I am doing a theoretical study about some transitions metals and I need to do a computational calculate. So I need to understand the electron configuration of Platinum and other transition metals, because of the parameters of charge and spin multipicity.
Elder, the electron configuration of heavy metals has a complex behavior. Using the Russel-Saunders coupling for atoms you will find all the terms for platinum energy levels. The Hund's rule will show that the ground state for this electron configuration is the 5d96d1.
We talk about a very special and typical electronic configuration of Pt where almost every argument is found wanting and we are in a very precarious situation as to how to explain the transfer the 6s electrons into 5d as Pt should be a 6s2 5d8 like Ni(4s2 3d8 ). But it is 6s1 5d9 .
[A]The ground state configurations of the elements is determined by the combined effect of n+l Energy Rule,Z*( effective nuclear charge),Higher exchange energy(Eex) associated with the half filled and completely filled subshell. But as the population of electrons becomes very large( atomic number more than 50 and especially in the third transition series), the factor like INTERELECTRONIC REPULSIONS OVERWEIGHS all the above named factors to make the ground state configurations of the higher atomic number elements different from those which are normally expected.
[B]As you go up, the energy difference among various shells becomes less:
En α 1/n^2------ [1]
So the 6s and 5d get closer together in energy and the transition of 6s electron to 5d is easily facilitated .
Two more factors[other than the overweighing effect of interelectronic repulsion] which make this transition easy are:
(i)It can be looked as that the 5d sub shell electrons are shielded better by the completely filled 4f sub shell electrons to make 5d to be lower energy than the 6s orbital.
(ii) This energy difference decreases all the more due to another effect called EINSTEIN’S RELATIVISTIC EFFECT.
It needs a bit more of explanation and I hope that my young friends would bear with it.
The effect is given by:
E=mc^2/(1-v^2/C^2)0.5-------[2]
Or
E=γmc^2------------------------[3]
So the dissipation of electonic energy should happen with change of velocity of the electron. As the electrons reach higher shells( 5th , 6th) ) ,the hold of the nucleus on these electrons decreases.So the energy dissipation becomes more and more and follows the order : s˃ p ˃ d due to their shapes.This makes the energy difference between 6s and 5d to decrease SOME WHAT MORE THAN what is expected from relation [I] . And thus there will be only a VERY little difference between excitation energies of the 5d and 6s to make the electronic transitions to occur easily from 6s to 5d.