Tabled Discussion

Chm 1311 Lectures for 15 & 19 June 2000

Your browser must support super- and sub-scripts and Greek letters (p). Previous lecture


Now that we have the rules that matter waves impose upon legitimate orbitals for electrons, we can blithely drop one electron after another into a gedanken atom (that's one we're thinking about) until we've added as many electrons as there are protons (so it's neutral). Then we look at the properties of the electrons "on the outside" (those in intimate contact with other atoms) to determine the Physics, Chemistry, Wit, and Wisdom of atoms in molecules.

We've seen that, due to its specification of increasing wave nodes, n plays the major role in determining the energy of the electrons in an atom. In fact, in hydrogen and the hydrogenic ions (those with only one electron like Li2+), n is exclusively responsible for the orbital energies. But due to electron-electron repulsion in non-hydrogenic atoms and ions, l makes small contributions to the energy as well.

Nature will demand that in a stable atom (or atomic ion), the electrons occupy the lowest energy orbitals available. The Pauli Principle prevents them from all tumbling down to the n=1 shell, so only the two that get there first can have that lowest energy orbital. The rest must content themselves with the upper floors.
For convenience when we speak of atomic orbitals, we need some sort of shorthand for the creature implied by ( n, l, etc. ). That shorthand takes the form of a number, the value of n, followed by a letter, a symbol for the value of l. Those symbols come from early spectroscopists who characterized spectral lines as sharp, principal, diffuse, or fundamental. So the first 4 symbols (for l=0,1,2,3) are s , p, d, and f. Since l can rise higher (all the way to n-1 and n can be indefinitely large), more symbols than just spdf are need. After f, the lettering just becomes alphabetical: spdfghiklmn... etc. (No, I didn't make a mistake; there is no j for, I'm sure, some very good reason...which I don't know. Maybe it's because j is used instead for total angular momentum? Whatever.)
So since energy rises with n and l < n, we can predict the ordering of the hydrogenic orbitals:

1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f < 5s = 5p = 5d = 5f = 5g < 6s ...

However, the shapes of the orbitals are such that the electrons are more confined as l increases, leading to slightly higher energies for multi-electron atoms (due to the electron-electron repulsions), and a different energy order:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p ...

which appears impossible to remember! (Although if you stare at it a while, it starts to have some sort of demonic order.) And how do we know that this order is correct? Well, one can do theoretical quantum mechanical calculations which show it, or you can just look at the periodic table! Keeping in mind that the maximum # of electrons in any suborbital is 2 (for ms=±½) :

lSymbolMax. # ml = 2l+1Max. # e- = 2×(2l+1)

2s2sDr. Parr's Colorful Periodic Table2p2p2p2p2p2p
3s3s 3p3p3p3p3p3p
4s4s3d3d3d3d3d3d3d3d3d3d 4p4p4p4p4p4p
5s5s4d4d4d4d4d4d4d4d4d4d 5p5p5p5p5p5p
6s6s5d4f 1-145d5d5d5d5d5d5d5d5d 6p6p6p6p6p6p
7s7s6d5f 1-146d6d6d6d6d6d6d6d6d 7p7p7p7p7p7p


