The spectroscopic notation for an element is given by $^{2S+1}L_{J}$. I want to use this notation for nitrogen with configuration


This is what I did. $l=1$ for p orbital. There are three electrons in p orbital, two electrons: spin up and spin down, and the remaining electron in spin up position, so that my $$S = \frac 12 + \left(-\frac{1}{2}\right) + \frac12 = \frac12$$ which gives


My query is can't the remaining electron in p orbital be in spin down position so that my $S$ becomes $-1/2$ and so that $J=l+s=1/2$.


Why not go one step further and make all the electrons have spin down? Then you should get $S = -3/2$, no?

In fact, what's happening is: you are confusing the projection of the spin quantum number, $M_S$, with its magnitude, represented by the spin quantum number $S$.

The $\pm1/2$ that you are summing are the individual projections of the electron spins, $m_s$. When you add up projections, you get a projection $M_S$. This projection can indeed be $-1/2$, and it can indeed be $-3/2$ as well, but the term symbol does not depend on $M_S$; it depends on $S$.

To obtain the total spin quantum number $S$ you have to use the Clebsch-Gordan series to couple different sources of angular momenta together. In this case we would first couple the spins of two electrons

$$S_{12} = s_1 + s_2, s_1 + s_2 - 1, \cdots, |s_1 - s_2|$$

Note that for an electron, it has spin $s = 1/2$. Again, the projection of the spin can be $+1/2$ or $-1/2$, but that is not what we are interested in. We are interested in the magnitude of its spin, which is only $1/2$, not $-1/2$.1 If we plug $s_1 = s_2 = 1/2$ into that equation above we get

$$S_{12} = 1, 0$$

and now if we couple this to the spin of the third electron we get

$$S = S_{12} + s_3, S_{12} + s_3 - 1, \cdots |S_{12} - s_3|$$

Again, $s_3 = 1/2$ only. But $S_{12}$ can take on two values so we will have to evaluate this sum two times, one time for each possibility:

$$\begin{align} S_{12} = 1 &\Longrightarrow S = \frac 32, \frac 12 \\ S_{12} = 0 &\Longrightarrow S = \frac 12 \end{align}$$

Note that the absolute value in the Clebsch-Gordan series forbids the possibility that $S < 0$. This makes sense since a magnitude of a spin cannot be negative.

In the end your two possibilities are $S = 3/2$ or $S = 1/2$. Due to Hund's first rule the $S = 3/2$ term is the ground state and the $S = 1/2$ terms are excited states.

Next, the $L$ in the term symbol is not equal to 1 just because the p orbitals have $l = 1$. Note that one is a small $l$ and the other is a big $L$. The big $L$ in the term symbol is obtained again by coupling the individual values of $l$ for each orbital. In this case all three p orbitals have $l =1$ and you would have to work through the Clebsch-Gordan series again.

$$L_{12} = l_1 + l_2 , l_1 + l_2 - 1, \cdots , |l_1 - l_2|$$

$$L = L_{12} + l_3, L_{12} + l_3 - 1, \cdots, |L_{12} - l_3|$$

and if you set $l_1 = l_2 = l_3$, should reach the conclusion that $L$ can take on the values of $3,2,1,0$. However you have to be careful about this, because the $L = 3$ value is forbidden by the Pauli principle, which is not taken into account by the Clebsch-Gordan series. A term with $L = 3$ would necessitate the existence of a microstate with $m_L = +3$, which would imply that all three electrons are in the p orbital with $m_l = +1$, which is not possible.

1 Technically, the spin is $\displaystyle \sqrt{\frac{1}{2}\left(\frac{1}{2} + 1\right)}\hbar$, but whatever. The correct phrasing is that the quantum number representing the magnitude of the spin can only be $1/2$.

  • $\begingroup$ I'm sorry, but I got a little confused by your statement A term with L=3 would necessitate the existence of a microstate with mL=+3, which would imply that all three electrons are in the p orbital with ml=+1, which is not possible.. Could you, please, explain it in a little more detail? I don't see the connection to $m_l$ state... $\endgroup$
    – Eenoku
    Nov 28 '18 at 14:12
  • 1
    $\begingroup$ @Eenoku (1) a term with $L = 3$ necessarily encompasses several states with $M_L$ ranging from $-L$ to $+L$. So, if that term is possible, it would imply the existence of states with $M_L = +3, +2, +1, 0, -1, -2, -3$. (2) $M_L$ is the direct sum of individual values of $m_l$. For $M_L = +3$ and three unpaired electrons in p-type orbitals (which have $m_l = +1, 0, -1$), the only way to accomplish this would be with three electrons in the $m_l = +1$ orbital, which is forbidden. Thus, the $L=3$ term cannot exist, which is why there is no $F$ term symbol for a $2p^3$ configuration. $\endgroup$
    – orthocresol
    Nov 28 '18 at 14:19

An algorithmic approach to writing down term symbols for an electron configuration can be found here. If you know group theory, that can help.

Anyway, if you follow the procedure described in the linked wikipedia article, then you will arrive at the following terms. $^4S,\, ^2D,\ \text{and } ^2P $.

Now, the rule's for determining the ground state term are:

The lowest energy term is that which has the greatest spin multiplicity. and For terms with the same spin multiplicity the term with the greatest orbital angular momentum is lower in energy

So in this case, the ground state term would be $^4S$. Now, J can take on values $J = |L+S|, |L+S-1|....|L-S|$

As an example, for the ground state term, $L = 0$ and $S = \frac{3}{2}$

So, $J = \frac{3}{2}$

The situation you describe, where one electron is spin up and the other two are spin paired is possible, and they give rise to the other two terms, and aren't the ground state term, which is what you were interested in? (I am guessing)


It can be spin down, but that would give you a doublet state, compared to a quartet if all electrons were spin up. The ground state is the one with the largest multiplicity.


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