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PROFESSOR: All right, let's
just take 10 more seconds.
All right, so someone
want to explain
why this is the correct answer?
And we have a
syringe highlighter.
You probably never had
something like this before.
AUDIENCE: OK, so the fourth
excited state is n equals 5.
And then IE is opposite of
the negative number shown.
So it would be a
positive reaction.
PROFESSOR: Right.
So IE is always
going to be positive.
And you have to pay attention
to what n equals when
you're in the excited state.
So we've been talking about
the hydrogen atom and binding
energies.
What comes out of the
Schrodinger equation?
We have the binding
energies that come out.
And we also have wave functions.
So today we're
going to be talking
about wave functions,
which are often
referred to as
orbitals in chemistry,
for the hydrogen atom.
So when you solve the
Schrodinger equation,
you get out this information
about wave functions.
And what comes out of it
is these quantum numbers.
And we already saw quantum
number n coming out.
But there are three
quantum numbers
that are going to come out
of the Schrodinger equation.
And those three
quantum numbers are
necessary to describe the
wave function or the orbital.
So we have n, the
principle quantum number.
We've already talked about that.
And we've already seen
that n is an integer.
So I'll just put that down here.
So n can equal 1, 2,
3, on to infinity.
So this describes the
energy level or the shell.
Then we have l, which we
haven't talked about yet.
So that's the angular
momentum quantum number.
So it tells you about
the angular momentum.
It also tells you
about the subshell
or the shape of the orbital.
And so l is related to n.
And it can be 0, 1, 2,
3, onward to n minus 1.
So its biggest
number is n minus 1.
Then we have m, the
magnetic quantum number.
And we often see this
also listed as m sub
l because m is
related back to l.
And this is equal to minus
l, dot, dot, dot, to 0,
dot, dot, dot to plus l.
And m describes the behavior
in a magnetic field.
It also describes
the orientation
of the orbital with
respect to an axes.
And it tells you about the
specific orbital in question.
So we need all three of these
to describe any orbital.
All right, so let's look at
this in a slightly other way.
So we're going to have lots of
different sort of nomenclatures
for the same thing.
So to describe an orbital,
we need those three quantum
numbers.
We need n, l, and m.
And this can also be expressed
as our wave function sub nlm.
And again, we talked
about this last time.
We're going to
talk more about it.
So our wave function
is also described
by r, the radius, and theta
and phi, which are two angles.
And we're going to talk
a lot about those today.
So the wave function
for the ground state
is abbreviated wave
function sub 1, 0, 0.
Because it's the ground state.
So n equals 1,
and l and m are 0.
So what you see down
here, the 1, 0, 0,
refers back to what is
n, what is l, what is m.
And this also has another name.
So in the terminology
of chemists,
we call the wave function 1,
0, 0 1s, or the 1s orbital.
So let's look again
now at the same things
we just talked about,
but going through kind
of chemistry lingo.
So again, n describes the
shell or the energy level.
Again, it's integers,
1, 2, 3, et cetera.
l in chemistry lingo,
the subshell or the shape
of the orbital.
And instead of
listing it this way,
we have another way to
list it if we're a chemist,
and that is s, p,
d, f, et cetera.
So chemists like numbers, but
we also throw in some letters
every once in a while.
And then m, again, designates
this orbital orientation
or the specific orbital.
So for s, there's only s.
It doesn't have any
other designation,
as we'll talk more about later.
But for p, we start
having suborbitals.
And there is a
difference in terms
of the orientation of this.
So we have px, py, pz.
So that's what m tells us about.
So if we have all
three of these numbers,
we get down to the
specific orbital,
we can say oh, that's
pz, for example.
So we need all of these three
numbers to define the orbital.
And this is in then
the chemistry lingo.
All right, also a little
bit more chemistry lingo.
So here we have l equals 0.
So that is the s orbital.
When l equals 1,
that's the p orbital.
l equals 2 is the d orbital.
And l equals 3 is the f orbital.
And frankly we don't
really go much beyond that.
And in this part of
the course, we're
really only going to be talking
mostly about s and p orbitals.
We get to d orbitals
around Thanksgiving time.
So you can look forward to that.
And pretty much we're not going
to really talk about f orbitals
very much at all.
You'll need to know
some things about them,
but we're not going to go into
them in any kind of detail.
All right, so if
we keep going then,
we can think about l
equals 1 or our p orbitals.
And then when l equals 1, then
m can equal 0 plus 1 or minus 1.
And when m equals 0, that's
by definition the pz orbital.
