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:


$S=3/2, 1/2$

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?


1 Answer 1


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 $J$.

  • $\begingroup$ 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? $\endgroup$
    – Deglupta
    Aug 21, 2019 at 15:22
  • 1
    $\begingroup$ Yes, pretty much. $\endgroup$
    – orthocresol
    Aug 21, 2019 at 16:49

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