Sciencemadness Discussion Board - Antibonding and Delocalization....

Antibonding and Delocalization....

Rich_Insane - 17-12-2009 at 21:46

I'm taking a Chemistry class right now, and we're on the chapter concerning bonding and the orbital arrangements in bonding. The chapter first covered Lewis dot structures briefly, then went onto hybridization, which I barely understood (taken electrons from the 2s orbital and putting them in p orbitals). Now this new idea of antibonding has scrambled my brains up.

So the first topic was about how antibonding orbitals affected paramagnetism in O2 but not N2 (I understand that this is because O2 has two unpaired electrons in the p orbital?) . I'm not sure what I don't understand. I really don't understand what exactly an antibonding orbital is, and what the significance of it is. the next topic was about antibonding in heteronuclear atoms. I'm not sure how it works, since I ddin't get the whole thing in the first place. Then finally, there was a lecture on delocalization -- I understand that this is where the electrons are moving throughout the molecule, but I do not get how this works with the pi orbitals (antibonding and bonding). I remember vaguely from an O Chem class I took (just to see how difficult it was) that on the section with allylic systems, we looked at how there were "figure eight" orbitals arranged in different manners, and the one that did not repulse the charges would best fit, but I'm not sure what that meant. Will someone kindly explain what all of this is about?

JohnWW - 17-12-2009 at 23:29

Frankly, I am suspicious of the term "anti-bonding" or "anti-bonding orbitals", simply because such orbitals, in which extra ("anti-bonding") electrons are supposed to be accommodated, simply do not appear as legitimate solutions of the Schrödinger Equation, as s, p, d, f, or (for superheavy elements) g orbitals which have their characteristic shapes. I suspect a "fudge". It is much more realistic to expect, instead, that such supernumary electrons would be accommodated in the vacant orbital (being one allowed to exist by the Schrödinnger Equation) that is next in energy above the highest-energy occupied orbital.

sonogashira - 18-12-2009 at 02:27

One can imagine that atomic orbitals combine constructively or destructively - rather like a light wave does/can - to give a bonding orbital (of lower energy than the 2 atomic orbitals)... and... to conserve energy, an antibonding orbital of higher energy (out of phase interaction).

Delocalization is more about p-orbitals (your 'figure 8' orbitals) being aligned like in benzene, dienes, other aromatics etc., where one may imagine that they constructively combine to produce a 'cloud' above and below the ring.

Electrons go in lowest energy orbitals for preference, of cause, but with energy (light for instance) they may be promoted to higher (antibonding) orbital which weakens the bond.

It is some time since I read this but really I think it is just a simple model of explaining the observed molecular orbital electron energy levels. I'm sure an organic chemistry book gives a simple explanation better than me... and better than a silly physical chemistry textbook... but that is my prejudice that nature does not conform to man's silly rules (mathematics).

kmno4 - 18-12-2009 at 08:14

Two remarks:
1.Dot(ted) structures of bondings are mainly for "children".
2. You cannot mess delocalization with "antibondig" orbitals.

Delocalization = orbital(s) more than two centered.
For example, benzene possess 6 delocalized molecular orbitals (of course multicenter)
http://www.cem.msu.edu/~reusch/VirtualText/react3.htm#rx8
"Antibonding" orbitals - molecular orbitals with higher energy than original (before bonding creation) orbitals. Maximum electron density is out of line joining bonded partners (cores, nucleus), on the contrary the to bonding orbitals. This simple picture complicates when there are more than 2 partners of bonding.
Sometimes they are called "virtual" (when are empty, no electrons) because they are only solutions of mathematical equations.

