# Give the number of core electrons for F-?

## A) 22 B) 75 C) 9 D) 4 E) 2 Please more detailed on how to get the answer. I'm really confused on this chapter of valance electrons and etc. Thank you so much!!!

Dec 11, 2017

They include the $1 s$ electrons in its electron configuration, so, there are two of them.

overbrace(1s^2)^"core" " "overbrace(2s^2 2p^6)^("valence")

I suppose I'll do a detailed review from particle properties all the way through quantum numbers and electron configurations.

FLUORINE ELECTRONS

$\text{F}$ as-written as the fluorine atom, with atomic number $9$, giving it $9$ protons by definition.

As a neutral atom, the number of protons it has must be equal to the number of electrons, as they are particles of opposite charge and a neutral atom has a total charge of zero.

Therefore, $\text{F}$ atom has $9$ electrons total and ${\text{F}}^{-}$ has $10$ electrons total. Note that these include both the core and valence electrons, the latter being the ones used most often to react.

FLUORINE ORBITALS AS RELATED TO QUANTUM NUMBERS

An atomic orbital is given as the symbol $n l$, which we will define later.

As $\text{F}$ is on the 2nd row of the periodic table, it has access up to (and including) the atomic orbitals belonging to principal quantum number $n = 2$, where we could potentially have

$n = 1 , 2 , 3 , . . .$

defining each energy level. These are the coefficients of atomic orbital symbols.

For each energy level $n$, the orbitals that exist have certain shapes given by $l$, the angular momentum quantum number, where

$l = 0 , 1 , 2 , . . . , n - 1$.

Since the maximum $l$ is $n - 1$, fluorine uses orbitals up to and including $l = 1$, which are found in the subshells it has access to. As it turns out,

$l = 0 \leftrightarrow s$ subshell
$l = 1 \leftrightarrow p$ subshell
$l = 2 \leftrightarrow d$ subshell
$l = 3 \leftrightarrow f$ subshell

and so on. This completes the orbital symbol given by $n l$.

Therefore, fluorine atom contains electrons within $1 s$, $2 s$, and $2 p$ subshells.

Furthermore, each subshell ($s , p , d , f , . . .$) contains a certain number of orbitals, each assigned a value of ${m}_{l}$, where ${m}_{l}$ is the magnetic quantum number:

$l = 0 \leftrightarrow {m}_{l} = 0$
$\to$ $1 \times s$ orbital for any $n$

$l = 1 \leftrightarrow {m}_{l} = \left\{- 1 , 0 , + 1\right\}$
$\to$ $3 \times p$ orbitals for any $n$

$l = 2 \leftrightarrow {m}_{l} = \left\{- 2 , - 1 , 0 , + 1 , + 2\right\}$
$\to$ $5 \times d$ orbitals for any $n$

etc.

This means we have one $1 s$, one $2 s$, and three $2 p$ orbitals.

"PUTTING" ELECTRONS INTO ORBITALS

Each of these orbitals (of a particular ${m}_{l}$) will contain certain numbers of those $10$ electrons we say ${\text{F}}^{-}$ has.

By the Pauli Exclusion Principle, no two electrons can be in the same orbital and have the same spin value ${m}_{s} = \pm \frac{1}{2}$ at the same time.

Since two electrons in the same orbital have the same $n$, $l$, and ${m}_{l}$, these two electrons must then have a spin quantum number of ${m}_{s} = \pm \frac{1}{2}$.

As a result, each orbital can only contain two electrons, maximum.

This then leads to the electron configuration of ${\text{F}}^{-}$:

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

where the superscripts denote how many electrons are in the subshell defined by $n l$.

CORE VS VALENCE ELECTRONS

Since helium has two electrons, it is convenient to define the noble gas shorthand $\left[H e\right] = 1 {s}^{2}$. Similarly, we could say $\left[N e\right] = 1 {s}^{2} 2 {s}^{2} 2 {p}^{6}$, and so on.

We also call this the noble gas core for a reason! It contains the core electrons (most of the time)!

As such, it should at this point be straightforward that there are $\boldsymbol{2}$ core electrons in ${\text{F}}^{-}$, and $\boldsymbol{8}$ valence electrons:

color(blue)(overbrace(1s^2)^"core" " "overbrace(2s^2 2p^6)^("valence"))