5.9 Heav­ier Atoms

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.

5.9.1 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 4.3. The to­tal en­ergy Hamil­ton­ian for the elec­trons of an el­e­ment with atomic num­ber $Z$ with is:

$$H = \sum_{i=1}^Z \Bigg[ - \frac{\hbar^2}{2m_{\rm e}} \na... ...\skew0\vec r}_i -{\skew0\vec r}_{\underline i}\vert} \Bigg] %$$ (5.34)

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:

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

Key Points

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

The Hamil­ton­ian for the elec­tron struc­ture has been writ­ten down.

5.9.2 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 4.3.

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:

$$\psi^Z=\pp1/{\skew0\vec r}_1//z1/\pp2/{\skew0\vec r}_2//z2/\ldots \pp Z/{\skew0\vec r}_Z//zZ/.$$

Sub­sti­tu­tion of this as­sump­tion into the eigen­value prob­lem $\sum_{i}h_i\psi^Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E\psi^Z$ and di­vid­ing by $\psi^Z$ pro­duces

$$\frac{1}{\pp1/{\skew0\vec r}_1//z1/} h_1 \pp1/{\skew0\vec r... ...w0\vec r}_2//z2/} h_2 \pp2/{\skew0\vec r}_2//z2/ + \ldots = E$$

since $h_1$ only does any­thing to the fac­tor $\pp1/{\skew0\vec r}_1//z1/$, $h_2$ only does any­thing to the fac­tor $\pp2/{\skew0\vec r}_2//z2/$, 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 $\pp1/{\skew0\vec r}_1//z1/$ it­self does, since none of the other terms in the equa­tion varies with those vari­ables. That means that

$$h_1 \pp1/{\skew0\vec r}_1//z1/ = \epsilon_1 \pp1/{\skew0\vec r}_1//z1/,$$

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 4.3; al­low­ing for spin, the pos­si­ble so­lu­tions are,

$$\psi_{100}({\skew0\vec r}_1){\uparrow}(S_{z1}),\, \psi_{10... ..., \psi_{200}({\skew0\vec r}_1){\downarrow}(S_{z1}),\, \ldots$$

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,

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

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 5.7. That will be done dur­ing the dis­cus­sion of the dif­fer­ent atoms that is next.

Key Points

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

The Hamil­ton­ian eigen­value prob­lem is too dif­fi­cult to solve an­a­lyt­i­cally.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

This is a very crude ap­prox­i­ma­tion.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

5.9.3 Hy­dro­gen and he­lium

This sub­sec­tion starts off the dis­cus­sion of the ap­prox­i­mate ground states of the el­e­ments. Atomic num­ber $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 cor­re­sponds to hy­dro­gen, which was al­ready dis­cussed in chap­ter 4.3. The low­est en­ergy state, or ground state, is $\psi_{100}$, (4.40), 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:

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

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 4.3.

For he­lium, with $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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 5.7):

$$\Psi({\skew0\vec r}_1,S_{z1},{\skew0\vec r}_2,S_{z2};t) = ... ...{\skew0\vec r}_2){\downarrow}(S_{z2}) \end{array} \right\vert$$ (5.35)

or, writ­ing out the Slater de­ter­mi­nant:

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

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 5.5.6.

Fig­ure 5.4 shows the ap­prox­i­mate prob­a­bil­ity den­sity for the first two el­e­ments, in­di­cat­ing where elec­trons are most likely to be found. In re­al­ity, the shield­ing ap­prox­i­ma­tion un­der­es­ti­mates the nu­clear at­trac­tion and the shown he­lium atom is much too big.

Fig­ure 5.4: Ap­prox­i­mate so­lu­tions for the hy­dro­gen (left) and he­lium (right) atoms.

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 13.7). As a re­sult, the sec­ond pro­ton is only partly shielded, and the elec­tron is held much more tightly than the analy­sis pre­dicts. See ad­den­dum {A.38.2} 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 {N.7}.

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

5.9.4 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, fig­ure 5.5, lithium starts a new row.

Fig­ure 5.5: Ab­bre­vi­ated pe­ri­odic ta­ble of the el­e­ments. Boxes be­low the el­e­ment names in­di­cate the quan­tum states be­ing filled with elec­trons in that row. Cell color in­di­cates ion­iza­tion en­ergy. The length of a bar be­low an atomic num­ber in­di­cates elec­troneg­a­tiv­ity. A dot pat­tern in­di­cates that the el­e­ment is a gas un­der nor­mal con­di­tions and wavy lines a liq­uid.