So when you see m equals
0, that's going to be pz.
And when m is plus
1 or minus 1, those
are the px or the py orbitals.
And this is just something
that you need to remember,
that z is the one
that's special.
It's the one that
has m equals 0.
All right, so we can take
all of the nomenclatures
now and use it to fill
in this awesome table.
So this will help
you kind of keep
track of all the
different ways you
can designate the same things.
And we'll fill this in.
So first, state label.
What do I mean by this?
By this I mean this one 1, 0, 0
to generate this wave function
where we have this 1, 0, 0
listed below the wave function
here.
And so now this is just
a little color coding.
But it's blank in your handout.
So n equals 1, so n is first.
l is the second number.
And m is the third here.
So 1, 0, 0, and what
kind of orbital is this?
You can just yell it out.
AUDIENCE: 1s
PROFESSOR: Yep, so
that's the 1s orbital.
And so the 1, n
equals 1, that's 1s.
And now we have our
binding energies again.
And so we can write those
in two different ways.
So we saw for the
hydrogen atom before what
comes out of the
Schrodinger equation,
that the binding energy of
the electron for the nucleus
is minus the Rydberg constant
RH, divided by n squared.
And here n is 1, so
divided by 1 squared.
So this is just the value
for the Rydberg constant,
the negative value.
And binding energies,
again, are always negative.
So we have our first one down.
So now for the
second, what number
am I going to write here
for the state label?
You can just yell it out.
Yep, 200 or 2, 0, 0.
And then you would
put it this way
where the state label
is by the wave function.
What orbital is this-- 2s.
And then we also know the
binding energies for this.
So here we have minus
RH over n squared
where n is 2, 2 squared.
And we saw this
number last time.
So we can keep going.
Now we have 2, 1, 1.
So we can write that down.
We can write it both ways.
What orbital is this?
AUDIENCE: [INAUDIBLE].
PROFESSOR: So it's a 2p.
And because n is plus 1 and
not 0, it's either x or y.
Do we have a different or
the same binding energy here?
We have the same, right, because
it's just over n squared.
We're still talking about
n equals 2, so 2 squared.
So it's the same value here.
Now we have m equals 0.
So we write 2, 1, 0.
And now what is that orbital?
AUDIENCE: [INAUDIBLE].
PROFESSOR: 2pz, right,
because that's m
equals 0, by the
definition I gave you.
So we know that one for sure.
And again, the energies
are going to be the same.
And then the last one, so
now we write 2, 1, minus 1.
And now it's again a 2p orbital.
And it's either y or x.
And the energies are
going to be the same.
So these are just
a table that kind
of interconverts
different ways that you
will see things written.
And you'll know if you
see it one way, what
orbital to put down.
And we can also think
about the binding energies
for those particular
orbitals, or for electrons
in those particular orbitals.
All right, so why don't you
try a clicker question on this?
10 seconds.
Ah, excellent.
Right.
So you're getting
the hang of this.
It's great.
Some things, it's
always nice when
there's some things that
are pretty straightforward.
So n equals 5.
l equals 1, which means
p orbital and m equals 0,
means pz.
So let's think now about
these orbitals again.
And we looked at
that table and saw
that if we were talking
about n equals 2,
they all seem to
have the same energy.
So for a hydrogen atom-- and
it will get more complicated
when we start
talking about things
with more than one electron.
But for a hydrogen atom,
orbitals that have the same n
value have the same energy.
So here we have n
equals 1, l equals 0.
This is our 1s.
We have n equals 2, our
2s, and our 2p orbitals.
n equals 3, we have
our 3s, 3p, and 3d.
And in this case,
all these orbitals
are what's known as degenerate
with respect to each other.
They have the same energy.
And so for any n with a hydrogen
atom, or any one electron
system, for n shells,
there n square degenerate--
or for any n there are n
squared generate orbitals.
So they're all going
to be the same energy.
And that changes when we go
to more complicated systems.
But for hydrogen, this holds.
So now I'm going
to tell you why you
should care a little about
these energy levels again.
And today you're going to
hear in their own words
from a graduate student in the
physical chemistry division.
[VIDEO PLAYBACK]
- My name is
Benjamin Ofori-Okai.
I'm entering my third
year of graduate school
in the chemistry
department here at MIT.
And the work that
I've been focusing
on for the last couple
of years involves
nanoscale magnetic resonance
imaging or nano MRI.
When you think of
typical MRI, what
comes to mind for most people
is the image of a brain
scan or a heart scan
or some sort of organ
scan inside the human body.
