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PROFESSOR: All right.
As everyone finishes getting settled in, why don't you take
10 more seconds on the clicker question here, and let's see
how you did on that this, this is very similar to the clicker
question that we had on Friday.
OK, so let's get started here.
It looks like we are doing a lot better.
We now have 77% getting the correct answer, we only had
about 30-something percent on Friday for a
very similar question.
So if you're not in this 77%, let's quickly go over why, in
fact, this is the correct answer, 0 .
9 times 10 to the negative 18 joules.
So I'm using the same kind of tricky language that we'd used
before, not to trick you, but so that you're not tricked in
the future.
So if we're talking about the fourth excited state, and we
talk instead about principle quantum numbers, what
principle quantum number corresponds to the fourth
excited state of a hydrogen atom.
STUDENT: Five.
PROFESSOR: Five.
OK.
So, hopefully that cleared up for some of you why you got
the wrong answer.
So we know that we're in the n equals 5 state, so we can find
what the binding energy is here.
The ionization energy, of course, is just the negative
of the binding energy.
We know that binding energy is always negative, we know that
ionization energy is always positive.
So hopefully, putting all those things together, if you
looked at this question again we'd get 100% on it, that our
only option here is 0 .
9, and that it's not the negative, it's the positive
version, because we're talking about how much energy we have
to put into the system in order to eject an electron.
All right.
And today we're going to mostly be talking about wave
functions of electrons, but before we get to that, I
wanted to review one last thing that's back on to
Friday's topic, which was when we were solving the
Schrodinger equation, or in fact, using the solution to
the Schrodinger equation for the energy, the binding energy
between an electron and a nucleus.
And when we talked about that, what we found was that we
could actually validate our predicted binding energies by
looking at the emission spectra of the hydrogen atom,
which is what we did as the demo, or we could think about
the absorption spectra as well.
And what we predict as an energy difference between two
levels, we know should correspond to the energy of
light that's either emitted, if we're giving off a photon,
or that's absorbed if we're going to take on a photon and
jump from a lower to a higher energy level.
So we came up with two formulas, which are similar to
the two that I'm showing here.
The formula tells us the frequency of the light that's
emitted or absorbed based on the energy difference between
the two levels that we're going between, that the
electron is transitioning between.
You'll notice that there's a little bit of a difference in
these equations here from the ones from the other day, which
is that you have this z squared value in there.
So these are both called Rydberg formulas for figuring
out the frequency of light emitted or absorbed, and
before we were looking at the Rydberg formula specifically
for the hydrogen atom, and now that we have this z squared
term in the formula here, we're now talking about
absolutely any one electron atom.
And it should make sense where we got this from, because we
know that the binding energy, if we're talking about a
hydrogen atom, what is the binding energy equal to?
Negative Rydberg over what?
Yes.
So, it's negative Rydberg constant over n squared.
But if we're talking more generally about any one
electron atom, now we have a more general equation for the
binding energy, which has this z squared term out in front of
it, right, so it's negative z squared times the Rydberg
constant all over n squared.
So, essentially when we're talking about these equations
up here, all we're doing is talking about the regular
Rydberg formulas, but instead we could go back and re-derive
the equation for any one electron atom, which would
just mean that we put that z squared term in the front.
So when you solve certain types of problems, such as
problems later on in the second half of your p-set, if
you need to talk about the frequency of light emitted or
absorbed for a one electron atom, such as lithium plus 2,
for example, then you would need to plug in z, and
remember the z value for lithium would just be 3.
The z value for hydrogen, of course, is 1, and that's why
this term falls out of that equation when we're talking
specifically about the hydrogen atom.
So, just to finish our review of what we talked about on
Friday, when we're thinking about transitions between two
different states, and we're talking about a situation
where the final state, the n final, is greater than n
initial, in this case, are we talking about absorption or
are we talking about emission?
Hearing a little bit of a mix here.
In fact, we're talking about absorption when n final is
greater than n initial.
We start at this lower energy state and go up -- that means
we need to absorb a photon, we have to take in energy.
Specifically, we have to take in this exact amount of energy
in order to bump the electron up to the higher energy level.
So that means that when instead we start high and go
low, we're dealing with emission where we have excess
energy that the electron's giving off, and that energy is
going to be equal the energy of the photon that is released
and, of course, through our equations we know how to get
from energy to frequency or to wavelength of the photon that
we're talking about.
All right.
