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PRINCIPLES OF ALCHEMY
Introduction to the Periodic Table |
OK. Here it is for the first 20 elements. These are the most important
elements in the universe, and you're already familiar with most
of them.
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Yeah, I see what you mean. All the elements in the first column
(hydrogen, lithium, sodium and potassium) have one electron in
their outer shell. The elements in the second column (beryllium,
magnesium and calcium) have two electrons in their outer shell.
Hey, wait a minute. Helium also has two electrons in its outer shell. Why isn't it placed next to hydrogen, right above beryllium? Why is it over with neon? Neon's got 8 electrons in its outer shell. This table is wrong! |
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Yes.
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I'll ignore that. I'll even let you in on my little secret
about how I keep track of the first 20 elements. Instead of memorizing
each element, I memorized a silly poem which reminds me about
them and their order. It goes like this:
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Ah, that there are a lot of new abbreviations to learn?
You'll pick up the abbreviations as we go along. The important ones are the ones that will stick in your head without really thinking about it. The rest you can look up later in a list.OK. You don't have to convince me not to study. |
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I think so. I suspect that Rb (rubidium), Cs (cesium) and Fr (francium),
will all form electrovalent bonds just like Li (lithium), Na (sodium),
or K (potassium), because they are in the same Group (Group I).
That's right! All the Group I elements have a single electron in their outer shell and would be happy to lose it in order to achieve a complete outer shell. So all the Group I elements will form cations (with a charge of +1). |
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Yes. Can you "guess" a few compound molecules you could make using group I and group VII elements?Sure. Sodium chloride (NaCl) is the one we've talked about a lot already. I suppose you could also make sodium bromide (NaBr), sodium iodide (NaI) and sodium astatinide (NaAt).I suppose you can use lithium as if it were sodium to make lithium fluoride (LiF), lithium chloride (LiCl), lithium bromide (LiBr), lithium iodide (LiI) and lithium astatinide (LiAt). Hey, I can see even more, like potassium fluoride (KF) and.... Yes. Yes! I think you've got it. There are many different compounds one can make from combinations of the Group I and Group VII elements. That, again, is the power of the Periodic Table. You can match any Group I element with any Group VII element to make an ionic compound containing one atom of each. |
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Very good and clear thinking Arthur.
OK. We talked about magnesium chloride (MgCl2) a lot last time.
All the Group II /VII compounds will be made the same way.
Each Group II atom will form electrovalent bonds to two Group
VII atoms. So you can have magnesium chloride (MgCl2), magnesium
bromide (MgBr2), magnesium iodide (MgI2), mag.......
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Alchemists use X-rays to figure out the distance between two atoms,
using a technique I don't want to get into right now. For example,
using X-rays an Alchemist measures the distance between the two
hydrogens in H2. From that "interatomic distance", we
can calculate the radius of each hydrogen atom.
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Carbons involved in covalent bonds will have smaller radii (plural of radius) than a pile of unbonded carbons. Can you tell me why?Could it be that when atoms share their outer electrons, to make covalent bonds, they are drawn closer together?Yes. Covalently bonded atoms not only share their outer electrons, they also share their entire outer shell. |
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When carbons are involved in covalent bonds they are able
to get much closer together by sharing electrons in their outer
shell.
Yes. The atom's covalent radius will be smaller than its atomic radius because the covalent radius is measured using covalently linked molecules of the atoms. |
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Why?Because lithium (Li) has an extra shell. Hydrogen has only a K-shell, but lithium has a full K-shell and a start on an L-shell. Lithium has an extra shell so it will be a little bit bigger than hydrogen.That's right and it is right for all atoms within a Group. As you go down a Group the atoms get bigger and bigger because they have more and more shells. Understand?I think so. Hydrogen is the smallest atom. The next largest in the Group is lithium, then sodium, then potassium, than rubidium, then cesium and finally francium is the largest! Right? |
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Why? What causes the shells to shrink as you move across a period?
The protons! As you move across the Period (from left to right), you add a proton to the nucleus. Right?Right. So what? You also add an electron to balance the charge.Yes. However, each extra proton draws the shell a tiny bit closer. It's that electrostatic attraction again.You mean, because beryllium has one more proton than lithium, it attracts the electrons in the L-shell closer to it? |
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It applies to ALL Periods in the table, not just the second Period,
and not just the typical elements. All of them!
But in ANY GROUP the atomic radius will increase as you move down
the group because another level of electron shell is added each
time you move down.
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What do you mean "relative size"? I just mean, how they compare to each of their neighbor elements.
Let's look at the size of the typical elements and you'll see
what I mean.
That's because of the effect of different numbers of protons
attracting the same shell differently.
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Yes, I do and here it is. Notice that the energies are lowest
at the lower left of the Table, near Cs, and largest at the upper
right, near He. The numbers change in a way similar, but OPPOSITE
to, the way they change for atomic radius.
Yeah, I think so. As you go down a Group it is easier to ionize
the next element. Is that because the atoms get bigger?
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Hey! Not so fast wizard.
Look, beryllium (Be) has a HIGHER ionization energy than boron (B)! And look, look! Magnesium (Mg) and calcium (Ca) do that too! Their neighbors to the right (Al and Ga) have lower ionization energies! Some of the Group II elements stick out. They are wrong. Well, they aren't "wrong", they just don't follow the "rule". (So it isn't a rule at all. It's a "trend".) There are some subtle factors involved which I don't want to go into because they require more advanced knowledge about quantum mechanics and electron behavior. The exceptions you've found are the only exceptions to this trend in first ionization energies. OK? |
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I see. If I understand this right, the Periodic Table shows that
I could use 13.6 eV to ionize a hydrogen atom. Could I use
it to ionize a lithium (Li) atom?
