# Why does it look like potassium has more than 8 valence electrons in its outer shell? Isn't its KLMN configuration 2, 8, 9???

May 24, 2016

It can be very confusing to think of it this way.

Potassium has access to its $1 s$, $2 s$, $2 p$, $3 s$, $3 p$, and $4 s$ orbitals. Its electron configuration is $\textcolor{g r e e n}{1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{1}}$.

Hence, your "KLMN" configuration would be:

$\textcolor{w h i t e}{\left[\left(\textcolor{b l a c k}{\underline{\text{e"^(-)"'s/subshell"))), (color(black)(2)), (color(black)(2,6)), (color(black)(2,6)), (color(black)(1))] = [(color(black)(ul("Total e"^(-)"'s}}}\right) , \left(\textcolor{b l a c k}{2}\right) , \left(\textcolor{b l a c k}{8}\right) , \left(\textcolor{b l a c k}{8}\right) , \left(\textcolor{b l a c k}{1}\right)\right] = \left[\left(\textcolor{b l a c k}{\underline{l}}\right) , \left(\textcolor{b l a c k}{0}\right) , \left(\textcolor{b l a c k}{0 , 1}\right) , \left(\textcolor{b l a c k}{0 , 1}\right) , \left(\textcolor{b l a c k}{0}\right)\right] = \left[\begin{matrix}\textcolor{b l a c k}{\underline{n}} \\ \textcolor{b l a c k}{1} \\ \textcolor{b l a c k}{2} \\ \textcolor{b l a c k}{3} \\ \textcolor{b l a c k}{4}\end{matrix}\right]}$.

i.e. You should have written $2 , 8 , 8 , 1$.

Thus, in your words, the 19th electron goes into the "4th shell". It does have access to $n = \left\{1 , 2 , 3 , 4\right\}$, being on the 4th period of the periodic table.

The "KLMN" configuration numbers arise from the number of ${m}_{l}$ values possible for all values of $l$ possible at a given $n$ (recall these are quantum numbers), combined with the maximum of two electrons of opposite spin (${m}_{s} = \pm \frac{1}{2}$) in the same orbital.

Recall that $l = 0 , 1 , 2 , . . . , n - 1$.

• For $n = 1$, we have $l = 0$ available.
• For $n = 2$, we have $l = 0$ or $1$.
• For $n = 3$, we have $l = 0 , 1$, or $2$.
• For $n = 4$, we have $l = 0 , 1 , 2$, or $3$.

For each of these cases, we thus have...

1S ORBITAL

For $n = 1$ and $l = 0$, we have ${m}_{l} = \left\{0\right\}$ and ${m}_{s} = \pm \frac{1}{2}$. Thus, for the $1 s$ orbital, 2 electrons can fill the set of subshells.

2S AND 2P ORBITALS

For $n = 2$ and $l = 0 , 1$, we have ${m}_{l} = \left\{0\right\}$ and ${m}_{l} = \left\{0 , \pm 1\right\}$, and ${m}_{s} = \pm \frac{1}{2}$. Thus, for the $2 s$ and $2 p$ orbitals combined, 2 + 6 = 8 electrons can fill the set of subshells. (Hence, we have the octet rule.)

3S, 3P, AND 3D ORBITALS

For $n = 3$ and $l = 0 , 1 , 2$, we have ${m}_{l} = \left\{0\right\}$, ${m}_{l} = \left\{0 , \pm 1\right\}$, and ${m}_{l} = \left\{0 , \pm 1 , \pm 2\right\}$, as well as ${m}_{s} = \pm \frac{1}{2}$. Thus, for the $3 s$, $3 p$, and $3 d$ orbitals combined, 2 + 6 + 10 = 18 electrons can fill the set of subshells.

4S, 4P, 4D, AND 4F ORBITALS

For $n = 4$ and $l = 0 , 1 , 2 , 3$, we have ${m}_{l} = \left\{0\right\}$, ${m}_{l} = \left\{0 , \pm 1\right\}$, ${m}_{l} = \left\{0 , \pm 1 , \pm 2\right\}$, and ${m}_{l} = \left\{0 , \pm 1 , \pm 2 , \pm 3\right\}$, as well as ${m}_{s} = \pm \frac{1}{2}$. Thus, for the $4 s$, $4 p$, $4 d$, and $4 f$ orbitals combined, 2 + 6 + 10 + 14 = 32 electrons can fill the set of subshells.