This Quantum World/Implications and applications/Beyond hydrogen: the Periodic Table

Beyond hydrogen: the Periodic Table

If we again assume that the nucleus is fixed at the center and ignore relativistic and spin effects, then the stationary states of helium are the solutions of the following equation:

$E\,{\partial \psi \over \partial t}=-{\hbar ^{2} \over 2m}\left[{\partial ^{2}\psi \over \partial x_{1}^{2}}+{\partial ^{2}\psi \over \partial y_{1}^{2}}+{\partial ^{2}\psi \over \partial z_{1}^{2}}+{\partial ^{2}\psi \over \partial x_{2}^{2}}+{\partial ^{2}\psi \over \partial y_{2}^{2}}+{\partial ^{2}\psi \over \partial z_{2}^{2}}\right]+\left[-{\frac {2e^{2}}{r_{1}}}-{\frac {2e^{2}}{r_{2}}}+{\frac {e^{2}}{r_{12}}}\right]\psi .$

The wave function now depends on six coordinates, and the potential energy $V$ is made up of three terms. $r_{1}={\sqrt {x_{1}^{2}+y_{1}^{2}+z_{1}^{2}}}$ and $r_{2}={\sqrt {x_{2}^{2}+y_{2}^{2}+z_{2}^{2}}}$ are associated with the respective distances of the electrons from the nucleus, and $r_{12}={\sqrt {(x_{2}{-}x_{1})^{2}+(y_{2}{-}y_{1})^{2}+(z_{2}{-}z_{1})^{2}}}$ is associated with the distance between the electrons. Think of $e^{2}/r_{12}$ as the value the potential energy associated with the two electrons would have if they were at $r_{1}$ and $r_{2},$ respectively.

Why are there no separate wave functions for the two electrons? The joint probability of finding the first electron in a region $A$ and the second in a region $B$ (relative to the nucleus) is given by

$p(A,B)=\int _{A}\!d^{3}r_{1}\!\int _{B}\!d^{3}r_{2}\;|\psi (\mathbf {r} _{1},\mathbf {r} _{2})|^{2}.$

If the probability of finding the first electron in $A$ were independent of the whereabouts of the second electron, then we could assign to it a wave function $\psi _{1}(\mathbf {r} _{1}),$ and if the probability of finding the second electron in $B$ were independent of the whereabouts of the first electron, we could assign to it a wave function $\psi _{2}(\mathbf {r} _{2}).$ In this case $\psi (\mathbf {r} _{1},\mathbf {r} _{2})$ would be given by the product $\psi _{1}(\mathbf {r} _{1})\,\psi _{2}(\mathbf {r} _{2})$ of the two wave functions, and $p(A,B)$ would be the product of $p(A)=\int _{A}\!d^{3}r_{1}\,|\psi (\mathbf {r} _{1})|^{2}$ and $p(B)=\int _{B}\!d^{3}r_{2}\,|\psi (\mathbf {r} _{2})|^{2}.$ But in general, and especially inside a helium atom, the positional probability distribution for the first electron is conditional on the whereabouts of the second electron, and vice versa, given that the two electrons repel each other (to use the language of classical physics).

For the lowest energy levels, the above equation has been solved by numerical methods. With three or more electrons it is hopeless to look for exact solutions of the corresponding Schrödinger equation. Nevertheless, the Periodic Table and many properties of the chemical elements can be understood by using the following approximate theory.

First,we disregard the details of the interactions between the electrons. Next, since the chemical properties of atoms depend on their outermost electrons, we consider each of these atoms subject to a potential that is due to (i) the nucleus and (ii) a continuous, spherically symmetric, charge distribution doing duty for the other electrons. We again neglect spin effects except that we take account of the Pauli exclusion principle, according to which the probability of finding two electrons (more generally, two fermions) having exactly the same properties is 0. Thus two electrons can be associated with exactly the same wave function provided that their spin states differ in the following way: whenever the spins of the two electrons are measured with respect to a given axis, the outcomes are perfectly anticorrelated; one will be "up" and the other will be "down". Since there are only two possible outcomes, a third electron cannot be associated with the same wave function.