What do you think?I think I could. According to the Table, lithium needs only 5.39 eV to ionize it. That is much less than 13.6 eV, so I would even have energy left over.That's right and that is because lithium lies below hydrogen in the first Group. It is always easier to ionize an atom lower on the periodic table.Hey, you know, if I had 13.6 eV and it takes only 5.39 eV to ionize lithium, I would have enough energy to ionize two electrons from the lithium. It would take twice the ionization energy.Look it takes only 10.78 eV to ionize two electrons from the lithium. (2 X 5.39 = 10.78 eV). I would even have 2.82 eV left over. Right? |
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Right. 10 eV will not ionize hydrogen.But it would ionize all the other Group I elements because they all have first ionization energies below 10 eV. Right?Absolutely. Go on.That's it. What do you mean "Go on."?Go on and tell me what would happen if you had more than just one atom of each Group I element.Oh, I see. (I thought you were trying to trick me into mistakenly using the first energies twice on the same atom. That would be wrong.)Well, to ionize TWO lithium atoms takes 10.78 eV (2Li X 5.39 eV per Li = 10.78 eV). So only one lithium atom can be ionized. I need 10.28 eV to ionize two sodium atoms (2Na X 5.14 eV per Na = 10.28 eV), so again I am left with only one ionized atom of sodium (Na+). |
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OK. Potassium (K) is a different story because its first ionization
is only 4.34 eV, much less than 5 eV. I need only 8.68 eV to ionize
two of them (2K X 4.34 eV = 8.68 eV). So that 10 eV will make
two ions of potassium (2K+) with a little bit of energy left over
(1.32 eV). Right?
Right!Rubidium (Rb) needs only 4.18 eV to ionize its first electron, so 10 eV could ionize two rubidiums (2Rb X 4.18 eV = 8.36 eV) with some energy left over (10 eV - 8.36 eV = 1.64 eV).Finally, that 10 eV could ionize two cesium atoms (2Cs X 3.89 eV = 7.78 eV) with some energy left over (10 eV - 7.78 eV = 2.22 eV). If I had a little more energy I could ionize a third cesium atom. Your Table doesn't have a value for the first ionization of francium (Fr), but I would guess that it has a lower value then rubidium. So I might be able to ionize three francium atoms with 10 eV Very good, Arthur. You made a good prediction. (And you're right!)
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So it is a measure of how easily an atom becomes an anion!
Yes, that's right! Elements with high electron affinity readily take electrons, when they can get them, to become anions. Those would be the elements close to fluorine in the Table.It's that greedy fluorine again stealing electrons every chance it gets!Yes, it is! Look at fluorine (F) on this Table! It has a value of 3.4 eV. Look here at chlorine (Cl), 3.62. What a monster!I sort of see some trends in this Table but it isn't real obvious. |
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That they are all bad at attracting electrons but good at releasing
them. The negative elements become cations. Right?
Well, you are half right. Negative values mean these elements are bad at accepting an electron. That much is right, but it is wrong to think that just because an atom has a negative electron affinity it will be good at releasing them.But they all form cations, don't they? I recall that magnesium forms the cation with two positive charges. Magnesium chloride is MgCl2 and the magnesium loses two electrons to become a cation. Right? |
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I see what you mean. So the negative electron affinity values
for the Group II elements don't have anything to do with their
ability to form cations. It has to do with their low affinity
for extra electrons.
Right. You would have to look at the (first and second) ionization energies to see how easily magnesium loses electrons.You know, even nitrogen (N) has a negative value, just barely.So what does that tell you? What does nitrogen's electron affinity tell you in comparison to other elements in the Table? |
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I can see why these two Tables are valuable. They tell you how
atoms treat electrons. Their own electrons as well as spare ones. It's electrons that make bonds.
Right you are! You can use the properties of ionization energy
and electron affinity to understand bonds better. Compounds too.
Sure. The atom with low ionization energy (Na) will readily lose
its electron, becoming the cation (Na+). The atom with the
high electron affinity (Cl) will readily accept an electron, becoming
the anion (Cl-).
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I see. Chlorine (Cl) is very electronegative, so it quickly picks
up an electron becoming an anion (Cl-). But sodium (Na) is not
very electronegative, so it readily gives up its electron becoming
a cation (Na+).
Both of them do that in order to obtain the electronic configuration of a noble gas (a full outer shell). That's right. Notice that sodium (Na) and chlorine (Cl) are on opposite sides of all the Tables. That's because they have opposite electronegativity and that displays a trend across the Periods. |
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Fluorine is a monster! No wonder it sucks up electrons every chance
it gets. Other elements don't stand a chance!
Yes, fluorine is an "electron monster"!
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OK. Carbon has an electronegativity of 2.55 and hydrogen has an
electronegativity of 2.20. So the difference in electronegativity
is 0.35. That's not much of a difference! It's about a tenth (1/10)
the difference in CsF. So I think the C-H bonds in methane are
covalent. The difference in their electronegativity is so low
that neither atom can give or take an electron from the other.
So, they share. They form covalent bonds.
Very good. You worked that out very well. Here's another one to
try.
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Right. Polarized bonds leave the atoms with very slight charges
which we call "delta minus" or "delta plus".
They are not complete ions! These polarized bonds are due to the
electronegativity of the two different elements which make the
bond.
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Yeah. The polarized bonds draw the electrons away from the hydrogens, so the hydrogen could be used to make hydrogen bonds. A polarized bond will cause the hydrogen to be slightly positive (delta plus) because its shared electron spends most of the time with the other, more electronegative, atom. That slightly positive charge on that hydrogen can attract and hold (slightly) any negatively charged atoms. That's a hydrogen bond! |
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PRINCIPLES OF ALCHEMY
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