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.