[chrisso80b]

Hi guys,
I dont want to start a topic where I simply ask; What are term symbols? or How do I calculate term symbols? but I've been searching around the internet, looking through all my books, and yet I still can't work out Russell-Saunders term symbols!
Is there an easy way to do it? I've used the ^2s+1L equation thing but that never seems to work. For example, in my book it says that the term symbol for a 3d² free ion is ³F.
My way of working it out is this:
d² means 2 electrons, so s= 1/2 + 1/2= 1
and 2s+1= 3
therefore is triply degenerate yes?
The to work out L. The electrons are in the d shell so l=3 which means L=D
So the term symbol should be ³D
But its not... my book says its ³F. I dont get why its F (and I dont really get why its triply degenerate either because I think im working out S wrong)
Please someone *delete me*
Thanks

[dipesh747]

Angular momenta couple.

The simplest scheme (appropriate for light atoms) is known as Russell-Saunders Coupling
Individual spin momenta couple => S = s1+s2, …, |s1-s2|
Individual orbital angular momenta couple => L = l1+l2, …, |l1-l2|
S and L now combine to give a total angular momentum=> J = L+S, …, |L-S|

The are 2S+1 values of J (L>S)=> the spin multiplicity provides an indication of the                number of J subcomponents

 Term symbols convey information about the configuration of a state

2P3/2 or 3D2

The term symbol gives 3 pieces of information
The letter indicates the total orbital angular quantum number, L.

L         0   1   2   3   4
letter   S   P   D   F   G

The left superscript gives the multiplicity, 2S+1.
The right subscript is the value of the total angular momentum quantum number, J.

Examples:
d2:
l1 = 2, l2 = 2
l1 + l2 = 4, |l1 - l2| = 0             => L = 4, 3, …, 0      => G, F, D, P, S.
p3:
l1 = 1, l2 = 1, l3 = 1
Couple l1 to l2       => l1 + l2 = 2, |l1 - l2| = 0             => L’ = 2, 1, 0 Couple L’ to l3      L’ = 2 =>    L = 2+1, …, 1    = 3, 2, 1      L’ = 1 =>    L = 1+1, …, 0    = 2, 1, 0      L’ = 0 =>    L = 1+0    = 1      =>   1 x F, 2 x D, 3 x P, 1 x S.


Na = [Ne] 3s1
single 3s electron
L = l = 0
S = s = ½
J = ½
   => Term symbol, 2S+1{L}J = 2S½


C = [He] 2s2 2p1 3p1
2p and 3p electrons
l1 + l2  = 2, |l1 - l2| = 0             => L = 2, 1, 0
s1 + s2 = 1, |s1 - s2| = 0      => S = 1, 0
Hence 3D, 1D, 3P, 1P, 3S & 1S
Hence term symbols:
3D: L = 2, S = 1 => J = 3, 2, 1    => 3D3, 3D2, 3D1
1D: L = 2, S = 0 => J = 2      => 1D2
3P: L = 1, S = 1 => J = 2, 1, 0    => 3P2, 3P1, 3P0
1P: L = 1, S = 0 => J = 1      => 1P1
3S: L = 0, S = 1 => J = 1      => 3S1
1S: L = 0, S = 0 => J = 0      => 1S0

[chrisso80b]

Ok thats really helped a lot, but still have a few queries.
Firstly, I now understand that each individual spin momenta couple, and individual orbital angular momenta couple (which I never actually realised) and you have written:
L= l1 + l2, ..., |l1-l2|
Im a little lost with this bit? what does the |l1-l2| mean?

Secondly, for my question about the d² state symbol, I now have this..
there are 2 electrons, each in a d orbital, therefore l1=2 and l2=2
Therefore, as L= l1 + l2, L=4
L= 4 corresponds to the letter G so shouldn't the term symbol be ³G?
Im obviously wrong just cant see how?
Thanks again sorry if im being slow!

[dipesh747]

2p and 3p electrons

l1 + l2  = 2, |l1 - l2| = 0             => L = 2, 1, 0

the bars mean modulus, which means if it is negative make it positive.

So for d2 system

d2:
l1 = 2, l2 = 2 (here you are defining what l1 and l2 are, as they are d they both = 2)
l1 + l2 = 4, |l1 - l2| = 0             => L = 4, 3, 2, 1,

So the letters = 4, 3 ,2, 1 and 0 which = G, F, D, P and S

These will combine with the s1 + s2 = 1, |s1 - s2| = 0      => S = 1,


The easiest way to follow this is if you work through the carbon example I gave you. When you get a bit stuck look at the answers and carry on working through. As soon as you understand that example you will understand this topic.

[chrisso80b]

Ok just spent the day going through it all, and I get it now! thanks so much for the help it was really annoying me!
Cheers