Sciencemadness Discussion Board - d-metal splitting, symmetry and momentum

d-metal splitting, symmetry and momentum

fluorescence - 22-10-2015 at 02:48

Hey Guys,

I was preparing for a short presentation on Crystal Field Theory and
took the time to go a bit deeper into the details. So this is probably too much for the presentation but I didn't come up with a solution and it kinda bothers me now so maybe someone has an answer.

So for a d-d transition in a d-metal complex there is the so called LaPorte's Rule which says that by excitation with photons there is also a change in the momentum meaning that l, the azimuthal quantum numbers has to change in +/- 1 in order to make the transition allowed.
So an s -> p would be possible since s = 0 and p=1 and that is +1 so possible. A d -> d tansition however does not work.

But this problem only arises in octahedral coordinations. I know that an octahedron has an inversion symmetry thus creating the t_2g and e_g ( If they are called like this in English, too). A Tetrahedron does not have an inversion symmetry and because of that doesn't have that g either.

I know octahedral complexes are in fact colored due to the ligand vibrations and small movements that are not synced and therefore distort the symmetry a bit but still it's actually a d -> d transition that is forbidden.

Now Tetrahedrons don't have that rule. And the question is why ?
Does the Inversion Symmetry cause the splitted orbitals to have the same azimuthal quantum number or in other words does the azimuthal quantum number change when it's not symmetrical and how does it change.

I already had the lecture on Theoretical Chemistry but we were not talking about Symmetry back then and I now have the lecture on Symmetry but due to time issues we don't talk about the quantum mechanical solutions. That's why I'm quite stuck at the moment...

I'd be glad if someone could explain that to me or show me a link or literature where this is mentioned.

Thank you all in advance,
Greetings.

DraconicAcid - 22-10-2015 at 07:16

I seem to recall that the transitions are "kinda" forbidden, so the colours are paler than those due to charge transfer transitions....

fluorescence - 22-10-2015 at 07:47

Yeah, CT-Transitions are quite colorfull since the electrons usually come frome a p-Orbital (Ligand) into the d-Orbitals and that is allowed and thus colorful and since I have no spin-forbidden transition it's colored, too. So CT-Complexes usually don't have these selection rules and therefore are quite colorful.

Tetrahedral complexes are colored a bit more intensense than octahedral ones due to the laporte/parity-rule.

blogfast25 - 22-10-2015 at 08:31

@fluorescence:

CT and field theory are very different.

I think I might know where your confusion comes from but before I enter into a long diatribe, have a look and see if this (below) sheds any light on your problem:

http://www.sciencemadness.org/talk/viewthread.php?tid=62973&...

I don't see how weak e-t transitions could occur in anything other than an octahedral ligand field, as this is what causes the energy splitting between e and t d electrons.

[Edited on 22-10-2015 by blogfast25]

fluorescence - 22-10-2015 at 08:57

I had a quite long discussion with my instructor because I focused on Hybrids like the d²sp³ or sp³d². Or sd³ for CN =4.
Problem is my presentation will only be on Ligand Fields. I had the Inner and Outer Shell Complexes from the VB-Model, too but I was told to unly use the Crystal Field Splitting.

Well has nothing to do with the problem. I think Riedel even says that anything from 2 Electrons upwards will have so many couplings that those simple rules for colours just don't help anymore. But I'm looking for the reason in symmetry. I think I found the answer. It could be that literature just has it wrong.

So I think that laporte and parity are something completely diferent. In Literature it says that the Inversion Symmetry is part of LaPorte. But I think all d- Transitions are LaPorte forbidden it's always a d-Orbital. Only Transitions from Ligand to Metal are LaPorte allowed.

That diference between Octahedron and Tetrahedron is not the Laporte Rule but the Parity Rule which says that in excitation of an electron the parity of an orbital has to change from g -> u or u -> g. And therefore since in an octahedron they are both g it doesn't work.

Still I'm not sure whether I get that concept of Partiy in Tetrahedrons... I think I will have to look at that again.
Does anybody know why that Parity rule occurs ? Like what causes that ? Some property like the momentum of the photon should actually cause that selection rule.

blogfast25 - 22-10-2015 at 09:34

Quote: Originally posted by fluorescence

So I think that laporte and parity are something completely diferent. In Literature it says that the Inversion Symmetry is part of LaPorte. But I think all d- Transitions are LaPorte forbidden it's always a d-Orbital. Only Transitions from Ligand to Metal are LaPorte allowed.

That diference between Octahedron and Tetrahedron is not the Laporte Rule but the Parity Rule which says that in excitation of an electron the parity of an orbital has to change from g -> u or u -> g. And therefore since in an octahedron they are both g it doesn't work.

No, no: LaPorte and parity (g and u) are very related.

But these transitions aren't so forbidden at all, see wiki e.g.:

Quote:

However, forbidden transitions are allowed if the centre of symmetry is disrupted, and indeed, such apparently forbidden transitions are then observed in experiments. Disruption of the centre of symmetry occurs for various reasons, such as the Jahn-Teller effect and asymmetric vibrations. Complexes are not perfectly symmetric all the time. Transitions that occur as a result of an asymmetrical vibration of a molecule are called vibronic transitions, such as those caused by vibronic coupling. Through such asymmetric vibrations, transitions that would theoretically be forbidden, such as a d → d transition, are weakly allowed.

I think you are really over-thinking this.

Re. your point about what causes that parity rule, I'll definitely look for that.

