# How do I know which term symbols are acceptable for an np^2 electron configuration? I found all of them, but which ones should I keep?

Jul 9, 2016

I think if you connect orbital configurations with each term symbol, you would have a clearer way to understand term symbols. All of them represent an electron configuration, and so you can relate them back to Hund's Rule and the Pauli Exclusion Principle.

As a note, we call ""^1 L_J a singlet state, which has all paired electrons, and we call ""^3 L_J a triplet state, which has two unpaired electrons.

We can start from the known configuration of carbon:

$1 {s}^{2} 2 {s}^{2} 2 {p}^{2}$

Now, recall that term symbols are defined as:

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

where:

• $2 S + 1$ is the spin multiplicity, and $S$ is the total spin due to adding up the ${m}_{s}$ values for each electron in the same subshell, i.e. $S = \sum {m}_{s}$.
• $L$ is the total orbital angular momentum, or $L = \sum {m}_{l}$, due to adding up the ${m}_{l}$ values for each orbital. Each electron in an orbital contributes one value of ${m}_{l}$, so if an ${m}_{l} = + 1$ orbital is fully occupied, then you have $L = 2$. Make sure you know the maximum possible $L$ before using it to calculate $J$.
• $J$ is the total angular momentum, or $J = \left\{| L - S | , | L - S + 1 | , . . . , 0 , . . . , | L + S - 1 | , | L + S |\right\}$. So, an example is $S = 1$ and ${L}_{\max} = 1$. Then $J = \left\{L - S , L - S + 1 , L + S\right\} = \left\{0 , 1 , 2\right\}$.

THE $\setminus m a t h b f \left(n {s}^{2}\right)$ CONFIGURATIONS

When we consider $n {s}^{2}$ configurations, we can have these:

$\uparrow \uparrow$, $\downarrow \downarrow$, $\uparrow \downarrow$, $\downarrow \uparrow$

• The first and second are not allowed according to the Pauli Exclusion Principle, which requires the ${m}_{s}$ for each electron to be opposite.
• The fourth is redundant, since electrons are indistinguishable (you can't tell whether one electron in particular is spin-up or if it's the other one). So we didn't need to consider that one, actually, but I wrote it out to show that it was something to think about.

For the third $n {s}^{2}$ configuration, we have that:

• $S = \sum {m}_{s} = - \frac{1}{2} + \frac{1}{2} = 0$, so $2 S + 1 = 1$.
• $L = \sum {m}_{l} = 0$.
• Since $L = 0$, and $0 \le S < 1$, $J$ only has one value, and $J = 0 \pm 0 = 0$.

Therefore, with $\sum {m}_{l} = {m}_{l} = \left\{0\right\}$ (only one $n s$ orbital exists for a given $n$), $l = 0$ and we have an color(blue)(""^1 S_0) term symbol for $\uparrow \downarrow$, corresponding to either the $1 {s}^{2}$ or $2 {s}^{2}$ configuration.

EXAMINING ALL $\setminus m a t h b f \left(n {p}^{2}\right)$ CONFIGURATIONS

This is quite a bit harder, mainly because of the number of possibilities. Work it out, and you'd get:

Utilizing ${m}_{l} = \left\{- 1 , 0 , + 1\right\}$ (including unfavorable configurations):

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$
$\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$

Utilizing ${m}_{l} = \left\{0 , + 1\right\}$ (including unfavorable configurations):

$\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$
$\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$
$\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\downarrow \textcolor{w h i t e}{\downarrow}}$

Utilizing paired electrons nonredundantly, and ${m}_{l} = \left\{- 1 , 0 , + 1\right\}$:

$\underline{\uparrow \downarrow}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\uparrow \downarrow}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$
$\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\uparrow \downarrow}$

(You could also have $\downarrow \uparrow$ configurations for the paired electrons, but again, those are redundant.)

That gives a total of $\setminus m a t h b f \left(15\right)$ possibilities, or microstates, counting 'unacceptable' configurations. There is also a mathematical way to do this, but it should be in your book. I think this visual way can help you make connections.

The allowed ones follow the Pauli Exclusion Principle, and Hund's rule asks for all electrons to be parallel spin when possible, meaning when one tries to maximize total spin.

Therefore, $n {p}^{2}$ configurations number $\setminus m a t h b f \left(1 , 2 , 9 ,\right)$ and $\setminus m a t h b f \left(13\right)$ (OR $\setminus m a t h b f \left(14\right)$, OR $\setminus m a t h b f \left(15\right)$) are acceptable. The rest are either redundant or don't follow Hund's rule or the Pauli Exclusion Principle. Configurations $13$, $14$, and $15$ are degenerate (same in energy), so only one is necessary to consider.

EXAMINING ALLOWED $\setminus m a t h b f \left(n {p}^{2}\right)$ CONFIGURATIONS

Now, considering each nonredundant, allowed configuration, we can determine their term symbols. We expect $5 - 1 = 4$ more configurations left to consider, since we have already counted ""^1 S_0.

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ ($L = - 1$)

$\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ ($L = 0$)

$\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ $\underline{\uparrow \textcolor{w h i t e}{\downarrow}}$ ($L = 1$)

• With ${m}_{l} = + 1$ and $0$ orbitals occupied, we would have the maximum possible $L = \sum {m}_{l} = + 1 + 0 = 1$, corresponding with a $P$ term.
• With two parallel electrons, $S = \sum {m}_{s} = \frac{1}{2} + \frac{1}{2} = 1$, so the spin multiplicity is $2 S + 1 = 3$.
• With $S > 0$, we can have more than one $J$ value. For $J$, we have three possible values. One way to do this is: $L + S = \textcolor{red}{2}$, $L + S - 1 = \textcolor{red}{1}$, and $L - S = \textcolor{red}{0}$.

Therefore, the second, third, and fourth term symbols here are $\textcolor{b l u e}{{\text{^3 P_0, ""^3 P_1, }}^{3} {P}_{2}}$. These are nearly-degenerate (very close together in energy) and nonredundant configurations.

$\underline{\uparrow \downarrow}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$ $\underline{\textcolor{w h i t e}{\uparrow \downarrow}}$

• Finally, for this one, two electrons in the ${m}_{l} = + 1$ orbital would generate the maximum possible $L = \sum {m}_{l} = 1 + 1 = 2$. That corresponds with a $D$ term, just like how $l = 2$ is a $d$ orbital (but this is for a $p$ orbital configuration!!).
• The electrons are paired, so $S = \sum {m}_{s} = \frac{1}{2} + \left(- \frac{1}{2}\right) = 0$.

Therefore, $J = \left\{L + 0\right\} = L = 2$, and our fifth term symbol is color(blue)(""^1 D_2).

So, to summarize our answer, all five allowed configurations correspond with the term symbols ""^1 S_0 (paired $n {s}^{2}$), $\left\{{\text{^3 P_0, ""^3 P_1, }}^{3} {P}_{2}\right\}$ (unpaired $n {p}^{2}$), and ""^1 D_2 (paired $n {p}^{2}$).

(They are pronounced "singlet S zero", "triplet P zero", etc.)

If you wish, you can identify the electron configurations that are physically represented by each term symbol that is not allowed, and you should find that those have some reason related to Hund's Rule or the Pauli Exclusion Principle.