So that's all I'm going to say today in terms of solving the
energy part of the Schrodinger equation, so what we're really
going to focus on is the other part of the Schrodinger
equation today, which is solving for psi.
So we're going to for psi, and before that, we're going to
figure out that instead of just that one quantum number
n, we're going to have a few other quantum numbers that
fall out of solving the Schrodinger equation
for what psi is.
We're also going to talk more about what psi actually means.
When we first introduced the Schrodinger equation, what I
told you was think of psi as being some representation of
what an electron is.
We'll get more specific here, more specific even than just
saying you can think of it as an orbital.
We'll really think about what psi means.
And in doing that, we'll also talk about the shapes of h
atom wave functions, specifically the shapes of
orbitals, and then something called radial probability
distribution, which will make sense when we get to it.
But, as I said before that, we have some more quantum numbers
to take care of, because it turns out that when you solve
the Schrodinger equation for psi, these other quantum
numbers have to be the defined.
When we talked about binding energy, we just had one
quantum number that came out of it.
And that quantum number was n, which was our principle
quantum number, and we know that n could be equal to any
integer value, so, 1, 2, 3, all the way up to infinity.
And this quantization that comes out of having n is what
gives us the quantization of different energy levels.
That's why we can't have a continuum of energy, we
actually have those quantized points.
So, it turns out that n is not the only quantum number needed
to describe a wave function, however.
There's two more that you can see come out of it.
And the first is l, and l is our angular momentum quantum
number, and it's called that because it actually dictates
the angular momentum that our electron has in our atom.
And when we talk about l it is a quantum number, so because
it's a quantum number, we know that it can only have discreet
values, it can't just be any value we want, it's very
specific values.
And unlike n, l can start all the way down at 0, and it
increases by integer value, so we go 1, 2, 3,
and all the way up.
But also unlike n, l cannot have just any value, we can't
go into infinity.
L is limited such that the highest value of
l is n minus 1.
We can't get any higher than that.
So, it would be a good question to ask why are we
limited -- clearly there's this relationship between l
and n, and we can't get any higher than n equals one.
We can actually think about why that is, and the reason is
because l is our angular momentum.
It describes the angular momentum of the electron.
So another way to think about that is just the rotational
kinetic energy of our electron.
And we know that n describes the total energy, that total
binding energy of the electron, so the total energy
is going to be equal to potential energy
plus kinetic energy.
So if we say that l is just talking about our kinetic
energy part, our rotational kinetic energy, and we know
that electrons have potential energy, then it makes sense
that l, in fact, can never go higher than n.
And, in fact, it can't even reach n, because then we would
have no potential energy at all in our electron, which is
not correct.
So, that's the second quantum number.
And the third one is called m, it's also m sub l.
This is what we call the magnetic quantum number, and
we won't deal with the fact of its being the magnetic quantum
number here -- that kind of tells us the shape of the
orbital or the way that the electron will behave in a
magnetic field, but what's more relevant to thinking
about the limits of this number is that it's also the z
component of the angular momentum.
So since it's a component of the angular momentum, that
means that it's never going to be able to go higher than l
is, so it makes sense that, for example, it could start at
0 and then go all the way up to l.
But since it is a component it can have a direction, too, so
can go up between negative l and positive l.
So the allowed values for m sub l are going to be negative
l, all the way up to 0, and then up to positive l.
So, if we think of just an example, we could say that 4 l
equals 2, what would be our lowest value of m sub l?
Yup.
So m sub l could equal negative 2,
negative 1, 0, 1 or 2.
So we could have five different values of m sub l.
So, those are our three quantum numbers.
So if, in fact, we want to describe a wave function, we
know that we need to describe it in terms of all three
quantum numbers, and also as a function of our three
positional factors, which are r, the radius, plus the two
angles, theta and phi.
So, we have now a complete description of a wave function
that we can talk about.
So, we can think about what is it that we would call the
ground state wave function.
We knew from Friday, when we talked about energy, that
ground state was that n equals 1 value, that was the lowest
energy, that was the most stable place for
the electron to be.
But now we need to talk about l and m as well.
So now when we talk about a ground state in terms of wave
function, we need to talk about the wave function of 1,
0, 0, and again, as a function of r, theta and phi.
So this is our complete description of the ground
state wave function.
So, a lot of you talked about different types of orbitals in
high school, I'm sure, or in previous courses, and it might
be less common that you actually talked about a wave
function that was labeled like this.
We're used to labelling orbitals as an s, or a p, or a
d, for example, but it turns out that these correlate to
those letters that we're more used to seeing.
