# Question #47537

##### 1 Answer

#### Answer:

#### Explanation:

The first thing to do here is to write the electron configuration of the iron(III) cation,

#"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.

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*, *magnetic quantum number*,

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 **two electrons** for which

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 -># thes subshell#l=1 -># thep subshell#l=2 -># thed 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 **electrons** as given by *Pauli's Exclusion Principle*.

Because the magnetic quantum number can take the value **every subshell** will have **one orbital** for which

#m_l + l = 0#

For the **s subshell** you have **every s subshell** will have one orbital, i.e. *maximum* *electrons*, for which

The same can be said for the other subshells. In the **p subshell**, **d subshell**,

Now, let's focus on the iron(III) cation. The cation has three completely filled **s subshells**, **one s orbital**, which in turn contains **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 thes subshells

Next, move on to the **p subshells**. The iron(III) cation has two completely filled p subshells, *one p orbital*, which in turn contains **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 thep subshells

Finally, move on to the **d subshell**. Notice that the iron(III) cation has a total of **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 **electron** when half-filled, for which

#m_l + l = 0#

#(-2) + 2 = 0#

You will thus have

#"no. of e"^(-) = "1 e"^(-) -># just from thed 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"^(-)#