# How many electrons in an iron(III) cation have n + l + m_l = 4 ?

Jan 28, 2015

Let's start by showing the electron configuration of a neutral $\text{Fe}$ atom, which has a total of 26 electrons.

$\text{Fe} : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 4 {s}^{2} 3 {d}^{6}$

${\text{Fe}}^{3 +}$ has 3 less electrons than $\text{Fe}$; these electrons will be removed starting from the outermost shell, which in $\text{Fe}$'s case will be $4 s$ and $3 d$. The first two will come from $4 s$ and the last one will come from $3 d$. So,

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

We've established that ${\text{Fe}}^{3 +}$ has $\text{26 - 23 = 26}$ electrons. I assume the $\text{m}$ you're asking for is actually ${\text{m}}_{l}$, the magnetic quantum number. Let's start counting the electrons that meet the $\text{n + l + m = 4}$ criterion.

When $\text{n=1}$, you only get one choice for $\text{l}$ and ${\text{m}}_{l}$, that is $\text{l = 0}$ and $\text{m"_l = 0}$. This clearly doesn't satisfy the criterion. Moving on.

When $\text{n=2}$, you now get multiple values for $\text{l}$ and ${\text{m}}_{l}$. If $\text{l} = 0$, ${\text{m}}_{l}$ is again equal to zero and your equation is $\text{2 + 0+ 0}$.

When $\text{l=1}$, ${\text{m}}_{l}$ can be equal to $\text{-1}$, $0$, or $+ 1$. As you can see, we've found our first match: "2 + 1 + 1 = 4. Since the spin quantum number can be either $\text{-1/2}$ or $\text{+1/2}$, we've found the first two electrons that match the equation.

When $\text{n=3}$, you've got multiple choices again, since $\text{l}$ can be either $0$, $1$ or $2$. When $\text{l = 0}$, ${\text{m}}_{l} = 0$ and we don't have a match.

Let's take the case when $\text{l=1}$. In this case, ${\text{m}}_{l}$ can once again be $\text{-1}$, $0$, or $+ 1$. When ${\text{m}}_{l}$ is equal to zero, we have another match: $\text{3 + 1 + 0 =4}$. Two more electrons meet the criterion.

If $\text{l} = 2$, which means that ${\text{m}}_{l}$ can be $\text{-2}$, $\text{-1}$, $\text{0}$, $\text{+1}$, and $\text{+2}$, we can see that we get one more option, $\text{m"_l "= -1}$. We have "3 + 2 + (-1) = 4.

Here's where it gets a little tricky. Notice that the $3 d$ subshell is only half-full. This means that each of the five 3d-orbitals will only hold 1 electron. As a result, just one electron meets the criterion this time.

Add up all the electrons and you'll get $\text{2 + 2 + 1 = 5}$ electrons satisfy the criterion given.

Jan 28, 2015

in $F {e}^{3 +}$ there are 5 electrons for which the value of $n + l + m = 4$

The electron structure of iron is :

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{6} 4 {s}^{2}$

When $F {e}^{3 +}$ forms the 4s are lost and one of the 3d to give:

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{6} 3 {d}^{5}$

$n$ is the principal quantum number and has integral values 1, 2, 3....etc
$l$ is the angular momentum quantum number and takes integral values of 0 up to $\left(n - 1\right)$
$m$ is the magnetic quantum number and takes integral values of $- l$..through to 0 up to $+ l$

Non of the 1s or 2s electrons can reach a sum of 4 for these numbers but for the $2 {p}^{6}$ electrons:

$n = 2$ , $l = 1$ and $m$ can be -1 , 0 or 1 (these are the ${p}_{x} , {p}_{y} \mathmr{and} {p}_{z}$ electrons)

So 2 of these electrons has $n = 2 , l = 1 \mathmr{and} m = 1$ i.e a sum of 4

For $3 {s}^{2}$ $n = 3$, $l = 0$ and $m = 0$ so non of them can add up to 4.

For $3 {p}^{6}$, $n = 3$, $l = 1$ and $m$ can be -1, 0 , or 1

So there are 2 electrons with $n = 3$, $l = 1$ and $m = 0$

For $3 {d}^{5}$, $n = 3$, $l = 2$ and $m$ can be -2, -1, 0, 1, 2

There is only one electron where $n + l + m = 4$ which is $n = 3 , l = 2 \mathmr{and} m = - 1$. This is because the 3d orbitals are singly occupied so the electrons take separate values of $m$.

So the total number of electrons which have $n + l + m = 4$ is 2 + 2 + 1 = 5.