# Term symbols for nitrogen: determining J

I'm trying to figure out what terms are possible for nitrogen with the electron configuration $$\ce{[He] 2s^2 2p^3}$$. There is an old question on StackExchange and the corresponding answer was a great help already but doesn't mention $$J$$ or specify what terms exactly are possible. I'll try to explain how I would determine the possible terms:

The s-electrons needn't be considered for they complete the s subshell, this means I'm left with three p-electrons. I do get the same results for the possible values of $$L$$ and $$S$$ as user orthocresol provided with his answer namely:

$$L=3,2,1,0$$

$$S=3/2, 1/2$$

Now the possible values of J range from $$L+S$$ to $$|L-S|$$:

For $$L=0$$ there are two S-Terms*: $$^4\!S$$ and $$^2\!S$$ giving

$$^4\!S_{3/2} \quad ^2\!S_{1/2}$$

$$L=1$$ gives two terms*: $$^4\!P$$ and $$^2\!P$$, the former term has the following possible J values: $$J = 5/2, 3/2, 1/2$$ and the latter: $$J=3/2, 1/2$$. The P-Terms of this electron configuration are:

$$^4\!P_{5/2} \quad ^4\!P_{3/2} \quad ^4\!P_{1/2} \quad ^2\!P_{3/2}\quad ^2\!P_{1/2}$$

Doing the same for $$L=2$$ there are two D-Terms*: $$^4\!D$$ and $$^2\!D$$, possible J values for the former: $$J = 7/2, 5/2, 3/2, 1/2$$ and for the latter: $$J = 5/2, 3/2$$. This gives six different terms.

I'm stopping at this point because I think I'm not doing it right. Using the NIST ASD I only see five terms with this electron configuration in the Grotrian diagram:

$$^4\!S_{3/2}, ^2\!D_{5/2}, ^2\!D_{3/2}, ^2\!P_{1/2}, ^2\!P_{3/2}$$

and according to the book "Physical Chemistry: A Molecular Approach" Table 8.4 for three p-electrons only $$^2\!P,^2\!D,^4\!S$$ Terms are possible. What's wrong with my approach that I'm that much off?

*Question on the side: I don't think it's correct to use the term "term" here because what I'm talking about is split into actual terms. Is there a way to refer to, for example, the $$^4\!D$$-Terms in general, maybe term system?

Firstly, regarding the extra question: Atkins' Molecular Quantum Mechanics (5th ed.) uses "term" for $$^2\!S$$, and "level" for $$^2\!S_{1/2}$$.

Back to the main question. It's been a long time since I did term symbols, so I am happy to be corrected, but if I am not wrong, your $$^4\!P$$, $$^4\!D$$, and $$^2\!S$$ terms are not allowed because of the Pauli exclusion principle. The argument is pretty similar to the one in the comments on my answer, which you're already aware of.

A term with $$S = 3/2$$ must have a set of states with $$M_S = +3/2, +1/2, -1/2, -3/2$$. Likewise, a $$P$$ term with $$L = 1$$ must also have a set of states with $$M_L = +1, 0, -1$$. Consequently, the total number of states associated with a single term $$^{2S+1}\!L$$ is $$(2S+1)(2L+1)$$. (The total number of states in a level, $$^{2S+1}\!L_J$$, is $$2J+1$$, and you can also show that the sum of this over the allowed values of $$J$$ is equal to $$(2S+1)(2L+1)$$.)

One of those twelve states in the purported $$^4\!P$$ term must have $$(M_S, M_L) = (+3/2, +1)$$. However, $$M_S = +3/2$$ can only be achieved if all three electrons have the same spin (spin up in this case), which in turn necessitates that the three electrons are all in different p-orbitals, such that $$M_L = \sum m_l = +1 + 0 + -1 = 0$$.

Ergo, a state with $$(M_S, M_L) = (+3/2, +1)$$ cannot exist, and the entire term $$^4\!P$$ cannot be possible. The same argument also applies to the $$^4\!D$$ term.

I suspect that showing that the $$^2\!S$$ term is forbidden is somewhat trickier. I had a similar issue before with carbon, which is an easier system to understand (only two electrons to consider instead of three), so maybe you will find this useful: Pauli-forbidden term symbols for atomic carbon.

Out of the three surviving terms ($$^2\!P$$, $$^2\!D$$, $$^4\!S$$) you already see from NIST that all the possible $$J$$ values (from $$|L-S|$$ to $$L+S$$) are allowed, so there is no issue with your calculation of $$J$$.

• So for each term there is a set of microstates and if one of these microstates can't be fulfilled without disregarding the Pauli principle than the whole term needs to be discarded? – Deglupta Aug 21 '19 at 15:22
• Yes, pretty much. – orthocresol Aug 21 '19 at 16:49