For more information, please see full course syllabus of Physical Chemistry

For more information, please see full course syllabus of Physical Chemistry

### Spin Quantum Number: Term Symbols III

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Spin Quantum Number: Term Symbols III 0:14
- Deriving the Term Symbols for the p² Configuration
- Table: MS vs. ML
- ¹D State
- ³P State
- ¹S State
- J Value
- Degeneracy of the Level
- When Given r Electrons to Assign to n Equivalent Spin Orbitals
- p² Configuration
- Complementary Configurations

### Physical Chemistry Online Course

### Transcription: Spin Quantum Number: Term Symbols III

*Hello, welcome back to www.educator.com, welcome back to Physical Chemistry.*0000

*Today, we are going to round out our discussion of term symbols, let us jump right on in.*0005

*For a change of pace, let us go ahead and do black.*0013

*I think blue was actually nicer.*0025

*In the last two lessons, we derived the term symbols for the P2 configuration by an explicit procedure.*0029

*We basically wrote out by hand every possible electron configuration with the lines and with the arrows.*0074

*We are going to go through the same example, except a little bit quicker.*0082

*We are simply to be doing the same thing, except we would use *0086

*a shorthand notation for all those configuration instead of drawing them all out.*0089

*We now go through the same example in a quicker way.*0094

*Now that we know how to handle these arrangements, how to handle the various microstates,*0122

*this is the general procedure for any electron configuration.*0143

*There were 2 electrons right, we have a P2 configuration.*0148

*There are 2 electrons, each is in a P orbital, l = 1.*0154

*For each electron, l = 1.*0172

*The sum of the individual L sub I, the L for each electron, *0178

*in this particular case I have 2 electrons and they are both in the P orbital.*0185

*L = 1 for 1 electron, L = 1 for the other electron.*0190

*I'm going to get 1 + 1 = 2, this is going to equal L.*0194

*M sub L is going to equal to 1, 0, -1, 2.*0201

*The sum of the S sub I, the individual spin angular momentum, we have ½ + ½ that = 1.*0209

*That = S, the M sub S = 1, 0, and -1.*0223

*We have about a series of values for M sub L.*0231

*We have a series of values for M sub S.*0234

*We are going to create a table with M sub S along the top row and all the values of M sub L down the left column.*0238

*I did not have to be like this, you can put the M sub S in the columns and the M sub L along the row.*0265

*The order does not matter, you are creating a matrix, you are creating a table.*0271

*You are going to end up with something that looks like the following.*0275

*I'm going to do 1, 0, -1.*0283

*I will give myself a little bit of room here 1, 0, -1 that is the M sub S.*0288

*I have 2, I have 1, I have 0, -1, -2.*0296

*Those are the values of the M sub L, something like this.*0301

*I have M sub S on the top row and the different values of the M sub L along the bottom row.*0306

*And I got those just by adding up the L for each individual electron configuration, getting my L value.*0312

*And from that, finding the values of M sub L then adding my S values, getting my total S,*0320

*and then getting my different values for the M sub S.*0327

*What I'm going to do is the entries of this table, the entries are going to be the microstates, *0332

*such that the sum of the individual M sub L = the M sub L in that particular row.*0353

*The sum of the individual M sub S is going to equal the M sub S in that column.*0373

*Let us go ahead and see what we mean by that.*0390

*We have 2 electrons, in this particular case we need the two M sub L values.*0392

*The m sub l PX PZ PY.*0399

*M sub L here = 1, M sub L = 0, M sub L = -1.*0410

*We are going to add the individual values to come up with, this particular entry here is what it is actually look like.*0419

*It is going to be 1 and 1.*0429

*M sub L 1 is, what I'm saying is I can put both electrons into this, the P sub X orbital.*0432

*I can give each one of them a positive spin.*0440

*Notice the 1 + 1 = 2, the positive and positive or the positive means positive ½ positive ½ = 1.*0444

