physical chemistry - How does one find the ground-state term symbol for a configuration that is exactly half-filled?

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.)

Answer

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 $^2P$