[MrHappy0]

I'm not fully understanding how to assign term symbols to a certain electronic configuration. Most resources seem to be heavy on the physics/math so it's hard to decipher.

For example the ground state of hydrogen is ²S_1/2 because 2S+1=2(1/2)+1, L=S, and J=L+S=0+(1/2) (Does S always equal n/2 where n=number of electrons?).

But now for hydrogen excited config (3d) it is ²D_1/2. Everything makes sense for this except for the the J. Why isn't J=L+S=3+(1/2)=7/2?

EDIT: using this site as a reference to what I'm doing here --> http://www.chem.ufl.edu/~itl/4412/lectures/ATermSym.html

[Corribus]

First, in case you didn't know, there are multiple terms (states) for each electron configuration. The 3d¹ configruation of hydrogen has 2 of them: ²D_5/2 and ²D_3/2.  J values can take on L+S, L+S-1, L+S-2...|L-S|. Here, L = 2 and S = 1/2, therefore J = 5/2 or 3/2. The reason the J = 3/2 state is the lowest energy is because the outermost subshell is less than half filled (Hund's Rules).

These questions should be in the physical chemistry forum by the way.

[MrHappy0]

Okay, I am studying for an Inorganic final but I will post there instead. I'm not really sure if this stuff is even necessary for what we studied. I'm trying to understand the Selection rules for UV/Vis spectroscopy (the previous post) and went back to states because I thought it was necessary to understand selection rules. Is it?

Now back to the question: Why does 3d¹ have 2? If it only has one electron how can it have 2 states? And I don't really even understand what S is. I just know that one electron gets 1/2. And if J=L+S, L+S-1, L+S-2....|L-S| why wouldn't J be 5/2, 3/2, 1/2?

[Corribus]

Okay, I am studying for an Inorganic final but I will post there instead. I'm not really sure if this stuff is even necessary for what we studied. I'm trying to understand the Selection rules for UV/Vis spectroscopy (the previous post) and went back to states because I thought it was necessary to understand selection rules. Is it?

Term symbols can help you quickly identify which electronic transitions are allowed, so yes it is very relevant.  You can google "Term Symbols and Selection Rules" to find some links about this.  If you have specific questions, I can help.

Now back to the question: Why does 3d¹ have 2? If it only has one electron how can it have 2 states? And I don't really even understand what S is. I just know that one electron gets 1/2. And if J=L+S, L+S-1, L+S-2....|L-S| why wouldn't J be 5/2, 3/2, 1/2?

There are two states because of spin-orbit coupling: there are two ways that the orbital and spin angular momenta of the single electron can interact.  In one state, they are interacting constructively (in the same dirrection, additive); in the other, they are interacting destructively. This is the point of the J quantum number.

[MrHappy0]

Each search seems to confuse me more. I need to answer this question below. I have the answer but I'm confused how to get to them. For example why is it ³A_2g and not just A_2g. What does the 3 indicate?

Rank the intensity following electronic transitions from least intense to most intense:
i. ³A_2g → ¹A_1g for [Ni(H2O)6]²⁺
ii. ³T₁ → ³T₂ for [Ni(H2O)4]²⁺
iii. ³A_2g → ³T_1g for [Ni(H2O)6]²⁺
iv. ³A_2g → ¹A1_g for [Pt(H2O)6]²⁺

Okay, I am studying for an Inorganic final but I will post there instead. I'm not really sure if this stuff is even necessary for what we studied. I'm trying to understand the Selection rules for UV/Vis spectroscopy (the previous post) and went back to states because I thought it was necessary to understand selection rules. Is it?

Term symbols can help you quickly identify which electronic transitions are allowed, so yes it is very relevant.  You can google "Term Symbols and Selection Rules" to find some links about this.  If you have specific questions, I can help.

Now back to the question: Why does 3d¹ have 2? If it only has one electron how can it have 2 states? And I don't really even understand what S is. I just know that one electron gets 1/2. And if J=L+S, L+S-1, L+S-2....|L-S| why wouldn't J be 5/2, 3/2, 1/2?

There are two states because of spin-orbit coupling: there are two ways that the orbital and spin angular momenta of the single electron can interact.  In one state, they are interacting constructively (in the same dirrection, additive); in the other, they are interacting destructively. This is the point of the J quantum number.

