Why platinum is stable with its actual electronic configuration?
this question is of 11th std
Answers
hi mate,
Answer : 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.
The initial configuration is : [Xe] 4f14 5d8 6s2, then one electron is transfered from 6s to 5d, so that all orbitals become stable, either through full filling or half filling, which is better then having one empty and unstable. ... It cannot be the other two because, in both of, one orbital is empty and unstable.
This seems odd to me that platinum is very inert. It seems like pulling off that 6s1 electron should not be difficult as compared to other metals.
It has to do with the relativistic effect that contracts the orbit of the 6s electron.
Platinum isn't unreactive. It is resistant to some specific reactions (oxidation or tarnishing) but is a very good catalyst for others (e.g. methanol oxidation, oxidation of residual hydrocarbons in catalytic converters, ammonia oxidation...). And it has a rich chemistry.
i hope it helps you.
Answer:
- The initial configuration is :
- [Xe] 4f14 5d8 6s2, then one electron is transfered from 6s to 5d, so that all orbitals become stable, either through full filling or half filling, which is better then having one empty and unstable.
- It cannot be the other two because, in both of, one orbital is empty and unstable.
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