Exam 1, Problem 2 | MIT 3.091SC Introduction to Solid State Chemistry, Fall 2010


Uploaded by MIT on 16.11.2010

Transcript:

The following content is provided under a Creative
Commons license.
Your support will help MIT OpenCourseWare continue to
offer high-quality educational resources for free.
To make a donation or view additional materials from
hundreds of MIT courses, visit MIT OpenCourseWare at
ocw.mit.edu.

Hi, I'm Jocelyn, and we're going to go over Fall 2009,
Exam 1, problem number 2.
So as with every problem, we want to first
read the full question.
In box notation, give the complete electron
configuration of each of the following gas-phase species:
calcium 2 minus and magnesium 4 plus.
So the first thing to note here, is you need to know what
electronic configuration is.
If you don't, you might want to go back and review that
material, but it is basically assigning
electrons to orbitals.
The next thing you need to know is what box notation is.
And Professor Sadoway introduced that in class, and
you'll see, as I go through this, a refresher of that, if
you forgot.

So to assign electrons to orbitals, we first need to
know how many electrons we have. So let's start with the
first part of this problem, which is calcium 2 minus.

We will look at our periodic table, and see that neutral
calcium has 20 electrons.
So calcium with no charge has 20 electrons.
The 2 minus tells us that it has 2 extra electrons.
Right?
Because electrons have a negative charge.
So we have a total of 22 electrons that we need to
assign into orbitals.
In order to actually do the assigning, we need to follow
the three rules that Professor Sadoway introduced in class.
So I will put that over here.

The first rule is that you need to fill them, at least
for a ground state species, in order of increasing energy.
That is, you want to fill up the lowest energy orbitals
first, and then move to the higher energy orbitals.
This makes sense because ground state species want to
have the lowest total energy possible.
So the first is--

Now, how do we figure that out?
Well, there's a few ways.
One way is a nice trick that you can do on the side of your
page or something.
And it goes like this.
So we have our s, p, d, and f orbitals.
Right?
We know that s can be in the first energy level up.

We know that p starts at the second energy level, d starts
at the third, and f starts at the fourth.

The tricky part is that you don't just fill up 1s, 2s, 2p,
3s, 3p, 3d.
There's a little bit of mixing up of the energy levels versus
the actual energy of the orbital.
So a nice side of the page trick is to draw diagonal
lines down through these numbers.

And if you follow these arrows, you'll fill up in
order of increasing energy.
For example, after you fill up the 2p, you're going to fill
up the 3s then the 3p, the 4s, and then you'll go to the 3d.
If that doesn't make sense, you might want to go and
review a little bit about the orbitals in your textbook, or
other resources.
The other way you can think about the ordering of the
energy of the orbitals, is to be familiar with
the periodic table.
So we're just going to have an aside over here.
And a general, very rough sketch of the periodic table
would look something like this.
OK?
You have your alkali, alkali earth metals over here.
You have aluminum, oxygen.
Your halides, your noble gases over there.
But the way I drew it here is in what we call blocks.
So this is your s block, and then we have our p block over
here, d block, and the the f block is down here.

So once you get more familiar with electron configurations,
you can actually looks towards the periodic table, and know,
depending upon where your element is in the periodic
table, what its electron configuration is.
Or at least the electron configuration
of the valence electrons.
OK.
So moving back to our trick, which we'll use now, we know
that we have 22 electrons, and we know the order that we need
to fill the orbitals.
So let's start writing out our boxes, which is how Professor
Sadoway wanted us to answer this problem.
So we have 1s, our 2s, 2p.
If you're unsure why I'm only drawing 1 orbital for s and 3
orbitals for p, again, you might want to go review that
beginning orbital material.
So we filled, we have 1s, 2s, 2p, 3s, then we're going to do
the 3s here, we're going to do the 3p.

And after the 3p, we do the 4s, and after the
4s, we do the 3d.
Which has 5 boxes, right?

So now that we know which orbitals we're going to fill,
we need to know some more rules about filling them.
So the second rule, over here, is the Pauli exclusion
principle, which says that no two electrons can have the
same quantum numbers.
For us, that means that we basically can only have two
electrons per orbital, and of opposite spin.
So that's something we need to remember for this.
The third is Hund's Rule.

And that is, we want to have as many unpaired
electrons as possible.
So that will come into play when we're filling up our
boxes in just a second.
If these are confusing, again, we want to look
at the lecture material.

