# Draw the electron configuration diagram for carbon?

Jul 29, 2016

Each row on the periodic table introduces a new value for the principal quantum number $n$, while $l$ goes as $0 , 1 , . . . , n - 1$.

Recall that $l = 0 \to s$ orbital, and $l = 1 \to p$ orbital. Therefore, we would have these orbitals available:

$\text{Row 1:}$ $1 s$
$\text{Row 2:}$ $2 s , 2 p$
$\text{Row 3:}$ $. . .$
$\text{Row 4:}$ $. . .$

Carbon has access to only $n = 2$ and $n = 1$, so its six electrons can only go into the $1 s$, $2 s$ and $2 p$ orbitals, from lowest to highest energy (Aufbau Principle) one at a time to maximize spin (Hund's Rule), with opposite spins when pairing up (Pauli Exclusion Principle).

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

$\textcolor{w h i t e}{\left[\begin{matrix}\textcolor{b l a c k}{\underline{\uparrow \textcolor{w h i t e}{\downarrow}}} \\ \textcolor{b l a c k}{2 {p}_{x}}\end{matrix}\right]} \textcolor{w h i t e}{\left[\begin{matrix}\textcolor{b l a c k}{\underline{\uparrow \textcolor{w h i t e}{\downarrow}}} \\ \textcolor{b l a c k}{2 {p}_{y}}\end{matrix}\right]} \textcolor{w h i t e}{\left[\begin{matrix}\textcolor{b l a c k}{\underline{\textcolor{w h i t e}{\downarrow} \textcolor{w h i t e}{\downarrow}}} \\ \textcolor{b l a c k}{2 {p}_{z}}\end{matrix}\right]}$

$\textcolor{w h i t e}{\left[\begin{matrix}\textcolor{b l a c k}{\underline{\uparrow \downarrow}} \\ \textcolor{b l a c k}{2 s}\end{matrix}\right]}$

$\text{ }$
$\text{ }$

$\textcolor{w h i t e}{\left[\begin{matrix}\textcolor{b l a c k}{\underline{\uparrow \downarrow}} \\ \textcolor{b l a c k}{1 s}\end{matrix}\right]}$

If you want to read more on it:
https://socratic.org/questions/how-do-you-draw-electron-orbital-diagrams