4.9 Heavier Atoms [Descriptive]

This section solves the ground state electron configuration of the atoms of elements heavier than hydrogen. The atoms of the elements are distinguished by their “atomic number” $Z$, which is the number of protons in the nucleus. For the neutral atoms considered in this section, $Z$ is also the number of electrons circling the nucleus.

A crude approximation will be made to deal with the mutual interactions between the electrons. Still, many properties of the elements can be understood using this crude model, such as their geometry and chemical properties, and how the Pauli exclusion principle raises the energy of the electrons.

This is a descriptive section, in which no new analytical procedures are taught. However, it is a very important section to read, and reread, because much of our qualitative understanding of nature is based on the ideas in this section.

4.9.1 The Hamiltonian eigenvalue problem

The procedure to find the ground state of the heavier atoms is similar to the one for the hydrogen atom of chapter 3.2. The total energy Hamiltonian for the electrons of an element with atomic number $Z$ with is:

$$H = \sum_{i=1}^Z \Bigg[ - \frac{\hbar^2}{2m_e} \nabl... ...ew0\vec r}_i -{\skew0\vec r}_{\underline i}\vert} \Bigg] %$$ (4.33)

Within the brackets, the first term represents the kinetic energy of electron number $i$ out of $Z$, the second the attractive potential due to the nuclear charge $Ze$, and the final term is the repulsion by all the other electrons. In the Hamiltonian as written, it is assumed that half of the energy of a repulsion is credited to each of the two electrons involved, accounting for the factor $\frac12$.

The Hamiltonian eigenvalue problem for the energy states takes the form:

$$H \psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2},\ld... ..._{z1},{\skew0\vec r}_2,S_{z2},\ldots,{\skew0\vec r}_Z,S_{zZ})$$

Key Points

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The Hamiltonian for the electron structure has been written down.

4.9.2 Approximate solution using separation of variables

The Hamiltonian eigenvalue problem of the previous subsection cannot be solved exactly. The repulsive interactions between the electrons, given by the last term in the Hamiltonian are too complex.

More can be said under the, really poor, approximation that each electron “sees” a repulsion by the other $Z-1$ electrons that averages out as if the other electrons are located in the nucleus. The other $Z-1$ electrons then reduce the net charge of the nucleus from $Ze$ to $e$. An other way of saying this is that each of the $Z-1$ other electrons “shields” one proton in the nucleus, allowing only a single remaining proton charge to filter through.

In this crude approximation, the electrons do not notice each other at all; they see only a single charge hydrogen nucleus. Obviously then, the wave function solutions for each electron should be the $\psi_{nlm}$ eigenfunctions of the hydrogen atom, which were found in chapter 3.2.

To verify this explicitly, the approximate Hamiltonian is

$$H = \sum_{i=1}^Z \left\{ - \frac{\hbar^2}{2m} \nabla_i^2 - \frac{e^2}{4\pi\epsilon_0} \frac{1}{r_i} \right\}$$

since this represents a system of noninteracting electrons in which each experiences an hydrogen nucleus potential. This can be written more concisely as

$$H = \sum_{i=1}^Z h_i$$

where $h_i$ is the hydrogen-atom Hamiltonian for electron number $i$,

$$h_i = - \frac{\hbar^2}{2m} \nabla_i^2 - \frac{e^2}{4\pi\epsilon_0} \frac{1}{r_i}.$$

The approximate Hamiltonian eigenvalue problem can now be solved using a method of separation of variables in which solutions are sought that take the form of products of single-electron wave functions:

$$\psi^Z=\pes1/{\skew0\vec r}_1/e/1/\pes2/{\skew0\vec r}_2/e/2/\ldots \pes Z/{\skew0\vec r}_Z/e/Z/.$$

Substitution of this assumption into the eigenvalue problem $\sum_{i}h_i\psi^Z=E\psi^Z$ and dividing by $\psi^Z$ produces

$$\frac{1}{\pes1/{\skew0\vec r}_1/e/1/} h_1 \pes1/{\skew0\ve... ...vec r}_2/e/2/} h_2 \pes2/{\skew0\vec r}_2/e/2/ + \ldots = E$$

since $h_1$ only does anything to the factor $\pes1/{\skew0\vec r}_1/e/1/$, $h_2$ only does anything to the factor $\pes2/{\skew0\vec r}_2/e/2/$, etcetera.

