# Question #f4b81

##### 1 Answer

Seven electrons.

#### Explanation:

Your starting point here will the **electron configuration** of a *neutral* iron atom,

Iron is located in period 4, group 8 of the periodic table and has an atomic number equal to **electrons**

#"Fe: " 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2#

Now, a set of **four quantum numbers** is assigned to each of those

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 **two electrons**, one having spin-up,

Starting with the electrons located on the *first energy shell*,

#n= 1" " implies" " l = 0 " "implies " "m_l = 0#

The two electrons located on the first energy level can have only **one set** of *spin quantum number*,

For the electrons located on the *second energy level*,

#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, " "l = 1, " "m_l = +1 " "implies" " 2 + 1 + 1 = 4#

This set of **two electrons** located on the second energy shell, in the *spin-up* and the other with *spin-down*.

For the electrons located on the *third energy level*,

#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, " " l=1, " "m_l = 0 " "implies" " 3 + 1 + 0 = 4#

This set of **two electrons** located on the third energy level, in the *spin-up* and the other *spin-down*.

Notice that you can also have

#n=3, " " l=2, " " m_l = -1 " " implies" " 3 + 2 + (-1) = 4#

This set of **one electron** located on the third energy level, in the *spin-up*.

This is the case because iron has a total of **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*.

If you assign **two electrons**, let's say **a single electron** in the orbital that corresponds to

Finally, for the electrons located on the *fourth energy level*,

#n=4 " "implies" "l=0 " "implies" "m_l = 0#

Notice that iron has only **two electrons** on its fourth energy level, both in the

These two electrons share the

#n=4, " " 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)|)))#