The way that MRI works
now, the way that you
take a picture of anything in
your body is you use water.
And the reason
that you use water
is because it's made up of
hydrogen atoms and oxygen
atoms.
And hydrogen atoms actually
generate a magnetic signal.
And so you can take
a picture of that.
The idea behind nano MRI is
that you want to take a picture.
You want to do the
same kind of imaging,
but on a considerably
smaller scale.
We have this probe
which is sensitive
to local magnetic fields.
And the way that
the probe works is
that you have these electrons.
There's a ground state
for these electrons
and two excited states for these
electrons, which are actually
degenerate with each other.
And degenerate means that they
just have the exact same energy
level.
As you move the probe
around, anything
that's in the environment that
generates a magnetic field
will change what
the energy levels
of these two excited states is.
So when you're far
away, there's no change
and they're exactly the same.
And as you get
closer and closer,
these levels start to split.
And what we actually
care about is
what is the splitting
between these two levels,
because that's what tells us
what the magnetic field is.
In traditional MRI,
the probe that we
use, the thing that
measures the fields,
itself is very, very big.
It's person sized.
The probe that we're
using in this nano MRI
is nanometer sized.
So this gives us the ability
to look at things that
are on the nanometer scale.
And to give you a sense of size,
that's like 1/10,000 the width
of a human hair.
So that includes viruses,
cells, parts of proteins,
not just the entire protein.
And on top of that, we'll be
able to look within objects.
So you're not just sensitive
to what's on the surface.
You can actually
see how are things--
what's the constitution?
What's the makeup of things
within the object that you
want to image?
So the long term
goal, the one thing
that I'd really love to see
this technology be able to do
is say, OK, we've
got this virus.
Let's just see how it works.
Let's watch it in real time.
Let's see if we can see
how it attaches to cells
and invades them and
ultimately kills them.
[END PLAYBACK]
PROFESSOR: OK, so I always think
this is a great time of year
to show this video because
pretty much viruses, I think,
start to be on people's minds.
Everyone has sinuses and colds
and other things going on.
And so understanding,
we're still very far away
from having a real cure
for the common cold.
So I think it's very
timely to be talking about,
talking about this research.
I'll also use this to
remind myself to tell you
that if you qualify for
extra time on the exam,
you should get me your
form for the exam.
And it reminded me to
say that because Ben,
who is a former
TA for this class,
always proctors the
extra time folks.
So you'll get to
meet him in real life
if you qualify for
extra time on exams.
So hydrogen is in
fact important.
I'm excited to get
on to elements that
have more than one electron.
But hydrogen actually does turn
out to be extremely important.
A lot of imaging, as you heard
from Ben, is based on hydrogen.
So we're spending a lot
of time on hydrogen,
but hydrogen really, really
is an important element.
So continuing on now,
what is the significance
of this wave function?
Why do we care about this?
And so really, we're
interested in trying
to understand not just how
tightly the electron is
bound to the nucleus, but kind
of how the electrons exist
around the nucleus.
And so the wave function
really gets at this.
It gets at the probability
density, the likelihood
that you'll find an electron
at a certain location,
the probability per unit volume.
And again, this is a
three dimensional problem.
So our wave function
depends on a radius r.
But it also depends on two
angles, the theta and phi.
And so you can kind of think of
those as latitude and longitude
if you will.
And so we want to know
what the probability is
that an electron will be at
a certain r, theta, and phi
position in a particular small
unit volume in that area.
How well can we understand
where the electron is?
And this gives rise to a lot of
the properties of the elements.
So probability density,
density per unit volume.
So really, when we're talking
about where electrons are,
we're thinking about
a shape of an orbital,
a shape of a probability density
of where that electron might
be.
So now we're going to
think about shapes.
So we can define a
wave function in terms
of two properties, a radial wave
function and an angular wave
function.
So again, the wave function
has these three things.
We are considered with a
radius and these two angles.
So we can rewrite
this, breaking up
these two different components--
the radial component that
depends on the radius-- so
that's easy to remember,
radial, radius-- and
the angular component
that depends on the angles.
So the nomenclature
here is pretty good.
All right, so we have
these two components.
So now I'm going to
show you a table that
is largely from your book.
Don't let it scare you.
You do not need to memorize
any of these things.
And I'm showing this
to you because I
want you to believe me about
certain properties of these two
functions.
So here they are solved.
You can look them up.
Actually I think we just
typed a new copy of this
so it was easier to see.
If you find any typos,
please let me know.
But there's a couple
of important points.
