# How would I determine all the possible term symbols for an s^1p^2 electron configuration (such as first-excited-state boron)?

## They are $\text{^4 P_"1/2}$, $\text{^4 P_"3/2}$, $\text{^4 P_"5/2}$, $\text{^2 D_"3/2}$, $\text{^2 D_"5/2}$, $\text{^2 P_"1/2}$, $\text{^2 P_"3/2}$, and $\text{^2 S_"1/2}$. I'll answer this question, for people who eventually may run into this topic.

Nov 24, 2016

DISCLAIMER: This is a long process! If you want to try this, set aside about 1-2 hours.

Let's say you wanted to find each possible term symbol for an ${s}^{1} {p}^{2}$ configuration. The general notation is:

bb(""^(2S + 1) L_J)

where

• $S$ is the total spin.
• $L$ is the total orbital angular momentum.
• $J$ is the total angular momentum, taking on the range $\left\{| L - S | , | L - S + 1 | , . . . , | L + S - 1 | , | L + S |\right\}$.
• $2 S + 1$ is the spin multiplicity.

For this, I would first identify all the possible values of ${m}_{l}$ and ${m}_{s}$ for the $s$ and $p$ electrons:

• ${s}^{1} : {m}_{l} = 0$, ${m}_{s} = \pm \frac{1}{2}$
• ${p}^{2} : {m}_{l} = \left\{- 1 , 0 , + 1\right\}$, ${m}_{s} = \pm \frac{1}{2}$

ELECTRON CONFIGURATION "OUTLINE"

To outline the possible electron configurations, let us list each possible electron configuration out. We call them microstates.

The way I think makes sense to organize them is doing all the spins for some lefthand ${m}_{l}$, and then restricting the lowest lefthand ${m}_{l}$.

• Without electron pairing, and with a spin-up $s$ electron (${L}_{\max} = {\sum}_{i} {l}_{i} = 0 + 1 = 1$): • Without electron pairing, and with a spin-down $s$ electron (${L}_{\max} = {\sum}_{i} {l}_{i} = 0 + 1 = 1$): • With electron pairing, with a spin-up or spin-down $s$ electron (${L}_{\max} = {\sum}_{i} {l}_{i} = 0 + 1 + 1 = 2$): That gives us a total of $30$ electron configuration "microstates".

CONSTRUCTING A MICROSTATE TABLE

Each microstate has its corresponding total spin angular momentum $S$ and total orbital angular momentum $L$ in the $z$ direction, which are called ${M}_{S}$ and ${M}_{L}$, respectively. These are defined as:

${M}_{L} = {\sum}_{i} {m}_{l} \left(i\right)$
${M}_{S} = {\sum}_{i} {m}_{s} \left(i\right)$

meaning the sum of the ${m}_{l}$ or ${m}_{s}$ values for electron $i$.

Earlier, we said that we had an ${L}_{\max}$ of $1$ or $2$. Well, that gives the allowed range of ${M}_{L}$ to be $\textcolor{g r e e n}{\left\{- 2 , - 1 , 0 , + 1 , + 2\right\}}$, just like how ${m}_{l} = \left\{- l , - l + 1 , \ldots , l - 1 , l\right\}$.

That will be the number of rows of our table.

Also, with $3$ electrons, the total spin could be $S = \frac{1}{2} , \frac{3}{2}$. Therefore, the range of ${M}_{S}$ is $\textcolor{g r e e n}{\left\{- \frac{3}{2} , - \frac{1}{2} , + \frac{1}{2} , + \frac{3}{2}\right\}}$.

That will be the number of columns of our table.

From this, the blank microstate table that organizes our electron configurations is:

${M}_{L} \uparrow \text{ "" } \leftarrow {M}_{S} \rightarrow$
$\underline{\text{ "" "" "" "" "" "" "" "" "" "" "" "" "" }}$
$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\text{),color(black)(-3/2),color(black)(-1/2),color(black)(+1/2),color(black)(+3/2)),(color(black)(+2),color(black)(""),color(black)(""),color(black)(""),color(black)("")),(color(black)(+1),color(black)(""),color(black)(""),color(black)(""),color(black)("")),(color(black)(0),color(black)(""),color(black)(""),color(black)(""),color(black)("")),(color(black)(-1),color(black)(""),color(black)(""),color(black)(""),color(black)("")),(color(black)(-2),color(black)(""),color(black)(""),color(black)(""),color(black)(}}\right)\right]}$

The outline we did above is how we can keep track of which ones we've accounted for already.

As an example of the notation we'll put into the table,

$\underline{\textcolor{w h i t e}{\uparrow \downarrow}} \text{ "ul(uarr color(white)(darr))" } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$\underline{\downarrow \textcolor{w h i t e}{\uparrow}}$

would be written as ${0}^{-} {0}^{+} {1}^{+}$, to indicate that:

• the $s$ electron went into an orbital of ${m}_{l} = 0$ as spin-down $\left(-\right)$
• a $p$ electron went into an orbital of ${m}_{l} = 0$ as spin-up $\left(+\right)$,
• a $p$ electron went into an orbital of ${m}_{l} = 1$ as spin-up $\left(+\right)$.

