The periodic table
We would like now to apply the theory of the hydrogen atom in an approximate way to get some understanding of the chemist’s periodic table of the elements. For an element with atomic number $Z$ there are $Z$ electrons held together by the electric attraction of the nucleus but with mutual repulsion of the electrons. To get an exact solution we would have to solve Schrödinger’s equation for $Z$ electrons in a Coulomb field. For helium the equation is  \begin{equation*} -\frac{\hbar}{i}\ddp{\psi}{t}=-\frac{\hbar^2}{2m} (\nabla_1^2\psi\!+\!\nabla_2^2\psi)\!+\!\biggl(\! -\!\frac{2e^2}{r_1}\!-\!\frac{2e^2}{r_2}\!+\!\frac{e^2}{r_{12}}\!\biggr)\psi, \end{equation*}   where $\nabla_1^2$ is a Laplacian which operates on $\FLPr_1$, the coordinate of one electron; $\nabla_2^2$ operates on $\FLPr_2$; and $r_{12}=\abs{\FLPr_1-\FLPr_2}$. (We are again neglecting the spin of the electrons.) To find the stationary states and energy levels we would have to find solutions of the form \begin{equation*} \psi=f(\FLPr_1,\FLPr_2)e^{-(i/\hbar)Et}. \end{equation*} The geometrical dependence is contained in $f$, which is a function of six variables—the simultaneous positions of the two electrons. No one has found an analytic solution, although solutions for the lowest energy states have been obtained by numerical methods.  With $3$, $4$, or $5$ electrons it is hopeless to try to obtain exact solutions, and it is going too far to say that quantum mechanics has given a precise understanding of the periodic table. It is possible, however, even with a sloppy approximation—and some fixing—to understand, at least qualitatively, many chemical properties which show up in the periodic table.  The chemical properties of atoms are determined primarily by their lowest energy states. We can use the following approximate theory to find these states and their energies. First, we neglect the electron spin, except that we adopt the exclusion principle and say that any particular electronic state can be occupied by only one electron. This means that any particular orbital configuration can have up to two electrons—one with spin up, the other with spin down. Next we disregard the details of the interactions between the electrons in our first approximation, and say that each electron moves in a central field which is the combined field of the nucleus and all the other electrons. For neon, which has $10$ electrons, we say that one electron sees an average potential due to the nucleus plus the other nine electrons. We imagine then that in the Schrödinger equation for each electron we put a $V(r)$ which is a $1/r$ field modified by a spherically symmetric charge density coming from the other electrons.  In this model each electron acts like an independent particle. The angular dependence of its wave function will be just the same as the ones we had for the hydrogen atom. There will be $s$-states, $p$-states, and so on; and they will have the various possible $m$-values. Since $V(r)$ no longer goes as $1/r$, the radial part of the wave functions will be somewhat different, but it will be qualitatively the same, so we will have the same radial quantum numbers, $n$. The energies of the states will also be somewhat different.  With these ideas, let’s see what we get. The ground state of hydrogen has $l=m=0$ and $n=1$; we say the electron configuration is $1s$. The energy is $-13.6$ eV. This means that it takes $13.6$ electron volts to pull the electron off the atom. We call this the “ionization energy”, $W_I$. A large ionization energy means that it is harder to pull the electron off and, in general, that the material is chemically less active.  Now take helium. Both electrons can be in the same lowest state (one spin up and the other spin down). In this lowest state the electron moves in a potential which is for small $r$ like a Coulomb field for $z=2$ and for large $r$ like a Coulomb field for $z=1$. The result is a “hydrogen-like” $1s$ state with a somewhat lower energy. Both electrons occupy identical $1s$ states ($l=0$, $m=0$). The observed ionization energy (to remove one electron) is $24.6$ electron volts. Since the $1s$ “shell” is now filled—we allow only two electrons—there is practically no tendency for an electron to be attracted from another atom. Helium is chemically inert.  