So on this side, we have
the radial wave function,
and over here we have the
angular wave function,
for various values of n and l.
So again, not an
exhaustive list here.
And a lot of these are
written in terms of a0,
which is the Bohr radius, which
is a constant, 52.9 picometers.
All right, so now let's just
consider the ground state.
So we'll start with
that lowest energy
state or most stable state,
the 1s orbital for the hydrogen
atom.
So we have our wave
function 1, 0, 0 here.
And this is 1s up here.
Again, n equals 1. l equals 0.
So that's 1s.
And z for hydrogen atom is 1.
So I've gotten rid of all z's
to make it a little simpler.
So here we have the
radial wave function
times the angular wave function,
which is listed up here.
And the thing that I
really want you to notice
is that for all of the s
orbitals, this is a constant.
So this is always the angular
component for all s orbitals.
And in fact, there are no
angular components in there.
So all 1s, 2s, 3s, all
have this same constant.
And that leads to a
very important property
of s orbitals, which
is that they're
spherically symmetrical.
In other words, they're
independent of those angles,
of theta and phi.
And so that means that
the probability of finding
the electron away
from the nucleus
is just going to depend on r.
There's only r in this equation.
The angles are not
part of the equation.
So s is spherically symmetrical.
The probability of
finding the electron
just depends on the radius.
So we can draw a picture,
or multiple pictures,
of what that could look like.
And these are
three common plots.
So I'll tell you
that on your handout,
the plots are
listed on one page,
and then the plots are
shown on the next page.
And I'm going to kind of go
back and forth between things.
So the plots-- don't
have to write this down.
They're on the other page.
But if you want to pay
attention to which kind of plot
goes with which plot.
So these are three different
ways to, quote, visualize.
And some people say, can you
give me another visualization?
We're really just trying to
think about probabilities
of finding electrons here.
And so you can't sort of
take a picture of an orbital.
So these are just
different ways to help
people think about that possible
distribution of electrons
around the nucleus.
All right, so one thing that
everyone's feeling pretty good
about is that it should
be spherically symmetric
hole for an s orbital.
And so we have a circle.
And so the probability
density, which
is shown in this plot-- and
the probability density parts
are basically just dots
where the more concentrated
the dots are, the
higher the probability
density for that
particular-- the probability
for that particular
volume exists.
So in here there are sort of
more dots and then less dots
as you come out.
And so that is a
circle, which is what?
It's symmetrical.
So you can always
recognize a 1s.
You have this symmetrical thing.
So this is the wave function
squared, is this probability
density plot.
Another kind of plot
that you can see
looks at the radial
wave function
plotted against the
distance r here,
distance from the nucleus.
And then a third kind of plot
is another probability plot,
like this one up here.
But instead of the dots
indicating the higher
probability density, you
have a radial probability
distribution.
And so at the nucleus,
at 0, well then
the probability goes up.
The electron is not going
to crash into the nucleus,
so it won't be right
on top of the nucleus.
But as you get out a
little bit farther away,
there's a high probability
that it's there.
And then that decreases again.
So the top one
and the bottom one
both talk about the probability
of finding an electron
in a particular unit.
And I'll give you just a
little more definition of this.
And this is on the same page
above those different plots.
So the radial
probability distribution
reports on the
probability of finding
an electron in the
spherical shell
at some little distance
dr from the origin.
And one thing that
comes out of this,
which is pretty important,
is the most probable value
for that distance
r, which is denoted
rmp, so most probable distance.
And for a hydrogen atom,
this is a0, the Bohr radius.
And you can see it expressed
in different units over here.
And from the plot, that will
be the top part of the plot,
the most probable distance.
In this case, that's the Bohr
radius for the hydrogen atom.
So we have now these
three different kinds
of plots that you'll see.
And I want to point out that
they're different plots.
Sometimes people are thinking
that there is sort of one plot
and they're trying to read one
of them as probability density,
and that's not what it is.
So we'll look at these again.
All right, so going
back and we'll just
look at them again now that
we sort of talked about what
all of them are,
again, we have our sort
of dot density, probability
density plot, our wave function
plot, and our radial
probability distribution plot.
And for 1s, we have the dots
closer to the nucleus here.
Probability goes
up and goes down.
And here, you're
thinking about this
as the amplitude of finding
an electron as you move away
from the nucleus.
So 1s is pretty simple.
And I think these plots are a
lot more meaningful when we go
on to look at other orbitals.
So let's think about
those other orbitals.
And we'll finish
the other plots.
So this is just-- you can
actually stay, in this case.