So,

• $\boldsymbol{{M}_{S}} = {\sum}_{i} {m}_{s} \left(i\right) = - \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \boldsymbol{+ \frac{1}{2}}$
• $\boldsymbol{{M}_{L}} = {\sum}_{i} {m}_{l} \left(i\right) = 0 + 0 + 1 = \boldsymbol{1}$

Therefore, it goes into the cell that is indicated by ${M}_{S} = + \frac{1}{2}$ and ${M}_{L} = + 1$.

Give yourself maybe half an hour to an hour, and you should get: SEPARATING INTO INDIVIDUAL MICROSTATE TABLES FOR EACH FREE-ION TERM

Now, to find each term symbol, we first make the table easier to manage by setting each microstate as $x$. That gives: Above, I've highlighted the microstates as follows:

1. Starting at the maximum number of ${M}_{L}$ rows, and then the maximum number of those ${M}_{S}$ columns, and choose the first term in each cell.
2. Then, decrease the range of $S$ symmetrically (thus going from 4 columns to 2 columns) and find the new maximum number of ${M}_{L}$ rows out of the available microstates.
3. Then, decrease the range of $L$ once you've reached the minimum number of ${M}_{S}$ columns.

Each color of $x$ is placed into a separate microstate table.

• The first table would be the $\textcolor{b l u e}{\text{blue}}$ $x$'s.
• The second would be the $\textcolor{red}{\text{red}}$ $x$'s.
• The third would be the $\textcolor{\mathmr{and} a n \ge}{\text{orange}}$ $x$'s.
• The fourth would be the $\textcolor{g r e e n}{\text{green}}$ $x$'s.

Here's a GIF illustrating how to do it: FINDING EACH FREE-ION TERM SYMBOL (NO J)

This is how I knew which free-ion term symbols to write for the above microstate tables:

• The number of ${M}_{L}$ rows is the range of $L$ in the $+ z$ and $- z$ directions, so $| {M}_{L , \max} | = {L}_{\max}$, which tells you what letter the term symbol is ($0 , 1 , 2 , 3 , 4 , \ldots \leftrightarrow S , P , D , F , G , \ldots$).
• The number of ${M}_{S}$ columns is the range of $S$ in the $+ z$ and $- z$ directions, so $| {M}_{S , \max} | = {S}_{\max}$, which tells you what the total spin for the term symbol is.

Once you work it out, you should confirm that your initial term symbols are:

• ${\text{^(2(3/2) + 1) (L = 1) = }}^{4} P$ (blue $x$'s)
• ${\text{^(2(1/2) + 1) (L = 2) = }}^{2} D$ (red $x$'s)
• ${\text{^(2(1/2) + 1) (L = 1) = }}^{2} P$ (orange $x$'s)
• ${\text{^(2(1/2) + 1) (L = 0) = }}^{2} S$ (green $x$'s)

FINDING EACH "MULTIPLET" TERM SYMBOL (INCLUDING J)

Finally, find $J$ by using the $L$ and $S$ values you have available. For each $L$ and $S$, take the largest $| {M}_{L} |$ and use each $| {M}_{S} |$, respectively:

""^4 P: L = 0,bb(1); S = 1/2,3/2
$\implies \textcolor{g r e e n}{J} = \left(1 - \frac{1}{2}\right) , \left(1 + \frac{1}{2}\right) , \left(1 + \frac{3}{2}\right) = \textcolor{g r e e n}{\frac{1}{2} , \frac{3}{2} , \frac{5}{2}}$

""^2 D: L = 0,1,bb(2); S = 1/2
$\implies \textcolor{g r e e n}{J} = \left(2 - \frac{1}{2}\right) , \left(2 + \frac{1}{2}\right) = \textcolor{g r e e n}{\frac{3}{2} , \frac{5}{2}}$

""^2 P: L = 0,bb(1); S = 1/2
$\implies \textcolor{g r e e n}{J} = \left(1 - \frac{1}{2}\right) , \left(1 + \frac{1}{2}\right) = \textcolor{g r e e n}{\frac{1}{2} , \frac{3}{2}}$

""^2 S: L = bb(0); S = 1/2
$\implies \textcolor{g r e e n}{J = \frac{1}{2}}$

So, we finally have:

$\textcolor{b l u e}{\text{^4 P_"1/2", ""^4 P_"3/2", ""^4 P_"5/2", ""^2 D_"3/2", ""^2 D_"5/2", ""^2 P_"1/2", ""^2 P_"3/2", ""^2 S_"1/2}}$