[Edited on 18-12-2009 by kmno4]

bbartlog - 18-12-2009 at 08:42

Delocalization isn't just about aromatic compounds, either. One of the simplest examples is the carboxyl group (COOH), which is generally drawn as if there were one -OH group and one double-bonded oxygen - but in reality, the bond lengths between carbon and the two oxygen atoms are equal; it's as if the two oxygens are sharing three electrons (from the carbon) and one (from the hydrogen) equally...

kmno4 - 18-12-2009 at 10:57

Quote: Originally posted by bbartlog

One of the simplest examples is the carboxyl group (COOH), which is generally drawn as if there were one -OH group and one double-bonded oxygen - but in reality, the bond lengths between carbon and the two oxygen atoms are equal (...)

You are wrong - these bonds differ markedly, see for example:
http://dx.doi.org/10.1021/ja01232a022
or even for solid acid
http://dx.doi.org/10.1107/S0365110X58001341

DJF90 - 18-12-2009 at 11:38

Yes what you say bbartlog is complete bullshit. What is close, is that the ANION of the carboxylic acid group ("carboxylates") have equal C-O bond distances, as there is delocalisation/resonance.

kmno4: Lewis structure representation can be very useful for working out oxidation states/formal charges. They are not just "for children".

Sonogashira: You are right about the constructive/destructive superposition. "Atomic orbitals" are really wavefunctions that describe the position of an electron. As "waves", they therefore can superimpose with eachother, forming "molecular orbitals. If you're interested further I suggest you look at LCAO (thats Linear Combination of Atomic Orbitals for those not in the know) for simple diatomic molecules (like nitrogen, oxygen, fluorine - these are probably the most interesting simple diatomics with respect to how the MO diagram changes). For construction of MO diagrams for more complcated molecules (by more complicated I mean stuff like water, ammonia, methane, SF6, you need a pretty thorough understanding of group theory (from a chemist's perspective) and use symmetry groups to combime the atomic orbitals.

You would be suprised as to how much information can be derived from an MO diagram; they can be far more useful that just predicting paramagnetic properties and bond orders...

Of course, in between the simple diatomics and the simple polyatomics, it would make sense to look at the first transition series and the orbital diagrams for complexes of different geometries. Alot of this can be done without knowledge of point groups and symmetry, and there are two models you could consider looking at. Crystal field theory is the simpler one, and is based upon purely electrostatic interactions. Ligand field theory is more complicated, and comprises covalency and pi-bonding between the metal atom/ion and the ligands.

Using MOs in this context, one can rationalise the spectroscopic series, and also aspects of organometallic chemistry (A good book if you are interested in this area of chemistry is Organometallics by Elschenbroich (3rd Edition.)

And JohnW: I'm pretty sure this isn't correct also. Using LCAO as the simplest approach to this problem, then the combination of the wavefunctions (atomic orbitals) can be addition or subtraction, leading to TWO molecular orbitals (this is a key principle in MO theory - combining a number of AOs gives an EQUAL number of MOs). One of these new wavefunctions describes the bonding orbital, and the other the antibonding orbital. The schrodinger equation cannot be solved exactly for anything other than a hydrogen atom (or more correctly a "hydrogenic atom" - anything with only one electron; He(+), Li(2+), etc...), but we can make some reasonable assumtions and solve it with good accuracy for multi-electron systems. So long as the wavefunctions that describe the MOs are eigenfunctions of the hamiltonian for the system then the schrodinger equation is satisfied. The wavefunctions that are found to satisfy the schrodinger equation (generally by trial and error) can the be mapped mathematically to determine the "shape" of the bonding and antibonding MOs.

[Edited on 18-12-2009 by DJF90]

kmno4 - 18-12-2009 at 15:17

Quote: Originally posted by DJF90

kmno4: Lewis structure representation can be very useful for working out oxidation states/formal charges. They are not just "for children".

"Oxidation state" or "formal charge" is misleading fiction.
It represents no physical feature of any really existing molecule.
What information does formal charge +1 for nitrogen in NH4(+) ion give me ? Gives me nothing, because real charge on N in ammonium ion is negative.

