This course will cover the topics of a full year, two semester General Chemistry course. We will use a free on-line textbook, Concept Development Studies in Chemistry, available via Rice’s Connexions project. The fundamental concepts in the course will be introduced via the Concept Development Approach developed at Rice University. In this approach, we will develop the concepts you need to know from experimental observations and scientific reasoning rather than simply telling you the concepts and then asking you to simply memorize or apply them. So why use this approach? One reason is that most of us are inductive learners, meaning that we like to make specific observations and then generalize from there. Many of the most significant concepts in Chemistry are counter-intuitive. When we see where those concepts come from, we can more readily accept them, explain them, and apply them. A second reason is that scientific reasoning in general and Chemistry reasoning in particular are inductive processes. This Concept Development approach illustrates those reasoning processes. A third reason is that this is simply more interesting! The structure and reactions of matter are fascinating puzzles to be solved by observation and reasoning. It is more fun intellectually when we can solve those puzzles together, rather than simply have the answers to the riddles revealed at the outset. Recommended Background: The class can be taken by someone with no prior experience in chemistry. However, some prior familiarity with the basics of chemistry is desirable as we will cover some elements only briefly. For example, a prior high school chemistry class would be helpful. Suggested Readings: Readings will be assigned from the on-line textbook “Concept Development Studies in Chemistry”, available via Rice’s Connexions project. In addition, we will suggest readings from any of the standard textbooks in General Chemistry. A particularly good free on-line resource is Dickerson, Gray, and Haight, "Chemical Principles, 3rd Edition". Links to these two texts will be available in the Introduction module.
CDS 4 Electron Shell Model of an Atom III
Loading...
Из курса от партнера Rice University
General Chemistry: Concept Development and Application
230 оценки
Из урока
Structure of an Atom and the Electron Shell Model
Proving the existence of atoms and knowing that they combine to form molecules does not provide a means to predict how or why these atoms might combine. This requires greater detail about the structure and properties of individual atoms. In this module, we extend our understanding of atoms by making observations which reveal the internal structure of the atom including a model for the arrangement of the electrons around the atomic nucleus.
Познакомьтесь с преподавателями
John Steven HutchinsonDean of Undergraduates and Professor of Chemistry
In this lecture, we're going to dig more deeply into our understanding of atomic
structure by looking more closely at this electron shell model that we have
developed for describing the arrangement of the electrons about the nucleus.
Let's reiterate a bit of what we've learned about this electron shell model.
We know that what we mean by an electron shell is a group of electrons
which are a certain distance away from the nucleus about equal to each other,
that there are multiple sub shells,
that each shell can only accommodate a certain number of electrons.
We can't put all the electrons into the lowest energy shell.
Once a shell is filled we have to add subsequent electrons to
shells which are larger.
And by larger, we mean at a radius farther away from the nucleus,
meaning they are less strongly attracted to the nucleus.
They're at a higher energy level.
At some point, we reach the highest energy shell for each atom, and we call that
shell the valence shell, what we mean by that is it's the most important shell.
The electrons in that shell are most important
to the chemistry of the each individual atom.
And we call those electrons the valence electrons.
One of the things we'll noticed that the atoms in the same group of the periodic
table has the same number of a valence electrons, for example,
say, fluorine and chlorine, which both have seven valence electrons.
And we imagine that that's the reason why they have very similar
kinds of chemical properties.
For example, somewhat similar ionization energies so that's for ionization.
All the remaining electrons that are not in the valence shell we're going to refer
to those as core electrons.
Now this model you recall, actually it grows out of examination of this data
having to do with the ionization energies of the elements.
And there were really two aspects to this data that we were focused on.
Aspect one, was this rather dramatic rise
in the ionization energies as we move across the periodic table.
We accounted for
this by saying the electrons are about the same distance away from the nucleus.
In other words, in the same shell.
But that there's an increased nuclear charge as we move from
lithium across the periodic table to oxygen, fluorine, and finally, neon.
And therefore, those electrons about the same distance from the nucleus are more
strongly attracted.
And therefore, are harder to ionize.
We also noted, this rather precipitous drop in the ionization
energies as we went from the farthest right element on the periodic table, neon,
to the next element, sodium, all the way on the left side of the periodic table.
And we accounted for this by saying that the next electron added when we increased
the nuclear charge on sodium, we increased the number of electrons,
that next electron must be in a new shell farther from the nucleus, and
therefore less strongly attracted.
But we should think a little bit more carefully about this.
Because, let's examine this particular data here,
this drawing that I've made to compare fluorine and neon.
If I take a little bit of a close examination here,
I noticed that neon has a higher ionization energy than fluorine does, and
our model accounts for that as follows.
Here is a drawing of the valence electrons for fluorine.
