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For this problem,

we are going to do the electron configuration of four different elements.

And then we're going to look at the valence shell and

do the quantum numbers for each at, electron in that valence shell.

Let's begin with boron.

If we look at a periodic table and find boron on the periodic table.

We see that we're going to element five and if we're going

to work our way through from the beginning, we do 1s2, 2s2.

I'll go ahead and write that down, 1s2, 2s2 and

then we come over here to 2p and 1.

A refresher, this is level one, two, three, four, five, six.

This is the S block, it's over 1s2, 2s2 and we come over this is the p block,

over here in purple and so we have one electron in that group.

So let's write that down, 1s2 2s2 2p1.

Now the valence shell are all electrons with the highest end.

So this is my valence shell, 2s2 2p1.

There are three electrons there, so we would need three sets of quantum numbers.

So here's one set, two set, three sets of quantum numbers.

The first number tells me n, the 2 right here is the n, so

we have a 2 for each one of those.

The s is a subshell and when s is for an l, or

s gives us an l equal to 0.

A p gives us an l equal to 1.

So for these two electrons, they're both in the 0 subshell.

They are both n the same order, orbital.

One spins in the positive one-half direction and

the other one in its opposite direction.

And we go to the next electron, which is in the p subshell, so we put a 1 here.

And I have three choices for m sub l, this is orbital it sits in.

We can have, when l equals 1, m sub l can be a negative 1, 0, or 1.

It's important that you know that it can be any one of those three numbers that I

can put next.

I generally go from left to right in my choices, so

I put a negative 1 first, but it is not necessary that you do so.

It just needs one electron in the p orbital of that p subshell.

And it doesn't matter which way it spins, so

I'll spin in it a positive one-half direction.

So that's a.

Let's go to b.

B is sulfur.

So once again, let's look at a periodic table.

Let's find sulfur.

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So we would start up here at the top and we're going to work our way to Sulfur.

It would be 1s2, then we'll come to here that's 2s2.

Coming across, we enter the 2p block.

So that's 2p and count all the way across, four to six.

Then we enter the three, that's 3s.

There's two.

We struggle across to here and we have the 3p.

And we count over one, two, three, four, that would be 3p4.

So let's write that out.

1s2 2s2 2p6 3s2 3p4.

Now quicker way to do that is to look at that periodic table and

find the noble gas that comes prior to the sulfur, that would be neon.

So we could write neon and

say, everything at this electron configuration is the same as neon.

And then we enter, come across and enter the 3s.

Put in the two electrons, come one across and enter the 2p.

I mean, sorry, the 3p and put in the four electrons.

So another way to write the electron configuration would

be neon in square brackets and then 3s2 3p4.

This is my valence shell.

So if I want to write the quantum numbers of the valence shell,

it's going to be all four of these electrons.

Let's do the s electrons, first.

That would be 3, 0 for the subshell, 0 for the orbital.

One spins in a positive direction and the other one spins in the negative direction.

Okay.

One direction, the other direction.

So that takes care of these two electrons.

Now lets go the the electrons in the p.

First of all, lets think about where those electrons are located.

So I'm going to draw an orbital diagram for these four electrons.

We have the 3p subshell with it's three orbitals.

We put the electrons in as follows one, two,

three, we spread them out first and then we go four.

So when I do the orbital diagrams of these four electrons, they are all in the third

shell, they're all in a p subshell, but they're in three different orbitals.

So I'm going to do them in the order that I wrote them down.

So I'm going to do this one, then this one, then this one.

And then I'm going to come back and to this one.

So the first one there.

It would be a 3 shell, 1 because it's a p subshell.

I'm going to call the first block a negative 1 and

I'm spinning it in the up direction, so I'm going to give it a positive one-half.

The next one, it's in the third shell.

It's still the p subshell, but it's a different orbital.

I'm going to call this one, 0.

It is spinning in the same direction as the first electron I laid down.

The next one is in the third shell, the p subshell, but in a different orbital.

So I'm going to call it plus 1 and it's still spinning in the same direction, so

plus one-half.

Coming back to that first, it should match up to that set of quantum numbers.

