# What is the electron configuration of an element with atomic number 8?

Jun 5, 2018

Well, once you get your periodic table out, spot the element with $Z = 8$... and you should find it is oxygen. It is on period 2, column 16.

$\text{O": " } 1 {s}^{2} 2 {s}^{2} 2 {p}^{4}$

DEFINING THE ORBITALS

The principal quantum number can only be $n = 1 , 2 , 3 , . . .$, and being a main group element, oxygen's highest-energy orbitals have $n = 2$; so,

• $n = 2$ contains the valence electrons.
• $n = 1$ contains the core electrons, since it must be lower in energy. In fact, it is MUCH lower in energy.

The angular momentum quantum number $l$ gives the shape of the orbital, and $l = 0 , 1 , 2 , . . . , n - 1$, so

• if $n = 1$, $l = 0$
• if $n = 2$, $l = 0 , 1$

And $l = 0 \leftrightarrow s , 1 \leftrightarrow p , 2 \leftrightarrow d , . . .$.

Therefore, oxygen has access to the $1 s , 2 s$, and $2 p$ orbitals at the moment.

FILLING THE ORBITALS

As per the Aufbau principle, we PREDICT that the orbitals are filled from lowest to highest energy, and from Hund's rule, we PREDICT that the orbitals are more stable when more electrons are unpaired.

And of course, the Pauli Exclusion Principle applies to electrons, only allowing two electrons per orbital.

Hence, we fill them as follows:

$\underbrace{\underline{\uparrow \downarrow} \text{ "ul(uarr color(white)(darr))" } \underline{\uparrow \textcolor{w h i t e}{\downarrow}}}$
$\text{ "" "" } \textcolor{w h i t e}{/} 2 p$

$\underline{\uparrow \downarrow}$
$2 s$

$\underline{\uparrow \downarrow}$
$1 s$

and we denote this as $\textcolor{b l u e}{1 {s}^{2} 2 {s}^{2} 2 {p}^{4}}$ to indicate the number of electrons in each orbital.