Question

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

Answer 1

Let's start by showing the electron configuration of a neutral `"Fe"` atom, which has a total of 26 electrons.

`"Fe": 1s^(2)2s^(2)2p^(6) 3s^(2)3p^(6)4s^(2)3d^(6)`

`"Fe"^(3+)` has 3 less electrons than `"Fe"`; these electrons will be removed starting from the outermost shell, which in `"Fe"`'s case will be `4s` and `3d`. The first two will come from `4s` and the last one will come from `3d`. So,

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

We've established that `"Fe"^(3+)` has `"26 - 23 = 26"` electrons. I assume the `"m"` you're asking for is actually `"m"_l`, the magnetic quantum number. Let's start counting the electrons that meet the `"n + l + m = 4"` criterion.

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

When `"n=2"`, you now get multiple values for `"l"` and `"m"_l`. If `"l"=0`, `"m"_l` is again equal to zero and your equation is `"2 + 0+ 0"`.

When `"l=1"`, `"m"_l` can be equal to `"-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 `"-1/2"` or `"+1/2"`, we've found the first two electrons that match the equation.

When `"n=3"`, you've got multiple choices again, since `"l"` can be either `0`, `1` or `2`. When `"l = 0"`, `"m"_l =0` and we don't have a match.

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

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

Here's where it gets a little tricky. Notice that the `3d` 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 `"2 + 2 + 1 = 5"` electrons satisfy the criterion given.

Answer 2

in `Fe^(3+)` there are 5 electrons for which the value of `n+l+m=4`

The electron structure of iron is :

`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(6)4s^(2)`

When `Fe^(3+)` forms the 4s are lost and one of the 3d to give:

`1s^(2)2s^(2)2p^(6)3s^(2)3p^(6)3d^(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 `(n-1)`
`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 `2p^(6)` electrons:

`n=2` , `l=1` and `m` can be -1 , 0 or 1 (these are the `p_x, p_y and p_z` electrons)

So 2 of these electrons has `n=2,l=1 and m=1` i.e a sum of 4

For `3s^(2)` `n=3`, `l=0` and `m=0` so non of them can add up to 4.

For `3p^(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 `3d^(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=2and 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.