Rich_Insane - 18-12-2009 at 15:42

So for clarification, antibonding orbitals are simply extra orbitals with very slightly larger energies than bonding orbitals? As the electrons are filled in a shell, the antibonding orbitals get filled last?

12AX7 - 18-12-2009 at 23:22

Antibonding orbitals mean what they say: they don't contribute to bonding. (I'm not sure if they have a repelling action, but they certainly don't have the attractive action of ordinary bonds.) If an electron is there, it would only be there due to electron affinity, charge balance or something like that. Presumably, there are more possible compounds which fill the bonding orbitals only, i.e., which attain a lower overall energy state, that is to say, this molecule is relatively unfavored among other possibilities: this suggests that molecules with antibonding electrons should be fairly reactive.

Tim

kmno4 - 19-12-2009 at 00:07

"Antibonding" orbitals may contribute to lowering bonding energy.
It is easy to see on MO diagram for N2 molecule (for example)
http://www.meta-synthesis.com/webbook/39_diatomics/diatomics...
The highest "antibonding" sigma<sub>s</sub> orbitals have lower energy than the lowest bonding sigma<sub>p</sub> orbitals. So "antibonding" orbitals are - in this case - bonding ones

.

[Edited on 19-12-2009 by kmno4]

DJF90 - 19-12-2009 at 08:41

kmno4 - Lewis structures are useful, and can be used for clarification when electrons need to be counted. For example, the molecule carbon monoxide. Using the Lewis structures allow us to see that there is a dative bond from the oxygen (as carbon would be electron deficient) and there is a formal -1 charge on carbon, which explains its behaviour as a nucleophile in reactions such as "ring stitching" when used in conjuction with thexyl borane or 9-BBN and a couple of other reagents (NaOH, H2O2, H2O, sometimes ethylene glycol also) to form ketones from dienes.

Another example of when Lewis structures are useful is for electron counting. For example, NO can be either a one electron or three electron donor as a ligand in transition metal complexes. The geometry of the complex is often (maybe always) different in each case, as the NO is bent or linear depending on how many electrons it provides. This can be determined by counting the electrons around the metal centre, and following the 18 electron rule (generally). Thus using Lewis structures to count electrons is not as "childish" as you make out; although the geometry about the metal centre is the same, the symmetry of the molecule is different, and thus it will belong to a different point group, which can only be predicted in this manner.

And your latest comment is incorrect. Antibonding orbitals are exactly that; ANTIBONDING. If you take the time to go back to the page you link to, and calculate the bond order for N2 in the way you are suggesting, you will find that it has a bond order of 5, which we all know is wrong as nitrogen has a TRIPLE BOND (bond order 3). If you are incapable of working out the bond order, use this formula:

Bond Order = (# bonding electrons - #antibonding electrons)/2

Now, if you do it the RIGHT way, counting the electrons in ANTIBONDING orbitals as ANTIBONDING ELECTRONS (as would make sense, no?) then you will get the RIGHT answer; a bond order of 3. Such an answer is expected.

Rich_Insane: If you look at the link kindly provided by kmno4 in the above post, you will see that on the left and the right are the atomic orbitals for each nitrogen atom, and in the middle is the molecular orbitals for the N2 molecule. If I can draw your attention to the 1s atomic orbitals. See how the 1sigma and 1sigma* (a slightly different notation to what is used on the page but is correct) molecular orbitals are vertically equidistant from the 1s atomic orbitals. This is primarily because energy is conserved.

If you go further up the page to the example for Helium, this will make more sense. The theoretical He2 molecule that the molecular orbital diagram is for has four electrons, a 1sigma and a 1sigma* molecular orbitals. As 2 electrons can enter each orbital (revise Aufbau principle, Pauli principle and Hund's rule of maximum multiplicity), both MOs are filled. Now as the lower (bonding) orbital is the same distance from the 1s atomic orbitals as the upper (antibonding) orbital, the average electron energy is that of the 1s atomic orbital, and so there is no energetical benefit for two He atom to bond to each other - this coincides with the fact that He is a MONOatomic gas. A further point to note is that if you try to calculate the bond order (using the equation I provided earlier), you find that it is zero; there is no bond between the He atoms, which again reinforces the fact that He is a monoatomic gas. Happy days.

