# Question #47537

Sep 16, 2016

${\text{11 e}}^{-}$

#### Explanation:

The first thing to do here is to write the electron configuration of the iron(III) cation, ${\text{Fe}}^{3 +}$

${\text{Fe}}^{3 +} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{5}$

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, $l$, 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

${m}_{l} = \left\{- l , - \left(l - 1\right) , \ldots , - 1 , 0 , 1 , \ldots , \left(l - 1\right) , l\right\}$

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

• $l = 0 \to$ the s subshell
• $l = 1 \to$ the p subshell
• $l = 2 \to$ 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, $1 {s}^{2}$, $2 {s}^{2}$, and $3 {s}^{2}$. 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

${\text{no. of e"^(-) = 2 + 2 + 2 = "6 e}}^{-} \to$ just from the s subshells

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

${m}_{l} + l = 0$

$\left(- 1\right) + 1 = 0$

you will have

${\text{no. of e"^(-) = 2 + 2 = "4 e}}^{-} \to$ 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

${m}_{l} + l = 0$

$\left(- 2\right) + 2 = 0$

You will thus have

${\text{no. of e"^(-) = "1 e}}^{-} \to$ just from the d subshell

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

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