# What is the ground-state term symbol for the aluminium atom in a magnetic field?

Aug 8, 2016

I got $\text{^2 P_"1/2}$ (read as "doublet-P-one-half").

Well, the ground-state electron configuration is:

$\left[N e\right] 3 {s}^{2} 3 {p}^{1}$

which looks like this:

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

$\underline{\uparrow \downarrow}$

If you recall, an atomic term symbol is written like so:

\mathbf(""^(2S + 1) L_J)

where:

• $S$ is the total spin angular momentum ($| \sum {m}_{s} |$) , and $2 S + 1$ is the spin multiplicity.
• $L$ is the total orbital angular momentum, which is the maximum sum of ${m}_{l}$ values ($| \sum {m}_{l} |$) of occupied orbitals, made into a positive quantity.
• $J = \left\{| L - S | , L - S + 1 , . . . , 0 , . . . , L + S - 1 , | L + S |\right\}$ is the total angular momentum, of which there can be more than one possible value if $S > 0$. The ground term though, has only one of these $J$ values (differing in integer increments).

Starting by knowing the ground-state electron configuration, finding the term symbols are a bit easier. Then we'll have to narrow the ground term down with Hund's Rule.

$\setminus m a t h b f \left(3 {s}^{2}\right)$ TERM SYMBOL

$\underline{\uparrow \downarrow}$

• Since the $3 s$ electrons are paired, $S = | \sum {m}_{s} | = | + \frac{1}{2} + \left(- \frac{1}{2}\right) | = \textcolor{g r e e n}{0}$, and therefore, $2 S + 1 = \textcolor{g r e e n}{1}$.
• $S = 0$, so there is only one value of $J$.
• Since the maximum $L$ possible would be if the electrons are in ${m}_{l} = 0$ (there is only one $3 s$ orbital), ${L}_{\max} = 0$. That corresponds to the letter $\textcolor{g r e e n}{S}$ in the term symbol, just as $l = 0 \to s$.
• Therefore, $J = L + S = \textcolor{g r e e n}{0}$

That means the $\setminus m a t h b f \left(3 {s}^{2}\right)$ (ground) term symbol is color(blue)(""^1 S_0). And in fact, for any configuration that has fully-occupied subshells ($n {s}^{2}$, $n {p}^{6}$, etc), the term symbol is ""^1 S_0.

$\setminus m a t h b f \left(3 {p}^{1}\right)$ TERM SYMBOLS

For this, we don't have to consider all $6$ possible microstates (spin-up in each orbital, and spin-down in each orbital; $\text{2 spins"xx"3 orbitals" = "6 microstates}$) unless we want all possible term symbols, and you only want the ground term symbol.

However, since the ground term requires that we understand how Hund's Rule works, we'll look at the two primary possibilities.

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

• Since there is only one electron, $S = | \sum {m}_{s} | = \textcolor{g r e e n}{+ \frac{1}{2}}$.
• That means $2 S + 1 = \textcolor{g r e e n}{2}$.
• The maximum $L$ is when the electron is in the ${m}_{l} = \pm 1$ orbital, so $L = | \pm 1 | = 1$, and the letter is $\textcolor{g r e e n}{P}$, just as $l = 1 \to p$.
• With $S > 0$, we have more than one $J$ value (differing in integer increments).
• $J = \left\{L - S , . . . , L + S\right\}$ cannot be $0$, and we should get:

J = L pm S = 1 pm "1/2" = color(green)("1/2, 3/2"). The value of $J$ due to $S$ cannot use a greater $S$ than $\frac{1}{2}$, and $J$ must be positive, so we have two values of $J$.

So, our $2$ term symbols (one of which is the ground term) are:

$\textcolor{\mathrm{da} r k b l u e}{\text{^2 P_"1/2}}$, $\textcolor{\mathrm{da} r k b l u e}{\text{^2 P_"3/2}}$

GROUND TERM SYMBOL

To find the ground term, we'll need to look at Hund's Rule. Hund's Rule (for Physical Chemists) states:

1. The ground term with the largest $\setminus m a t h b f \left(S\right)$ is most stable, unless all have the same $S$.
2. If all terms have the same $S$, the largest $\setminus m a t h b f \left(L\right)$ is the most stable, unless they all have the same $L$ as well.
3. For terms with the same $S$ AND $L$, the ground term for a configuration where the subshell is less than half-filled has the lowest $J$, or vice versa.
4. If the subshell is EXACTLY half-filled, then $L = 0$, and thus, $J = {S}_{\text{max}}$.

Since $\text{^2 P_"1/2}$ and $\text{^2 P_"3/2}$ have the same $S$ and $L$, and since ${p}^{1}$ is less than half-filled (half-filled is ${p}^{3}$, which is $\text{3/6}$ electrons), the ground term has $J = \text{1/2}$:

$\implies \setminus m a t h b f \left(\textcolor{b l u e}{\text{^2 P_"1/2}}\right)$