I don't really get your point about tetrahedral complexes: show me one that shows colour explained by FLT?

See also the spectrum of Ti(+3) hydrate, no lines:

The involved energy levels must therefore be quite diffuse.

[Edited on 22-10-2015 by blogfast25]

fluorescence - 22-10-2015 at 10:10

I think I mentioned somewhere above that in fact due to vibrations there isn't a perfect symmetry.

The problem is that you need to change the azimuthal quantum number by +-1. Does that symmetry thing change
anything on the a.q.n. of a d-orbital ? If not it shouldn't be LaPorte. But on the other hand I'm not sure about that parity concept either. I mean what is a tetrahedron ? An Octahedron will get 2 times a "g" because of the symmetry. But a tetrahedron doesn't have that. Does it mean that one is "g" and the other "u" ? See the rule says "g->g" is forbidden so I think g->u is allowed. But if there isn't any symbol like that like in the tetrahedron "t2 and e", the regular Mulliken Symbols is that now parity allowed or not ?

See either one of these two rules must be diferent here. Either LaPorte and Parity are the same rule and a splitted d-Orbital has a diferent a.q.n. which I don't believe happens or they are something diferent in that case and I have to find which one of them causes the diference.

The problem is that I will have to mention that there is a diference between Oct and Tet in the LaPorte Rule and if someone asks why I need a quick summary. Like I dunno a wave function, equation, something where I can point out a reason why it's happening.

blogfast25 - 22-10-2015 at 10:29

LaPorte is about parity inversion in transitions: it states that for an electronic transition to occur the quantum state must change parity. But in octahedral fields it's not strictly observed and low probability g to g transitions occur. I'll try and expand a little more later on, on LaPorte.

Can you please give me an example of a tetrahedron you have in mind? g to u or u to g would of course be allowed and give highly probable transitions (strong emissions).

[Edited on 22-10-2015 by blogfast25]

fluorescence - 22-10-2015 at 21:29

I don't have any specific one in mind, perhaps [CoCl4]2- would be an example as it is an intense blue colored compound while the Aqua Complex as Octahedron is only slightly pink. Might be a good comparison for intensities.

But I think LaPorte is pretty clear to me or at least the reason why. I don't understand why the parity has to change. Parity Operation was the behaviour of the Wave Function when the Coordinate System is inverted to be either "g" when -psi = + or "U" when -psi = -

But does that affect the a.q.n., too ? Are all numbers inverted between a g and a u Orbital ? That would be the reason for the Parity rule then that for photon excitation you will need to change the a.q.n. - no wait if one was actually 2 and the other one -2 that would change it by delta = 5 so not allowed either. But what causes this rule then ?

fluorescence - 23-10-2015 at 03:26

Update:

So I asked my PC II Prof today and he explained why this happens.

For an excitation you need a transition dipol moment. Later your two functions
for the original orbital and the excited one are together with µ integrated over
the whole space. Now µ is x or odd and your parity can either be odd or even.

So

g u g = u
u u g = g

If the function formed is odd like for a s-orbital

x² x x² = x^5 -> odd the graph for it will have the same area in the negative part as it has in the positive one
and by integrating over the whole space you will eventually get 0. And therefore it's not allowed.

So the reason I was looking for has nothing to do with the photon but rather with the fact that the
transition dipol moment is odd, too.

[Edited on 23-10-2015 by fluorescence]

blogfast25 - 23-10-2015 at 06:16

Quote: Originally posted by fluorescence

Update:

So I asked my PC II Prof today and he explained why this happens.

For an excitation you need a transition dipol moment. Later your two functions
for the original orbital and the excited one are together with µ integrated over
the whole space. Now µ is x or odd and your parity can either be odd or even.

So

g u g = u
u u g = g

If the function formed is odd like for a s-orbital

x² x x² = x^5 -> odd the graph for it will have the same area in the negative part as it has in the positive one
and by integrating over the whole space you will eventually get 0. And therefore it's not allowed.

So the reason I was looking for has nothing to do with the photon but rather with the fact that the
transition dipol moment is odd, too.

[Edited on 23-10-2015 by fluorescence]

Interesting. Thanks for posting it.

fluorescence - 23-10-2015 at 07:21

Yeah he said we will have that topic in some weeks and I guess I'll just update that then. He just gave me a brief summary.

His example was the s-Orbital like exciting an electron from 1s to 2s Orbital is Parity Forbidden since a s-orbital is a sphere and therefore x²+y²+z² so if I only look at like the symmetry of the x-Component for example you will see that it's

< Psi | Hamiltonian | Psi > -> Int ( x² x x² dx (d tau)) = Int (x^5) and that function is symmetric so the area would be 0 and therefore the probability of <...> = 0

blogfast25 - 23-10-2015 at 07:43

Simply put, for hydrogenic atoms, the selection rules for transitions are:

Δl=+/-1, Δj=0, +/-1, Δm=0, +/-1

BTW, an s-orbital isn't really a sphere but it is radially symmetric. The wave functions can be found here (hydrogenic atoms):

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.htm...

One would not use Cartesian (x,y,z) coordinates here, only spherical coordinates.

[Edited on 23-10-2015 by blogfast25]

fluorescence - 23-10-2015 at 08:31

And µ isn't x it's just to show the symmetry easier. I mean there are certain factors in there, too but like you said there is much more to the color of a compound than these simple rules. I just wanted to understand the basics and I guess that example is something easy I could show on my presentation if someone asked.