Okay but how do you determine how many interactions there are? The formula for J seems to indicate there are more than 2. Say you have 3 electrons in a p-orbital.

[MrHappy0]

Bare with me, Corribus but I think if I can get through some of the details that are confusing me I will get this. So I attached an answer key and look at the microstates for [Ar]3d⁸.  I almost fully see the pattern of how this chart was filled out. Here are the things that are confusing me: Why when tabling the microstates in the top row where it says M_S +1,0,-1 when on a different document for the nitrogen the table has +3/2,+1/2,-1/2,-3/2 for the M_S values? My second question is, why does it then remove ¹G then ³F then ¹D, then ³P then ¹S? Final question, what do the numbers actually imply for these terms, e.g. what does the 3 actually imply in ³F?

[Corribus]

MrHappy, I will be happy to answer these questions, but I must delay until tomorrow morning. Please check back in about 10-12 hours.

[Corribus]

OK, here we go:

Each search seems to confuse me more. I need to answer this question below. I have the answer but I'm confused how to get to them. For example why is it 3A2g and not just A2g. What does the 3 indicate?

The 3 represents the spin multiplicity of the state, given by the simple formula M = 2S + 1, where M is the multiplicity and S is the total spin angular momentum of all the electrons in the system.  You might wonder why we use multiplicity rather than just S.  The reason is because the multiplicity not only includes the total spin but also the degeneracy of the state.  A triplet (M = 3) state actually is three degenerate states. Degenerate means they all have the same energy, so are virtually indistinguishable in a normal laboratory frame.  However, in a magnetic field, the states will split, with one having slightly higher energy, one having slightly lower, and one unchanged. You can actually observe this in a spectroscopic experiment and it's called the Zeeman effect.

Let's take an example. Let's say you have a system with two electrons in which they aren't both in the same orbital. The spin of each electron is ± 1/2. Since you have two electrons, they can either be both directed up (positive Z-axis), both directed down (negative Z-axis), or one up one down.  In the third case it can be down-up or up-down - both combinations give us a total spin of zero. This gives us a total of 4 possible combinations, which we call microstates. The two-up combination gives you a total spin of +1 and the two-down combination gives you a total spin of -1. There's a weird quantum mechanical thing now that happens with the up-down and down-up combinations: we actually take linear combinations of them, so we call one microstate

2^-1/2(down-up + up-down)

and the other

2^-1/2(down-up - up-down)

This is called taking linear combinations of the two microstates.  The 2^-1/2 factor ensures the wavefunctions remain normalized.

Anyway, point is that when you have four spin microstates.  We can group them like so:

|++>     
|+-> + |-+>
|-->

and

|+-> - |-+>

Here I'm using Dirac notation because it's easier.  If you don't know this notation, don't worry too much. It just means that the "plus" combination of the up-down/down-up state is grouped with the up-up and down-down combinations and the "minus" combination is grouped alone. (A common question that comes up is why the "plus" combination is grouped with the ++ and -- microstates and the "minus" combination grouped by itself; there are some complicated mathematical reasons which are probably better off left ignored here.)

It shouldn't be too hard to see that the upper grouping contains three microstates and the lower grouping contains one. The upper grouping is called the triplet state, the lower grouping the singlet.  The upper grouping contains microstates with spin quantum numbers (M_S) of +1, 0 and -1, for a total spin quantum (S) of 1 and a multiplicity of 3: hence the triplet.  The lower grouping contains microstates with spin quantum number (M_S) of 0, for a total spin (S) of 0 and a multiplicity of 1: hence the singlet.

Collectively these represent two unique states with different energies, which is why the multiplicity is included in term symbol. You should be able to see now that the triplet "state" actually includes three degenerate states. In the absence of a magnetic field, a system with the appropriate energy can be in any one of these three microstates at a given time. In a magnetic field, the degeneracy is broken. This is because the magnetic field will have a vector and depending on the alignment of the system with the field, the up-up state will have a different interaction energy with the field than the down-down state. 