So we're going a lot of
back and forth today.
So here is your table
that we had before.
And here's 1s.
Here's 2s.
Here's 3s.
These terms are in fact
different, as you can see.
But the angular term,
as we mentioned before,
is still the same.
So that means 2s and 3s
are still symmetrical.
So we're still thinking
about the probability
of finding an electron
in some volume
as just going out
as a distance of r.
So let's look now at the three
plots, and compare those plots.
And this is the one on
your handouts we looked at.
I showed you this.
And now we have all of these
three plots together here.
And in the comparison
of these three,
I think it helps
differentiate what
you're seeing in these plots.
So important point,
they're spherical.
1s, 2s, 3s, they're
all spherical.
And here we see the
dot density increase.
And then the dot
density goes to 0.
And that's known as a node.
So a node is a value
of r or theta or phi
for which the wave
function and wave function
squared, or the
probability density, is 0.
And in this particular case, the
type of node that we're seeing
is a radial node.
And so that's a value of r
for which the wave function,
wave function squared
probability density is 0.
So it goes to 0.
We have a node, a radial node.
Then there's more probability.
And then it increases, and
then starts decreasing again.
And so if you plot this with the
radial wave function versus r,
you see it go down.
And it crosses the
zero line here.
And that's the node.
And that's at 2a0.
And then it goes back up.
And this plot often
bothers people.
They're saying, what, there's
now negative probability?
No, these are not the
probability diagrams.
This is thinking
about the amplitude
of finding an electron.
So we don't have to worry.
It can have a positive or
a negative phase to it.
And if you look at this
plot, the radial probability
distribution plot,
then you'll see
that actually the radius,
the most probable radius
is in this region over here.
And you see that this is
concentrated dots up here.
So if we think about these two,
which are really probability
distribution diagrams, we're
thinking about the probability
of finding an electron.
You have a probability in
here close to the nucleus.
Then you get a node.
And then you have another
probability, high probability
of finding the electron.
In fact that's the most
probable radius here for 2s.
And then it decreases.
So this line shows
you what a radial node
looks like in all three plots.
In this probability
diagram, wave function
squared plot, it looks like
there's just an empty space,
no dots at all.
Down here, it's where
it crosses the line.
And in the bottom plot, it
is where you go up and down
and again touches the
line before going back up.
So you should be able
to look at these plots
and think about what they mean.
For 3s, we see the same thing.
But now we have an intense spot
in the middle near the nucleus.
That is indicated down here.
There is probability of finding
the electron near the nucleus.
Then there's a node.
And that's in this plot
where it crosses the line
and in this plot where
you have the empty space.
Then you have more probability
of finding the electron.
You have another bump here.
And then we have another node,
indicated by touching the zero
line here, touching here.
That's at 7.1a0.
And then we have
more probability
of finding the electron.
And this is where the most
probable radius is at 11.5.
So again, you need to be able
to look at these diagrams
and recognize what
constitutes a radial node.
And a node is a
place where there
is no probability that you're
going to find an electron.
So now let's think about
how many nodes, or radial
nodes you should
have when you have
different types of orbitals.
And this is just a similar
diagram to what I just showed.
This is the wave function
squared, probability diagram.
And now instead of blue
you have orange dots,
but otherwise should be the
same-- so for 1s, for 2s,
and 3s.
So for the 1s orbital,
we can calculate
how many radial
nodes that we should
have by using this handy
formula, n minus 1 minus l.
So for 1s we have 1 minus 1.
And l is 0.
So we have zero radial nodes.
And we can see that
from that diagram
there are zero radial nodes.
2s now-- 2, n is 2.
Minus 1, minus 0-- so
that's one radial node.
And the radial node, again,
in this kind of diagram
is the empty space.
And that radial node is at 2a0.
For 3s, we have n equals 3 minus
1 minus l, which is still 0.
So we have two radial nodes.
And so again, the empty
space here at 1.9a0 and then
at 7.1a0.
So why don't you
give this a try now
and tell me what kind of radial
nodes you would expect for 4p.
OK, 10 seconds.
These are pretty fast.
Yep.
So again, we have to do
n, which is 4, minus 1.
And then what is l
in this case-- 1.
So that gives you 2.
All right, so 4 minus 1
minus 1 or 2 radial nodes.
All right, don't put
your clickers away.
Let's try something else.
So now tell me which of these
is correct both in terms
of the indicated
number of radial nodes
and in terms of the
plot for a 5s orbital.
All right, let's just
do 10 more seconds.
