This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.
2.03 Effective Nuclear Charge and atomic and ionic radii
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来自 University of Kentucky 的课程
Chemistry
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从本节课中
Week 2
Once we understand the electronic structure, we will be able determine how the periodic table can be used to compare the elements according to atomic size and various energies. We will conclude by seeing how these trends can be used to predict chemical properties of the elements.
与讲师见面
Dr. Allison SoultLecturer
Chemistry
Dr. Kim WoodrumSr. Lecturer
Chemistry
In our third learning objective, we're going to be developing an understanding of
effective nuclear charge, and we're going to see the role that
effective nuclear charge has, in the size of the atoms, or size of ions.
To understand effective nuclear charge, we must begin by understanding shielding.
Here's the definition, the effect,
in which electrons can reduce the electrostatic attraction,
between the protons of the nucleus, and other electrons.
So what we are considering here, is we have a nucleus, and
we know that in this nucleus there are protons.
There's neutrons as well, but that nucleus has got a charge.
And we have electrons that are being attracted to those protons.
That attraction is opposite charges, and it's called an electrostatic attraction.
Shielding is an ability for electrons to block a little bit,
the feel of that outer electron, from the nucleus.
And that's what shielding is.
The effect in which electrons can reduce the electrostatic attraction.
So, with the idea, shielding, we can now define effective nuclear charge,
it's the nuclear charge felt by an electron, okay?
So, it is that electrostatic attraction, but
it's not going to feel the full necessarily,
the full positive charge of the nucleus, because of the shielding effect.
So, the effective nuclear charge, will be the actual nuclear charge, okay?
So, that would be the number of protons, in the nucleus.
That's, it's nuclear charge.
And then, this sigma that we see here, is the symbol for shielding.
So, the shielding by other electrons keeping from,
I, an electron that we're interested in, from filling that nucleus.
So, each electron has an effective nuclear charge.
When it senses that nucleus as being, the stronger that effective nuclear charge,
the more tightly the electron will be held onto that nucleus.
The lower that effective nuclear charge,
the further away that electron will be from the nucleus,
coz it is not getting the same tug from that electrostatic attraction.
So, what kind of shielding are there?
There are basically three very,
three degrees of shielding that we're going to look at.
The first one is, electrons are shielded very effectively,
by electrons in subshells that are closer to the nucleus.
So, if we redraw the picture that I had drawn earlier,
where we had the protons in the nucleus, and we had an electron out here.
If we have an electron between the, this electron and its nucleus, this electron
here is very effective at shielding the outer electron, from the nucleus.
If you have an electron that's in the same shell, okay, so
it's the same relative distance, so let's put an electron here.
This electron is competing for the positive charge of the nucleus, with this,
this electron is competing with this electron, so
there's a little bit of shielding, but not extensively.
This one certainly would do it a lot more,
because it's between the outer electron that we're considering, and the nucleus.
Now, if there were an electron sitting way out here,
this one would not be effective at all, at shielding this electron from its nucleus.
Okay? So,
outer electrons beyond the one that we're interested in, don't shield.
So, what does this have to do with anything?
Why do we care about effective nuclear charge, why do we care about shielding?
Well the reason, is we can use this to learn trends in atomic radius,
and then eventually in ionic radius.
How do we define atomic radius?
Well, we know what a radius is.
A radius would be the distance from the nucleus to the outside of the circle.
However, you can't determine where the outside of that circle is, very well.
Take our half the distance between two nuclei.
It's very easy to find nuclei.
We've got means by which we can do that.
So if we look at this nucleus, and
this nucleus, and determine the distance between that nucleus.
Cut it in half, we can assume that, that would be the atomic radius.
If we had a diatomic molecule, like maybe this was Cl2, again,
we would assume that these two are bumping next to each other, and we would define
the atomic radius as one half the distance between those two nuclei.
So, that's how it's defined.
Now let's see how affective nuclear charge and
shielding, define the size of the atom.
Now, before we do that,
I want you to understand that, it is never an issue of crowding.
Where, because you'd have more electrons, you'd need more space for
them, to move about in.