Does anyone know what the 1, 0, 0 orbital is also called?
Yeah.
And specfically it's the 1 s, so not just the
s, but the 1 s orbital.
So, using the terminology of chemists, which is a good
thing to do, because in this course we are all chemists, we
want to make sure that we're not using just the physical
description of the numbers, but that we can correlate it
to what we understand as orbitals, and instead of 1, 0,
0, we call this the 1 s orbital.
The reason that we do this is because this is another way to
completely describe it.
The n designates the shell, so that's what this number is
here, we're in the first shell.
The l is what we call the sub shell.
And instead of having a 0 there, what we
have here is an s.
So, if we look at what the other sub shells are called,
essentially we're just converting
the number to a letter.
L equals 0 is s, what is l equals 1?
Um-hmm, it's the p.
What about 2? d, and 3?
Yup, so 3 is f.
So these names, they don't really make any sense if we're
looking at them why they're called past s p and f, and it
turns out that it comes from spectroscopy terms that are
pre-quantum mechanics where, for example, this is called
the sharp line, I think the principle, the diffuse, and
the fundamental.
It doesn't even make sense now, they're not used in
spectroscopy anymore, but this is where the names originally
came from and they did stick.
So, we being chemists, we'll call that 1 s instead of 1, 0.
In addition to having another name to denote l, we also have
another name for the m designation here.
So, for example, when l is equal to 0, we're going to
find that we have to call -- we have to specify
what m is as well.
All right.
So, when we have, for example, l equal to 1, what kind of
orbital is this?
The p orbital.
And for example, we could also in this case, have
m is equal to 0.
If m is equal to 0, in this case we would call it the p z
orbital, so we would have the subscript z here.
Similarly, if m is equal to either plus 1 or minus 1, we
would in turn call it the p y orbital, or the p x orbital.
So you should know that any time m is equal to zero when
we are talking about p orbitals, that it's the p z.
The p y and the p x are actually a bit more
complicated, they're linear combinations of the m plus 1,
and the m minus 1 orbital, where 1 is the positive linear
combination, and 1 is the negative linear combination.
You're not responsible for that, you're not responsible
for correlating plus 1 to y, minus 1 to x.
Just know that you have plus or minus 1, for our class, you
can call it either x or y, either is fine, because it's a
little bit more complicated than just the 1:1 translation
between, for example, m equals 0 and having a p z orbital.
All right.
So let's look at some of these wave functions and make sure
that we know how to name all of them in terms of orbitals
and not just in terms of their numbers.
Once we can do that we can go on and say okay, what actually
is a wave function, but first we need to know how to
describe which ones were talking about.
So we saw that our lowest, our ground state wave
function is 1, 0, 0.
We can call that psi 1, 0, 0 is how we write
it as a wave function.
We said that's the 1 s orbital.
We also know how to figure out the energy of this orbital,
and we know how to figure out the energy using this formula
here, which was the binding energy, which is negative r h,
and instead of n, we can plug it in because n equals 1, so
over 1 squared, and the actual energy is here.
So, our next level up that we can go is going to be the n
equals 2 energy level, but we also have an l and an m value,
so our lowest l is going to be a 0 there.
So we'll call that psi 2, 0, 0 wave function.
What will we call that in terms of orbitals?
Yup, so that's the 2 s orbital.
So something I actually wanted to point out that I forgot to
here is you'll notice that there's no subscript to the s.
We said we have a subscript to the p, for example, that
describes what m is equal to.
The reason that we have no subscript to the s, is because
the only possibility for m when you have an s orbital is
that m has to be equal to 0.
So we just assume it, you don't actually have to write
it because there is, in fact, only one possibility.
We can also figure out the energy of this orbital here,
and the energy is equal to the Rydberg constant.
The negative of the Rydberg constant now
divided by 2 squared.
So we can go on and do this for any orbital or any state
function that we would like to.
So, for example, if we talk about the 2, 1, 1 state label,
that's just psi 2, 1, 1.
What, in this case, would be our orbital?
2 p what?
OK, good, I heard mixed answers, which is correct.
So you can either write 2 p x or 2 p y, whichever one you
want is fine.
And again, you'll notice that our energy is absolutely the
same for an electron in that 2 p x orbital
and in the 2 s orbital.
So that's true for a hydrogen atom, it doesn't matter if
you're in a p or an s orbital, their energies are the same.
Then we can also talk about the 2, 1, 0 state function,
which would be psi 2, 1, 0.