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 {N.8}.

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.

$$\frac{a_1}{\sqrt{6}} \left\vert \begin{array}{ccc} \psi_{100}({\s... ...S_{z3}) & \psi_{200}({\skew0\vec r}_3){\uparrow}(S_{z3}) \end{array}\right\vert$$

$${} + \frac{a_2}{\sqrt{6}} \left\vert \begin{array}{ccc} \psi_{100... ...{z3}) & \psi_{200}({\skew0\vec r}_3){\downarrow}(S_{z3}) \end{array}\right\vert$$ (5.36)

Fig­ure 5.6: 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 $\vert E_2\vert$ $\vphantom0\raisebox{1.5pt}{$=$}$ 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 $4\times4$ Slater de­ter­mi­nant;

$$\frac{a}{\sqrt{24}} \left\vert \begin{array}{cccc} \psi_... ...{\skew0\vec r}_4){\downarrow}(S_{z4}) \end{array} \right\vert$$ (5.37)

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 5.7 shows the ap­prox­i­mate so­lu­tion in which the $\psi_{210}$, or 2p$_z$ state is oc­cu­pied. It may be re­called from fig­ure 4.11 that this state re­mains close to the $z$-​axis (which is hor­i­zon­tal in the fig­ure.) As a re­sult, the wave func­tion be­comes di­rec­tional.

Fig­ure 5.7: 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 re­al­ity, 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 4.3, 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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

With the third el­e­ment, lithium, 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.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

Con­versely, he­lium will not take on a third elec­tron.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

5.9.5 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 fig­ure 5.5 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 at the same the­o­ret­i­cal en­ergy level, 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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

5.9.6 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 chromium and cop­per 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.

Key Points

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

Af­ter that, the tran­si­tion met­als fill the skipped 3d states be­fore the old logic re­sumes.

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

$\begin{picture}(15,5.5)(0,-3) \put(2,0){\makebox(0,0){\scriptsize\bf0}} \put(12... ...\thicklines \put(3,0){\line(1,0){12}}\put(11.5,-2){\line(1,0){3}} \end{picture}$

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

5.9.7 Full pe­ri­odic ta­ble

A com­plete pe­ri­odic ta­ble of the el­e­ments is shown in fig­ure 5.8. The num­ber in the top left cor­ner of each cell is the atomic num­ber $Z$ of the el­e­ment. The num­bers to the left of the ta­ble in­di­cate the pe­ri­ods. The length of the pe­ri­ods ex­pands from 2 to 8 el­e­ments in pe­riod 2 when $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, p, states must be filled with elec­trons. Then in pe­riod 4, a de­layed fill­ing of $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2, d, states ex­pands the pe­ri­ods by an­other 10 el­e­ments. Fi­nally, in pe­riod 6, a de­layed fill­ing of $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3, f, states adds an­other 14 el­e­ments per pe­riod.

Fig­ure 5.8: Pe­ri­odic ta­ble of the el­e­ments. Cell color in­di­cates ion­iza­tion en­ergy. Boxes in­di­cate the outer elec­tron struc­ture. See the text for more in­for­ma­tion. [pdf]

The top part of the shown ta­ble is called the main group. For some rea­son how­ever, hy­dro­gen is not in­cluded in this term. Note that com­pared to the pre­vi­ous ab­bre­vi­ated pe­ri­odic ta­ble, hy­dro­gen and he­lium have been moved to the fi­nal columns. The idea is to com­bine el­e­ments with sim­i­lar prop­er­ties to­gether into the same columns. He­lium is a no­ble gas like the group VIII el­e­ments with filled elec­tron shells. He­lium has ab­solutely noth­ing in com­mon with the group II al­ka­line met­als that have two elec­trons in an oth­er­wise empty shell. Sim­i­larly, hy­dro­gen be­haves much more like a halo­gen with 1 elec­tron miss­ing from a filled shell than like an al­kali metal with 1 elec­tron in an oth­er­wise empty shell. How­ever, hy­dro­gen is still suf­fi­ciently dif­fer­ent that it should not be con­sid­ered an ac­tual halo­gen.