Remember when we talked about the relative size of
the nucleus versus the size of the atom.
And we said to imagine that a marble was suspended in the middle of a big
football stadium, or
soccer stadium, and we had that marble suspended above the 50 yard line.
The rest of that stadium would be the size of the atom.
Electrons are way tinier than the nucleus, and they're moving about in that space, so
the rest of the stadium, it's never going to be a crowding issue.
What defines the size of these atoms, is going to
be the attractive force that the nucleus is, is pulling, on those electrons.
The more those outer electrons are feeling that nuclear charge, the higher that
effective nuclear charge, the tighter it's going to draw those electrons in.
The lower that effective nuclear charge, the bigger the atom's going to be.
So, never think, well I've got more electrons, therefore it has to be bigger.
That's not the argument.
It is the attractive force that those electrons are feeling that nuclear,
nuclear charge, okay, effective nuclear charge.
So let's consider two atoms here.
I have on the left, an atom of lithium, okay?
With this atom of lithium, I want to consider this outer electron right here,
and what it is feeling, that nuclear charge as.
Now, we know the nuclear charge is a full plus 3, but
it's not going to fill that full plus 3, because, it has two electrons between it.
So, these are in the 1s orbital.
This is the 2s orbital.
We know that 2s is further away, and
those 1s can shield it effectively, from the nucleus.
Now, we can use as a rough estimate of shielding, works pretty well for
arguments, just how many electrons are between that outer shell and
the nucleus, okay?
So, we have got the effective nuclear charge being the nuclear charge,
minus the two electrons that stand in the way, between the outer electron and
the nucleus.
So, this number two, is because of these two electrons.
And that'll give me an effective nuclear charge of plus 1.
So, this electron that is sitting out here in the outer shell,
is going to feel a plus 1 pull.
Now let's go to the next atom on the periodic table.
Which would be beryllium, okay?
In beryllium, we have one more electron, and
therefore we have one more proton in the nucleus.
So, we have got four protons in the nucleus, and
the nuclear charge would be four.
If we consider the pool of this electron,
it's effectively being shielded by this electron and this electron.
So, that it's being effectively shielded by only two.
Now, this is a little bit of shielding, but not enough to really consider.
So, when we work on the affect of nuclear charge of this, we see that we
have a nuclear charge of four because there are four protons in the nucleus.
The two is the shielding, okay, and that is coming from these two
inner electrons, and that is going to give us an effective nuclear charge of plus 2.
So, if I were to consider this electron right here.
It is filling a plus 2 pull, by the nucleus.
Now, if it is feeling a plus 2 pull, and an electrostatic attraction of a plus 2
with a minus 1, it is going to pull it in more tightly.
An therefore, the atomic radius of beryllium is going to be smaller.
Now, if we continue to work our way across the periodic table, so
we work across the periodic table, from lithium to beryllium to aluminum.
And actually, well boron is next, and carbon an nitrogen.
As we work our way across that periodic table, what is happening?
We're continuing to add one elec, nu, one proton into the nucleus.
So, as we go across the periodic table, this keeps going up.
But we're continuing to add electrons into the same shell, where n is equal to 2.
We'll put 'em in the s, and then we'll put 'em in a p,
but we're putting 'em all in the same general distance from the nucleus.
Therefore, the shielding is not changing as we go across the periodic table.
The shielding is staying at two.
So, as we go across the periodic table, okay, Z keeps going up by one.
Keeps plus 1, plus 1, as we go across.
But, the shielding is staying constant.
And so, the effective nuclear charge, is going to keep on going up by one.
So what is that going to do to our size trend?
Well, let's look at it.
So, here is just the representative elements.
Actually, the main group elements represented, so
we don't see the transition elements there.
And we can see as we follow a group across, I mean,
not group, but a period across.
And we see this decrease in size as we move across.
And there's some minor variations to that, but for
the most part, we see that decreasing occurring.
So, as the effective nuclear charge is going up, because the sil,
shielding is staying constant across there, we see that decrease in size.
Now, all of these are measured in picometers, just so, so you know.
All right, so, we're adding a proton, an electron as we go across.
