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Grade: 12th pass
        
Why does Lanthanum violates aufbau principle? 

and how does moving to 5d instead of 4f orbital imparts stability to Xenon core? 
 
Please explain!
2 years ago

Answers : (1)

Arun
23368 Points
							
Dear student
 
 
That is a good question. And to answer it, we will have to start off by discussing the building up of electrons in atoms in the first place.In the hydrogen atom, the s,p,d, and f subshells within any principal quantum shell are all of equal energy. This is because hydrogen has only one electron and the interaction of the one electron with the one proton of the nucleus is always direct. There aren't any other electrons to get in the way.With Helium, we have two electrons present. If each electron occupied a side of the atom, then each would see the full nuclear charge - a full +2. But, of course, the shape and radial distribution of electron density in the 1s orbital means that the electrons interact with each other. That is, sometimes one electron gets between the second and the nucleus with the result that there is repulsion and not attraction. Of course, both electrons are continually moving and both interfer with the other but only some of the time, so the net result is that each electron only feels some of the nuclear charge. This is called the "effective nuclear charge" because it is what the electron actually experiences versus the number of protons in the nucleus.With atoms more complicated than Helium, things get pretty mucky, pretty fast. With each successive shell, there is a shape factor (the angular component of the wavefunction) and a radial factor. These combine to both localize electrons in regions of space and ensure that there will be interactions. The "effective nuclear charge" is very sensitive to which orbitals are occupied and what is the overall nuclear charge. And this is manifested by a shielding factor that takes into account all of the electrons in all of the orbitals in an atom that are "inside" the orbital of interest. It is the shielding of outer electrons by inner electrons that results in the decrease in overall nuclear charge to give the effective nuclear charge.Now, the energy of an orbital in hydrogen is dependent upon the nuclear charge, but for more complex atoms, a more accurate description would be to realize that it is dependent upon the "effective nuclear charge". That is, the energy expression for the electrons in any orbital is a function of the nuclear charge actually "seen" by electrons in that orbital. This is - loosely speaking - the source of the structure that is the periodic table and is manifest in the Aufbau principle. The effective nuclear charge for the 4sorbital is larger than for the 3d, and hence it is lower in energy - until you start to fill the 3d, and then the increase in nuclear charge results in a lowering of the 3d orbitals below the 4s. This is why transition metals all preferentially lose their 4s electrons and the chemistry of the ions is the chemistry of the 3d electron configurations.So, with all of that as back ground, the answer to your question is related to the relative shielding factors. Once barium is reached, the energy of the 4f, 5d, and 6s subshells (the effective nuclear charge for an electron in each) has converged. As a consequence, the lanthanide series - after lanthanum - is formed by filling the 4f shell. But for lanthanum, itself, the energy balance is tipped slightly in favour of having that single 5d electron. In essence, by adding one more proton to the nucleus and shifting one more electron into a d-orbital that does not penetrate the atomic core nor interact with the f-orbitals, the effective nuclear charge felt by the 4f is enhanced enough that it becomes the orbital of choice for incoming electrons. Another way to think about this is that the size of the nucleus is enhanced enough to pull the 4f orbitals closer and therefore decrease their energy relative to the 5d. It is not really a "violation" of the Aufbau principle but rather a natural consequence of the "true way" that electrons and the nucleus interact. And, of course, if you look at the electron configurations of all of the lanthanides, it is quickly apparent that the is a bouncing of the electron in and out of the 5d shell with each additional measure of nuclear charge - that is, the filling of the 4f series. This is much more so in the case of the actinides Regards
Arun (askIITians forum expert)
2 years ago
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