The el­e­ments in the pe­ri­odic ta­ble are clas­si­fied as met­als, met­al­loids, and non­metals. Met­al­loids have chem­i­cal prop­er­ties in­ter­me­di­ate be­tween met­als and non­metals. The band of met­al­loids is in­di­cated by dark red cell bound­aries in fig­ure 5.8. It ex­tends from boron to polo­nium. The met­als are found to the left of this band and the non­metals to the right. Hy­dro­gen and he­lium are most def­i­nitely non­metals and their shown po­si­tion in the ta­ble re­flects that.

The color of each cell in­di­cates the ion­iza­tion en­ergy, in­creas­ing from bluish to red­dish. The length of the bar be­low the atomic num­ber gives the elec­troneg­a­tiv­ity. In the top right cor­ner wavy lines in­di­cate that the el­e­ment is a liq­uid un­der nor­mal con­di­tions, and dots that it is a gas. A dag­ger in­di­cates that the atomic nu­cleus is ra­dioac­tive (for every iso­tope, chap­ter 14). If the dag­ger is fol­lowed by an ex­cla­ma­tion mark, the ra­dioac­tiv­ity causes the nu­cleus to de­cay fast enough that there are no us­able quan­ti­ties of the el­e­ment found in na­ture. These el­e­ments must be ar­ti­fi­cially pre­pared in a lab.

The boxes be­low the el­e­ment names in­di­cate the s, p, d, and f shells be­ing filled in that pe­riod of the ta­ble. The shells al­ready filled in the no­ble gas at the end of the pre­vi­ous pe­riod re­main filled and are not shown. Note that the fill­ing of $n$d states is de­layed one pe­riod, to pe­riod $n+1$, and the fil­ing of $n$f states is de­layed two pe­ri­ods, to pe­riod $n+2$.

Be­sides el­e­ment name, sym­bol, and ra­dioac­tiv­ity, pe­ri­odic ta­ble fig­ure 5.8 lim­its it­self to data for which the pe­ri­odic ta­ble arrange­ment is mean­ing­ful. Many other pe­ri­odic ta­bles also list the av­er­age atomic mass for the iso­topic com­po­si­tion found on earth. How­ever, for pur­poses of un­der­stand­ing atomic masses phys­i­cally, graphs in chap­ter 14 on nu­clei, like fig­ures 14.2 and 14.4, are much more use­ful.

It should be noted that pe­ri­odic ta­ble fig­ure 5.8 de­vi­ates in a num­ber of as­pects from the nor­mal con­ven­tions. Fig­ure 5.8 is what seems the sim­plest and most log­i­cal. If you put his­tor­i­cal odd­i­ties and a few com­mit­tees in charge, you get some­thing dif­fer­ent.

Most promi­nently, most pe­ri­odic ta­bles leave hy­dro­gen in group I in­stead of mov­ing it to the top of group VII. But if you move he­lium to group VIII be­cause of its sim­i­lar­ity with the other no­ble gases in that group, then it is lu­di­crous to leave hy­dro­gen in group I. Hy­dro­gen has vir­tu­ally noth­ing in com­mon with the al­kali met­als in group I. Like the light halo­gens, it is a di­atomic gas un­der nor­mal con­di­tions, not a solid metal. Even at the ex­tremely low tem­per­a­tures at which hy­dro­gen so­lid­i­fies, it is a non­con­duct­ing mol­e­c­u­lar solid, not a metal. The melt­ing, boil­ing, and crit­i­cal points of hy­dro­gen form a log­i­cal se­quence with those of the halo­gens. They are to­tally in­con­sis­tent with those of the al­kali met­als. Hy­dro­gen has the ion­iza­tion en­ergy of oxy­gen and the elec­troneg­a­tiv­ity of phos­pho­rus. A ionic com­pound like NaH is a di­rect equiv­a­lent of NaCl, salt, with the hy­dro­gen as the neg­a­tive ion.

