# Question #db9e4

Apr 29, 2016

The answer is (3) a fluorine atom in an excited state

#### Explanation:

Right from the start, you should be able to look at the given electron configuration and say that it cannot possibly belong to an atom or ion in the ground state.

Why? Because it's in violation of the Aufbau Principle.

As you know, the Aufbau Principle states that electrons must first completely fill an energy level before moving on to fill a higher energy level.

In your case, the given electron configuration is

$1 {s}^{2} \textcolor{red}{2} {s}^{2} \textcolor{red}{2} {p}^{4} 3 {s}^{1}$

Now, the second energy level, denoted by $n = \textcolor{red}{2}$, contains two subshells, the 2s-subshell and the 2p-subshell.

The important thing to remember about p-subshells in general is that they can hold a maximum of $6$ electrons, $2$ for each of the three orbitals, ${p}_{x}$, ${p}_{z}$, and ${p}_{y}$, they contains.

When completely filled, the second energy level should look like this

$\text{second energy level: } 2 {s}^{2} 2 {p}^{6} \to$ completely filled

According to the Aufbau Principle, you cannot have an electron in the third energy level, $n = 3$, before the second energy level is completely filled with electrons. Therefore, you can say for sure that

$\textcolor{red}{\cancel{\textcolor{b l a c k}{1 {s}^{2} \textcolor{red}{2} {s}^{2} \textcolor{red}{2} {p}^{4} 3 {s}^{1}}}} \to$ violates the Aufbau Principle

This means that you're dealing with an atom or ion in an excited state. Here one of the electrons that in the ground state would be located in the 2p-subshell was promoted to the third energy level, in the 3s-subshell.

$\textcolor{w h i t e}{a}$

Now, it's worth noting that an electron that was promoted to a higher energy level by absorbing a photon can return to its ground state by emitting a photon of the same frequency.

This means that you can demote this excited electron back to the 2p-subshell to get the ground state configuration of the atom.

You will have

$\text{ground state: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{5} 3 {s}^{0}$

which is of course

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{ground state: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{5}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This electron configuration corresponds to a neutral atom of fluorine, $\text{F}$

which means that the electron configuration given to you corresponds to a fluorine atom in an excited state.