Destabilizing Molecules Through Electron Rearrangement?

[jaketodd]

By changing the arrangement of electrons in a molecule's energy levels (without adding or removing any electrons), can the molecule become destabilized or come apart? An example or two would really help.

Thanks,

Jake

 

[chill_factor]

sure.

in a previous post in the chemistry forum i repeated a derivation of the hydrogen molecule ion, H2+. it took a while and can't find it anymore.

long story short: it turns out that from the 2 atomic 1s orbitals on the 2 hydrogens, in the molecule you get 2 molecular orbitals, one of which is symmetric, and the other is antisymmetric.

the symmetric orbital has lower energy than the antisymmetric orbital. thus, with a H2+ ion, the single electron goes into the symmetric orbital, and the entire molecule is stable. This also explains the neutral hydrogen molecule, since the symmetric orbital can hold 2 electrons of different spin states.

but if one of the electrons was excited, say by a laser, the next higher molecular orbital up is the antisymmetric one with a node for the probability distribution between the 2 protons, and the molecule gets destabilized.

 

[sophiecentaur]

Sounds to me like you're talking Allotropes, here.

 

[chill_factor]

Allotropes means there is at least one metastable phase that is not the most stable thermodynamic phase, but is kept stable at room temperature by kinetics.

for example, diamond is thermodynamically less stable than graphite, but since the relaxation time is unimaginably long, it is essentially stable, but the constituent molecules are the same.

this has little to do with molecular electronic structure, but the OP asked about electronic effects.

 

[jaketodd]

sure.

in a previous post in the chemistry forum i repeated a derivation of the hydrogen molecule ion, H2+. it took a while and can't find it anymore.

long story short: it turns out that from the 2 atomic 1s orbitals on the 2 hydrogens, in the molecule you get 2 molecular orbitals, one of which is symmetric, and the other is antisymmetric.

the symmetric orbital has lower energy than the antisymmetric orbital. thus, with a H2+ ion, the single electron goes into the symmetric orbital, and the entire molecule is stable. This also explains the neutral hydrogen molecule, since the symmetric orbital can hold 2 electrons of different spin states.

but if one of the electrons was excited, say by a laser, the next higher molecular orbital up is the antisymmetric one with a node for the probability distribution between the 2 protons, and the molecule gets destabilized.

So what are the requirements for destabilization when promoting electrons up energy levels please? I am not understanding the symmetric, antisymmetric, the nodes, etc. Please help.

Thanks,

Jake

 

[chill_factor]

So what are the requirements for destabilization when promoting electrons up energy levels please? I am not understanding the symmetric, antisymmetric, the nodes, etc. Please help.

Thanks,

Jake

sure.

lets say you have 2 protons and an electron. this is a hydrogen molecule ion: H2+.

let the distance between the protons be R and apply the Born-Oppenheimer approximation such that the protons do not move. The chemical dynamics of the molecule is determined by the electron motion alone. We write this as the electronic Hamiltonian

He = -h^2/2m * ∇^2 - e^2/rA - e^2/rB + e^2/R

if you let R approach infinity and look at one of the protons (lets call this one proton A) the potential looks very much like a hydrogen atom. Likewise, if you look at the other proton called B, the potential also looks like a hydrogen atom.

ψA = β*e^rA/a0
ψB = β*e^rB/a0

where β is a constant of normalization. so for the case that R = infinity, the assertion that the molecular wavefunction ψM is the normalized sum of the 2 hydrogenic wavefunction is correct. if R is reduced to some finite number, here's the thing: while ψM = C1*ψA ± C2*ψB is no longer exactly true, it is a good approximation. Indeed it turns out to capture the correct qualitative behavior, although energy calculations require much more sophistication.

ψM = C1*ψA ± C2*ψB

we know that the nuclei are identical, so the electron distribution must be symmetric. The only way for this to happen is the absolute values of C1 and C2 are the same, otherwise the electron would spend more time at one of the protons despite both protons being identical. So |C1| = |C2|.

So the molecular wavefunction is actually 2 wavefunctions:

ψM+ = C1(ψA + ψB) and ψM- = C1(ψA - ψB)

For ψM+ the wavefunction is very low far away from both protons and hit a peak at the nuclei. But it is nowhere zero and the same applies for its absolute square. That's the bonding orbital.

For ψM- note that the wavefunction has a - sign. That means it crosses zero somewhere. When you graph its absolute square, it turns out that it crosses zero between the protons. Therefore, when you square it, there is a point in the graph that is a ZERO (node) and has NO ELECTRONS there. That is the anti-bonding orbital.

When you promote an electron up from a bonding orbital to the anti-bonding orbital, you're giving existence to a system state that includes a node between the 2 protons, which separates them.