So I forgot my boxes here.
Now let's start the fun part of filling up the elections.
So we fill up 1s, spin up and down, the
2s, spin up and down.
So that's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
16, 17, 18, 19, 20, and we have 2 more.
So in the 3d, we're going to do 21, 22.
And I separate them out because of Hund's rule.
We want to maximize the number of unpaired spins.
They both spin down, or both spin up, but they will be the
same spin, and they will be in different orbitals.
So this was the answer for this part of the problem.
Now let's move to the second part.
And maybe you can try by yourself, and then
we can do it here.
All right.
Now we're going to do the second part of this first part
of the problem number 2.
So it asks us for the electron configuration in box notation
of magnesium 4 plus.
So again, we're going to determine how many electrons
we have. Which, neutral magnesium has 12 electrons.
But the 4 plus signifies that we have 4 less electrons than
neutrality, and so we subtract 4, and we have 8.
With 8 electrons to work with, we again want to write down
our boxes using the energy level tricks that we have. And
now we can fill them up.
The Pauli Exclusion Principle tells us 1 spin up and 1 spin
down per orbital, and Hund's rule tells us that when we
fill up this 3p, we're going to try to maximize the number
of unpaired electrons, and then, when we have to, we pair
up the spins.
So that's another main mistake that people make, is they pair
up too many spins, violate the Pauli exclusion principle,
putting 2 electrons of the same spin in one orbital, or
something of that nature.
So this again is the correct answer, and would-- oh!
I'm sorry.
This is 2p.
I'm sure you guys caught that.
And would have awarded you full points in this part.
All right.
Now we're going to start the second part of this
problem, part B.
So first we want to read what the question is asking.
Give the chemical identities of the species with these
ground state electron configurations.
So now we're working in the opposite direction.
In part A, we were given a chemical species and asked for
the electron configuration.
Now we are given the electron configuration, and asked for
the chemical species.
So the first one has the electron
configuration of xenon.
So this is a noble gas configuration, plus what turns
out to be 27 electrons.
One way you can do this, is to go to xenon, which is in the
fifth row of the Periodic Table, and count through 27
electrons, remembering that when you get to the d on the
sixth and seventh row, you need to go to the f block,
which is signified in this problem by
having f block electrons.
Count through the f block, the d block,
and end up at thallium.
Or if you are familiar with the Periodic Table enough that
you can recognize the valence electrons are in 6p.

And the 6p only has one electron in it.
So that tells me, xenon is in row 5.
This element is going to be in row 6.
I go to the first column of the p block, which signifies
that there's one electron in that 6p orbital, and I see
that it's thallium.
That's the quicker way to do it.
On an exam, you might want to be that familiar with the
periodic table to be able to do that, because you're
certainly time-crunched, a lot of times.
So because that's a neutral atom, that method will work
very easily.
The second part is a little trickier, because it has a
charged species.
So we have a net charge of 4 plus, an electron
configuration of argon and 3d5.

This may seem a little confusing at first glance,
because we always fill the 4s before the 3d.
But when you're ionizing an atom, it turns out you take
electrons from the 4s first, even though
it's of lower energy.
So that's a little subtlety that's not too important to
this problem, but it may have tripped you up when you were
looking at this, and certainly caused some problems for the
people in this class.

Again, we go to the periodic table and we see where argon
is, and then we go to the next row.
The way I like to think about this, is this is the electron
configuration, missing 4 electrons.
To find the element that has this charge, I'm just going to
add back in the electrons, go to that position in the
periodic table, and that is my elemental species.
So we want argon plus, instead of 5 electrons--
I'm sorry, the problem says this is a 3--

instead plus 3 electrons, we are going to go to argon plus
7 electrons.
And going through the periodic table, you'll see that that
leaves you with manganese.

Adding the 4 plus, because that is part of this chemical
identity, we get the answer of the problem.
So again, the charged species make this a little more
complicated.
The easiest way is to go back and add in the missing
electrons, because that is your chemical species, and
then remember to signify that there actually is that charge
for your answer.
So moving on to part C, it asks us, write the quantum
numbers of one of the 3d and one of the 4s
electrons in iron.
So this requires a knowledge of what quantum numbers are,
and what values they can have.
So let's start with the 4s, because that's a little more
straightforward.
He tells us that we have an n quantum number, l quantum
number, m and s.
And, hopefully, we all know that n is the principal
quantum number.
It's basically your energy level.

l gives you what type of orbital you're in.
Right?
If you're in the s, the p, the d.
So it just tells you your orbital shape, basically.

m gives you the angular identity of the
orbital you're in.
We didn't really talk about that in depth.
We just need to know, it tells you which of the p orbitals
you're in, or which of the d orbitals you're in.
So I'll say, which orbital.

And the s is your spin.
Right?
So for a 4s electron, it doesn't actually matter what
the identity of our species is.
So that he tells us is iron is kind of extra information.
A 4s electron already has 3 of its quantum numbers set.
So we know that n equals 4, for s, l equals 0, and m,
which orbital, s only has 1 orbital.
So m always equals 0 for s, and your spin.
And I'm going to make that a script, just so it's less
confusing, is plus or minus 1/2.
So the answer to this problem could have been 4, 0, 0, 1/2,
or 4, 0, 0, minus 1/2.

Three of them are set by the fact that
this is the 4s orbital.
The spin can either be plus or minus 1/2.
If you're unsure of where I got these numbers, and why
they have to be such, you may want to review
the lecture on that.
So moving onto the 3d quantum numbers, we know that n is 3,
l for d is 2, m is negative 2, negative 1, 0, 1, or 2, right?
Because there's 5 different orbitals in a d set.
And then again, s can be plus or minus 1/2.
So picking one set of those, you could have said 3, 2,
minus 2, plus 1/2, or any combination of 3, 2, and
the m and the s.
You don't have to get the exact same answer as your
neighbor or someone else you're-- well, you shouldn't
be looking your neighbor's paper, or a classmate, but
these do signify that you're in a 3d electron.