For more information, please see full course syllabus of Physical Chemistry
For more information, please see full course syllabus of Physical Chemistry
Discussion
Download Lecture Slides
Table of Contents
Transcription
Related Books
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 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 www.educator.com.2280
We will see you next time.2282
Start Learning Now
Our free lessons will get you started (Adobe Flash® required).
Sign up for Educator.comGet immediate access to our entire library.
Membership Overview