Here's the +9 nuclear charge on fluorine, the two electrons filling the first shell,
and then seven subsequent electrons in the valence shell of fluorine.
Here's neon with its +10 nuclear charge.
There are ten electrons all together, two in the inner core,
and eight in the valence shell.
And what we've been imagining here is that, for example,
this electrons attraction to the +9 nucleus is not as great as this electrons,
attraction to the +10 nucleus because of the greater nuclear charge.
But when we stare at these diagrams,
there should be something else that jumps out at us.
Electrons are attracted to positive charges, but
they're repelled by negative charges.
Let's count the numbers of negative charges.
For this neon electron here, it's attracted to a +10 nucleus.
But it gets repelled by, one, two, three, four, five,
six, seven, eight, nine, other electrons.
And, if I look at the fluorine electron, this one here,
it's attracted to a +9 nuclear charge that is weaker.
But how many electrons is it being repelled by?
One, two, three, four, five, six, seven, eight, not nine.
In other words, adding the extra nuclear charge, of course,
also means that we have added an extra electron for
repulsion amongst the electrons in the valence shell.
So the question is, then, as the atomic number increases,
why isn't it the case that the increase in the attraction to the nucleus
is all set by an increase in electron-electron repulsion.
We need an answer to these question that we're going to be able to understand
our ionization energy data.
And thereby, understand our shell model.
To do that, we need some more data.
Here's the data we're going to look at.
We're going to call these successive ionization energies.
In our previous data, we have examined only the first ionization.
We're moving a single electron, and
we'll call that energy the first ionization energy, IE1.
The energy required to remove the first electron.
But of course, we could remove a second electron.
That's this process here, taking electron out of the ion which is already formed.
The energy required to remove the electron from the ion already formed
will call the second ionization energy.
Core sector remove the third electron, fourth electron, fifth electron etc., and
measure all of those energies.
Let's analyze that data by taking a look at sodium, and
thinking about what the ionization would just going to look like for sodium.
And to do this, now, let's take our data, and
here are our successive ionization energies.
That's a lot of data in this table.
So we need to analyze it slowly.
First, let's make sure we know what we're looking at here.
These are the elements sodium through argon.
Those are all elements in the third row of the periodic table.
Let's pop up the periodic table here.
You can see them all in the same row, sodium going across to argon there.
And then in this column, to the left,
what we see is in fact the successive ionization energies starting with IE1,
and working our way down all the way we've removed maybe as many as seven electrons.
What do we see in this data?
Well, I mentioned we're going to focus on sodium first, so
let's analyze this data here.
What do we see?
Two things,
one is that each successive ionization energy is larger than the one before it.
But secondly, that one particular comparison here is particularly important.
We see as we go from ionizing the first electron to ionizing the second electron,
a gigantic increase a factor of almost ten, actually about nine.
Subsequently, if we ionize a third electron,
the comparison of the third ionization energy to the second ionization energy,
yes, it goes up, but it only goes up by about 50%.
It doesn't go up by a factor of nine.
So there's something unique in sodium about ionizing the first electron.
But our diagram for sodium, back over here, explains this.
We have sodium, which has a plus 11 nuclear charge, and
has 11 electrons as a neutral atom.
Two in the center core here, eight in the next shell in the core, and
then one valence electron out here.
If I remove that one valence electron, its ionization energy is going to be
determined by its attraction to the +11 nucleus.
And by its repulsion from the ten other electrons which are here.
If I remove a second electron, that electron is now closer to the nucleus and
we expect it to have a higher ionization energy, and that's exactly what we see.
So the fact that there's a significant jump as we go from the first
ionization to the second ionization tells me that there's only one valence electron
in the sodium atom.
We could actually use the same kind of analysis to find the number of valence
electrons in each of these elements, look for the large increases, there is a large
increase, there is a large increase, here is the large increase, here, and here.
And in each circumstance for example, what I can see is that the large increase for
magnesium occurs here between the second ionization energy and
the third ionization energy.
That means after I've removed two electrons from magnesium,
it's much harder to remove the third.
That tells me there are two valence electrons in magnesium.
Working over here to sulfur, the large increase
occurs between the sixth ionization energy, and the seventh ionization energy.
That means that sulfur has six valence electrons.
So that all looks pretty darn good, except if we analyze this data again,
and notice the changes here now.
And notice the changes,
that change is quite a bit larger than we might have expected.
Go back and look at our diagram here again.
As they go from this electron and removing it, to removing this electron.
Yes, it is closer to the nucleus, but we wouldn't expect an increase of a factor
of nine as we move one step closer into the nucleus.
That's two grade of change.
What's accounting for this has to be
the repulsion of this election by all of the other electrons as well.
That is, if we consider what the attractions are,
the attraction is the attraction of the electron 2 a +11 nucleus.