Do note that, all other things being equal, the triplet state is lower in energy than the singlet state. Also, any time you have a triplet state, you also have a singlet state associated with it. The converse is not, however, true. A singlet state arising from a closed shell does not have the possibility of a triplet state, because two electrons cannot be in the same orbital with the same spin (Pauli exclusion principle).

Rank the intensity following electronic transitions from least intense to most intense:
i. 3A2g → 1A1g for [Ni(H2O)6]2+
ii. 3T1 → 3T2 for [Ni(H2O)4]2+
iii. 3A2g → 3T1g for [Ni(H2O)6]2+
iv. 3A2g → 1A1g for [Pt(H2O)6]2+

What are the selection rules?

Okay but how do you determine how many interactions there are? The formula for J seems to indicate there are more than 2. Say you have 3 electrons in a p-orbital.

The number of interactions essentially depends on the number of possible spin states. Let's go back to the triplet state again. Here S = 1, corresponding to three microstates with spin angular momenta of +1 (both electrons up), 0 (one up and one down), and -1 (both electrons down). Now the electrons are also in orbitals with associated angular momenta L.  Basically, the oribtal angular momenta create their own electric fields which can interact positively or negatively with the spins.  So the L interacts with the positive spin microstate to a different degree than it does with the negative spin microstate and the zero-spin microstate.  We define a quantum number J to represent this interaction; the interaction is simply vectoral and we really only care about one direction; thus it is simply additive between the the L and S quantum numbers. In the case of a triplet, therefore, we have J values of L + (S = 1), L + (S = 0), and L + (S = -1). More generally we can define total momentum J as equal to values ranging from L + S to L + S - 1 to ... |L-S|. 

(E.G.: If S = 2, a doublet state, then M_S = ± 1/2 and J is equal to L + (S = 1/2) and L + (S = -1/2).  If S = 1, a singlet state, then M_S = 0 and J = L.  If S = 4 (quartet), as in the case of three unpaired electrons, then M_S = 1.5, 0.5, -0.5, -1.5 and J is equal to L + (S = 3/2), L + (S = 1/2), L + (S = -1/2) or L + (S = -3/2).)

Bare with me, Corribus but I think if I can get through some of the details that are confusing me I will get this. So I attached an answer key and look at the microstates for [Ar]3d8.  I almost fully see the pattern of how this chart was filled out. Here are the things that are confusing me: Why when tabling the microstates in the top row where it says MS +1,0,-1 when on a different document for the nitrogen the table has +3/2,+1/2,-1/2,-3/2 for the MS values?

If you have eight electrons distributed among five d-orbitals, six of the electrons must be paired.  This leaves two unpaired electrons, which can both be up (M_S = +1), both be down (M_S = -1), one-up-one-down (M_S = 0), or one-down-one-up (M_S = 0). This is essentially the same situation we discussed above, almost identical to a p² configuration. The difference here is that the total orbital angular moment of the d-orbitals is different, so the term symbols will be different two, even if the spin components are the same.  In nitrogen, the configuration is p³ - three electrons in the p-orbitals. Here there are no pairing restrictions because the three available p-orbitals can easily accomodate three electrons (one in each orbital).  Therefore we have possibilities of all three up (M_S = 1.5), all three down (M_S = -1.5) and a bunch where some are up and some are down (which must either have M_S values of +0.5 or -0.5. One of the results is a quartet, but you'll also probably have some doublets in there as well, just as you have a triplet and singlet for a 2-unpaired electron configuration.

My second question is, why does it then remove 1G then 3F then 1D, then 3P then 1S?

There's no strict rules about the way you do this. If you consult five books, you'll probably see five different ways to tabulate all the possible microstates and then pull out the states.  In most cases it is easier to pull out the states in decreasing order of "bigness" from the table, and the higher the total orbital momentum, the more microstates are involved.  You can pull out smaller ones first, but I think you'll be more likely to make an error that way.

Final question, what do the numbers actually imply for these terms, e.g. what does the 3 actually imply in 3F?

I've already discussed above what the preceding superscript means.  In this case ³F means that it is a triplet state (S = 1) which has a total orbtial angular momentum (L) of 3.  It should be obvious from those tables why the S value is 1: because there are M_S values of +1, 0 and -1 that range over this group where M_L ranges from 3 to -3.

If you have more questions, let me know.