0.3 Quantum energy levels in atoms  (Page 9/9)

A simple way to deal with the effect of electron-electron repulsion is to examine the shell structure ofthe atom. The two $n=1$ electrons in beryllium are in a shell with a comparatively short average distance from the nucleus. Therefore,the two $n=2$ electrons are in a shell which is, on average, "outside" of the $n=1$ shell. The $n=1$ electrons are thus the "core" and the $n=2$ electrons are in the valence shell. This structure allows us to see in a simple way the effectof electron-electron repulsion on the energies of the $n=2$ electrons. Each $n=2$ electron is attracted by the +4 charge on the tiny beryllium nucleus, but is repelled by the two -1 charges fromthe inner shell formed by the two $n=1$ electrons. Net, then, an $n=2$ electron effectively "sees" roughly a +2 nuclear charge. We refer to this +2 as the "core charge" sinceit is the net charge on the core resulting from the balance of attraction to the nucleus and repulsion from the core electrons.The nucleus is partially "shielded" from the valence electrons by the core electrons.

This shielding effect does not seem to account for the difference in ionization energies between 2s and 2p or forthe lower ionization energy of boron compared to beryllium, since, in each atom, the valence electrons are in the $n=2$ shell. However, the shielding effect is not perfect. Recall that we only know the probabilities for observing the positions of the electrons. Therefore, we cannot definitely state that the $n=2$ electrons are outside of the $n=1$ core. In fact, there is some probability that an $n=2$ electron might be found inside the $n=1$ core, an effect called "core penetration." When an $n=2$ electron does penetrate the core, it is no longer shielded from the nucleus. In this case, the $n=2$ electron is very strongly attracted to the nucleus and its energy is thus lowered. What is the extentof this penetration? We must consult quantum theory. The answer is in , which shows the probability of finding an electron a distance $r$ away from the nucleus for each of the 1s, 2s, and 2p orbitals. We can see that there is a greater probability (thoughsmall) for the 2s electron to penetrate the core than for the 2p electron to do so.

As a result of the core penetration, an electron in a 2s orbital feels a greater "effective nuclearcharge" than just the core charge, which was approximated by assuming perfect shielding. Thus the effective nuclear charge for a2s electron is greater than the effective nuclear charge for a 2p electron. Therefore, the energy of an electron in the 2s orbital inberyllium is lower than it would be in the 2p orbital.

A detailed analysis from quantum mechanics gives the following ordering of orbitals in order of increasingenergy:

This ordering can be rationalized on the basis of effective nuclear charge, shielding, and corepenetration.

Review and discussion questions