It is true that hy­dro­gen can also form pos­i­tive ions in chem­i­cal re­ac­tions, more or less, some­thing that the halo­gens sim­ply do not do. But do not ac­tu­ally ex­pect to find bare pro­tons when other atoms are around. Also the ion­iza­tion en­ergy and elec­troneg­a­tiv­ity of hy­dro­gen are quite a bit out of line with those of the other halo­gens. Hy­dro­gen is cer­tainly not a true halo­gen. But if you or­der the el­e­ments by prop­er­ties, there is no doubt that hy­dro­gen be­longs in group VII, not I. If you want to re­fer to the quan­tum-me­chan­i­cal shell struc­ture, the term s block can still be used to in­di­cate the al­kali and al­ka­line met­als along with hy­dro­gen and he­lium. The re­main­der of the main group is the p block. These names in­di­cate the quan­tum states be­ing filled.

The term tran­si­tion met­als may not in­clude the el­e­ments in group IIB of the d-block, for rea­son re­lated to the fact that their s and d shells have been com­pletely filled. The f-block el­e­ments are some­times re­ferred to as the in­ner tran­si­tion met­als.

Fur­ther, ac­cord­ing to the 2005 IU­PAC Red Book the lan­thanides and ac­tinides should be more prop­erly called the lan­thanoids and acti­noids, since ide usu­ally means neg­a­tive ion. Since oid” means “-like, ac­cord­ing to IU­PAC the lan­thanoids should not re­ally in­clude lan­thanum, and the acti­noids should not in­clude ac­tinium. How­ever, the IU­PAC does in­clude them be­cause of com­mon us­age. A rare tri­umph of sci­en­tific com­mon sense over lousy ter­mi­nol­ogy. If lan­thanum and ac­tinium are to be in­cluded, the lan­thanides and ac­tinides should of course sim­ply have been re­named the lan­thanum and ac­tinium groups, or equiv­a­lent, not lan­thanoids and acti­noids.

More sig­nif­i­cantly, un­like fig­ure 5.8 sug­gests, lutetium is in­cluded in the lan­thanoids and lawren­cium in the acti­noids. The term rare-earth met­als in­clude the lan­thanoids, as well as scan­dium and yt­trium as found im­me­di­ately above lutetium.

Also, both lutetium and lawren­cium are ac­cord­ing to IU­PAC in­cluded in the f-block. That makes the f-block 15 columns wide in­stead of the 14 col­umn block shown at the bot­tom of fig­ure 5.8. Of course, that does not make any sense at all. The name f-block sup­pos­edly in­di­cates that an f-shell is be­ing filled. An f-shell holds 14 elec­trons, not 15. For lutetium, the f-shell is full and other shells have be­gun to fill. The same is, at the time of writ­ing, be­lieved to be true for lawren­cium. And while the first f-shell elec­trons for lan­thanum and ac­tinium get tem­porar­ily bumped to the d-shell, that is ob­vi­ously a mi­nor er­ror in the over­all logic of fill­ing the f-shell. (Ap­par­ently, there is a long-stand­ing con­tro­versy whether lan­thanum and ac­tinium or lutetium and lawren­cium should be in­cluded in the f-block. By com­pro­mis­ing and putting both in the f-block of their 2007 pe­ri­odic ta­ble, the IU­PAC got the worst of both worlds.)

A nice re­cent ex­am­ple of a more con­ven­tional pe­ri­odic ta­ble by an au­thor­i­ta­tive source is from NIST. This also in­cludes the lat­est up­dates on var­i­ous data, un­like the pe­ri­odic ta­ble in this book. An ear­lier ver­sion can be found at the web lo­ca­tion of this doc­u­ment. The hi­ero­glyphs found in the NIST ta­ble are ex­plained in chap­ter 10.7.1.

Pe­ri­odic ta­ble fig­ure 5.8 was based on data from var­i­ous sources. Shell fill­ings and ion­iza­tion en­er­gies agree with the NIST list­ing and ta­ble. The un­cer­tain shell fill­ings at atomic num­bers 103 and 104 were left out. The clas­si­fi­ca­tion whether the el­e­ments must be ar­ti­fi­cially pre­pared was taken from the NIST pe­ri­odic ta­ble. The elec­troneg­a­tiv­i­ties are based on the Paul­ing scale. They were taken from Wikipedia use val­ues, that were in turn taken from We­bEle­ments, and are mostly the same as those in the 2003 CRC Hand­book of Chem­istry and Physics, and the 1999 Lange’s Hand­book of Chem­istry. Dis­crep­an­cies be­tween these sources of more than 10% oc­cur for atomic num­bers 71, 74, 82, and 92.