The lithium nucleus has a charge of $3$. The electron states will again be hydrogen-like, and the three electrons will occupy the lowest three energy levels. Two will go into $1s$ states and the third will go into an $n=2$ state. But with $l=0$ or $l=1$? In hydrogen these states have the same energy, but in other atoms they don’t, for the following reason. Remember that a $2s$ state has some amplitude to be near the nucleus while the $2p$ state does not. That means that a $2s$ electron will feel some of the triple electric charge of the Li nucleus, but that a $2p$ electron will stay out where the field looks like the Coulomb field of a single charge. The extra attraction lowers the energy of the $2s$ state relative to the $2p$ state. The energy levels will be roughly as shown in Fig. 19–8—which you should compare with the corresponding diagram for hydrogen in Fig. 19–7. So the lithium atom will have two electrons in $1s$ states and one in a $2s$. Since the $2s$ electron has a higher energy than a $1s$ electron it is relatively easily removed. The ionization energy of lithium is only $5.4$ electron volts, and it is quite active chemically.  So you can see the patterns which develop; we have given in Table 19–2 a list of the first $36$ elements, showing the states occupied by the electrons in the ground state of each atom. The Table gives the ionization energy for the most loosely bound electron, and the number of electrons occupying each “shell”—by which we mean states with the same $n$. Since the different $l$-states have different energies, each $l$-value corresponds to a sub-shell of $2(2l+1)$ possible states (of different $m$ and electron spin). These all have the same energy—except for some very small effects we are neglecting.  Beryllium is like lithium except that it has two electrons in the $2s$ state as well as two in the filled $1s$ shell.  Boron has $5$ electrons. The fifth must go into a $2p$ state. There are $2\times3=6$ different $2p$ states, so we can keep adding electrons until we get to a total of $8$. This takes us to neon. As we add these electrons we are also increasing $Z$, so the whole electron distribution gets pulled in closer and closer to the nucleus and the energy of the $2p$ states goes down. By the time we get to neon the ionization energy is up to $21.6$ electron volts. Neon does not easily give up an electron. Also there are no more low-energy slots to be filled, so it won’t try to grab an extra electron. Neon is chemically inert. Fluorine, on the other hand, does have an empty position where an electron can drop into a state of low energy, so it is quite active in chemical reactions.  With sodium the eleventh electron must start a new shell—going into a $3s$ state. The energy level of this state is much higher; the ionization energy jumps down; and sodium is an active chemical. From sodium to argon the $s$ and $p$ states with $n=3$ are occupied in exactly the same sequence as for lithium to neon. Angular configurations of the electrons in the outer unfilled shell have the same sequence, and the progression of ionization energies is quite similar. You can see why the chemical properties repeat with increasing atomic number. Magnesium acts chemically much like beryllium, silicon like carbon, and chlorine like fluorine. Argon is inert like neon.  You may have noticed that there is a slight peculiarity in the sequence of ionization energies between lithium and neon, and a similar one between sodium and argon. The last electron is bound to the oxygen atom somewhat less than we might expect. And sulfur is similar. Why should that be? We can understand it if we put in just a little bit of the effects of the interactions between individual electrons. Think of what happens when we put the first $2p$ electron onto the boron atom. It has six possibilities—three possible $p$-states, each with two spins. Imagine that the electron goes with spin up into the $m=0$ state, which we have also called the “$z$” state because it hugs the $z$-axis. Now what will happen in carbon? There are now two $2p$ electrons. If one of them goes into the “$z$” state, where will the second one go? It will have lower energy if it stays away from the first electron, which it can do by going into, say, the “$x$” state of the $2p$ shell. (This state is, remember, just a linear combination of the $m=+1$ and $m=-1$ states.) Next, when we go to nitrogen, the three $2p$ electrons will have the lowest energy of mutual repulsion if they go one each into the “$x$,” “$y$,” and “$z$” configurations. For oxygen, however, the jig is up. The fourth electron must go into one of the filled states—with opposite spin. It is strongly repelled by the electron already in that state, so its energy will not be as low as it might otherwise be, and it is more easily removed. That explains the break in the sequence of binding energies which appears between nitrogen and oxygen, and between phosphorus and sulfur.  After argon, you would, at first, think that the new electrons would start to fill up the $3d$ states. But they don’t. As we described earlier—and illustrated in Fig. 19–8—the higher angular momentum states get pushed up in energy. By the time we get to the $3d$ states they are pushed to an energy a little bit above the energy of the $4s$ state. So in potassium the last electron goes into the $4s$ state. After this shell is filled (with two electrons) at calcium, the $3d$ states begin to be filled for scandium, titanium, and vanadium.  