The first term in the equation above must be some constant $\epsilon_1$; it cannot vary with ${\skew0\vec r}_1$ or $S_{z1}$ as $\pes1/{\skew0\vec r}_1/e/1/$ itself does, since none of the other terms in the equation varies with those variables. That means that

$$h_1 \pes1/{\skew0\vec r}_1/e/1/ = \epsilon_1 \pes1/{\skew0\vec r}_1/e/1/,$$

which is an hydrogen atom eigenvalue problem for the single-electron wave function of electron 1. So, the single-electron wave function of electron 1 can be any one of the hydrogen atom wave functions from chapter 3.2; allowing for spin, the possible solutions are,

$$\psi_{100}({\skew0\vec r}_1){\uparrow}(S_{z1}),\, \psi_{... ... \psi_{200}({\skew0\vec r}_1){\downarrow}(S_{z1}),\, \ldots$$

The energy $\epsilon_1$ is the corresponding hydrogen atom energy level, $E_1$ for $\psi_{100}{\uparrow}$ or $\psi_{100}{\downarrow}$, $E_2$ for any of the eight states $\psi_{200}{\uparrow}$, $\psi_{200}{\downarrow}$, $\psi_{211}{\uparrow}$, $\psi_{211}{\downarrow}$, $\psi_{210}{\uparrow}$, $\psi_{210}{\downarrow}$, $\psi_{21{-1}}{\uparrow}$, $\psi_{21{-1}}{\downarrow}$, etcetera.

The same observations hold for the other electrons; their single-electron eigenfunctions are $\psi_{nlm}{\updownarrow}$ hydrogen atom ones, (where ${\updownarrow}$ can be either ${\uparrow}$ or ${\downarrow}$.) Their individual energies must be the corresponding hydrogen atom energy levels.

The final wave functions for all $Z$ electrons are then each a product of $Z$ hydrogen-atom wave functions,

$$\psi_{n_1l_1m_1}({\skew0\vec r}_1){\updownarrow}(S_{z1}) \... ...dots \psi_{n_Zl_Zm_Z}({\skew0\vec r}_Z){\updownarrow}(S_{zZ})$$

and the total energy is the sum of all the corresponding hydrogen atom energy levels,

$$E_{n_1} + E_{n_2} + \ldots + E_{n_Z}.$$

This solves the Hamiltonian eigenvalue problem under the shielding approximation. The bottom line is: just multiply $Z$ hydrogen energy eigenfunctions together to get an energy eigenfunction for an heavier atom. The energy is the sum of the $Z$ hydrogen energy levels. However, the electrons are identical fermions, so different eigenfunctions must still be combined together in Slater determinants to satisfy the antisymmetrization requirements for electron exchange, as discussed in section 4.7. That will be done during the discussion of the different atoms that is next.

Key Points

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The Hamiltonian eigenvalue problem is too difficult to solve analytically.

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To simplify the problem, the detailed interactions between electrons are ignored. For each electron, it is assumed that the only effect of the other electrons is to cancel, or “shield,” that many protons in the nucleus, leaving only a hydrogen nucleus strength.

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This is a very crude approximation.

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It implies that the $Z$-electron wave functions are products of the single-electron hydrogen atom wave functions. Their energy is the sum of the corresponding single-electron hydrogen energy levels.

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These wave functions must still be combined together to satisfy the antisymmetrization requirement (Pauli exclusion principle).

4.9.3 Hydrogen and helium

This subsection starts off the discussion of the approximate ground states of the elements. Atomic number $Z=1$ corresponds to hydrogen, which was already discussed in chapter 3.2. The lowest energy state, or ground state, is $\psi_{100}$, (3.22), also called the “1s” state, and the single electron can be in the spin-up or spin-down versions of that state, or in any combination of the two. The most general ground state wave function is therefore:

$$\Psi({\skew0\vec r}_1,S_{z1}) = a_1 \psi_{100}({\skew0... ..._1) \Big(a_1{\uparrow}(S_{z1}) + a_2{\downarrow}(S_{z1})\Big)$$ (4.34)

The “ionization energy” that would be needed to remove the electron from the atom is the absolute value of the energy eigenvalue $E_1$, or 13.6 eV, as derived in chapter 3.2.

