# What are the quantum numbers of the seventh electron in fluorine?

Jun 5, 2018
• $n = 2$
• $l = 1$

Neither ${m}_{l}$ nor ${m}_{s}$ cannot be determined without further information on the spin of the electron or the orientation of its orbital.

#### Explanation:

The quantum numbers of an electron pinpoint its location in an atom. It would thus be necessary to determine the orbital containing this particular electron before deducing its quantum numbers. Start by writing the electron configuration of a fluorine atom.

Fluorine is the $9$th element on the periodic table and is located in the $p$ block at the right end of the second period. As a result, an atom of fluorine should contain $9$ electrons; among all occupied atomic orbitals, $2 p$ is the one of the highest potential energy.

$1 {s}^{2} \textcolor{w h i t e}{l} 2 {s}^{2} \textcolor{w h i t e}{l} 2 {p}^{5}$

The $1 s$ and $2 s$ orbitals of fluorine hold the first $2 + 2 = 4$ of its $9$ valence electrons. The rest $5$ electrons- the $5$th to the $9$th- are located in the partially-filled $2 p$ orbital. The $7$th electron should thus be a $\textcolor{g r e e n}{2} \textcolor{p u r p \le}{p}$ electron.

Each electron in an atom should have a set of four quantum numbers- $\textcolor{g r e e n}{n}$, $\textcolor{p u r p \le}{l}$, ${m}_{l}$, and ${m}_{s}$- that is- by the Pauli Exclusion Principle- unique to the particular electron.

The principal quantum number $\textcolor{g r e e n}{n}$ is a positive integer dependent on the index number of the main energy level containing the electron. The $\textcolor{g r e e n}{n}$ number for an electron in the second atomic energy level is $\textcolor{g r e e n}{2}$.

The angular quantum number $\textcolor{p u r p \le}{l}$ resembles the shape of the orbital that contains the electron. It is supposed to be an integer between $0$ and $\textcolor{g r e e n}{n} - 1$. For example, given $n = 2$, $\textcolor{p u r p \le}{l} = 0$ corresponds to atomic orbital $2 s$ and $\textcolor{p u r p \le}{l} = 1$ indicates a $2 p$ orbital. The electron in question is located in a $2 p$ orbital$\mathmr{and} \therefore h a s$color(purple)(l)=1# as its angular quantum number.

The value of ${m}_{l}$ and ${m}_{s}$ is dependent on the geometric orientation (e.g., in the direction of the $x$, $y$, or $z$ axis) and the spin of the electron (up or down) and cannot be derived from the given information.

Reference
"Quantum numbers (electronic)", Chemistry Libretext, Ed Vitz et. al, https://chem.libretexts.org/Textbook_Maps/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/05The_Electronic_Structure_of_Atoms/5.08%3A_Quantum_Numbers