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For this problem,

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

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

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

Â Let's begin with boron.

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

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

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

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

Â then we come over here to 2p and 1.

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

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

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

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

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

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

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

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

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

Â we have a 2 for each one of those.

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

Â s gives us an l equal to 0.

Â A p gives us an l equal to 1.

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

Â They are both n the same order, orbital.

Â One spins in the positive one-half direction and

Â the other one in its opposite direction.

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

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

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

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

Â can put next.

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

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

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

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

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

Â So that's a.

Â Let's go to b.

Â B is sulfur.

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

Â Let's find sulfur.

Â 2:53

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

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

Â Coming across, we enter the 2p block.

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

Â Then we enter the three, that's 3s.

Â There's two.

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

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

Â So let's write that out.

Â 1s2 2s2 2p6 3s2 3p4.

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

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

Â So we could write neon and

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

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

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

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

Â So another way to write the electron configuration would

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

Â This is my valence shell.

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

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

Â Let's do the s electrons, first.

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

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

Â Okay.

Â One direction, the other direction.

Â So that takes care of these two electrons.

Â Now lets go the the electrons in the p.

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

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

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

Â We put the electrons in as follows one, two,

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

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

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

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

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

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

Â So the first one there.

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

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

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

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

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

Â I'm going to call this one, 0.

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

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

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

Â plus one-half.

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

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

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

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

Â that would give you all six valence electrons.

Â 6:28

Okay. Let's go to c.

Â C is vanadium.

Â So let's find vanadium on our periodic table.

Â Here it is.

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

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

Â I'm sorry.

Â We're going to do the shorthand electron configuration for

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

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

Â That would be argon.

Â So the electron configuration of vanadium would be argon.

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

Â then we enter the d block.

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

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

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

Â So we, 3d3.

Â So 4s2 3d3, let's write that down.

Â 7:33

Now, the valence shell is everything past the argon.

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

Â that would be the 3d.

Â So I have these five electrons to consider.

Â For the s, we have the fourth shell,

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

Â One is spinning in the positive one-half.

Â The other electron is in the same orbital, but

Â spinning in the negative one-half direction.

Â So if two electrons are in the exact same orbital,

Â they have the same first three numbers, but

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

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

Â So I can think about where they're located.

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

Â So these three electrons are in different orbitals, but

Â spinning in the same direction.

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

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

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

Â I can choose any one I want, but

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

Â I'm going to spin it in one direction or

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

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

Â The next electron is in the same shell and

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

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

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

Â 2 here, but I've chosen the 0.

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

Â these have to be parallel spins.

Â They have to be spinning in the same direction.

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

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

Â So let's look at periodic table and

Â let's get the electron configuration of chromium.

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

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

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

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

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

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

Â that this is not exactly what happens.

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

Â that sub, 3d subshell.

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

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

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

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

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

Â So we're going to go back over here.

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

Â So we know that in this case,

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

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

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

Â Choose a spin and I choose plus one-half.

Â All other spins will spin in the same direction.

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

Â That's a 3 and a 2 and

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

Â A minus 1 spinning in the positive direction.

Â A 0 spinning in the positive direction.

Â What's next?

Â A 1 spinning in the positive direction.

Â And what's left?

Â Last one has to be in a different orbital,

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

Â So what other variations could I have?

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

Â it would still be correct.

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

Â