The energies of the $3d$ and $4s$ states are so close together that small effects can shift the balance either way. By the time we get to put four electrons into the $3d$ states, their repulsion raises the energy of the $4s$ state just enough that its energy is slightly above the $3d$ energy, so one electron shifts over. For chromium we don’t get a $4$, $2$ combination as we would have expected, but instead a $5$, $1$ combination. The new electron added to get manganese fills up the $4s$ shell again, and the states of the $3d$ shell are then occupied one by one until we reach copper.  Since the outermost shell of manganese, iron, cobalt, and nickel have the same configurations, however, they all tend to have similar chemical properties. (This effect is much more pronounced in the rare-earth elements which all have the same outer shell but a progressively filling inner shell which has much less influence on their chemical properties.)  In copper an electron is robbed from the $4s$ shell, finally completing the $3d$ shell. The energy of the $10$, $1$ combination is, however, so close to the $9$, $2$ configuration for copper that just the presence of another atom nearby can shift the balance. For this reason the two last electrons of copper are nearly equivalent, and copper can have a valence of either $1$ or $2$. (It sometimes acts as though its electrons were in the $9$, $2$ combination.) Similar things happen at other places and account for the fact that other metals, such as iron, combine chemically with either of two valences. By zinc, both the $3d$ and $4s$ shells are filled once and for all.  From gallium to krypton the sequence proceeds normally again, filling the $4p$ shell. The outer shells, the energies, and the chemical properties repeat the pattern of boron to neon and aluminum to argon.  Krypton, like argon and neon, is known as “noble” gas. All three are chemically “inert.” This means only that, having filled shells of relatively low energy, there are few situations in which there is an energy advantage for them to join in a simple combination with other elements. Having a filled shell is not enough. Beryllium and magnesium have filled $s$-shells, but the energy of these shells is too high to lead to stability. Similarly, one would have expected another “noble” element at nickel, if the energy of the $3d$ shell had been lower (or the $4s$ higher). On the other hand, krypton is not completely inert; it will form a weakly-bound compound with chlorine.  Since our sample has turned up most of the main features of the periodic table, we stop our examination at element number $36$—there are still seventy or so more!  We would like to bring up only one more point—that we not only can understand the valences to some extent but also can say something about the directional properties of the chemical bonds. Take an atom like oxygen which has four $2p$ electrons. The first three go into “$x$,” “$y$,” and “$z$” states and the fourth will double one of these states, leaving two—say “$x$” and “$y$”—vacant. Consider then what happens in H$_2$O. Each of the two hydrogens are willing to share an electron with the oxygen, helping the oxygen to fill a shell. These electrons will tend to go into the “$x$” and “$y$” vacancies. So the water molecule should have the two hydrogen atoms making a right angle with respect to the center of the oxygen. The angle is actually $105^\circ$. We can even understand why the angle is larger than $90^\circ$. In sharing their electrons the hydrogens end up with a net positive charge. The electric repulsion “strains” the wave functions and pushes the angle out to $105^\circ$. The same situation occurs in H$_2$S. But because the sulfur atom is larger, the two hydrogen atoms are farther apart, there is less repulsion, and the angle is only pushed out to about $93^\circ$. Selenium is even larger, so in H$_2$Se the angle is very nearly $90^\circ$.  We can use the same arguments to understand the geometry of ammonia, H$_3$N. Nitrogen has room for three more $2p$ electrons, one each for the “$x$,” “$y$,” and “$z$” type states. The three hydrogens should join on at right angles to each other. The angles come out a little larger than $90^\circ$—again from the electric repulsion—but at least we see why the molecule of H$_3$N is not flat. The angles in phosphene, H$_3$P, are close to $90^\circ$, and in H$_3$As are still closer. We assumed that NH$_3$ was not flat when we described it as a two-state system. And the nonflatness is what makes the ammonia maser possible. Now we see that also that shape can be understood from our quantum mechanics.  The Schrödinger equation has been one of the great triumphs of physics. By providing the key to the underlying machinery of atomic structure it has given an explanation for atomic spectra, for chemistry, and for the nature of matter. 