What is this orbital?
Yup.
And there's only one correct answer here,
which is to 2 p z.
Is the energy going to be the same or different as up here?
It's going to be the same energy.
Again, the reason for that is because the energy only
depends on the n value here, it doesn't
depend on l or on m.
So finally, if we talk about our last example of when n is
going to be equal 2, we can have 2, 1 for l and
then minus 1 for m.
We can re-write this as psi 2 1 negative 1.
And then our orbital is going to be just the opposite of
whatever we said it was up here.
So if you said 2 p x the first time, say 2 p y this time.
And again, our energy is going to be the same where we again
only depend on the n value.
All right.
So hopefully we're pretty comfortable naming any type of
wave function using the chemist terminology.
Let's switch to a clicker question and just confirm that
that is, in fact, true.
So what's the corresponding orbital if we talk about this
state, 5, 1, 0?
And you can go ahead and give 10 seconds on that.
OK.
All right, 77%.
So, that's OK, you don't have to memorize things as I speak,
you just need to go back and look at this and make sure you
understand how to name it and that you'll be able to, for
example, by next class, get a similar clicker question
correct, and good job to the 77% that did get it.
So I think we're safe to move on here.
And I just want to point out that now we have these three
quantum numbers.
The reason there are three quantum numbers is we're
describing an orbital in three dimensions, so it makes sense
that we would need to describe in terms of three different
quantum numbers.
And the complete description, as I said, is from n l and m.
And when you talk about n for an orbital, it's talking about
the shell -- that shell is kind of what you picture when
you think of a classical picture of an atom where you
have 1 energy level, the next one is further out, the next
one's further away.
That's kind of your shell that we're discussing.
L is the sub shell here, and then we have m, which is
finally the complete description of the orbital.
And what you can see is that for any n that has an l equals
0, you can see here how there's only one possibility
for and orbital description, and that's why we don't need
to include the m when we're talking about and s orbital.
The other thing that we know, which is what we were just
discussing when we were going through the table is how this
all relates to energy.
And I want to really highlight here we're talking about for a
hydrogen atom -- orbitals with the same n value
have the same energy.
Some of you might be saying in your heads, wait a second, I
happen to know, I happen to remember from high school,
that p orbitals have different energies then,
for example, s orbitals.
And that is not true for one electron atoms. We're going to
get to more complicated atoms eventually where we're going
to have more than one electron in it, but when we're talking
about a single electron atom, we know that the binding
energy is equal to the negative of the Rydberg
constant over n squared, so it's only depends on n.
So, for example, if we're talking about the n equals 2
state, all of these four orbital descriptions are going
to have the same energy.
And we can generalize to figure out, based on any
principle quantum number n, how many orbitals we have of
the same energy, and what we can say is that for any shell
n, there are n squared degenerate orbitals.
And the word degenerate simply means same energy, so you have
n squared orbitals that are of equal energy when they're
degenerate.
So, let's look at where this comes from with an energy
level diagram here.
So what you can see is again, we've got this ground state.
So if we go to the ground state, what you see is we're
at that lowest energy level, and we only have one
possibility for an orbital, because when n equals 1,
that's all we can do.
So that's the 1 s orbital -- we have n squared or 1
degenerate orbitals.
When we talk about the n equals 2 state, we now have 2
squared or 4 degenerate same energy orbitals, and those are
the 2 s orbital.
And then we also have the l being equal to 1 orbital, so
those are going to be the 2 p x, the 2 p z,
and the 2 p y orbital.
All four of these orbitals have the same energy, they're
degenerate.
And as we go up the next energy level, which is based
on n equals 3 principle quantum number, well now we
have again the s, so we have the 3 s orbital, we're going
to have three 3 p orbitals, right, so we'll have 3 p x, 3
p z, and 3 p y, and now we're actually also going to have
five different possible l equals 2 orbitals.
Does anyone remember the l equals 2?
Yes, everyone remembers.
Good.
So we have five possible d orbitals.
We'll call these here the 3 d x y, as the subscript, the 3 d
y z, the 3 d z squared, the 3 d x z, and the 3 d x squared
minus y squared.
So, what do you need to know here?
What you need to know is that when m equals
0, it's 3 d z squared.
That's it.
Again, these other p -- or the d x y, d y z, those are going
to be those more complicated linear combinations, you don't
need to worry about them.
Eventually you will, at least, need to know the labels and
know a little bit more about them.