*If I want I can put ½ there, I just decided to use a positive and negative signs.*0456

*It is pretty standard to use the positive and negative signs, that is all it means.*0460

*You are going to add the individual M sub L values for where the electron can go, *0465

*to add up to this number, that is the numbers.*0470

*In these things, you are going to add up the individual M sub S says, the spins, the + or -1/2 to get that number.*0474

*1 + 1 is 2, positive and positive, ½ + ½ is 1.*0482

*Over here, I'm going to have 1 and 1, and 1 and 1, because 1 + 1 they have to equal this number.*0486

*As far as the spins, I can go positive and negative.*0496

*That is if I did 1 - and 1 +, it is actually the same thing, because these are equivalent orbitals, they are all in the P orbital.*0504

*It does not matter whether if you just switch the order,*0511

*in other words put this 1 - over here and 1 + over here, you have not really changed anything.*0516

*It is the same thing, you do not have to double that up.*0521

*Over here, it is going to be 1 - and 1 -.*0524

*I’m going to add the individual M sub L values, the m sub l values to add up to the M sub L values in the row and column.*0527

*Over here, we are going to have 1, 0.*0538

*This is going to have to be + +.*0542

*We are just doing a shorthand notation for the microstates.*0545

*In other words, I’m saying I’m going to put 1 electron with a positive spin here and*0550

*I’m going to put in other electron with a positive spin here in the M sub L = 1, the M sub L = 0.*0556

*Positive spin positive spin that is a + +, these numbers add to that number.*0567

*These numbers add up to this number.*0572

*Over here, we are going to just go ahead and fill them in.*0575

*We have 1 + 0 -, we have 1 -0 +, I will explain all of these in just a minute.*0579

*1 -1 - and I have got 0 + 0 +, I have got 1 + -1 +, I have 1 + -1 -, I have got 1 - -1 +, and I also have 0 + 0 -.*0589

*Over here, I have 0 - 0 -, and I have 1 - -1 -.*0620

*Here I have got 0, -1 = -1, I get a + + gives me that, + ½ + ½ give us what the + means.*0630

*This means + ½ or - 1/2.*0641

*Over here, I have 0 + -1 - and I have got 0 - -1 +, over here I have got 0 - -1-, I got -1 + -1 +.*0645

*I got -1 + -1 -, I got -1 - -1 -.*0665

*Once again, the entries are going to be in the microstates such that the sum of the m sub l, *0675

*in this particular case is going to be 1, 0, -1 add up to that particular M sub L in that row.*0681

*And the sum of the individual spin values, the M sub S, the spin quantum numbers + ½ -1/2, *0690

*they are going to equal the particular value of the M sub S in that particular column.*0698

*That is all I’m doing, these are all the microstates.*0704

*This is a shorthand notation for what we did explicitly, when we drew out every single one in the previous lessons.*0708

*Let us see, this table is just a shorthand for each microstate.*0719

*Perhaps, it is not quicker.*0736

*The truth this when you write out all these microstates, you do have to be careful.*0738

*This is what actually is going to take the most time, writing out each microstate.*0742

*For each microstate, we did explicitly before.*0746

*A couple of things I want to talk about.*0758

*Let us take a look at the second row, second column, this right here.*0761

*The other ones that have multiple entries.*0767

*Here is where I have multiple entries.*0771

*In this particular case, we need the angular momentum, the M sub L to equal 1.*0774

*It can be 1, 0, this 1, 0.*0780

*This is 1 positive spin, 0 negative spin, that corresponds to this 1 positive spin and 0 negative spin.*0783

*This one over here is this one.*0797

*1 negative spin 0 suborbital positive spin, these are not the same.*0800

*If I switch the order, if I just said 1 + 0 - and 0 -1 +, if I just switched the order that would be the same.*0807

*Those I do not have to write but in this particular case, here the spin is up, here the spin is down.*0816