It's 3, it's 1, it's minus 1, but it's spinning in the opposite direction.

So there's the four numbers, the four sets of quantum numbers for

the 3p with the two sets of quantum numbers for the 2s and

that would give you all six valence electrons.

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Okay. Let's go to c.

C is vanadium.

So let's find vanadium on our periodic table.

Here it is.

Now we're just going to do the shorthand nom nomenclature for

this one, because I don't want to go.

I'm sorry.

We're going to do the shorthand electron configuration for

this, because I don't want to start at 1s and work my way all the way to that.

Let's find the noble gas that comes just before vanadium.

That would be argon.

So the electron configuration of vanadium would be argon.

Then we come across and we enter the 4s block, that's 4s2, and

then we enter the d block.

And when you count these, you start with number three.

So this is 3d and this is 4d and this is 5d.

So we're in 3d range, 3d and we count over to get to vanadium, one, two, three.

So we, 3d3.

So 4s2 3d3, let's write that down.

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Now, the valence shell is everything past the argon.

It's the highest n in any inner shell that's incompletely filled, so

that would be the 3d.

So I have these five electrons to consider.

For the s, we have the fourth shell,

0 is the subshell, we only have that choice for orbital.

One is spinning in the positive one-half.

The other electron is in the same orbital, but

spinning in the negative one-half direction.

So if two electrons are in the exact same orbital,

they have the same first three numbers, but

they have to spin in different directions that's the Pauli Exclusion Principle.

Now for the d, it's going to be helpful for us to think about those five orbitals.

So I can think about where they're located.

This is a 3d subshell and I only have three electrons in there.

So these three electrons are in different orbitals, but

spinning in the same direction.

So we have a 3 for the n, because of this 3.

When you have a d subshell, that is an l of 2.

My choices for n sub l are a negative 2 up to a positive 2.

I can choose any one I want, but

I'm going to choose it in order from a negative 2 up to a positive 2.

I'm going to spin it in one direction or

the other, it doesn't matter which one I choose.

But generally speaking, when you do an up arrow, you choose the positive one-half.

The next electron is in the same shell and

subshell, but in a different orbital but spinning in the same direction.

The third one is in the same shell and subshell, but a different orbital.

Again, I can choose any of the remaining ones, 0, 1, or

2 here, but I've chosen the 0.

But it has to spin in the positive one-half direction,

these have to be parallel spins.

They have to be spinning in the same direction.

So those are the five electrons of the valence shell of vanadium.

The last one we're going to do here is chromium.

So let's look at periodic table and

let's get the electron configuration of chromium.

Put a little cloud around it here, so we can see it.

We'll find the level gas that comes before chromium and that's argon, so it's argon.

And then we come to across and we're working our way to chromium,

so we do 4s2 and 3d, count over four, 3d4.

Now any time we have a four or a nine sitting right here,

we have to consider that there's an exception to the octet rule,

that this is not exactly what happens.

We're going to promote one of these electrons up to completely half filled,

that sub, 3d subshell.

So that's going to give me instead, argon 4s1 3d5.

So we've maximized the number of parallel spins by putting in a little bit of

energy to promote an electron up to the 3d subshell has higher energy.

But when that happens and you have that maxed out parallel spins,

it actually brings all the energies down and will actually lower energy.

So we're going to go back over here.

We'll write that electron configuration, it was argon 4s1 3d5.

So we know that in this case,

all your electrons are completely spread out and in different orbitals.

So we're going to do the six valence electrons for the chromium.

The first one that's in the 4s subshell would be a 4, a 0,0 and spin it.

Choose a spin and I choose plus one-half.

All other spins will spin in the same direction.

So now, we're going to the 3d subshell.

That's a 3 and a 2 and

then I have a choice of a negative 2 spinning in the positive direction.

A minus 1 spinning in the positive direction.

A 0 spinning in the positive direction.

What's next?

A 1 spinning in the positive direction.

And what's left?

Last one has to be in a different orbital,

it's in the 2 spinning in the positive one-half direction.

So what other variations could I have?

I could have every atom spinning in the negative one-half direction,

it would still be correct.

But otherwise, I have no other choice for this one.