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For instance, the ground-state configuration of N atom is a $p^3$ configuration of all parallel spins and one electron in each $2p$ orbital, which has:

  • Total spin angular momentum, $S = 3 * \frac{1}{2} = \frac{3}{2}$
  • Spin multiplicity, $2S + 1 = 2(\frac{3}{2}) + 1 = 4$
  • Total orbital angular momentum, $L = (-1) + (0) + (1) = 0$

and so, $J = {|L - S|, |L - S + 1|, . . . , |L + S - 1|, |L + S|} = {\frac{1}{2}, \frac{3}{2}}$. Thus, I've found that the two candidates are ${}^{4} S_{1/2}$ and ${}^{4} S_{3/2}$.

But my textbook (Pilar, from the '60s!) says that the ground state is ${}^{4} S_{3/2}$. How could I figure something like this out?

Hund's Rules from my text (as far as I can tell---this text is hard for me to read) don't specify how we treat exactly half-filled subshells, only less-than-half-filled, or more-than-half-filled, assuming equal $L_{max}$ and $S_{max}$ already.

(As a side note, my usual McQuarrie textbook is not with me this week, so I cannot simply refer to McQuarrie.)

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The full calculation is laid out below. Start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number $s = 1/2$ and magnetic quantum number $m_s = \pm 1/2$. Orbitals have angular momentum of $s=0, p=1, d=2, f=3$ etc.

The total spin angular momentum is the series of values $$S = |s_1 + s_2|\cdots|s_1 − s_2|$$ where $s_1 , s_2 , s_3 =+1/2$ are the quantum numbers of each electron. The total angular momentum is $$ L=|l_1+l_2|\cdots|l_1−l_2|$$ and total angular momentum $$J=|L+S|\cdots|L−S|$$ Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by $1$.

The term symbol has the form $^{2S+1}L_J$. The super-prefix is the spin multiplicity, for spin angular momentum S this is $2S+1$ or in general $2X+1$ for angular momentum X.

When there are three 3 electrons the the spin and orbital angular momentum terms have to be added in two parts. First as a pair, with the equations above then again with each of the values in the series with the last electron or orbital.

For the spin the S values are
$S = |s_1 + s_2|\cdots|s_1 − s_2| = |1/2+1/2|\cdots|1/2-1/2| = 1, ~ 0 $ . We call these values $S_1$ and $ S_0$ To calculate the total with the third electron gives $$S_{s_1}=|S_1+1/2|\cdots|S_1 - 1/2| = 3/2 , ~ 1/2$$ and for the $S_0$ there is one value $S_0=1/2$. Thus for the spin the values are $3/2$ and $1/2$. This means that the multiplicity produces quartet and doublet states, thus so far we have $^4L_J$ and $^2L_J$.

The same method is followed for the orbital angular momentum. The first two p orbitals give $$ L=|l_1+l_2|\cdots|l_1−l_2|= |1+1|\cdots|1-1| = 2,~1,~0 $$ then for each of these values called $L_1,L_2,L_3$ we combine with the last orbital angular momentum, for example $$ L_{L_1}=|L_1+l_1|\cdots|L_1−l_2|= |2+1|\cdots|2-1| = 3,~2,~1 $$ and so forth for the other values. The L values are $3,2,1,0$

Putting all these values into a table gives $$\begin{matrix} L & S & J & \text{term symbol}\\ 3 & 3/2 & 9/2, 7/2, 5/2, 3/2 & ^4F_{9/2, 7/2, 5/2, 3/2} \\ 2 & 3/2 & 7/2, 5/2, 3/2,1/2 & ^4D_{ 7/2, 5/2, 3/2,1/2}\\ 1 & 3/2 & 5/2, 3/2,1/2 & ^4P_{5/2, 3/2,1/2}\\ 0 & 3/2 & 3/2 & ^4S_{3/2}\\ 3 & 1/2 & 7/2, 5/2 & ^2F_{7/2,5/2}\\ 2 & 1/2 & 5/2, 3/2 &^2D_{5/2, 3/2}\\ 1 & 1/2 & 3/2, 1/2 & ^2P_{3/2, 1/2}\\ 0 & 1/2 & 1/2 & ^2S_{1/2}\\ \end{matrix}$$ Finally to find the lowest energy term use Hund’s rules, i.e. of the terms given by equivalent electrons, i.e. having the same L values, the one with the greatest multiplicity is lowest in energy and of these the lowest is that one with the greatest L value. The only terms remaining after considering Pauli exclusion are $^4S$ which is lowest in energy and next $^2D$ then $^2P$

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  • $\begingroup$ Ah, so that's what you were going for. Yeah, it's more rigorous than I would have done for this kind of problem, since it was only concerning the ground state, but it makes sense to me. I did something similar over here for $s^{1} p^{2}$: socratic.org/questions/… $\endgroup$ – timaeus222 Nov 25 '16 at 23:19
  • $\begingroup$ How did you apply Pauli exclusion principle to remove the extraneous term symbols? $\endgroup$ – Acnologia Oct 6 '17 at 23:01
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*sigh *

Face-palm moment: I've already chosen an $S_{max}$, and when the subshell is exactly half-filled, $L_{max} = 0$, so $J = S_{max}$. And that would be why it's ${}^{4}S_{3/2}$, rather than ${}^{4}S_{1/2}$.

But if anyone has anything more to add, feel free to edit this answer.


Here's what I would do if I started this question from scratch. I would write out the electron configuration of the ground state in accordance with general chemistry principles (maximize parallel spins and minimize coulombic repulsions by distributing electrons evenly).

$\underline{\uparrow \text{ } }$ $\text{ }$ $\underline{\uparrow \text{ } }$ $\text{ }$ $\underline{\uparrow \text{ } }$

$m_l = $ {$-1,0,+1$}

$m_s = +\frac{1}{2}$

In order to determine the ground-state term symbol, we note the usual format:

$\mathbf{{}^{2S+1} L_J}$

where:

  • $S$ is the total spin angular momentum.
  • $L$ is the total orbital angular momentum.
  • $J = |L - S|, |L - S + 1|, . . . , |L + S - 1|, |L + S|$ is the total angular momentum when one considers spin-orbit coupling to determine multiplet terms.

By definition:

  • $L = \sum_i m_{l}(i) = -1 + 0 + 1 = 0$.
  • $S = \sum_i m_{s}(i) = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2}$

So the spin multiplicity is $2S+1 = 2(\frac{3}{2}) + 1 = 4$, and the term symbol letter is $S$ since $L = 0$. Therefore, we have a quadruplet $S$ state.

Since $J$ takes on each positive term starting from $S = \frac{3}{2}$ in integer increments, $J = \frac{1}{2}, \frac{3}{2}$. So, the possible term symbols without narrowing them down yet are ${}^{4} S_{1/2}$ and ${}^{4} S_{3/2}$.

Since the subshell is exactly half-filled, $L = 0$, so $J = S$. From Hund's Rules, we first try to maximize $S$, then $L$, then we choose our $J$ depending on whether the subshell is less-than, more-than, or exactly half-filled.

Since we would have already gone through and chosen $S_{max} = 3/2$ and $L_{max} = 0$, $J = S_{max} = 3/2$, and our ground-state term symbol is:

$\mathbf{{}^4 S_{3/2}}$

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