Here's a wiki article that explains this in a better way and adds the energy calculations to boot.

http://en.wikipedia.org/wiki/Holstein–Herring_method

 

[jaketodd]

sure.

lets say you have 2 protons and an electron. this is a hydrogen molecule ion: H2+.

let the distance between the protons be R and apply the Born-Oppenheimer approximation such that the protons do not move. The chemical dynamics of the molecule is determined by the electron motion alone. We write this as the electronic Hamiltonian

He = -h^2/2m * ∇^2 - e^2/rA - 2^2/rB + e^2/R

if you let R approach infinity and look at one of the protons (lets call this one proton A) the potential looks very much like a hydrogen atom. Likewise, if you look at the other proton called B, the potential also looks like a hydrogen atom.

ψA = β*e^rA/a0
ψB = β*e^rB/a0

where β is a constant of normalization. so for the case that R = infinity, the assertion that the molecular wavefunction ψM is the normalized sum of the 2 hydrogenic wavefunction is correct. if R is reduced to some finite number, here's the thing: while ψM = C1*ψA ± C2*ψB is no longer exactly true, it is a good approximation. Indeed it turns out to capture the correct qualitative behavior, although energy calculations require much more sophistication.

ψM = C1*ψA ± C2*ψB

we know that the nuclei are identical, so the electron distribution must be symmetric. The only way for this to happen is the absolute values of C1 and C2 are the same, otherwise the electron would spend more time at one of the protons despite both protons being identical. So |C1| = |C2|.

So the molecular wavefunction is actually 2 wavefunctions:

ψM+ = C1(ψA + ψB) and ψM- = C1(ψA - ψB)

For ψM+ the wavefunction is very low far away from both protons and hit a peak at the nuclei. But it is nowhere zero. That's the bonding orbital.

For ψM- note that the wavefunction has a - sign. That means it crosses zero somewhere. When you graph it, it turns out that it crosses zero between the protons. Therefore, when you square it, there is a point in the graph that is a ZERO (node) and has NO ELECTRONS there. That is the anti-bonding orbital.

When you promote an electron up from a bonding orbital to the anti-bonding orbital, you're giving existence to a system state that includes a node between the 2 protons, which separates them.

Here's a wiki article that explains this in a better way and adds the energy calculations to boot.

http://en.wikipedia.org/wiki/Holstein–Herring_method

I really appreciate all you've laid out there, but I don't really follow. Maybe I could reduce my question to this: Let's start with an atom that has three electrons in its valence shell. Let's say these three electrons are bonded with three additional atoms. Now, if you were to excite an inner-shell electron to an energy level above the three bonded electrons level, it seems to me that there would be a new valence level defined, and the three bonds below it would break apart because they are no longer in the valence shell. Is this correct? Thank you!

 

[chill_factor]

I really appreciate all you've laid out there, but I don't really follow. Maybe I could reduce my question to this: Let's start with an atom that has three electrons in its valence shell. Let's say each of these three electrons is bonded with three additional atoms. Now, if you were to excite an inner-shell electron to an energy level above the three bonded electrons level, it seems to me that there would be a new valence level defined, and the three bonds below it would break apart because they are no longer in the valence shell. Is this correct? Thank you!

OK, I think I understand your question now. This depends.

The LUMO (lowest unoccupied molecular orbital) in a complicated molecule is not necessarily anti-bonding! I used hydrogen ions as an example, but almost all molecules aren't hydrogen ions. For some molecules, like oxygen, there are unoccupied bonding orbitals right above the occupied bonding orbitals.

 

[jaketodd]

OK, I think I understand your question now. This depends.

The LUMO (lowest unoccupied molecular orbital) in a complicated molecule is not necessarily anti-bonding! I used hydrogen ions as an example, but almost all molecules aren't hydrogen ions. For some molecules, like oxygen, there are unoccupied bonding orbitals right above the occupied bonding orbitals.

How about for main group elements? It says here: "For a main group element, a valence electron can only be in the outermost electron shell."

 

[chill_factor]

From the wiki:

"In chemistry, a valence electron is an electron that is associated with an atom, and that can participate in the formation of a chemical bond; in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair. The presence of valence electrons can determine the element's chemical properties and whether it may bond with other elements..."

Note that it says atoms and elements. You can only talk about valence electrons for elements and atoms. It is not a useful concept for molecules and solids. For molecules we talk about molecular (as opposed to atomic) orbitals and for solids we talk about bands.

http://en.wikipedia.org/wiki/Molecular_orbital

 

[jaketodd]

From the wiki:

"In chemistry, a valence electron is an electron that is associated with an atom, and that can participate in the formation of a chemical bond; in a single covalent bond, both atoms in the bond contribute one valence electron in order to form a shared pair. The presence of valence electrons can determine the element's chemical properties and whether it may bond with other elements..."

Note that it says atoms and elements. You can only talk about valence electrons for elements and atoms. It is not a useful concept for molecules and solids. For molecules we talk about molecular (as opposed to atomic) orbitals and for solids we talk about bands.

http://en.wikipedia.org/wiki/Molecular_orbital

I see. Is the bonding for a particular element, within a molecule, specific to the bonds that are connected to that particular element, or do you have to take every element in the molecule (even ones that aren't bonded to the particular atom) into account?