MO diagrams are useful for predicting several properties of molecules. Two examples follow:

If you find the MO diagram for HF on the following page: http://butane.chem.uiuc.edu/pshapley/312/Lectures/L5/index.h...
Then you will notice two things (I advise you read what it says too...):
1) The 1s orbital of the hydrogen atom is interacting with the 2pz (by convention it is the z axis) of the fluorine atom. The page explains why this is.
2) There are non-bonding electrons residing in the remaining 2p orbitals of the fluorine atom. This can be used to explain why there is an electric dipole moment in the HF molecule; there are more electrons on the fluorine than on the hydrogen.

You could also compare the MO diagrams of N2 and CO (to be found in the links provided by myself and kmno4). These two molecules are isoelectronic, yet the MO diagrams will reveal that whilst N2 is paramagnetic (unpaired electrons), CO is diamagnetic (no unpaired electrons). This is due to the ordering of the 1pi and 2sigma (labelled as sigma3 in the diagrams) orbitals (the page I provided is clearer on this for CO), which is caused by the s-p mixing in carbon monoxide, as being a heteronuclear diatomic, the 2p orbitals on oxygen are of an appropriate energy to mix with the carbon 2s orbital (Recall that increase in electronegativity of an atom brings its atomic orbitals lower in energy - this can be seen in the diagram).

Now when there is conjugation in a molecule, antibonding orbitals cannot be considered. The best way to see conjugation, is to draw the atoms singly bonded, and then when there are double bonds, place p-orbitals appropriately. If there continuous string of more than two p-orbitals, there is conjugation.

Antibonding orbitals have their use in organic chemistry when discussing mechanisms. When a pair of electrons is placed in an antibonding orbital, the associated bond is then broken. A simple example of this is the Sn2 mechanism, which is shown well in "Organic Chemistry" by Clayden et al. Another example of this is the anomeric effect; I'll leave you to figure out this one.

sonogashira - 19-12-2009 at 10:10

Wow! Revising for exams?

The best way to learn is to teach - it is true! Best of luck for you

Rich_Insane - 19-12-2009 at 12:23

Woah, that's quite a bit to swallow. I sort of understand what's going on now. I understand how the orbitals interact in bonding, or I think I do. I know that Sigma bonding occurs with 2pz orbitals or 1s orbitals.

So the antibonding orbital is simply formed when 2s orbitals are combined, but the orbitals have a region of zero electron density within, right? How is it determined whether antibonding orbital or bonding MO is formed? Is it just that they are added to alternatively according to the Aufbau principle? From the diagram, you see that antibonding Sigma MOs are higher in energy, therefore you see a more antibonding effect rather than a bonding one. I understand that O2 is paramagnetic because there are two unpaired electrons, which allows aligning with a magnetic field. I actually recall learning about HOMO and LUMO in my brief preview of O Chem, but will someone tell me what those acronyms mean again?

DJF90 - 19-12-2009 at 12:42

Sonogashira - unfortunately not, although I should get to it at some point. I have to have a full time job this xmas vacation to support myself financially due to an unforseen situation, and so have little time to do my vac work (the tutors here are quite liberal

) let alone revise.

Rich_insane: You are right, there is a region of zero electron density between the nuclei in a sigma* MO. Note that the "lobes" on each of the nuclei are also out of phase (this is represented as one being shaded, the other left alone). When two atomic orbitals (of energy a) come together, two molecular orbitals are formed, one at energy (a+b) and the other at energy (a-b). By convention b takes a negative value. The bonding orbital is the MO formed that is lower in energy (i.e. the one at (a+b)), and the antibonding orbital is the MO formed that is of higher energy (i.e. the one at (a-b)), in comparison to the constituent atomic orbitals. The distance between the two energy levels is thus equal to 2b.

