WEBVTT chemistry/physical-chemistry/hovasapian
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Hello, and welcome to www.educator.com, welcome back to Physical Chemistry.
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As I stated in the last lesson, this particular lesson is a continuation of the example that was started in the last lesson which was,
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we are developing term symbols for the electron configuration for carbon 1S2, 2S2, 2P2.
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The P2 electron configuration, we want to find the term symbols for that.
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Let us go ahead and continue that example.
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We started with 21 possible microstates.
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Let me do this in light blue.
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We started with 21 possible microstates.
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And by microstates, we just mean the arrangements of electron in a particular orbital.
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We started with 21 microstates and 6 of these, 6 violated the exclusion principle.
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That parallel spins in the same suborbital, it is not going to happen.
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6 violated the exclusion principle.
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We found that 5 of them were in the D1 level, which is what we wanted.
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We want to find the term symbol representing that particular electron configuration.
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One of the term symbols is D1.
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We have 10 configurations left, we have 10 microstates left, and we need to group those.
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Let us do the same thing.
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I’m going to ask you actually to look back over that particular page of microstates.
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It is a good idea to copy that page, make a screenshot of it so that you can keep referring back and forth to it.
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Looking back at our page of microstates of the 10 remaining, when you actually chose it, when you actually grouped a certain number of them,
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like when we took the 5 and we found that D1, let us go ahead and cross them out, so you do not confuse them with any others.
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Looking back at our page of microstates.
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Of the 10 remaining stases, the next largest value of M sub L is equal to 1.
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And for this value of M sub L, the largest M sub S is also equal to 1.
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What we have is the following.
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The fact that M sub L is equal to 1 it implies that L is equal to 1.
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The possible values of M sub L that are going to be represented are 1, 0, -1.
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M sub S equaling 1 it implies that S is equal to 1.
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The possible values, we will find a better way, M sub S is also 1, 0, and -1.
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L1, L equal to 1, you remember the correlation, when L is equal to 0 that is the S term.
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When L is equal to 1 that is the P term.
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S equal to 1 means that 2S + 1 = 2 × 1 + 1 that is equal to 3.
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Together, we have the P3 term.
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We found a term of the next set of microstates that fall into a given level, it is the P3 level.
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The number of the M sub L × the number of the M sub S is equal to 3 × 3 = 9.
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All those 10 microstates, 9 of those microstates fall into the P3 level.
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We actually do not say 3P, we say triplet P.
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We will talk a little bit more about what we actually call these.
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For example, the 1 D it is actually called singlet D.
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If you were let us say 2P, they would be doublet P.
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In this case, it is 3P or it is triplet P state.
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There are 9 microstates in the triplet P level.
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All those 9 states, they have the same energy.
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When I look back, this is what I find.
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I'm finding all combinations of M sub L and M sub S.
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The M sub L values are 1, 0, -1.
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The M sub S values are 1, 0, -1.
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I’m going to look back and in my microstate, I’m going to find the M sub L, M sub S combination 1 1, 1 0, 1- 1.
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I’m going to find 0 1, 0 0, 0 -1.
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I’m going to find -1 1, -1 0, -1 -1.
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These 9 microstates belong to the triplet P level.
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They are as follows, we have 1, 2, 3, 4, 5, 6, 7, 8, 9.
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And we have 123, 123, 123.
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And we have 123, 123, 123.
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We have up spin up spin, here M sub L = 1 and M sub S = 1.
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That is the 1 1 combination, down spin and up spin.
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M sub L = 1 M sub S =0, that is the 1 -0 combination, down spin down spin.
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M sub L = 1 M sub S = -1, 1 1, 1 0, 1 -1, 1 1, 1 0, 1 -1 that is a microstate, a microstate.
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We have this and this, let me just go ahead and draw them all out.
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That and that, that and that, we have up and up, we have down and up, we have down and down here.
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M sub L = 0 M sub S = 1, M sub L = 0 M sub S = 0, M sub L = 0 and M sub = -1.
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This is the 0 1, 0 0, 0 -1, combination.
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And here we have, M sub L =-1 M sub S = 1, we have M sub L = -1 and M sub S = 0.
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And we have M sub L =-1 and M sub S is equal to -1.
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These 9 belong to the triplet P level, these 9 have the same energy.
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I apologize, it is an old habit of mine to always say 3P, 1D.
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I really should say triplet P, singlet D, doublet, things like that, quadruplet, quintuplet, whatever.
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Just old habits or just for odd reason had not been able to shake things up.
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We have got 9 microstates of the 10 remaining.
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This accounts for 9 of them, that leaves 1 microstate.
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On our page of microstates, there is only one left, its ML value is equal to 0 and it is MS value is equal to 0.