We're varying it up
in terms of the plots.
So maybe someone want to say
what the right answer is here?
Yeah?
AUDIENCE: So by the
formula we just did,
that has four radial nodes.
And if you look at the
graph of one, there's three,
and then there's another
one at the origin.
So that's four radial nodes.
Right?
Right?
PROFESSOR: Actually,
I just realized
that-- let me count here.
So this answer here, we
should have four radial nodes.
That is correct because
we have n minus 1 minus l.
Actually, I think this
is going to this--
this should be going
to this answer,
because if we count 1, 2, 3, 4.
Sorry, the new plot
is highly confusing.
I have to count.
So the one at the origin
should actually not count.
AUDIENCE: It doesn't count?
PROFESSOR: This is not a node.
So we have 1, 2, 3, 4, should
be our four radial nodes.
Because that's a nucleus,
and there isn't one there.
But that doesn't
count as a node.
So this should be here.
I guess that's-- right.
But thank you very much,
and [INAUDIBLE], here.
You were brave enough to answer.
Yeah, there's a question?
AUDIENCE: Should there also
be a certain number of peaks
in the graph as well as nodes?
PROFESSOR: Yeah.
So if you look at the peaks,
these are really hard to draw.
And I think that's partly
what the problem is.
But when we look
later in the handout
where they're drawn a
little bit more carefully,
it does increase.
So there are different numbers.
So we'll have nodes
going down here.
But then we'll have
more distributions.
But often the ones
as you go along,
it does indicate where the
most probable radius is
as the taller ones, and that
it's usually drawn at the end.
So we have some plots and
I'll point this out later.
We're going to look at
more plots, don't worry.
So if anyone's good
at drawing those,
let me know, because
they're really hard to draw.
So a lot of them are
copied from the book,
but then they don't
copy very well.
So let's consider
other kinds of nodes.
And we're going to come
back to radial nodes.
All right, so what
about p orbitals?
So here we have our table again.
These are our p
orbitals over here.
And we have our n equals 2
cases here and our l equals 1.
So these are x, y, and z--
so our 3p orbitals over here.
And the important point
is not to memorize
what these values are.
But now all of a sudden we
have dependence on angles.
So we're going to have an
angular component to these.
And that means the
probability density
as you go out from the
nucleus doesn't just
depend on r anymore.
It depends on theta
and phi, which
are sort of the equivalent
to latitude and longitude,
if you're thinking
about geography.
All right, so let's see
what that looks like.
So that means then
the p orbitals
are not spherically symmetric,
because it depends on angle.
So you just don't go out and
have the probability depend
on the radius and
it's symmetrical
in all the different directions.
And here are what some
of them look like.
These figures are
in your handouts.
Here are some other figures.
So the orbitals
consists of two lobes.
So you could view this as a lobe
up here and a lobe down here.
Or you have these lobes as these
two different colors over here.
And the lobes are
separated by a nodal plane.
And the nodal plane is a
plane on which the probability
of finding the electrons is 0.
So in the top drawing, the
nodal plane is drawn as a plane.
And in the bottom drawings,
you don't see a plane.
You just see an empty
space between the lobes.
So empty space here, empty
space here, empty space there.
And so if it helps
you to kind of think
about an actual plane
in between, that's good.
Or you can just
think that there's
a break between these nodes.
And again, the
nodal plane, there's
no probability of finding an
electron in the nodal planes.
And the nodal planes
are at the nucleus.
Therefore, there
is zero probability
of finding a p electron
at the nucleus.
s can get pretty
close to the nucleus.
But with a p orbital,
there's a nodal plane there.
No electrons are going
to be at the nucleus.
So now if you're going out from
the nucleus, the probability
of an electron,
finding it, if you're
going out in this
direction, you're
not going to do very well.
If you're going
in this direction,
you should do a lot better.
So here the angular
components really matter.
That defines the
shape of the orbital.
And where you're going,
what direction you're
going in, what
angles you're going
in matters in terms of
whether you're going
to find that electron or not.
So another way to
think about this
in sort of these nodal
planes-- so here we'll
just define what plane it is.
So we have our pz orbital.
That's a nodal plane
then in x and y.
And so x and y are over here.
Our px orbital is
going to be in--
or the nodal plane is going to
be in yz plane, so over here.
And py will be in xz plane.
So again, these
nodal planes, there's
no electron density there.
And these arise from
these angular nodes
in the wave function.
So angular nodes
then or these angular
nodal planes are
values of theta and phi
for which the wave function,
wave function squared are 0.
