Question #f4b81

1 Answer
Jun 22, 2016

Answer:

Seven electrons.

Explanation:

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

#"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 #26# electrons.

figures.boundless.com

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 = +1/2#, and the other having spin-down, #m_s = -1/2#.

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

#n= 1" " 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 = +- 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, " "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 #color(blue)(2p_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, " " 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 #color(purple)(3p_z)# orbital, one having 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 #n#, #l#, and #m_l# quantum numbers belongs to one electron located on the third energy level, in the #color(brown)(3d_(yz))# 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.

#color(white)()#
#color(white)()#
http://chemistry.tutorvista.com/inorganic-chemistry/oxidation-states.html
#color(white)()#

If you assign #m_l=-2# to the 3d-orbital that contains two electrons, let's say #3d_(xz)#, 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 #3d_(xz)# orbital.

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

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

Notice that iron has only two electrons on its fourth energy level, both in the #color(red)(4s)# 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, " " 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)|)))#