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M sub L = 0 that implies that L is equal to 0, which means that all the possible M sub L values are just 0.
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M sub S = 0 implies that S is equal to 0.
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Those are the ones that we are going to use for term symbol, that implies that all the values for M sub S are also just plain old 0.
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L equaling 0 means the S term.
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S = 0 means 2 × 0 + 1 is equal to 1.
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We have a singlet S level.
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This microstate is this only one left.
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For every value, it is the microstate with ML = 0 MS = 0.
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One value of ML, one value of MS, 1 × 1 is 1.
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This microstate has M sub L = 0 and M sub S = 0 and it is the microstate with the electrons in the Z orbital paired spin.
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We have a singlet D state, we have a triplet P state, and we have a singlet S state.
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15 viable microstates, 15 viable arrangements of 2 electrons in 6 spin orbital arrange themselves in 3 different energies.
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5 of them are here in the singlet D, 9 of them are going to be the triplet P and 1 of them is going to be in the singlet S.
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This is how we derive the term symbols.
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Let us go ahead and finish these.
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Let us finish the term symbols by finding J, the right subscript, the total angular momentum.
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Let us go ahead and do the singlet D state first.
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In this particular case, D means that L is equal to 2.
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The one here, 2S + 1 = 1, 2S = 0, that means that S =0.
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This primary term symbol can actually give you the value of L and S.
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We have used to say configuration to find the term symbol.
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But if you are just given the term symbol, you can get these numbers.
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If L is equal to 2, M sub L is equal to 2, 1, 0, -1, and 2, and M sub S is equal to 0.
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We are going to calculate the M sub J values.
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We do not need to do the M sub J on that page where we are given the microstates, but we did it anyway, it is not a problem.
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M sub J, you are going to take all the values of M sub L.
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Remember, we said M sub J was M sub L + M sub S.
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We are going to add 2 + 0, 1 + 0, 0 + 0, -1 + 0, -2 + 0.
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When we do that, we are going to get the 5 values.
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We add each value of ML with each value of MS to get 2, 1, 0, -1, -2.
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This sequence 2, 1, 0, -1, -2 for the M sub J means this sequence, that the number J itself is equal to 2.
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The singlet D state has a total angular momentum equal to 2.
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The term symbol, the complete term symbol is singlet D2.
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Notice the correlation that we had.
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With the hydrogen atom we had L and we had M sub L which takes on the values from -L to + L passing through 0.
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L, these values of M sub L take on all the values from +L to -L passing through 0.
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S, all the values of M and S run from -S to + S.
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We have J, all the values of M sub J run from -J to +J passing through 0.
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It is that same correlation that goes through all of this.
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Our complete term symbol is singlet D2, it has 5 microstates in that energy level.
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Let us go ahead and take care of the P3 or the triplet P state.
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In this particular case, the P means that the L is equal to 1, the 3 here means 2S + 1 = 3, 2S = 2, S = 1.
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If L is equal to 1 that means the M sub L values are 1, 0, -1.
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S is equal to 1 means that the M sub S values are equal to 1, 0, and -1.
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The M sub J is equal to M sub L + M sub S, I'm going to add each one of these to each one of these.
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1 + 1, 1 + 0, 1 + -1, I get 2, 1, 0.
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Next one, 0 + 1, 0 + 0, 0 + -1, I get 1, 0, -1.
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-1 + 1, -1 + 0, -1 + -1, I get 0, -1, and 2.
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All of these matter, I look in here and I see a sequence of 2, 1, 0, -1, 2.
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There is a sequence 2, 1, 0, -1, 2, for the M sub J.
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That means that J is equal to 2, for one of those.
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We have a triplet P2 state.
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However, that just takes care of that.
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There is also a sequence of 1, 0, -1.
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There is also a sequence of the M sub J =1, 0, -1.
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This implies that J is equal to 1.
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We also have a triplet P1 state.
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Notice the triplet P states actually have more sub states.
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What is left over, there is a N sub J = 0 leftover, that implies that J is equal to 0.
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We also have a triplet P0 state.
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The 9 microstates in the triplet P level are subdivided as follows.
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5 of the microstates are in triplet P2.
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3 microstates in triplet P1 and 1 microstate in triplet P0.
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If we want to know where this 5, 3, and 1 came from, they are actually degeneracy of each individual fully complete term.
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These come from 2J + 1.
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For a complete term symbol, one that actually includes the J value, the degeneracy is 2J + 1.
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In this case, J is equal to 0, 2 × 0 + 1, 1 microstate falling in that term symbol energy level.
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Here J is equal to 1, 2 × 1 + 1 is equal to 3, there are 3 microstates in that particular term symbol energy level.