Please note that with heteronuclear bonds this does not apply: the bonding orbital is closer the lower lying atomic orbital, and the antibonding orbital is closer to the atomic orbital of higher energy. See the HF MO diagram for an example.

sonogashira - 20-12-2009 at 03:26

^ Sorry to hear that.. but Sir if that is your knowledge without revision then there is not a need for very much!! And many good men have taken xmas employment.. ever see miracle on 34th street?!

Best!

DJF90 - 20-12-2009 at 05:45

As great as that knowledge sounds, it is but the tip of an iceberg for what is required where I am

And as well as having to revise, I also have further stuff to learn, so revision is on hold for the moment (and probably, knowing my luck, right up to a couple weeks before the exam!). Fortunately I am much better at organic chemistry, which should require much less revision, allowing me to concentrate on the physical and inorganic chem.

Rich_Insane - 20-12-2009 at 14:21

This is just General Chem for me.... I'm really bad at this atomic level stuff, and I like the actual reactions in Organic Chemistry better, but sadly, I must know all this crap to be able to understand reaction mechanisms.....

DJF90 - 20-12-2009 at 14:28

Just be thankful you arent stuck doing quantum mechanics like me

kmno4 - 21-12-2009 at 03:56

Quote: Originally posted by DJF90

kmno4 - Lewis structures are useful, and can be used for clarification when electrons need to be counted. For example, the molecule carbon monoxide. Using the Lewis structures allow us to see that there is a dative bond from the oxygen (as carbon would be electron deficient) and there is a formal -1 charge on carbon, which explains its behaviour as a nucleophile in reactions such as "ring stitching" when used in conjuction with thexyl borane or 9-BBN and a couple of other reagents (NaOH, H2O2, H2O, sometimes ethylene glycol also) to form ketones from dienes.

As for me, it is very bad example because you can write structure of CO equally good with double bond as with triple or single bond.
But it experience does says that it cannot be double or single bond and on this basis you can count your formal charges. Electronic structure of CO was investigated many times, especially in theoretical context and it was definitely not simple counting electrons. It was investigated so many times because is it just unusual molecule.

"Antibonding" orbitals are what they are - orbitals with higher energy than original ones.
And as I have clearly shown these orbitals can have lower energy than so called bonding orbitals. Of course it is true for different sets of input orbitals. And in this context they are BONDING, because they lower electronic energy of molecule.
If you would stop using words "bonding" and "antibonding" for describing these orbitals, then nothing would change

If you have electron on "bonding" orbital and lower energy "antibonding" orbital is vacant then electron will go down to lower energy "antibonding" orbital - it is rather of course....

DJF90 - 21-12-2009 at 12:29

That is just the order of filling, it matters not whether an antibonding orbital is below a bonding orbital, it is still ANTIBONDING (otherwise they'd call it a BONDING orbital... or even there would be no point in differentiating between the two) You have to consider it in terms of bond order (this appears to be the only way you will understand, and even then you're having problems...).

An antibonding orbital is one which, when filled, will reduce the bond order (a measure of the strength of attraction (bond) between the two atoms), i.e. the bond is made weaker.

A bonding orbital is one which, when filled, will increase the bond order, i.e. the bond is made stronger.

This really is a simple concept and you seem intent on making it difficult for this fellow who started the thread. Do him (and yourself) a favour and go think about it.

CO can be drawn however you like, but again it should be relatively simple. Make as many covalent bonds as you can, until one of the atoms has a full valence shell (8 electrons in this case). Then see what possibilities you have for making the other atom satisfied. In this case, you have C=O, and a lone pair on the oxygen can be shared (dative cavalency) with the carbon so it too has a full octet.

There are many other examples I could give where electron counting is useful (like VSEPR theory- thats Valence Shell Electron Repulsion theory for those of you not familiar with it).