Aside from never knowing just where to put H and He (so they're left blushing pink ... embarrassed), that shape should be now be familiar to you even in your sleep! In fact, by now you could probably place over a dozen atoms in their correct positions. So instead of element names, I've put corresponding orbital names to show you how the energy ordering scheme places the elements correctly. But far easier, since you'll always have a Periodic Table in front of you (tatooed on the inside of your eyelids?), you can use the organization of the table to rationalize how the electronic orbitals can be "filled up" (aufbau, in German) to produce each new element by adding one more electron to the last set.
1st Row Although only a single electron goes into a 1s (no nodes) orbital to make hydrogen, we have to specify that two electrons share that 1s (with ms spins +½ and -½, respectively) to make helium. So while the electronic configuration of H is given as 1s, that for He is 1s². Now the "first shell" has petered out 'cause we could only have l=0. Thus the first row of the Periodic Table is complete!
2nd Row But with n=2, we can use both l=0 and 1 (s and p) orbitals. And when we put the requisite number of electrons in each, we close the 2nd shell at 8 electrons as shown.
Now comes the kicker. We'd expect all the n=3 orbitals to show up in the 3rd row, but 4s < 3d in energy, so the 3d (transition metals) don't show up until row 4! Indeed, the d electrons all show up one row "late" because of e--e- repulsions acting differently in differently-shaped suborbitals.
Since l<3 for n=3, we didn't expect any 3f electrons in row 3, 4, or anywhere! But 4f were expected, and they didn't even show up in row 5 with the 4d. Instead 4f all fall into row 6 and constitute the Lanthanides. (Actually, La itself is that first 5d element; it's the rest of the "Lanthanides" that bear f electrons.
MS Encarta '97
The 5f Actinides (all 14 of 'em) show up similarly 2 rows late, but we don't get to see the 5g elements or, for that matter, any beyond the 6d since no one has succeeded in manufacturing them; they'd be incredibly radioactive nuclei with lifetimes measured below milliseconds! Indeed, the "7p" show up in the table with a pallid white background because they haven't been confirmed either; same reason.

You have to admit that it's impressive being able to say that the very latest elements to be discovered, Z=116 and 118, are going (for sure) to have the Periodic Table's very first 7p electrons! So we now know why Mendele'ev was onto something so profound: the chemistry of the elements progresses logically through the Periodic Table due to the periodicity of sub-orbital electronic energies! Although Mendele'ev didn't have this wonderful perspective known to him (or anyone else at the time) to rationalize his great achievement on the foundation of progressive matter waves.


So we aufbau our way merrily along the Periodic Table from
H: 1s1 to He: 1s2
Li: 1s22s1 to Ne: 1s22s22p6
Na: 1s22s22p63s1 to Ar: 1s22s22p63s23p6
K: 1s22s22p63s23p64s1 to ...
and we suddenly wish for an easier symbology because our HTML typing fingers are wearing thin! Fortunately, current practice is to replace each configuration ending in np6 with the noble gas element name associated with it. So that progression becomes the easier read:
H: 1s1 to He: 1s2 (OK, technically it doesn't end in a p.)
Li: [He] 2s1 to Ne: [He] 2s22p6
Na: [Ne] 3s1 to Ar: [Ne] 3s23p6
K: [Ar] 4s1 to V: [Ar] 4s23d3
when a hiccup occurs. We know from Hund's Rule that electrons prefer to occupy separate suborbitals if there's no energy penalty, but this means that up to 3d5 electrons could singly occupy on 3d each. And all their spins would be aligned; if any of them were ms=½, then they'd all be ms=½, and the atom would be paramagnetic.
Why did they align their spins again? Don't magnets alternate north-south with south-north? Yes, but compasses all line up north 'cause they're responding to a global (literally) magnetic field. Same thing happens to atomic electrons; they're responding to an atomic magnetic field created by their spins and their orbital motions. And those fields get higher the more unpaired spins one has, and that alignment gets more and more energetically favorable. Until it's such a good deal to unpair spins that Cr, the next atom after vanadium, will even pull the "wrong" spin electron out of its 4s2 and flip it to create not 4s23d4, as expected, but rather 4s13d5, half-filling its d subshell, and gaining another aligned spin from the remaining 4s1! Six spin-aligned electrons!! No wonder chromium makes such a good magnetic material.