This approximate theory yields stationary wave functions $\psi _{nlm}(\mathbf {r} )$ called orbitals for individual electrons. These are quite similar to the stationary wave functions one obtains for the single electron of hydrogen, except that their dependence on the radial coordinate is modified by the negative charge distribution representing the remaining electrons. As a consequence of this modification, the energies associated with orbitals with the same quantum number $n$ but different quantum numbers $l$ are no longer equal. For any given $n\geq 1,$ obitals with higher $l$ yield a larger mean distance between the electron and the nucleus, and the larger this distance, the more the negative charge of the remaining electrons screens the positive charge of the nucleus. As a result, an electron with higher $l$ is less strongly bound (given the same $n$), so its ionization energy is lower.

Chemists group orbitals into shells according to their principal quantum number. As we have seen, the $n$-th shell can "accommodate" up to $n^{2}\times 2$ electrons. Helium has the first shell completely "filled" and the second shell "empty." Because the helium nucleus has twice the charge of the hydrogen nucleus, the two electrons are, on average, much nearer the nucleus than the single electron of hydrogen. The ionization energy of helium is therefore much larger, 2372.3 J/mol as compared to 1312.0 J/mol for hydrogen. On the other hand, if you tried to add an electron to create a negative helium ion, it would have to go into the second shell, which is almost completely screened from the nucleus by the electrons in the first shell. Helium is therefore neither prone to give up an electron not able to hold an extra electron. It is chemically inert, as are all elements in the rightmost column of the Periodic Table.

In the second row of the Periodic Table the second shell gets filled. Since the energies of the 2p orbitals are higher than that of the 2s orbital, the latter gets "filled" first. With each added electron (and proton!) the entire electron distribution gets pulled in, and the ionization energy goes up, from 520.2 J/mol for lithium (atomic number Z=3) to 2080.8 J/mol for neon (Z=10). While lithium readily parts with an electron, fluorine (Z=9) with a single empty "slot" in the second shell is prone to grab one. Both are therefore quite active chemically. The progression from sodium (Z=11) to argon (Z=18) parallels that from lithium to neon.

There is a noteworthy peculiarity in the corresponding sequences of ionization energies: The ionization energy of oxygen (Z=8, 1313.9 J/mol) is lower than that of nitrogen (Z=7, 1402.3 J/mol), and that of sulfur (Z=16, 999.6 J/mol) is lower than that of phosphorus (Z=15, 1011.8 J/mol). To understand why this is so, we must take account of certain details of the inter-electronic forces that we have so far ignored.

Suppose that one of the two 2p electrons of carbon (Z=6) goes into the $m{=}0$ orbital with respect to the $z$ axis. Where will the other 2p electron go? It will go into any vacant orbital that minimizes the repulsion between the two electrons, by maximizing their mean distance. This is neither of the orbitals with $|m|{=}1$ with respect to the $z$ axis but an orbital with $m{=}0$ with respect to some axis perpendicular to the $z$ axis. If we call this the $x$ axis, then the third 2p electron of nitrogen goes into the orbital with $m{=}0$ relative to $y$ axis. The fourth 2p electron of oxygen then has no choice but to go — with opposite spin — into an already occupied 2p orbital. This raises its energy significantly and accounts for the drop in ionization from nitrogen to oxygen.

By the time the 3p orbitals are "filled," the energies of the 3d states are pushed up so high (as a result of screening) that the 4s state is energetically lower. The "filling up" of the 3d orbitals therefore begins only after the 4s orbitals are "occupied," with scandium (Z=21).

Thus even this simplified and approximate version of the quantum theory of atoms has the power to predict the qualitative and many of the quantitative features of the Period Table.