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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

Transcription: Spin Quantum Number: Term Symbols III

Hello, welcome back to, 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 use0086

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 and0550

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 asked0900

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 symbol1788

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

We will see you next time.2282