One proton, one electron.
The shielding is not changing, so it keeps getting smaller and smaller.
But what happens as we go from neon to the next element?
It increases drastically in size.
It goes from 70 picometers to 186 picometers.
Now, what would cause that to happen?
Well, let's considering what's happening between the neon and the sodium.
In the neon, we have in the first shell, two electrons.
Okay, that would be the heliud, helium core.
When we move out to the second shell, as we're moving across this period that we
see here, we're putting our electrons in the second shell, okay?
So we're putting those electrons here,
until we get eight valence electrons in that shell, and then the next one for
the sodium, it needs to move out to the third shell.
So, it is going in to the three, it'll be a 3S1 for that last electron.
What, what happens?
Well this is the outer electron, what is suddenly happening to the shielding?
We've gone from, for neon, a shielding equaling to two, to sodium,
where our shielding is the two electrons we see here, plus,
the eight electrons that we see here, and so, the shielding has just jumped to ten,
versus the two that we had for the sodium.
So, suddenly there's lots more shielding, the effective nuclear charge drops down,
and that means that sodium will be much larger.
So, the trend down a group, is that it will increase in size,
because you're putting it out into the next available shell, and
increasing the amount of shielding that there are.
So, everyone of those,
without exception, is going to increase in size as you go down.
So, increasing size from right to left, and
from top to bottom, on the periodic table.
Here's another way of looking at that.
And one of the things that we see here, is that across the this area right here,
is going across the transition elements.
And as we go across the transition elements,
the size trend doesn't change very much.
It is, levels out quite a bit.
And the reason for this is, the electrons that you are placing as you go
across the d, are not being placed in an outer shell.
They're being placed into an inner shell.
Remember, for example let's do, titanium.
The electron configuration is (Ar)4s2 3d2.
And we're putting those, the electrons as we go, into an inner shell.
That's not the outer shell, that's not the outermost shell.
The outermost shell is this.
So, as we go across the transition elements, we are increasing the nuclear
charge by one, but we're also continually increasing the shielding by one.
So, we see a somewhat leveling out of the size as we
go across those transition element areas.
So, that's our atomic radius.
We know that for atoms, when you're comparing atoms,
they get smaller as we go across the periodic table,
they get larger as we move down the perio, periodic table, okay?
It levels off somewhat across the transition elements, and
that trend is the way it, the way it is,
the way it's laid out because of the electronic structure of the atoms.
I want you to not only know the trend, but understand the why behind the trend.
Understand the shielding effect,
and the effect of nuclear charge that defines those.
Now let's move on to ionic radius, okay?
So, ionic radius would be just the radius of an anion or cation.
And in order to consider trends of these radii, let's
start with cations, and let us compare the metal, to its cation, okay?
So, we're going to start, and we're going to consider lithium the element,
as an atom, verses lithium the cation.
When your're comparing an atom to an ion, and comparing their sizes,
it's very helpful to look at the number of protons, verses the number of electrons.
So, if we compare these two,
we know that they both will certainly have the same number of protons.
They both have three protons.
However, lithium, as a atom is going to have three electrons, but
lithium as a cation, is only going to have two.
We've removed one electron.
So, in this scenario, we have the same nuclear charge, okay?
In this scenario,
we've got three electrons being held on by those three protons, and
here we only have two electrons being held on by those three protons.
Plus, what did we remove?
We removed a valence electron.
So, we are considering electrons that are on the inner shell.
Okay?
So, think about those things, and answer the question, question.
Which one would have the larger radius?
Well, if you picked lithium, the element, the atom, you would be correct.
The atom is always going to be larger than the cation that you make from that atom,
because you are removing an electron, and the nuclear charge is
being holding on to fewer electrons, and can draw them in more tightly.
Now, if we consider for anions, let's consider an atom versus its ion.
And let's do it for fluorine, versus the fluoride ion.
So we'll construct a similar table here.
The number of protons as we go to the periodic table,
we see that the atomic number is nine.
So they both have nine protons.
For fluorine the atom, it would have nine electrons.
For fluoride the ion, it would have ten.
So, we have got nine protons in both ca, situations.