So this is very
different from the s case
where we only had radial nodes.
But now, when in
the p orbitals where
the angular component matters,
they're angular nodes as well.
So we can think about how to
calculate the angular nodes.
So total nodes is going
to be equal to n minus 1.
The angular nodes is l.
And as we saw before, the radial
nodes are n minus 1 minus l.
So let's have more practice
in calculating these.
And then we'll look
at some more diagrams.
So for 2s, total nodes-- and
you can just yell this out.
Total nodes will be what?
AUDIENCE: 1
PROFESSOR: 1-- 2 minus 1 or 1.
Angular nodes are?
AUDIENCE: 0
PROFESSOR: 0.
For 1s, there is none.
And if you forget,
l equals 0 there.
Radial nodes is going to be?
AUDIENCE: 1
PROFESSOR: Right, 2
minus 1 minus 0, or 1.
All right, let's try 3--
or sorry, 2p is next.
Total nodes?
1 again, so 2 minus 1 or 1.
Angular nodes?
1-- l equals 1 here.
And radial node?
Right, 2 minus 1 minus 1, or 0.
So since there's
only one total node,
if you figured out there
was one angular node,
you could even realize that
there had to be zero there.
It's a way to check
maybe your equations.
All right, so let's
try for 3d now.
How are we doing?
All right, let's just
do 10 more seconds.
And let's just work
that out over here.
So total nodes for 3d,
we have 3 minus 1 or 2.
Angular nodes, l equals 2 for d.
So radial nodes, we have
3 minus 1 minus 2, or 0.
All right, so bring these
handouts on Wednesday
because we need to go back and
look at more radial probability
diagrams.
And talk more about nodes.
All right, let's just
do 10 more seconds.
OK, good job everyone.
Let's look through
this a little bit.
And you can sort of--
everyone can help.
Yell out some responses.
So this was 2s.
And that was the correct answer.
Which type of
orbital is this-- 2p.
And if you couldn't read
this information here,
you should have been able
to read the information
about the nodes.
What equation is that for nodes?
Yeah, n minus 1 minus l,
for what kind of nodes?
AUDIENCE: Radial.
PROFESSOR: Radial nodes, right.
So if you know what it means if
l equals 0 versus l equals 1,
and you knew this
was l, then you
could tell if it was an
s orbital or a p orbital.
And then whether it was
2 or 3p is from the n.
So even if you
couldn't read this,
if you knew that expression,
then you were OK.
What kind of orbital
was in plot C?
This was a 3s.
l equals 0.
And then this is a what, 3p and?
l equals 2.
Louder.
D, right?
So do 3px, 3py, and 3pz
have different plots?
No, they wouldn't
have different plots.
So we'll continue
to look at this.
And we're going to be
starting with the handout
from last time.
And so let's
continue with Monday
and continue with these radial
probability distributions.
So this is again
from Monday, page 6.
We're talking
about orbital size.
And we've already looked
at this a little bit today.
So we should be able
to go through this now
in a little bit more detail.
You've already thought about it.
So here we have the 2s orbital.
And we're going to
have one node using
our equation that you just
told me, n minus 1 minus l.
And when we go from 2s to 2p,
here we have no radial nodes.
And we can look
at r and p, which
is the radius of the maximal
probability of finding
an electron.
And you can note that when
you go from the 2s to the 2p,
the radius actually decreases.
So the most probable radius
for 2p is less than that of 2s.
Now let's consider
the 3, n equals 3.
So we have the 3s
situation over here.
And so l equals 0.
We have two nodes here.
And now if you look at the
radius, the axis over here,
you'll see that the
most probable for 2s
is close to 5a0, where
a0 is the Bohr radius.
And over here you're
talking between 10 and 15.
So we see an increase
in size going this way.
And then when we go from 3s to
3p-- so here we have 3 minus 1
minus l, which is 1.
So we have one node, down to 3d,
3 minus 1 minus 2, zero nodes.
And you see that there
is a decrease here
in the most probable radius.
So, OK, interesting.
All right, so 3d has the
smallest, next 3p, next 3s.
So there's two different
trends we're seeing.
One, as we increase l
within the same n number,
and one going from a smaller
value of n to a larger value,
and then again within
the 3, within the n value
as we change l.
So again, to say the same
thing in a different way,
as n increases from 2 to 3, the
radius, most probable radius
or the size increases.
So from here to here we
have an increase in size.
I just want to make
sure people have
time to kind of get all of this
down, but it should be good.
I have a little
picture that just shows
they're very different in size.