How much to you weigh at the center of the Earth? If gravity increases the closer you get to a body, you can't get any closer than the middle! Does that mean you weigh lots? No. You don't weigh anything at all! That's because all the stuff that's attracting you is symmetrically distributed around you, pulling in all directions at once and canceling out!
Better still, half way to the center of the Earth, you only weigh half as much as on the surface. Only 1/8 of the Earth is at an altitude below you [remember V=(4/3) p r3], and you are twice as close to it as you were on the surface. Since gravitational force varies as r -2, that 1/8 makes you weigh (1/8)×(1/2)-2=½ as much. But what about the 7/8 distributed non-symmetrically about you at higher altitudes? Because of the 1/r² law, the closer (but lesser) earth directly above you pulls with exactly equal and opposite force to the further (but greater) earth opposite the center! [The 1/r² force is canceled by the r² surface area factor the further from your position it goes.]
"And what has this to do with atoms?", you ask (profoundly). The electrons are held and repelled by charge-charge (called Coulombic) forces which also vary as 1/r²! (Gauss showed that.) So the same nonsense applies. No symmetrically distributed charge at a radius above yours attracts or repels you! At the same time, All symmetrically distributed charge at a radius below yours attracts or repels you as if it were all at the symmetry center, in our case, as if it were at the nucleus!
So electrons in orbitals well beneath yours, are 100% effective in canceling out their number of protons! Conversely, you, in your lofty height above the nucleus are completely ineffective in cancelling the nuclear charge for those well below you. They see it all and deny you its attraction (because they repel you with the same force as each proton). That means when we fill up a shell, and increase n by one to use the next higher shell's s electron for the next element, that poor sucker is tossed high above the "core" electrons beneath him (because the extra node stretches out his matter wave)!
That lonely outer s looks back toward the nucleus and sees what? Z protons and (Z-1) electrons. The one extra pull over the pushes means that the effective charge that outer s responds to is only Zeff = 1. He is not tightly bound. He can be easily encouraged to leave. As a result, he can conduct electricity, and the alkali metals are metals.
Same Shell
In contrast, your buddy electrons in the same shell are only sometimes below you and sometimes above you; a matter wave is a broad three-dimensional distribution, after all. So same shell electrons are definitely not 100% effective in shielding one another from nuclear charge. When we walk across a row, we're adding more and more ineffective cancellations, but that means that the steadily increasing positive charge, Z, on the subsequent nuclei is more and more strongly felt. That causes outer electrons to be steadily more strongly bound to their atoms, increasing the ionization potential across the rows from left to right.


The same accumulating forces which bind the electrons more strongly across the Periodic Table, bind them into tighter and tighter geometries. Increasing effective charge from left to right causes atoms to shrink in size across the table. So the tightest smallest atom in a row should be the noble gas at its end!
In contrast, moving one element beyond a noble gas causes virtual abandonment of the next electron. Remember that it gets tossed into a bloated (one node larger) s orbital. And it sees an effective charge much, much less than the (n-1)p electron just beneath it. But if each shell is thus puffed up relative to the one beneath it, the trend in binding should be that it is less strong as one descends a Periodic Table column. After all, all alkali metals, for example, see the same Zeff=1, but as we go down Group 1, the outermost s orbitals get bigger and bigger, casting the outermost electron further and further from that effective +1 charge. Since the attraction falls off with distance, the 6s1 electron in Cs should be less strongly held than the 2s1 in Li.
So going down the table, atoms get bigger and ionization energies get smaller.
Ions? And the same must be true for ions. Imagine Ba: [Xe] 6s2 being ionized; that 6s electron is more strongly held than the single one in Cs: [Xe] 6s1 because Ba has one more nuclear charge, but that second 6s electron is not terribly effective in canceling it. But what now if we rip one off and make Ba+: [Xe] 6s1? That ion is "isoelectronic" (has the same number of electrons) with Cs, but there's the extra + charge. So Ba's 2nd ionization potential is quite a bit larger than Cs's 1st, even though it's the "same" electron.
But that difference begins to pale compared to what happens when you try to pull one more electron off Ba2+: [Xe]. All of those core electrons that are left are very strongly bound, and so barium's 3rd ionization potential is sky high; there are no more bloated "valence shell" electrons left to pull! By valence shell, we mean the outermost shell, but "valence" used to refer to the electrons which can engage in chemical bond formation! And the reason that's at all possible is that valence shell electrons are the least strongly-bound ones in the atom.
Atoms see one another only by their electrons? Right...and furthermore, only by their valence electrons! Ahhhhh. What a segue into chemical bonding (next chapter).

Next Lecture
Last modified 19 June 2000. Chris Parr