*Here the spin is down, here the spin is up.*0825

*I have to count for every single possibility and that is what is important.*0828

*I cannot just switch the order.*0833

*I hope that makes sense.*0840

*Notice that there are 21 microstates.*0848

*If you count all of these, we are going to have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, *0850

*11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 21 microstates just like we have before.*0856

*The general procedure, the first thing we did was cross out those microstates that violate the exclusion principle.*0862

*Personally, I myself I like to actually draw these out.*0872

*I do not particularly like this particular procedure, for some odd reason it tends to be a little harder.*0875

*I do not like shorthand, I never did.*0881

*I like to see my electrons, my orbitals, and everything.*0884

*When you are counting a bunch of microstates, you might get 15 of them, 20 of them, 25 of them, *0889

*30 of them, especially when you get into higher electron configurations.*0896

*Fortunately, the problems that you deal with, you are not going to be asked *0900

*to find something for like the D 7 state, simply because it is just there to many microstates to deal with.*0902

*But this is what we are doing here.*0911

*We want to cross out the ones that violate the exclusion principle.*0915

*In other words, same orbital, same parallel spin.*0918

*1 + 1 + is not going to work.*0924

*0 + 0 + is not going to work.*0928

*-1 + -1 + no, 1 -1 – no, 0 -0 – no, -1 - -1 – no, these violates the exclusion principle.*0931

*They have 2 electrons of the same spin in the same orbital, that is not going to work.*0944

*We have crossed all 6 of them, like we did before.*0949

*We are left with 15, 15 viable microstates.*0952

*We take the largest value of M sub L having a viable microstate.*0956

*Let me do this in red.*0964

*Here we have a viable microstate here, the largest value of M sub L = 2.*0966

*For that value of 2, the largest value of M sub S is 0.*0973

*Let me go to the next.*0980

*The largest value of M sub L having a viable microstate is M sub L = 2.*0989

*For this M sub L = 2, M sub S is actually equal to 0.*1014

*M sub L = 2 implies that L = 2.*1022

*M sub S = 0 implies that S = 0.*1029

*Both of these give the singlet D state.*1035

*L = 2 that is D.*1038

*S = 0, 2 × 0 + 1 is 1.*1040

*We find the D1 state.*1044

*L = 2 implies that M sub L = 2, 1, 0, -1, -2.*1051

*S = 0 implies that M sub S = just 0.*1061

*For every value of M sub L M sub S, 2 0, 1 0, 0 0, -1 0, -2 0, I'm going to pick a microstate from that column.*1072

*I’m going to pick a microstate having these values.*1087

*In the column for M sub S = 0, choose one microstate for each value of M sub L and cross it out.*1095

*Those that you cross out, those are the ones that belong to the singlet D state.*1125

*The same thing that we did in the previous example.*1130

*The first lesson where we discussed term symbols.*1132

*Cross it out, these belong to the singlet D state.*1138

*What you are left with when you have crossed them out is the following.*1148

*What you have left is, I can do it here 1, 0, -1, 2, 1, 0, -1, -2.*1159

*We have M sub S here, M sub L here, what we are left with is this.*1179

*You have a 1 + 0 +, you have a 1 -0 +, when you choose one, notice in the column where the MS was equal to 0.*1192

*In the previous table, there are some entries.*1205

*The second row has 2 entries, the third row has 3 entries, it does not matter which one you choose.*1208

*Just choose one, but only one to cross out.*1213

*What you are left with when you cross out certain ones are 1 -0 -.*1217

*You have got 1 + -1 +, 1 - -1 +.*1224

*You got 0 + 0 -, 0 -0 - and the -1 row you have 0 + -1 + you have 0 - -1 +, then you have 0 - -1 -.*1235

*The largest value of M sub L has a viable microstate is 1.*1255

*And the largest value of M sub S that has a viable microstates is 1.*1261

*Now, all of them are viable.*1269

*The largest M sub L with a viable microstate is M sub L = 1, that implies that L = 1.*1281

