Question

Question #47537

Answer 1

`"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"^(-)`