Talk:Laporte rule

Comments
I avoid applying those cheesy tags, but you chem wonks must realize that this is unnecessarily opaque. The first things to add are "where g represents..." and "where u represents." In expanding this, just imagine you are explaining this to an intelligent friend of your family. --Wetman 04:16, 16 January 2007 (UTC)

I agree with the above poster. I could probably add a little more information but don't quite have time now.

g = gerade u = ungerade

A designation of g means there is symmetry with respect to an inversion center. That is, if all the atoms (or orbitals) are inverted across the metal center, the resulting compound would look exactly how it did before having inversion applied to it (this includes same orientation in space). A designation of u means otherwise, namely that there is no inversion center.

Moreover, it's very important to note that these rules can be relaxed. This rule basically says that having inversion symmetry "forbids" transitions. Spectroscopy indicates that this isn't the case (citation would probably be HKK Inorganic textbook that I don't have at hand now). Due to asymmetric vibrations, complexes are not perfectly symmetric all the time. This "relaxes" the Laporte rule and allows for transitions that would theoretically be forbidden. 24.7.108.102 08:57, 28 February 2007 (UTC)


 * No, 'u' doesn't mean there isn't an inversion centre, in fact there has to be an inversion centre for it to be designated such! g comes from the German gerade which translates as even, and u is ungerade which means uneven. Basically therefore if the orbital is symmetric about the inversion centre then it is assigned g (like the function cos(x)), and if it is antisymmetric (changes mathematical sign) upon inversion it is u (like the function sin(x)).


 * These signs are actually from group theory. The reason this rule exists is because the intensity of a transition is proportional to the square of the integral of the complex conjugate of the upper state multiplied by the operatr for dipole moment acting on the lower state. Now as the dipole operator is mathematically antisymmetric (u) the integral is only non-zero (and hence a transition occurs) when either the lower or upper state is u and the other is g, so that overall the integrand is g, and hence once integrated has a value (if this doesn't make sense think about integrating an even function between two points -x and x and it having a value, while an odd function would have an integral of zero). When the transition is g -> g or u -> u the integrand is overall u and the integral of any antisymmetric function is necessary zero (e.g. integral of sin(x) [which is u] from minus to positive infinities). I don't normally post on these things but that was a blatent misunderstanding that had to be cleared up.85.211.247.232 16:02, 7 June 2007 (UTC)

ACC Request
This request came in to ACC:  D u s t i *Let's talk!* 21:35, 28 August 2013 (UTC)


 * I fixed this today. Yes, it should be ANTIsymmetry. Dirac66 (talk) 02:13, 15 February 2014 (UTC)