And in the second half of this course, Professor Drennen's
going to talk to us about transition metals in depth,
and that's when we'll really delve into d orbitals.
For right now, you can kind of put the d orbitals in the back
of your head.
You need to know how to think about them in the same way we
think about s and p orbitals, but for example, you don't yet
need to know what all of the names are except for this 3 d
z squared here.
So we'll wait on that until we start talking more
specifically about atoms where the d orbital becomes very
significant.
So, what we can see is this degeneracy.
So what we know now is we can start thinking about the next
step because we can fully describe the energy of
orbitals, and we can fully describe a complete orbital in
terms of its three quantum numbers, and its three
positional variables, r, theta, and phi.
So next we can think about okay, what is actually a wave
function, and for example, what might the shape of
different wave functions be.
So essentially, what we're asking for here is the
physical interpretation of psi, of the value of psi for
an electron.
And it turns out that the answer to can we have this
physical interpretation of thinking about what psi means,
the answer is really no, that we can't.
There's no classical way to think about what a wave
function is.
There's no classical analogy that explains oh, this is what
you can kind of picture when you picture a wave function.
And that's somewhat inconvenient because we're
working with wave functions, but it's a reality that comes
out of quantum mechanics often, which is that we're
describing a world that is so much different from the world
that we observe on a day-to-day basis, that we're
not always going to be able to make
those one-to-one analogies.
But luckily we don't have to worry about how we're going to
picture all this, now that I said that, because even though
there's no physical interpretation for what a wave
function is, there is a physical interpretation for
what a wave function squared means.
So when we talk about a wave function squared, we're taking
the square of the wave function, any one that we
specify between n, l and m, at any position that we specify
based on r, theta, and phi.
And if we go ahead and square that, then what we get is a
probability density, and specifically it's the
probability of finding an electron in a certain small
defined volume away from the nucleus.
So it's a probability density.
The important point here is it's not just a probability,
it's a density, so we know that it's a probability
divided by volume.
And the person we have to thank for actually giving us
this more concrete way to think about what a wave
function squared is is Max Born here.
And actually after the Schrodinger equation first was
put forth, people had a lot of discussions about how is it
that we can actually interpret what this wave function means,
and a lot of ideas were put forth, and none of them worked
out to match up with observations until Max Born
here came up with the idea that we just square the wave
function, and that's the probability density of finding
an electron in a certain defined volume.
And it's very helpful because it gives us a way
to think about it.
We can't actually go ahead and derive this equation of the
wave function squared, because no one ever derived it, it's
just an interpretation, but it's an interpretation that
works essentially perfectly.
Ever since this was first proposed, there has never been
any observations that do not coincide with the idea, that
did not match the fact that the probability density is
equal to the wave function squared.
So, also about Max Born, just to give you a little bit of a
trivial pursuit type knowledge, he not only gave us
this relationship between wave function squared, he also gave
us Olivia Newton-John.
This is her grandfather, I don't know if you can see from
the eyes, I feel like there's a little bit of
a resemblance there.
So, I don't know what she grew up hearing about when she went
to her grandparents' house, but it might have been wave
function squared.
So, a little tidbit of knowledge for you that's
somewhat trivial.
Then back to the non-trivial knowledge that is not trivial
at all, in fact, is OK, how do we think about this
probability density now that we have a little
bit more of an idea.
We know that it's a density, it's not an actual
probability.
So, one way we could look at it is by looking at this
density dot diagram, where the density of the dots correlates
to the probability density.
So, what you see is near the nucleus, the density is the
strongest, the dots are closest together.
As you get far away from the nucleus, the dots get farther
and farther apart, meaning the probability density at those
volumes far away from the nucleus is going to be quite
low, eventually going to almost zero, although it turns
out that it never goes to exactly zero, so if we're
talking about any orbital or any atom, it never actually
ends, it never goes to zerio.
But it turns out the probability is only
significant within one angstrom.
So you can either say that electrons are very, very tiny
or that they're never ending, and both are pretty accurate
ways to think about what an atom is.
So, that's probability density, but in terms of
thinking about it in terms of actual solutions to the wave
function, let's take a little bit of a step back here.
I have yet to show you the solution to a wave function
for the hydrogen atom, so let me do that here, and then
we'll build back up to probability densities, and it
turns out that if we're talking about any wave
function, we can actually break it up into two
components, which are called the radial wave function and
angular wave function.
So, essentially we're just breaking it up into two parts
that can be separated, and the part that is only dealing with
the radius, so it's only a function of the radius of the
electron from the nucleus.
