

Oh, lots of folks. But mostly, Kit Cummins, my PhD supervisor at MIT. He taught me how inorganic chemists really think outside of the boundaries of traditional rules and can make anything they put their minds to.

Questions or comments?Contact us or email catalysthelp@uw.edu

Copyright © 1998-2012 Learning & Scholarly Technologies | Privacy | Terms