For helium, with $Z=2$, in the ground state both electrons are in the lowest possible energy state $\psi_{100}$. But since electrons are identical fermions, the antisymmetrization requirement now rears its head. It requires that the two states $\psi_{100}({\skew0\vec r}){\uparrow}(S_{z})$ and $\psi_{100}({\skew0\vec r}){\downarrow}(S_{z})$ appear together in the form of a Slater determinant (chapter 4.7):

$$\Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2};t) ... ...ew0\vec r}_2){\downarrow}(S_{z2}) \end{array} \right\vert$$ (4.35)

or, writing out the Slater determinant:

$$a \psi_{100}({\skew0\vec r}_1) \psi_{100}({\skew0\vec r}_2... ...w}(S_{z2})-{\downarrow}(S_{z1}){\uparrow}(S_{z2})}{\sqrt{2}}.$$

The spatial part is symmetric with respect to exchange of the two electrons. The spin state is antisymmetric; it is the singlet configuration with zero net spin of section 4.5.6.

Figure 4.4 shows the approximate probability density for the first two elements, indicating where electrons are most likely to be found. In actuality, the shielding approximation underestimates the nuclear attraction and the shown helium atom is much too big.

Figure 4.4: Approximate solutions for hydrogen (left) and helium (right).

It is good to remember that the $\psi_{100}{\uparrow}$ and $\psi_{100}{\downarrow}$ states are commonly indicated as the “K shell” after the first initial of the airline of the Netherlands.

The analysis predicts that the ionization energy to remove one electron from helium would be 13.6 eV, the same as for the hydrogen atom. This is a very bad approximation indeed; the truth is almost double, 24.6 eV.

The problem is the made assumption that the repulsion by the other electron “shields” one of the two protons in the helium nucleus, so that only a single-proton hydrogen nucleus is seen. When electron wave functions overlap significantly as they do here, their mutual repulsion is a lot less than you would naively expect, (compare figure 9.13). As a result, the second proton is only partly shielded, and the electron is held much more tightly than the analysis predicts. See chapter 10.1.2 for better estimates of the helium atom size and ionization energy.

However, despite the inaccuracy of the approximation chosen, it is probably best to stay consistent, and not fool around at random. It must just be accepted that the theoretical energy levels will be too small in magnitude {A.30}.

The large ionization energy of helium is one reason that it is chemically inert. Helium is called a “noble” gas, presumably because nobody expects nobility to do anything.

Key Points

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The ground states of the atoms of the elements are to be discussed.

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Element one is hydrogen, solved before. Its ground state is $\psi_{100}$ with arbitrary spin. Its ionization energy is 13.6 eV.

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Element two is helium. Its ground state has both electrons in the lowest-energy spatial state $\psi_{100}$, and locked into the singlet spin state. Its ionization energy is 24.6 eV.

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The large ionization energy of helium means it holds onto its two electrons tightly. Helium is an inert noble gas.

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The two “1s” states $\psi_{100}{\uparrow}$ and $\psi_{100}{\downarrow}$ are called the “K shell.”

4.9.4 Lithium to neon

The next element is lithium, with three electrons. This is the first element for which the antisymmetrization requirement forces the theoretical energy to go above the hydrogen ground state level $E_1$. The reason is that there is no way to create an antisymmetric wave function for three electrons using only the two lowest energy states $\psi_{100}{\uparrow}$ and $\psi_{100}{\downarrow}$. A Slater determinant for three electrons must have three different states. One of the eight $\psi_{2lm}{\updownarrow}$ states with energy $E_2$ will have to be thrown into the mix.

This effect of the antisymmetrization requirement, that a new state must become “occupied” every time an electron is added is known as the Pauli exclusion principle. It causes the energy values to become larger and larger as the supply of low energy states runs out.

The transition to the higher energy level $E_2$ is reflected in the fact that in the so-called “periodic table” of the elements, table 4.1, lithium starts a new row.

Table 4.1: Abbreviated periodic table of the elements, showing element symbol, atomic number, ionization energy, and electronegativity.

For the third electron of the lithium atom, the available states with theoretical energy $E_2$ are the $\psi_{200}{\updownarrow}$ “2s” states and the $\psi_{211}{\updownarrow}$, $\psi_{210}{\updownarrow}$, and $\psi_{21-1}{\updownarrow}$ “2p” states, a total of eight possible states. These states are, of course, commonly called the “L shell.”

Within the crude nuclear shielding approximation made, all eight states have the same energy. However, on closer examination, the spherically symmetric 2s states really have less energy than the 2p ones. Very close to the nucleus, shielding is not a factor and the full attractive nuclear force is felt. So a state in which the electron is more likely to be close to the nucleus has less energy. Those are the 2s states; in the 2p states, which have nonzero orbital angular momentum, the electron tends to stay away from the immediate vicinity of the nucleus {A.31}.