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Here J is equal to 2, 2 × 2 + 1 is equal to 5.
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There are 5 microstates in the triplet P2 level.
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Our last one, we have our singlet S.
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In this case, L is equal to 0 and S is equal to 0, that means the M sub L is equal to 0, that means M sub S is equal to 0.
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M sub J is equal to M sub L + M sub S = 0 + 0 is equal to 0, that implies that J is equal to 0.
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We have singlet S0, that is the complete term symbol.
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There is actually a quicker way, once you have the initial grouping, the singlet D, the triplet P, singlet S.
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There is a quicker way to find the different values of J for a given basic term symbol.
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The basic term symbol is just the left subscript and it is the 2S + 1 and is the L value, without the J.
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There is a quicker way to find the different values of J for a given basic term symbol.
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The allowed values of J are, you start by adding L + S and then you do L + S -1, L + S -2, L + S -3, and you keep going.
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The lowest number is going to be the absolute value of L – S.
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Let us do the J values.
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For the singlet D state, we have L is equal to 2 and we have S is equal to 0.
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L + S is equal to 2 + 0 it is equal to 2.
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The absolute value of L - S is the absolute value of 2 -0, which is 2.
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2 down to 2, only 2.
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Singlet D2 state, that is how we get that one.
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Let us do the triplet P state, L is equal to 1, S is equal to 1.
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L + S = 1 + 1 is equal to 2.
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The absolute value of L - S = absolute value of 1 -1 which is 0.
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We have 2, 1, 0, that gives us the triplet P2, triplet P1, triplet P0.
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Nice and quick.
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Singlet S state, L = 0 S = 0 L + S 0 + 0 is equal to 0.
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The absolute value of L - S is 0, that means that 0 is the only one.
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We have a singlet S0.
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Let us see, where are we?
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We have the triplet P2, we have the triplet P1, we have the triplet P0, and we have the singlet S0.
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All of those 15 viable microstates are distributed among these.
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And these are the terms symbols that represent the energy of those.
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It is giving us more information than just a normal electron configuration, 1S2 2S2 2P2.
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This is telling us something about the spin angular momentum, the orbital angular momentum, and the total angular momentum.
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All of these have different energies.
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The question is which one was the ground state?
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Which do we call the ground state, the state of lowest energy.
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The one with the lowest energy is the ground state.
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Which level has the lowest energy, that is the question we are asking.
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Which level, which term level has the lowest energy?
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The rules that we use to find that are Hund’s rules.
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They are the empirical rules that decide which is the lowest energy state for a particular term.
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Hund’s empirical rules for specifying the term symbol for the ground electronic state.
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The first rule says this, the state with the largest value of S is the lowest in energy.
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In other words, the most stable.
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And as S decreases, stability decreases.
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Energy goes up, stability decreases.
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For states that have the same value of S like we just saw second ago.
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The triplet P has 3 states, triplet P2, triplet P1, triplet P0.
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How do we decide?
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For states having the same value of S, the state with the largest value of L is lowest in energy.
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In other words, the most stable.
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For states having the same values of S and L, break it down a little bit further.
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We have part A, if the shell is less than half filled show smallest J, smallest J is the lowest energy.
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And B, for more than half filled shell show it is the largest J, it is lowest in energy.
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For our 1S2, 2S2, 2P configuration, we have a singlet D2, triplet P2, a triplet P1, a triplet P0, and we have a singlet S0.
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Hund’s first rule, largest S means the triplet P state.
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The second rule did not tell us anything because the L value is the same for each PPP, L = 1.
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The second one, L was the same for all 3 triplet P states.
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Let me make this 3 a little bit clearer here.
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Our third one, 2P2 it is a less than half filled shell that means less than half filled.
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Therefore, we pick the smallest J between the 2, 1, 0, we choose the 0.
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Less than half filled, the triplet P0 state is the ground state.
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The triplet P0 state, we said that there were 5 in this one, 3 in this one, 1 in this one.
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One particular configuration, that is going to be the ground state.
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That takes care of term symbol but we are not exactly finished discussing it.
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But for this particular example, given a configuration in particular the P2 state,
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we were able to elucidate 5 term symbols, singlet D2, triplet P2, triplet P1, triplet P0, and a singlet S0.
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And using Hund’s rules, we were able to actually find out that the ground state is actually the triplet P0.
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These term symbols give us more information than just the basic electron configuration.
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They tell us about the spin angular momentum.
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They tell us about the orbital angular momentum.
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And it tells us about the total angular momentum.
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That is all, thank you so much for joining us here at www.educator.com.
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We will see you next time for a further discussion of term symbols.
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Take care.