# What is the ground state electron configuration for "O"^(2-) ion?

Oct 26, 2016

Based on the periodic table, $\text{O}$ is atomic number $8$, which means it has $8$ electrons. The first few atomic orbitals are $1 s$, $2 s$, and $2 p$.

Each orbital can hold $2$ electrons maximum, and there are $2 l + 1$ of each type of orbital ($s , p , d , f , g , \ldots$), where $l = 0$ corresponds to an $s$ orbital, $l = 1$ means $p$ orbital, and so on.

So, the configuration for neutral $\text{O}$ atom is:

$1 {s}^{2} 2 {s}^{2} 2 {p}^{4}$

where the $2 p$ orbitals hold the $4$ electrons like this:

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

When oxygen gains two electrons, it acquires a charge of $2 -$. Therefore, the two electrons, which go into the highest-energy atomic orbitals, give you a new configuration of

$\textcolor{b l u e}{1 {s}^{2} 2 {s}^{2} 2 {p}^{6}}$

for ${\text{O}}^{2 -}$, which is the same electron configuration as $\text{Ne}$.

That is known as a noble-gas-like configuration, and in general it is stable if oxygen is like that in a compound (having $8$ electrons in its $2 s$ and $2 p$ orbitals combined).