And we abbreviate that by calling it r, which is
specified by two quantum numbers, and an l as a
function of little r, radius.
And we have the angular wave function, which is specified
by l and m, and it's a function of the two angles
when we're describing the position of the electron, so
theta and phi.
So, let's look at what this actually is for what we're
showing here is the 1 s hydrogen atom.
If you look in your book there's a whole table of
different solutions to the Schrodinger equation for
several different wave functions.
So this is the 1 s, you can look it up if you're
interested for the 2 s, or 3 s, or 5 s, or whatever you're
curious about.
But what I'm going to show you here is the 1 s solution.
So you can see there's this radial part here, and you have
the angular part, you can combine the two parts to get
the total wave function.
And what you can see is we have this new constant that we
haven't seen before.
So what do you see in there that is new?
Yeah.
This a sub nought.
That's a new constant for us in this course.
This is what's called the Bohr radius, and we'll explain --
hopefully we'll get to it today where this Bohr radius
name comes from, but for now what you need to know is just
that it's a constant, just treat it like a constant, and
it turns out to be equal to 52 .
9 pekameters or about 1/2 an angstrom.
The more important thing that I want you to notice when
you're looking at this wave equation for a 1 s h atom, is
the fact that if you look at the angular component of the
wave function, you'll notice that it's a constant.
It doesn't depend on theta, it doesn't depend on phi.
No matter where you specify your electron is in terms of
those two angles, it doesn't matter the angular part of
your wave function is going to be the same.
So, what does that mean for us?
Well, essentially what that tells is that these s orbitals
are spherically symmetrical.
That should make sense, right, because they're only
dependent on r.
How far you are away from the nucleus in terms of a radius,
they don't depend at all on those two angles, they're
independent of theta and they're independent of phi.
So, what I'm showing in this picture here is just an
electron cloud that you can see.
Think of it as a probability density plot.
And what here is just a graph of the 1 s wave function going
across some radius defined this way, and you can see that
the probability -- well, this is the wave function, so we
would have to square it and think about the probability.
So this squared at the origin is going to be a very high
probability, and it decays off as you get farther and farther
away from the nucleus or from the center, and that's
independent of the angle.
So, let's look at these probability plots of different
s orbitals here, and up top here, we have the probability
density plot and what you can see is what I just said, a
very high probability density in the nucleus,
decays as you go out.
And what is plotted below is the actual wave function, so
you can see it starts very high and then the decays down.
More interesting is to look at the 2 s wave function.
So, if we look at the bottom here and the actual plot of
the wave function, we see it starts high, very positive,
and it goes down and it eventually hits zero, and goes
through zero and then becomes negative and then never quite
hits zero again, although it approaches zero.
So, at this place where it hits zero, that means that the
square of the wave function is also going to be zero, right.
So we can see if we look at the probability density plot,
we can see there's a place where the probability density
of finding an electron anywhere there is actually
going to be zero.
So we can think of a third case where we have the 3 s
orbital, and in the 3 s orbital we see something
similar, we start high, we go through zero, where there will
now be zero probability density, as we can see in the
in the density plot graph.
Then we go negative and we go through zero again, which
correlates to the second area of zero, that shows up also in
our probability density plot, and then we're positive again
and approach zero as we go to infinity for r.
So, what this means is that when we're looking at an
actual wave function, we're treating it as a wave, right,
so waves can have both magnitude, but they can also
have a direction, so they can be
either positive or negative.
So, for example, if we were looking at the actual wave
function, we would say that these parts here have a
positive amplitude, and in here we
have a negative amplitude.
And when we're looking at the probability density graphs, it
doesn't make a difference, it's okay, It has no meaning
for our actual plot there, because we're squaring it, so
it doesn't matter whether it's negative or positive, all that
matters is the magnitude.
But when we're thinking about actual wave behavior of
electrons, it's just important to keep in the back of our
head that some areas have positive amplitude and some
have negative.
So we'll talk about this more we get into p orbitals and
bonding is where it's going to become an issue.
So I just want to kind of introduce that idea here.
Because if we think about wave behavior of electrons and
we're forming bonds, then what we have to do is have
constructive interference of 2 different electrons, right, to
form a bond, we want to and together those probabilities.
So we want to have constructive interference to
form a bond, whereas if we had destructive interference, we
would not be forming a bond.
So that's where you have to think about whether it's
positive or negative.
You don't have to think about it right now, but you might
have heard in high school talking about p orbitals, the
phase, sometimes you mark a p orbital as being a plus sign
or negative sign.