Within the assumptions made, there is no preference with regard to the spin direction of the 2s state, allowing two Slater determinants to be formed.

$$\frac{a_1}{\sqrt{6}} \left\vert \begin{array}{ccc} \psi_{10... ... & \psi_{200}({\skew0\vec r}_3){\uparrow}(S_{z3}) \end{array} \right\vert$$

$${} + \frac{a_2}{\sqrt{6}} \left\vert \begin{array}{ccc} \... ... \psi_{200}({\skew0\vec r}_3){\downarrow}(S_{z3}) \end{array} \right\vert$$ (4.36)

Figure 4.5: Approximate solutions for lithium (left) and beryllium (right).

It is common to say that the “third electron goes into a $\psi_{200}$” state. Of course that is not quite precise; the Slater determinants above have the first two electrons in $\psi_{200}$ states too. But the third electron adds the third state to the mix, so in that sense it more or less “owns” the state. For the same reason, the Pauli exclusion principle is commonly phrased as “no two electrons may occupy the same state”, even though the Slater determinants imply that all electrons share all states equally.

Since the third electron is bound with the much lower energy $\vert E_2\vert$ instead of $\vert E_1\vert$, it is rather easily given up. Despite the fact that the lithium ion has a nucleus that is 50% stronger than the one of helium, it only takes a ionization energy of 5.4 eV to remove an electron from lithium, versus 24.6 eV for helium. The theory would predict a ionization energy $\vert E_2\vert=3.4$ eV for lithium, which is close, so it appears that the two 1s electrons shield their protons quite well from the 2s one. This is in fact what one would expect, since the 1s electrons are quite close to the nucleus compared to the large radial extent of the 2s state.

Lithium will readily give up its loosely bound third electron in chemical reactions. Conversely, helium would have even less hold on a third electron than lithium, because it has only two protons in its nucleus. Helium simply does not have what it takes to seduce an electron away from another atom. This is the second part of the reason that helium is chemically inert: it neither will give up its electrons nor take on additional ones.

Thus the Pauli exclusion principle causes different elements to behave chemically in very different ways. Even elements that are just one unit apart in atomic number such as helium (inert) and lithium (very active).

For beryllium, with four electrons, the same four states as for lithium combine in a single $4 \times 4$ Slater determinant;

$$\frac{a}{\sqrt{24}} \left\vert \begin{array}{cccc} \... ...ew0\vec r}_4){\downarrow}(S_{z4}) \end{array} \right\vert$$ (4.37)

The ionization energy jumps up to 9.3 eV, due to the increased nuclear strength and the fact that the fellow 2s electron does not shield its proton as well as the two 1s electrons do theirs.

For boron, one of the $\psi_{21m}$ “2p” states will need to be occupied. Within the approximations made, there is no preference for any particular state. As an example, figure 4.6 shows the approximate solution in which the $\psi_{210}$, or “2p$_z$” state is occupied. It may be recalled from figure 3.5 that this state remains close to the $z$-axis (which is horizontal in the figure.) As a result, the wave function becomes directional.

Figure 4.6: Example approximate solution for boron.

The ionization energy decreases a bit to 8.3 eV, indicating that indeed the 2p states have higher energy than the 2s ones.

For carbon, a second $\psi_{21m}$ state needs to be occupied. Within the made approximations, the second 2p electron could also go into the 2p$_z$ state. However, in actuality, repulsion by the electron already in the 2p$_z$ state makes it preferable for the new electron to stay away from the $z$-axis, which it can do by going into say the 2p$_x$ state. This state is around the vertical $x$-axis instead of the horizontal $z$-axis. As noted in chapter 3.2, 2p$_x$ is a $\psi_{21m}$ combination state.

For nitrogen, the third 2p electron can go into the 2p$_y$ state, which is around the $y$-axis. There are now three 2p electrons, each in a different spatial state.

However, for oxygen the game is up. There are no more free spatial states in the L shell. The new electron will have to go, say, into the p$_y$ state, pairing up with the electron already there in an opposite-spin singlet state. The repulsion by the fellow electron in the same state reflects in an decrease in ionization energy compared to nitrogen.

For fluorine, the next electron goes into the 2p$_x$ state, leaving only the 2p$_z$ state unpaired.

For neon, all 2p electrons are paired, and the L shell is full. This makes neon an inert noble gas like helium: it cannot accommodate any more electrons at the $E_2$ energy level, and, with the strongest nucleus among the L-shell elements, it holds tightly onto the electrons it has.