Thanks,

Jake

 

[jaketodd]

Let me ask a better question: Take an element such as Phosphorus, with its outer electrons all bonded in a big complex molecule. Now, let's leave Phosphorus' bonds as they are, but excite some of its more inner electrons (who are not participating in bonding) to the highest possible energy level (but not ionizing).

Would that destabilize the Phosphorus' bonds to the molecule, anyone?

Thanks!

 

[chill_factor]

Let me ask a better question: Take an element such as Phosphorus, with its outer electrons all bonded in a big complex molecule. Now, let's leave Phosphorus' bonds as they are, but excite some of its more inner electrons (who are not participating in bonding) to the highest possible energy level (but not ionizing).

Would that destabilize the Phosphorus' bonds to the molecule, anyone?

Thanks!

it depends on the specific molecule. it might, it might not, but get this: exciting its inner electrons would require huge energy. that energy can't be directed so specifically towards ONLY the inner electrons. it would definitely start ionizing stuff, and the bonds would break anyways. That's why x-rays are used to sterilize stuff; even though they come from core electrons they can still knock out valence electrons.

 

[jaketodd]

it depends on the specific molecule. it might, it might not, but get this: exciting its inner electrons would require huge energy. that energy can't be directed so specifically towards ONLY the inner electrons. it would definitely start ionizing stuff, and the bonds would break anyways. That's why x-rays are used to sterilize stuff; even though they come from core electrons they can still knock out valence electrons.

Oh I see, thank you. You sure know a lot.

 

[Khashishi]

But, if you very monochromatic light, you can target inner electrons, since there's a specific transition energy to go from an inner electron to some excited state. It works better if the molecules are cold to reduce Doppler broadening.

 

[ZapperZ]

Let me ask a better question: Take an element such as Phosphorus, with its outer electrons all bonded in a big complex molecule. Now, let's leave Phosphorus' bonds as they are, but excite some of its more inner electrons (who are not participating in bonding) to the highest possible energy level (but not ionizing).

Would that destabilize the Phosphorus' bonds to the molecule, anyone?

Thanks!

Please note that this is done all the time in many techniques, such as Auger spectroscopy. In x-ray photoemission, one is also exciting core-level electrons.

None of these do any "harm" to the either the molecules, or the solid.

Zz.

 

[jaketodd]

Please note that this is done all the time in many techniques, such as Auger spectroscopy. In x-ray photoemission, one is also exciting core-level electrons.

None of these do any "harm" to the either the molecules, or the solid.

Zz.

None of this stuff destabilizes molecules? Now I'm confused.

 

[chemisttree]

Please note that this is done all the time in many techniques, such as Auger spectroscopy. In x-ray photoemission, one is also exciting core-level electrons.

None of these do any "harm" to the either the molecules, or the solid.

Zz.

I would disagree. The absorption of xrays followed by inner shell ionization and equilibration results in the fragmentation of organic molecules.

 

[ZapperZ]

I would disagree. The absorption of xrays followed by inner shell ionization and equilibration results in the fragmentation of organic molecules.

You understood me wrong. I didn't say ALL types of x-ray absorption. I specifically cited those two phenomena as examples to invalidate the idea that ANY inner level excitation changes the material!

Unless, of course, you are arguing that both Auger and XPS change the material.

Zz.

 

[ZapperZ]

None of this stuff destabilizes molecules? Now I'm confused.

Do a search on those two techniques. If they destroy the original material, then we can't do repeated measurement on that same material. Yet, we do.

Zz.

 

[chemisttree]

You understood me wrong. I didn't say ALL types of x-ray absorption. I specifically cited those two phenomena as examples to invalidate the idea that ANY inner level excitation changes the material!

Unless, of course, you are arguing that both Auger and XPS change the material.

Zz.

I'm not quite sure what you mean by all types of x-ray absorption. Both Auger and XPS (ESCA) techniques produce local radiation damage. In the case of an organic molecule, any process that produces an ion can result in the molecule fragmenting, rearrangement and reduction and so forth. Yes, XPS and Auger experiments change organic materials.

 

[ZapperZ]

I'm not quite sure what you mean by all types of x-ray absorption. Both Auger and XPS (ESCA) techniques produce local radiation damage. In the case of an organic molecule, any process that produces an ion can result in the molecule fragmenting, rearrangement and reduction and so forth. Yes, XPS and Auger experiments change organic materials.

So then you are saying that the XPS and Auger experiments done in material science are all invalid. After all, when I do an XPS, I shoot that x-ray several hundred times on the same spot on the sample, and then I average out the signal that I get at the end. If merely ONE x-ray shot onto the sample destroys or change the nature of the sample itself, then my averaging technique is invalid.

So are you willing to tell the XPS and Auger community that their experiments are really suspect?

Zz.