Did any of you do that in high school at all?
A little bit, yeah.
So, that's having to do with the actual wave function.
So, that'll become more relevant later, bonding
actually, a couple lectures down the road.
But I just want to introduce it here while we do, in fact,
have the wave function plots up here.
But a real key in looking at these plots is where we, in
fact, did go through zer and have this
zero probability density.
We call that a node, and a node, more specifically, is
any value of either r, the radius, or the two angles for
which the wave function, and that also means the wave
function squared or the probability density, is going
to be equal to zero.
So, we can see in our 1 s orbital, how
many nodes do we have?
There's no nodes, yeah.
It looks like we hit zero, but we actually don't -- remember
that we never go all the way to zero, so there's these
little points if we were to look really carefully at an
accurate probability density plot, it would never
actually hit zero.
And then, for example, how many nodes do we have in the 3
s orbital? two.
That's correct.
So we have two nodes in the 3 s orbital.
We can actually specify where those nodes are, which is
written on your notes.
For the 2 s orbital, at 2 a nought, so it's just 2 times
that constant a nought, which is the Bohr radius.
And for the 3 s, we have one at 1 .
9 a nought, and one at 7 .
1 a nought.
We can also specify what kind of node we're talking about.
We'll introduce in the next course angular nodes, but
today we're just going to be talking about radial nodes,
and a radial node is a value for r at which psi, and
therefore, also the probability psi squared is
going to be equal to zero.
So, when we're talking about an s orbital, since there is
no angular dependence, and it only depends on r, every
single one of our nodes is actually going to specifically
be a radial node, right, because these are, for
example, this 2 a nought is a value of r, a value of the
radius, no matter which way you go around at which there's
going to be a node at which there is zero probability
density of finding an electron there.
So, it's very easy to calculate, however, the number
of radial nodes, and this works not just for s orbitals,
but also for p orbitals, or d orbitals, or whatever kind of
work of orbitals you want to discuss.
And that's just to take the principle quantum number and
subtract it by 1, and then also subtract from that your l
quantum number.
So what you can do for a 1 s is just take 1 minus 1 and
then l is equal to 0, so you have zero radial nodes.
And that matches up with what we saw.
If we try this for the 2 s, we have 2 minus 1 minus 0.
So what we should expect to see is one radial node, and
that is what we see here in the probability density plot.
And then if we think about 3 s, we want to start with 3, we
subtract 1, again l is equal to 0, so minus 0 and we have
two radial nodes.
So, this should be pretty straight forward, let's see if
we can get close to a 100% on this one, which is how many
radial nodes does a 4 p orbital have?
And let's give 10 seconds on that, make
you think fast here.
OK, so most people were correct, or well, the
majority, at least, were correct.
And seeing that it's a 4 p has two nodes -- let's just write
this out since not everyone did get it correct.
So, if we're talking about a 4 p orbital, and our equation is
n minus 1 minus l, the principle quantum number is 4,
1 is 1 -- what is l for a p orbital?
STUDENT: 1.
PROFESSOR: 1.
So, I tricked you a little, I guess I didn't put an s up
there and that's what we had been talking about, so that
was probably the issue.
But what we find is that we have two radial nodes.
All right.
So we can switch back to our notes here.
So, doing those probability density dot graphs, we can get
an idea of the shape of those orbitals, we know that they're
spherically symmetrical.
We're not going to talk about p orbitals today, we're going
to talk about p orbitals exclusively on Friday, and as
I said, d orbitals you'll get to with Professor Drennen.
But we can also think when we're talking about wave
function squared, what we're really talking about is the
probability density, right, the
probability in some volume.
But there's also a way to get rid of the volume part and
actually talk about the probability of finding an
electron at some certain area within the atom, and this is
what we do using radial probability
distribution graphs.
And what that is the probability of finding an
electron in some shell where we define the thickness as d
r, some distance, r, from the nucleus.
So, think about what we're saying here.
We're saying the probability of finding an electron at some
distance from the nucleus in some very thin shell that we
describe by d r.
So if you think of a shell, you can actually just think of
an egg shell, that's probably the easiest way to think of
it, where the yolk, if you really maybe make it a lot
smaller might be the nucleus.
And let's also make our egg perfectly symmetric and
perfectly round.
But still, when we're talking about the radial probability
distribution, what we actually want to think about is what's
the probability of finding the electron in that shell?
Think of it as that egg shell part.