On the other hand, the previous element, fluorine, has a nucleus that is almost as strong, and it can accommodate an additional electron in its unpaired 2p$_z$ state. So fluorine is very willing to steal an electron if it can get away with it. The capability to draw electrons from other elements is called “electronegativity,” and fluorine is the most electronegative of them all.

Neighboring elements oxygen and nitrogen are less electronegative, but oxygen can accommodate two additional electrons rather than one, and nitrogen will even accommodate three.

Key Points

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The Pauli exclusion principle forces states of higher energy to become occupied when the number of electrons increases. This raises the energy levels greatly above what they would be otherwise.

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With the third element, beryllium, one of the $\psi_{200}{\updownarrow}$ “2s” states becomes occupied. Because of the higher energy of those states, the third electron is readily given up; the ionization energy is only 5.4 eV.

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Conversely, helium will not take on a third electron.

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The fourth element is lithium, with both 2s states occupied.

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For boron, carbon, nitrogen, oxygen, fluorine, and neon, the successive $\psi_{21m}$ “2p” states become occupied.

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Neon is a noble gas like helium: it holds onto its electrons tightly, and will not accommodate any additional electrons since they would have to enter the $E_3$ energy level states.

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Fluorine, oxygen, and nitrogen, however, are very willing to accommodate additional electrons in their vacant 2p states.

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The eight states $\psi_{2lm}{\updownarrow}$ are called the “L shell.”

4.9.5 Sodium to argon

Starting with sodium (natrium), the $E_3$, or “M shell” begins to be filled. Sodium has a single 3s electron in the outermost shell, which makes it much like lithium, with a single 2s electron in its outermost shell. Since the outermost electrons are the critical ones in chemical behavior, sodium is chemically much like lithium. Both are metals with a “valence” of one; they are willing to sacrifice one electron.

Similarly, the elements following sodium in the third row of the periodic table 4.1 mirror the corresponding elements in the previous row. Near the end of the row, the elements are again eager to accept additional electrons in the still vacant 3p states.

Finally argon, with no 3s and 3p vacancies left, is again inert. This is actually somewhat of a surprise, because the $E_3$ M-shell also includes 10 $\psi_{32m}{\updownarrow}$ states. These states of increased angular momentum are called the “3d” states. (What else?) According to the approximations made, the 3s, 3p, and 3d states would all have the same energy. So it might seem that argon could accept additional electrons into the 3d states.

But it was already noted that the p states in reality have more energy than the s states, and the d states have even more. The reason is the same: the d states stay even further away from the nucleus than the p states. Because of the higher energy of the d states, argon is really not willing to accept additional electrons.

Key Points

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The next eight elements mirror the properties of the previous eight, from the metal sodium to the highly electronegative chlorine and the noble gas argon.

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The states $\psi_{3lm}{\updownarrow}$ are called the “M shell.”

4.9.6 Potassium to krypton

The logical continuation of the story so far would be that the potassium (kalium) atom would be the first one to put an electron into a 3d state. However, by now the shielding approximation starts to fail not just quantitatively, but qualitatively. The 3d states actually have so much more energy than the 3s states that they even exceed the energy of the 4s states. Potassium puts its last electron into a 4s state, not a 3d one. This makes its outer shell much like the ones of lithium and sodium, so it starts a new row in the periodic table.

The next element, calcium, fills the 4s shell, putting an end to that game. Since the six 4p states have more energy, the next ten elements now start filling the skipped 3d states with electrons, leaving the N-shell with 2 electrons in it. (Actually, this is not quite precise; the 3d and 4s energies are closely together, and for copper and chromium one of the two 4s electrons turns out to switch to a 3d state.) In any case, it takes until gallium until the six 4p states start filling, which is fully accomplished at krypton. Krypton is again a noble gas, though it can form a weak bond with chlorine.

Continuing to still heavier elements, the energy levels get even more confused. This discussion will stop while it is still ahead. However, a complete periodic table is shown in figure 4.7. Note how the table expands horizontally because there are more and more angular momentum states for increasing values of the principal quantum number $n$. The shown periodic table is from NIST; an alternate link can be found in the web version of this document.

Figure 4.7: Periodic table, from NIST.

Key Points

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Unlike what the approximate theory says, in real life the 4s states $\psi_{400}{\updownarrow}$ have less energy than the $\psi_{32m}{\updownarrow}$ 3d states, and are filled first.

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After that, the transition metals fill the skipped states before the old logic resumes.

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The states $\psi_{4lm}{\updownarrow}$ are called the “N shell.” It all spells KLM Netherlands.

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The substates are of course called “s,” “p,” “d,” “f,” ...