# Question f4b81

Jun 22, 2016

Seven electrons.

#### Explanation:

Your starting point here will the electron configuration of a neutral iron atom, $\text{Fe}$.

Iron is located in period 4, group 8 of the periodic table and has an atomic number equal to $26$. This means that the electron configuration of a neutral iron atom must account for a total of $26$ electrons

$\text{Fe: } 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{6} 4 {s}^{2}$

Now, a set of four quantum numbers is assigned to each of those $26$ electrons. Your goal here is to figure out how many electrons have

$n + l + {m}_{l} = 4$

It's important to realize that for every set of $n$, $l$, and ${m}_{l}$ quantum numbers that match your criteria, you get two electrons, one having spin-up, ${m}_{s} = + \frac{1}{2}$, and the other having spin-down, ${m}_{s} = - \frac{1}{2}$.

Starting with the electrons located on the first energy shell, $n = 1$, you will have

$n = 1 \text{ " implies" " l = 0 " "implies " } {m}_{l} = 0$

The two electrons located on the first energy level can have only one set of $n$, $l$, and ${m}_{l}$ numbers. The difference between the two electrons is made by the spin quantum number, ${m}_{s} = \pm \frac{1}{2}$.

For the electrons located on the second energy level, $n = 2$, you will have

n=2 " "implies" "{ (l=0 " "implies" "m_l = color(white)(-1) 0 color(white)( +1)), (l=1 " "implies" " m_l = {-1, 0, 1}) :}

This time you can have

$n = 2 , \text{ "l = 1, " "m_l = +1 " "implies" } 2 + 1 + 1 = 4$

This set of $n$, $l$, and ${m}_{l}$ quantum numbers belongs to two electrons located on the second energy shell, in the $\textcolor{b l u e}{2 {p}_{y}}$ orbital, one having spin-up and the other with spin-down.

For the electrons located on the third energy level, $n = 3$, you will have

n=3 " "implies" "{ (l=0 " "implies" "m_l = color(white)(-1) 0 color(white)( +1)), (l=1 " "implies" " m_l = {-1, 0, 1}), (l=2 " "implies" " m_l = {-2, -1, 0, +1, +2}) :}

This time you can have

$n = 3 , \text{ " l=1, " "m_l = 0 " "implies" } 3 + 1 + 0 = 4$

This set of $n$, $l$, and ${m}_{l}$ quantum numbers belongs to two electrons located on the third energy level, in the $\textcolor{p u r p \le}{3 {p}_{z}}$ orbital, one having spin-up and the other spin-down.

Notice that you can also have

$n = 3 , \text{ " l=2, " " m_l = -1 " " implies" } 3 + 2 + \left(- 1\right) = 4$

This set of $n$, $l$, and ${m}_{l}$ quantum numbers belongs to one electron located on the third energy level, in the $\textcolor{b r o w n}{3 {d}_{y z}}$ orbital, having spin-up.

This is the case because iron has a total of $6$ electrons in its 3d-subshell. This implies that only one of the five 3d-orbitals is completely filled, the other four being only half-filled.

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

If you assign ${m}_{l} = - 2$ to the 3d-orbital that contains two electrons, let's say $3 {d}_{x z}$, you will end up with a single electron in the orbital that corresponds to ${m}_{l} = - 1$, which would be color(brown)(3d_(yz), the one following the $3 {d}_{x z}$ orbital.

Finally, for the electrons located on the fourth energy level, $n = 4$, you will have

$n = 4 \text{ "implies" "l=0 " "implies" } {m}_{l} = 0$

Notice that iron has only two electrons on its fourth energy level, both in the $\textcolor{red}{4 s}$ orbital, so you don't need to worry about other values of $l$.

These two electrons share the $n$, $l$, and ${m}_{l}$ numbers for which

$n = 4 , \text{ " l=0, " " m_l = 0 " "implies" } 4 + 0 + 0 = 4$

So, the total number of electrons for which

$n + l + {m}_{l} = 0$

is equal to

"no. of e"^(-) = overbrace(color(white)(a)2color(white)(a))^(color(blue)("located in 2p"_ y)) + overbrace(color(white)(a)2color(white)(a))^(color(purple)("located in 3p"_ z)) + overbrace(color(white)(a)1color(white)(a))^(color(brown)("located in 3d"_ (yz))) + overbrace(color(white)(a)2color(white)(a))^(color(red)("located in 4s")) = color(green)(|bar(ul(color(white)(a/a)color(black)("7 e"^(-))color(white)(a/a)|)))#