*For this M sub L, the largest M sub S = 1.*1302

*That implies that S = 1.*1314

*L = 1, that is P.*1319

*S = 1, 2 × 1 + 1, we have a triplet P state.*1322

*L = 1 means that N sub L = 1, 0, -1.*1332

*S equal to 1 implies that M sub S = 1, 0, -1.*1341

*For each value of M sub S, we choose a value of M sub L in that table.*1349

*Choose it, it is going to belong to the triplet P state, we cross that out.*1357

*For M sub L 1 M sub S 1, 1 0, 1 -1.*1364

*For 0 1, 0 0, 0 -1, -1 1, -1 0, -1 -1, we choose a microstate in that row and column.*1368

*For each value of M sub S choose a microstate, the table that we just had in the previous page.*1382

*Choose a microstate for each value of M sub L.*1400

*There are 9 of them, of course there 9 of them.*1410

*There are 3 here and 3 here.*1412

*3 × 3 is 9.*1414

*There are 9 of them, cross them out.*1417

*If there is column called multiple entries, just choose one.*1427

*It does not matter which, cross these out because they belong to the triplet P state.*1430

*What is left is the following.*1443

*What is left is 1, 0, -1, 2, 1, 0, -1, -2 M sub S M sub L, what we are left with is the 0 0 state which is going to be 0 + 0 -.*1452

*It is going to be in a 0 row, it is going to be in the 0 column.*1475

*The largest value of M sub L is 0.*1483

*The largest M sub L = 0 which implies that L = 0.*1490

*For this ML = 0, the largest M sub S = 0.*1506

*It implies that S = 0.*1513

*These together L = 0 is the S term.*1516

*S= 0, 2 × 0 + 1 is 1.*1519

*That gives us our singlet S state.*1521

*This last microstate belongs to the singlet S state.*1528

*We find the J values.*1533

*For the J values, these are really quick.*1539

*J values, what we are left is the basic term symbol like singlet S, triplet P, quadruplet D, quadruplet F, whatever.*1544

*The J values are easy to find.*1551

*For singlet D, L = 2.*1554

*S = 0.*1558

*D L =2, 2S + 1 = 1 that means that S = 0.*1560

*That means that L + S = 2.*1566

*The absolute value of L - S = 2.*1572

*J starts at 2 all the way down to 2.*1578

*J = only 2, we have a singlet D2 state.*1584

*For the triplet P state, L = 1.*1592

*S = 1, L + S = 2.*1596

*The absolute value of L - S = 1 -1 which is 0.*1602

*J has the value of 2 all the way down to 0, 2, 1, 0.*1608

*We have triplet P2, we have triplet P1, and we have triplet P0.*1614

*For singlet S, L = 0, S = 0.*1627

*L + S = 0, the absolute value of L - S = 0.*1634

*Therefore, J = 0 and we are left with singlet S0 state.*1639

*There you go.*1646

*Let us go ahead and write.*1649

*For each complete term symbol which is L 2S + 1 sub J, the degeneracy at each level.*1653

*In other words, the number of microstates that have the same energy.*1680

*They belong to that particular term symbol.*1686

*The degeneracy of the level is 2J + 1.*1689

*In the case of the singlet D2 state, we have 2 × 2 + 1.*1706

*We have 5 microstates belonging to the singlet D2 state.*1714

*5 microstates have that energy, whatever the energy happens to be.*1721

*For the triplet P2 state, we have 2 × 2 + 1 = 5 microstates in the triplet P2 state.*1727

*P31 J = 1, 2 × 1 + 1 = 3.*1738

*There are 3 microstates in the triplet P1 state.*1750

*And for the triplet P0 state, we have 2 × 0 + 1, there is 1 microstate.*1754

*For the singlet S0 state, 2 × 0 + 1 there is 1 microstate.*1762

*5, 10, 13, 14, 15, a total of 15 viable microstates divided among all of these particular energy levels.*1771