But if we consider what the repulsions are,
then the repulsions actually repulsion from the electron 2.
Let's see, one, two, three, four, five, six, seven,
eight, nine, ten core electrons.
Now what apparently must be happening is that the attraction of the electron
to the nucleus is being at least partially offset by repulsion from the core
electrons.
Consequently, we're going to describe something now that we're going to call
the core charge.
The core charge is a number which attempts to take into account simultaneously,
the attraction of the electron to the nucleus,
and the repulsion of the electrons from the inner shell electrons.
In the next slide here,
we describe the core charge as being the difference between the nuclear charge +11,
and the number of core electrons, 10 in the case of sodium.
So we'll describe the core charge now as being +1.
That means this electron is effectively feeling an attraction to
the nucleus of only net +1 representing its attraction to the plus 11,
and it's repulsion from the -10.
Well, now let's go back and look at our data again.
Our data tell us, that we get a dramatic increase,
here, when we try to remove the second electron.
Let's think about what might be causing that, by looking at a model for
what the sodium ion looks like, instead of the sodium atom.
Here's a model of the sodium ion.
Notice, I've removed the electron out here which was a valence electron,
leading behind the ten core electrons.
But now we recognize that the sodium ion,
the outer most shell is a shell closed rim.
Now let's imagine what we have here.
We have the attraction of the electron, let's say that one,
to the nucleus is due to the attraction of the electron to the +11 nuclear charge.
The repulsion, if we follow the same model that we used a little while ago,
is the electron to now only two core electrons.
Notice for this electron here, as it looks in towards the nuclear area,
there are only two core electrons.
Like go back and I looked at my definition of the core charge,
then the core charge for the electrons in the valence shell of the sodium ion,
are +11- 2+9.
An enormous increase in the attraction of the electrons
in this new shell of the sodium ion to the nucleus.
This accounts for a gigantic increase in the ionization energy.
Let's go back again and look at the data.
And in this data, what do we see?
We see this factor of nine here,
as we go from the first ionization to the second ionization.
Why then don't we see such a precipitous increase as we go to the third ionization?
Well, the answer is, that this electron, let's say I've already pulled this one
out, this electron is still in the same shell, it still feels the same +11 charge,
it still feels the same repulsion from the two core electrons.
The core charge for
pulling that second electron out is a +9, the same as it was before.
As a consequence, we don't expect to see a dramatic
increase in the ionization energy, because the core charge has not gone up.
So, this idea of a core charge here actually can account for
an enormous amount of what we see in the variations.
There is still one more bit of data to analyze, though.
Which is if we look at this data little bit more closely as well, we'll notice
that even once we have gone past ionizing the valence electrons, in each case,
there continued to be significant increases in the ionization energy.
It's just not as great as the big jump we get when we removed the last valence
electron.
What's causing that?
Well again, it makes sense to us
that if this electron were missing from the sodium ion, the number of total
electrons-electron repulsions felt by this electron is now fewer.
Let's think about how many electrons this one is repelled by.
One, two, three, four, five, six, seven, eight, nine.
How many repulsions are there from this one, once this electron has been removed.
The answer is eight, one, two, three, four, five, six, seven, eight.
Consequently, electron-electron repulsion within the valence shell also matters and
accounts for the increases that we see, for example, once we remove
all of the valence electrons in sodium, or for example, in silicon here.
Those increases are arising from valence electron-valence electron repulsion.
That also make sense.
Once we've remove an electron from an atom what remains behind is a positively
charged ion.
It should be harder to remove a negatively charged electron from a positively charged
ion than from a neutral atom.
And the greater the positive charge, in other words the more electrons we've
already removed, the harder it will be to remove any further ones
because of greater attraction of the electron to the remaining positive charge.
So the effective electron-electron repulsion in our shell model here
is two-fold.
One, is repulsion of the valence electrons by the core electrons which we
take into account with this core charge model such as we've applied to sodium.
But, there's also, we should not forget,
electron-electron repulsion amongst the valence electrons, and
that has a lot to do with the properties of the atoms as well.
In the end, then, we've got a model, now, which really well accounts for
the variation in the ionization energies of the elements,
by understanding this idea of the valence shell, and
the core charge felt by that electrons when they are in the valence shell.
The ionization energies could be understood in terms of this.
And it turns out as we will see over the course of the rest of this semester,
much of chemistry can be understood by understanding the ionization interests
of the atoms because they're related
to how strongly attracted the electrons are to each atom.
And we'll pick that up a bit further in the next lecture.
Ознакомьтесь с нашим каталогом
Присоединяйтесь бесплатно и получайте персонализированные рекомендации, обновления и предложения.
Coursera делает лучшее в мире образование доступным каждому, предлагая онлайн-курсы от ведущих университетов и организаций.
© Coursera Inc., 2018 Все права защищены.