Here the nine protons is drawing in nine electrons, attracting nine electrons.
This one is trying to hold on to ten electrons,
one more electron than the nuclear charge itself.
So, with that information, maybe you could think through which one would
have the larger radius, the fluorine or the fluoride.
If you chose the fluoride, you would be correct.
And the truth of the matter is,
every time you compare an atom to it's anion that's, so you take a non metal, and
create anion from it, the anion will always be larger than the atom itself.
Now, just to show you a relative size change between the atoms and the ions as
we go from lithium to lithium plus, we know that, that's decreasing in size.
As we go from fluorine to fluoride, it is increasing in size.
So, by transferring one electron, one electron is coming from here,
and being placed on here, it makes a very big difference in the size of those ions.
So we've looked at atom tresnds.
We've looked at a comparison of an atom to its ion, now let's compare rad,
the radii of ions and atoms, in an isoelectronic series.
Remember, if they are isoelectronic,
that means they have the same electron configuration.
To have the same electron configuration, they must have the same number,
of electrons.
Now, once again, it's helpful to compare the relative number of protons and
electrons, when we do these isoelectronic series.
Since they're isoelectronic, the number of electrons is not going to be changing, but
the number of protons will.
So, we're going to look at a series of these four ions,
we've got sulfur, chloride, potassium ion, and calcium ion,
and we'll work our way through a table again, so let's construct another table.
Here we've got these four ions, and I did not put the nobel
gas that they are all isoelectronic with, argon is somewhere in the middle there.
But these are all isoelectronic, with argon.
They all will have an argon electron configuration.
Now, if you go to the periodic table,
you can determine the number of protons each of these will have.
Okay, so we have 16, 17, 19 and 20 protons in each of these.
Now, in terms of electrons, lets start with sulfur 2 minus, sulfur 2 minus.
We gain two electrons over the number it had to begin with,
which as an atom would be 16, so it's going to have 18.
And as we do that for each one of these, we'll say that they all have 18.
They're all isoelectronic with argon.
Argon has 18.
So, you're going to consider these two questions.
Number one, which ion would be the largest?
And which ion would be the smallest?
I want you to think about that, and
you're going to be answering that question here in just a moment, but
before I change to that question, what are you thinking about?
You're thinking about the pull that the protons are going to have for
the electrons, so it's drawing those electrons in.
Which one do you think will be more effective at pull, holding onto its 18
electrons, an atom that adds 16 electrons, or an atom that has 20 electrons?
So, the largest and
the smallest, is certainly going to be one end or the other.
Let's see what you can do with this.
Okay, so, if you answered the question, did you choose number four?
Then you'd be correct.
The largest one would be the one with the fewest number of protons,
holding on to 18 electrons.
And that was the sulfide.
The smallest one,
would be the one that had the most protons holding onto to those 18 electrons.
So number four is the correct answer.
And this is just showing relative sizes of these ions.
And I see students very often who will be asked to compare
two things on the periodic table,
maybe they're going to compare potassium plus and Cl minus on their size,
and these are isoelectronic, and so they'll say, okay, I know trends.
Trends are, they get smaller as we move to the top,
they get smaller as we go across the periodic table.
Therefore, since chlorine is further up, and
further to the right, it's going to be smaller.
Well, that certainly is not the case, is it?
It is larger.
So, the thing that you can't do is use the trends when you're comparing ions.
Having said that however, we can do a little bit of comparison.
Let me erase everything I have on here, and let's focus on a family.
If you're comparing the same charge for a group, okay?
So we're looking at the halogens, and we're looking at the halides, these ions.
It got bigger as we went down the family, or
down the group, when we were talking about atomic radius.
If they all have the same charge, then that is also going to be true,
as we go down with a its ions.
Okay? So, that trend does hold true when we're
looking at the periodic table, no matter which group we choose.
Okay, so this ends our learning objective number three.
We've talked about shielding, its role in effective nuclear charge.
We've looked at effective nuclear charge role in atomic radius.
And then we've compared ions to their atoms, and
we've compared isoelectronic ions to each other, in terms of their radius.
