What is the electron configuration of F^-?

Jun 23, 2016

${\text{F}}^{-} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$

Explanation:

A good starting point for when you must find the electron configuration of an ion is the electron configuration of the neutral atom.

In your case, you must find the electron configuration of the fluoride anion, ${\text{F}}^{-}$, so start by writing the electron configuration of a neutral fluorine atom, $\text{F}$.

Fluorine is located in period 2, group 17 of the periodic table and has an atomic number of $9$. This tells you that the neutral fluorine atom has a total of $9$ electrons surrounding its nucleus.

Its electron configuration will be

$\text{F: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{5}$

Now, the ${\text{F}}^{-}$ anion is formed when $1$ electron is added to a neutral fluorine atom.

Notice that the 2p-subshell of the neutral atom contains $5$ electrons. Its maximum capacity is actually $6$ electrons, two electrons for each p-orbital.

This means that the $\textcolor{red}{\text{extra electron}}$ will be added to one of the three 2p-orbitals, let's say to $2 {p}_{y}$.

The 2p-subshell will now be completely filled, i.e. it will hold $6$ electrons.

The electron configuration of the fluoride anion will thus be

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{F}}^{-} : \textcolor{w h i t e}{a} 1 {s}^{2} \textcolor{w h i t e}{a} 2 {s}^{2} \textcolor{w h i t e}{a} 2 {p}^{6}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Notice that the fluoride anion has a total of $8$ electrons in its second shell, the outermost shell. This tell you that the anion has a complete octet.

Because the fluoride anion is isoelectronic with neon, $\text{Ne}$, you can write its electron configuration using the noble gas shorthand notation as

"F"^(-): ["Ne"]

Here

$\left[\text{Ne}\right]$ - the electron configuration of neon