# What are the quantum numbers for the 9th electron in flourine?

Nov 22, 2017

$n = 2 , l = 1 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2}$

#### Explanation:

You know that fluorine has an atomic number equal to $9$, which means that the $\text{9th}$ electron in an atom of fluorine is the last one added when "constructing" the neutral atom of fluorine.

Fluorine is located in period $2$ of the Periodic Table, which implies that the $\text{9th}$ electron in fluorine is located on the second energy level.

Consequently, you can say that this electron will have a principal quantum number, $n$, which describes the energy level on which an electron is located in an atom, equal to

$n = 2$

Now, fluorine is located in group $17$ of the Periodic Table, which means that fluorine is part of the $p$ block. This implies that the angular momentum quantum number, $l$, which describes the energy subshell in which the electron is located, will be equal to

$l = 1$

This is the case because you have

• $l = 0 \to$ describes the $s$ subshell
• $l = 1 \to$ describes the $p$ subshell
• $l = 2 \to$ describes the $d$ subshell
$\vdots$

and so on. Next, for the $p$ subshell, the magnetic quantum number, ${m}_{l}$, which tells you the orientation of the orbital in which the electron is located, will be equal to

${m}_{l} = \left\{- 1 , 0 , + 1\right\}$

By convention, you have

• ${m}_{l} = - 1 \to$ describes the ${p}_{y}$ orbital $\to$ corresponds to groups $13$ and $16$
• ${m}_{l} = \textcolor{w h i t e}{-} 0 \to$ describes the ${p}_{z}$ orbital corresponds to groups $14$ and $17$
• ${m}_{l} = + 1 \to$ descrbes the ${p}_{x}$ orbital corresponds to groups $15$ and $18$

Since the $\text{9th}$ electron corresponds to group $17$, you can say that it will have

${m}_{l} = 0$

This electron will be added to the $2 {p}_{z}$ orbital, which is the ${p}_{z}$ orbital located on the second energy level.

Finally, the spin quantum number, ${m}_{s}$, which tells you the spin of the electron, can take two possible values

${m}_{s} = \left\{+ \frac{1}{2} , - \frac{1}{2}\right\}$

Now, the $\text{9th}$ electron will have

${m}_{s} = - \frac{1}{2}$

because, by convention, electrons added to a half-filled orbital are assigned spin-down. The ${p}_{z}$ orbital is half-filled because of Pauli's Exclusion Principle, which states that all three $p$ orbitals must be half-filled before any of them can be filled completely.

You can thus say that the full quantum number set that describes the $\text{9th}$ electron in an atom of fluorine will be

$n = 2 , l = 1 , {m}_{l} = 0 , {m}_{s} = - \frac{1}{2}$

The electron is located on the second energy level, in the $2 p$ subshell, in the $2 {p}_{z}$ orbital, and has spin-down