So we'll go back to this again.
And then as I also said, as
l increases for a given n--
so from l equals 0
to l equals 1 here,
then we have a
decrease in the size.
So you can see the most
probable radius moves over.
And then here is
another within n.
And n equals 3.
We see, again, this decrease.
So those are the
two trends that you
observe when you look at
these radial probability
distributions.
So for exam one next
week, you should
be able to draw
distributions like this.
You should be able to
tell me how many radial
nodes you have for
different types of orbitals.
And you should know
these trends in size.
So I think in the
exam instructions
it says up to a 5 case.
You don't have to go on forever
to be able to draw them,
but you should be
able to look at these
and tell what kind of orbital
it is and where the nodes are,
be able to draw where the
nodes are-- one node here,
one, two, one node here.
This kind of thing will
be on the exam next week.
So there's something that's a
little counterintuitive when
it comes to this size issue.
And that has to do with
how this correlates
to the amount of shielding,
and as we see later,
to the energy levels.
So only electrons
in the s state here
really have any kind of
substantial probability
that they'll be
close to the nucleus.
So we have this
little blip over here
that is close to the nucleus,
that at are very small
radii, very small values of r.
Even though the most
probable is out here,
if we compare 3s to 3p and
look at where the electrons are
that are closest to the
nucleus, they're quite a bit
farther away than in the 3s.
Or there's more
probability that there's
going to be some closer here.
And then the closest probability
over here for these electrons
is quite a bit farther away.
So we see these circles
kind of move out.
So even though the
overall radius,
the sort of size of the
whole thing is decreasing,
the probability
that there are going
to be electrons really
close is actually
going in the opposite direction.
And so what this means
is that s electrons
are the least shielded
because there's
higher probability that they'll
be some close to the nucleus.
There's more penetration
close to the nucleus.
So s electrons are
the least shielded.
And we're going to
come back to this when
we move on to today's handout.
This is really important
in terms of thinking
about the energy levels.
And I'm going to
have these diagrams
on the handout for today.
So we'll see them again.
All right, so before we
move to that handout,
we've got to finish
our quantum numbers
and talk about electron spin.
So the fourth quantum
number describes
the spin on the electron.
And we already saw the
magnetic quantum number m.
We saw m sub l.
And now we have m sub s.
And the s stands for spin.
So there's some nomenclature
that actually makes sense.
So there are two possible
spin values for an electron.
And s can equal plus 1/2, spin
up, or minus 1/2, spin down.
And here are some
little pictures of that.
So this ms term, this spin
magnetic quantum number,
completes the description
of the electron.
But it's not dependent
on the orbital.
To describe an
orbital completely,
you only need three
quantum numbers.
But to describe the
electron, you need four.
And that is shown, again,
here on this picture,
or on this slide.
You need three quantum numbers.
You need n, l, and m sub l to
describe the quantum number,
describe the orbital completely.
But you need a fourth
one, this m sub
s to describe the electron.
So if you see wave
function n, l, m sub l,
you say that's telling
me what the orbital is.
And if we add the m sub
s, then you look at that
and say oh, that's
going to tell me
all the way to the
electron what is going on.
So this final quantum
number led to what
we know as Pauli's
exclusion principle, which
is that no two electrons can
have the same four quantum
numbers.
They can't have the same--
no two electrons can have
the same spin, in other words.
So if we are drawing a
configuration for neon
with 10 electrons,
we are going to have
with one electron being
up spin, the next one
is going to be down.
Because if we had two
of these both going up,
they would have the same
four quantum numbers.
And that's not allowed by
Pauli's exclusion principle.
So when you have two
here, one spin up,
one spin down in
an orbital, then
we say that those
electrons are paired.
And an important thing that
kind of comes out of all of this
is that one orbital can't
hold more than two electrons.
If it did, there'd
be another electron
that would have the same
four quantum numbers.
Because you need three
quantum numbers to describe
the electron, or the orbital.
We need three to describe, say,
that it's n equals 1, and then
its s state.
So we need those other ones to
describe the orbital and then
the fourth one to
describe the spin.
So if we add another
electron, you'd
have two that were spin up, say.
And that just wouldn't work.
So you cannot have more than two
electrons in the same orbital.
And this makes a
lot of sense when
you think about why you would
be putting electrons in orbitals
that are higher energy.
Why not just keep putting him
in the low energy orbital?
And it's because
you can't do that.
You can't put more
than two electrons in.
And so therefore once you've
filled a lower energy orbital,
you've got to move up to the
next lowest energy orbital.