*On Hund’s rule, we found that the ground state was this one.*1782

*When we talk about 1S2, 2S2, 2P2 configuration, we are talking about the term symbol *1788

*that gives a total spin angular momentum.*1796

*The total spin angular momentum S = 1 L =1, the total angular momentum is going to be 0.*1799

*That is what is going on here.*1816

*In general, if you are given R electrons to assign to N equivalent spin orbitals, *1819

*when we say equivalent it means those belonging to the same sub level.*1846

*P, 2P, equivalent spin orbital, that means the same L value.*1857

*In general, given R electrons to assign to N equivalent spin orbitals, the same L value, there are N choose R viable microstates.*1877

*Let me think for a second here.*1925

*In general, given R electrons to assign to N equivalent spin orbital which means the same L value, *1938

*there are N choose R viable microstates.*1942

*On your calculator, you will also see this symbol NCR.*1947

*Remember the number of combinations, the number of ways of assigning R things to impossible.*1952

*The definition is N!/ R! × N – R!. *1959

*In the P2 configuration, we had 2 electrons to distribute among 6 spin orbitals that means 6 choose 2 and 6!/ 2! × 4!.*1973

*6 × 5 × 4 × 3 × 2 × 1/ 2 × 1 × 4 × 3 × 2 × 1.*2015

*4321, 4321, 6 × 5 is 30, 30 divided by 2 = 15.*2025

*15 viable microstates.*2031

*I’m not sure if this number actually is altogether that important to me.*2033

*The fact of the matter is the best way to approach this, even though it is longer,*2035

*it is just to actually write out every single microstate that is available.*2040

*Notice that you can see every microstate that is available.*2043

*In case, the question is just how many viable microstates are there for this configuration.*2047

*You can just do 6 choose 2, something like that.*2053

*Viable microstates, notice that we actually had 21 microstates.*2059

*The 6 who violated the exclusion principle.*2065

*This N choose R is binomial coefficient, this combination coefficient, *2068

*and combinatorial coefficient only talks about the viable options, the viable microstates.*2075

*We already talked about how we refer to these levels.*2084

*In the case of 1D2, this is a singlet D2, in the case of 3P2 this is triplet P2.*2087

*This is triplet P1, triplet P0, and singlet S0.*2098

*The 4 would be quadruplet, 5 would be quintuplet, and so on.*2105

*One final thing before we let you go.*2113

*I will write it as some things to know.*2116

*The following configurations are called complimentary.*2125

*P1 P5, P2 P4, D1 D9, D2 D8, D3 D7, D4 D6, these are called complimentary.*2137

*They are called complimentary because P1 P5, 5 + 1 is 6, 2 + 4 is 6.*2177

*In the case of the D orbits, we have 10 electrons in there.*2190

*9 + 1 is 10, 2 8, 3 7.*2194

*The P1 configuration, if you are to work out the P5 configuration, the term symbols for that is exactly the same.*2197

*The P2 configuration, if you would work out the P4 configuration, you get exactly what we got.*2204

*You get the 1D, 3P, 3P1, S0.*2208

*Complementary configurations are exactly the same.*2213

*If you are asked to find a configuration for let us say D9, do not worry about it.*2216

*Just go ahead and find the configuration for D1, it will be the same.*2220

*That saves a lot of work, believe me.*2224

*You do not want to distribute 9 electrons in the D orbital.*2226

*That is why I choose the D orbitals and assign 1 electron to those D orbitals and the term symbols will be the same.*2230

*This is true in general, complimentary configurations have the same term symbols.*2243

*Thank you so much for joining us here at www.educator.com.*2280

*We will see you next time.*2282

## Start Learning Now

Our free lessons will get you started (Adobe Flash

Sign up for Educator.com^{®}required).Get immediate access to our entire library.

## Membership Overview

Unlimited access to our entire library of courses.Learn at your own pace... anytime, anywhere!