Question #47537

1 Answer
Sep 16, 2016

Answer:

#"11 e"^(-)#

Explanation:

The first thing to do here is to write the electron configuration of the iron(III) cation, #"Fe"^(3+)#

#"Fe"^(3+): 1s^2 2s^2 2p^6 3s^2 3p^6 3d^5#

Now, we can use four quantum numbers to describe the position and spin of an electron inside an atom.

figures.boundless.com

Your goal here is to find the number of electrons for which

#m_l + l = 0#

In other words, you need to find how many electrons for which the angular momentum quantum number, #l#, and the magnetic quantum number, #m_l#, add up to give zero.

As you can see, the value of the magnetic quantum number depends on the value of the angular momentum quantum number as given by

#m_l = {-l, -(l-1), ..., -1, 0, 1, ..., (l-1), l}#

You can thus say that every subshell, i.e. every value of #l# will have two electrons for which #m_l + l = 0#.

Here's why that it the case.

The angular momentum quantum number gives you the subshell in which the electron is located. You can have

  • #l=0 -># the s subshell
  • #l=1 -># the p subshell
  • #l=2 -># the d subshell

and so on. The magnetic quantum number tells you the exact orbital in which the electron is located. As you know, an orbital can hold a maximum of #3# electrons as given by Pauli's Exclusion Principle.

Because the magnetic quantum number can take the value #-l#, you can say that every subshell will have one orbital for which

#m_l + l = 0#

For the s subshell you have #l=0# and #m_l = 0#, and so every s subshell will have one orbital, i.e. maximum #2# electrons, for which #m_l + l = 0#.

The same can be said for the other subshells. In the p subshell, #l=1# allows for #m_l = -1#, in the d subshell, #l=2# allows for #m_l = -2#.

Now, let's focus on the iron(III) cation. The cation has three completely filled s subshells, #1s^2#, #2s^2#, and #3s^2#. Since each s subshell contains one s orbital, which in turn contains #2# electrons when completely filled, for which

#m_l + l = 0#

#0 + 0 = 0#

you will have

#"no. of e"^(-) = 2 + 2 + 2 = "6 e"^(-) -># just from the s subshells

Next, move on to the p subshells. The iron(III) cation has two completely filled p subshells, #2p^6# and #3p^6#. Since each p subshell contains one p orbital, which in turn contains #2# electrons when completely filled, for which

#m_l + l = 0#

#(-1) + 1 = 0#

you will have

#"no. of e"^(-) = 2 + 2 = "4 e"^(-) -># just from the p subshells

Finally, move on to the d subshell. Notice that the iron(III) cation has a total of #5# electrons in its d subshell.

The trick here is to realize that these electrons are unpaired, i.e. each occupies a d orbital as given by Hund's Rule. This means that the d subshell contains one orbital, which in turn contains #1# electron when half-filled, for which

#m_l + l = 0#

#(-2) + 2 = 0#

You will thus have

#"no. of e"^(-) = "1 e"^(-) -># just from the d subshell

The total number of electrons that satisfy the given condition will be

#"total no. of e"^(-) = overbrace("6 e"^(-))^(color(blue)("from s subshells")) + overbrace("4 e"^(-))^(color(darkgreen)("from p subshells")) + overbrace("1 e"^(-))^(color(purple)("from d subshell")) = "11 e"^(-)#