So, we can do that by using this equation, which is for s
orbitals where the radial probability distribution is
going to be equal to 4 pi r squared times the wave
function squared, d r.
That should make sense to us, because when we talk about a
wave function, we're talking about a probability divided by
a volume, because we're talking about
a probability density.
So if we actually go ahead and multiply it by the volume of
our shell, then we end up just with probability, which is
kind of a nicer term to be thinking about here.
So, of course, if we're talking about a perfectly
spherical shell at some distance, thickness, d r, we
talk about it as 4 pi r squared d r, so we just
multiply that by the probability density.
We can graph out what this is where we're graphing the
radial probability density as a function of the radius.
And what you see is that at zero, you start at zero.
And so, the radial probability density at the nucleus is
going to be zero, even though we know the probability
density at the nucleus is very high, that's actually where is
the highest. The reason in our radial probability
distributions we start -- the reason, if you look at the
zero point on the radius that we start at zero is because
we're multiplying the probability density by some
volume, and when we're not anywhere from the nucleus,
that volume is defined as zero.
So, it's a little bit artificial that we're seeing
that zero point there.
So, actually I want you to go ahead in your notes and circle
that zero point and write "not a node." This is not a node
because a node is where we actually have
no probability density.
So this, where we start at zero is not a node, is the
first thing to point out.
And as we get further and further from the radius, the
volume we're multiplying it by actually gets bigger and
bigger, because you can see how the volume of that little
thin shell is going to get larger and larger as you get
further away.
So there's some distance where the probability of actually
finding an electron there is going to be your maximum
probability.
And that's what we label as r sub m p, or your
most probable radius.
This is the point at which your probability is highest
for finding an electron.
This is equal to a sub nought for a hydrogen atom, and we
remember that that's just our Bohr radius, which is 0 .
5 2 9 angstroms. And basically, what that means is
you can actually find an electron anywhere going away
from the nucleus, but you're most likely to find that you
have the highest probability at a distance of a sub nought,
or the Bohr radius.
So, I said I'd tell you a little bit more about where
this Bohr radius came from, and it came from a model of
the atom that pre-dated quantum mechanics, and Neils
Bohr is who came up with the idea of the Bohr radius, and
here is hanging out with Einstein, so he had some
pretty good company that he kept.
And what you need to remember when we're thinking about this
model of the atom is that in 1911 it had already been
discovered that we have an electron, and we have a
nucleus, and there needs to be some way that those two hang
together, but it was not for another 15 years that we
actually had the Schrodinger equation that allowed us to
understand the interaction fully between the electron and
the nucleus.
So all that Bohr, for example, had to go on at this point was
a more classical picture of the atom, as you can see on
the left side of the screen there, which is the idea that
the electrons actually somehow just orbiting the nucleus.
And even though he could figure out that this wasn't
possible, he still used this as a starting point, and what
he did know was that these energy levels that were within
hydrogen atom were quantized. and he knew this the same way
that we saw it in the last class, which is when we viewed
the difference spectra coming out from the hydrogen, and we
also did it for neon, but we saw in the hydrogen atom that
it was very discreet energy levels that we could observe.
He knew the same thing that had been
observed by that point.
So, what he did was kind of impose a quantum mechanical
model, not a full one, just the idea that those energy
levels were quantized on to the classical picture of an
atom that has a discreet orbit.
And what he came out with when he did some calculations is
that there's the radius that he could calculate was equal
to this number a sub nought, which is what we call the Bohr
radius, and it turns out that the Bohr radius happens to be
the radius most probable for a hydrogen atom.
And the reason we won't talk any more about this Bohr model
is because, of course, it's not correct.
So we're not going to spend too much time on it here.
But we can see, for example, one reason or one way in which
is not correct.
Because what it tells is that we can figure out exactly what
the radius of an electron and a nucleus are
in a hydrogen atom.
That's a deterministic way of doing things, that's what you
get from classical mechanics.
But the reality that we know from our quantum mechanical
model, is that we can't know exactly what the radius is,
all we can say is what the probability is of the radius
being at certain different points. so, that's a more
complete quantum mechanical picture of
what is going on here.
So if we superimpose our radial probability
distribution onto the Bohr radius, we see it's much more
complicated than just having a discreet radius.
We can actually have any radius, but some radii just
have much, much smaller probabilities of actually
being significant or not.
So, I think we're a little bit out of time today, but we'll
start next class with thinking about drawing radial
probability distributions of more than
just the 1 s orbital.